Class 12

Balbharti Maharashtra State Board Class 12 Psychology Solutions Chapter 4 Cognitive Processes Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Psychology Solutions Chapter 4 Cognitive Processes

1. Choose the correct option and complete the following statements.

Question 1.
When we select few stimuli from the environment and bring them into the center of our awareness, it is called ……………
(a) attention
(b) perception
(c) thinking
Answer:
(a) attention

Question 2.
Ivan Pavlov is known for explaining learning by …………..
(a) classical conditioning
(b) operant conditioning
(c) observation
Answer:
(a) classical conditioning

Question 3.
When we organise the information from the environment, group them and give some meaning, that process is called ……………..
(a) attention
(b) perception
(c) thinking
Answer:
(b) perception

Question 4.
In operant conditioning the experiment was conducted on ……………
(a) dog
(b) rat
(c) monkey
Answer:
(b) rat

Question 5.
The change in behaviour due to practice or experience is known as ……………
(a) perception
(b) thinking
(c) learning
Answer:
(c) learning

2. Answer the following questions in 35 – 40 words each.

Question 1.
What is fluctuation of attention?
Answer:
Attention is the mental process of bringing few stimuli into the centre of awareness out of the many stimuli present. It is difficult to attend to a single stimulus for a long period of time. Our attention shifts from the original stimulus to another stimulus for a fraction of time and then comes back to the original stimulus. This may be due to factors like fatigue, decreased interest, etc. In most cases, attention fluctuates due to the limitations of human attention and hence cannot be totally eliminated.

Question 2.
Give three examples of top-down processing in perception.
Answer:
Perception is defined as the process of assigning meaning to information received about the environment based on the past experiences. We make use of both top-down processing and bottom- up processing while perceiving a stimulus. When we utilize top-down processing, our ability to understand information is influenced by the context in which it appears.

Some examples of top-down processes in perception are-
(a) Mr. Kumar lives in Flat A not in Flat
The circled item will be perceived as the letter B as the brain has used the context of the sentence.

(b) If you are asked to identify (a) the rat (b) the man, your perception will be likewise.
(c) the huamn mnid deos not raed ervey lteter by istlef but the wrod as a whole.

Most of you would be able to correctly read the above sentence as “The human mind does not read every letter by itself but the word as a whole”.

Question 3.
What is meant by thinking?
Answer:
Thinking is the mental activity that uses various cognitive elements and processes that involves the manipulation of information.
The core elements of thinking are-

1. Mental representation – It is a coded internal sensation acquired by direct experiences through the sense organs or by indirect experiences such as narrations, pictures, etc.
2. Concepts – They form the basis for all cognitive processes, acting as building blocks by connecting with each other to form more complex concepts.
3. Schema – It is an internal representation that organizes knowledge about related concepts and relationships among them.
4. Language – Mental representations, concept and schema are represented by language, for e.g., the word ‘cat’ is a symbol for a ‘cat’.

Question 4.
Describe any two processes of learning.
Answer:
Learning is defined as a relatively permanent change in behaviour that occurs due to experience or practice. The characteristics of learning are : (i) It involves some relatively permanent change in behaviour, (ii) The change in behaviour is due to experience or practice, (iii) Change in behaviour may be in knowledge or in skill or in muscular movements, (iv) Learning is an inferred process. The main processes of learning are:
(i) Learning by classical conditioning – It was first explained by Ivan Pavlov. It is learning by forming associations and stimulus substitution. In daily life, we learn many things by classical conditioning for e.g. a child is given an injection by the doctor and begins to cry in pain. He soon makes the connection between ‘doctor’, ‘injection’ and ‘pain’ and begins crying as soon as he is taken to a doctor.

(ii) Learning by operant conditioning – Learning by operant conditioning was first explained by B. F. Skinner. He said that learning takes place to gain positive consequences or to avoid negative consequences.

3. Classify the following into Mental Representation, Concepts and Schemata.

(1) Image of your mother in your mind
(2) Building
(3) Tree
(4) School
(5) Theater
(6) Table
(7) Neil Armstrong
Answer:

4. Find the odd one out.

Question 1.
Schema, Perception, Attention, Thinking
Answer:
Schema

Question 2.
Searching solution, Defining problem, Incubation, Implementation of the solution
Answer:
Incubation

Question 3.
Reasoning, Judgement, Decision making, Relaxing
Answer:
Relaxing

Question 4.
Concept, Schema, Solution, Mental Representation
Answer:
Solution

5. Identify if the following behaviours are due to learning or not. Give reasons for your answers.

Question 1.
After much hard work you are able to perform a dance step properly.
Answer:
It is due to learning as it involves a relatively permanent behavioural change due to practice.

Question 2.
In spite of getting hurt, Shayana, a five year old child, continues to play with knife.
Answer:
It is not learning as in spite of an unpleasant experience, there is no change in behaviour.

Question 3.
Once Hanif had to pay fine for breaking the signal while driving a car. Now he makes sure to stop the car at red signal.
Answer:
It is due to learning as Hanif has made an association between breaking the signal and paying a fine, i.e., negative consequences.

6. Answer the following questions with the help of given points.

Question 1.
Explain the stages of problem solving.
Points:
(i) Defining problem
(ii) Generating alternative solutions
(iii) Selecting a solution
(iv) Implementing and taking follow up of the solution.
Answer:
Problem solving is a type of thinking. It refers to the process of finding appropriate solutions to problems encountered in life. The main steps in problem solving are:
(i) Defining the problem – We need to identify and define the problem correctly. Otherwise, it will be difficult to reach a solution, for e.g., when a baby cries due to colic/stomach ache and if the mother interprets it as due to hungry, the problem will remain unresolved.

(ii) Generation alternative solutions – The person searches for possible solutions to the problem. Some solutions may be effective while other solutions have to be discarded, for e.g., Sunil finds maths difficult. He may think of solving more sums or of taking tuitions or of dropping the subject, etc.

(iii) Selecting a solution – The person selects what he considers the most effective solution, based on reasoning. This helps to resolve the problem in a realistic manner.

(iv) Implementing the solution and follow up on the solution – The person tries out the selected solution and evaluates the outcome i.e. whether it has helped to solve the problems. If the problem remains unresolved, the person may need to carry out the problem – solving procedure all over again.

Question 2.
Explain the stages of creative thinking.
Points:
(i) Preparation
(ii) Incubation
(iii) Illumination
(iv) Verification
Answer:
Creative thinking is characterised by the ability to perceive the world in new ways, to find hidden patterns, etc. It is a way of looking at problems or situations from a fresh/novel perspective. In involves divergent thinking that focuses on exploration of ideas and generating many possibilities. It is referred to as “thinking outside of the box.” Researchers, poets, designers, film makers always think in a creative way.

The four stages in creative thinking are-
(i) Preparation – This involves formulating the problem and gathering information about it. Many tentative solutions are tried out and discarded. No progress seems to be made by the person.

(ii) Incubation – If the person does not get the required results, he/she may focus on things unrelated to that problem. The period helps to work out the problem without consciously thinking about it. Incubation period appears to be non-fruitful.

(iii) Illumination – After incubation, suddenly the correct solution appears to the person, due to which he/she experiences excitement. It seems that a solution has fallen into place.

(iv) Verification – The new solution may need to be evaluated a number of times. In most cases, minor changes will be required but sometimes it may demand a major overhaul of the entire process.

For e.g., A teacher asks the students to think of unusual uses for a bottle.

• The students will begin brainstorming.
• The students will come up with many uses for e.g., to store liquids, to hold plants, etc. But these are not unusual uses. They will give up and focus on something else.
• The students will suddenly find novel uses for the bottle e.g., in juggling etc.
• The students will verify these solutions with their teacher.

7. Answer the following questions in detail.

Question 1.
Explain the laws of perceptual organization.
Answer:
Perception is defined as the process of assigning meaning to information received about the environment based on the past experiences. Our brain has the tendency to organize our sensations as a meaningful whole. Max Wertheimer first explained this tendency in the form of laws of perceptual organization.

Some laws of perceptual organization are as follows:
(i) Law of proximity – Stimuli that are near to each other are perceived as together, rather than stimuli that are far away from each other.

In the above figure, we perceive pairs of dots in each line because the dots which are near to each other are perceived together. So instead of perceiving a line of 8 dots, generally a line of four pairs of dots is perceived.

(ii) Law of similarity – Stimuli that are similar to each other are perceived together than stimuli that are distinct from each other.

In the above figure, we perceive 4 alternate vertical lines each of circles and crosses as similar stimuli are perceived together. Generally, we do not perceive 4 horizontal lines each having circles and crosses in alternate sequence.

(iii) Law of continuity – There is a tendency to perceive a stimulus in continuation according to its established direction.

In the above figure, a straight vertical line and a straight horizontal line are perceived together as a letter ‘L’ and a cutting line is perceived separately as a line following the smoothest path. Generally, we do not perceive here four different lines going in different directions.

(iv) Law of closure – There is a tendency to fill in the gaps in an incomplete stimulus so as to perceive it as a meaningful figure.

In the above figure, we fill in the gaps and perceive it as a triangle and square. Generally, we do not perceive here the three or four separate lines going in different directions.

Question 2.
Explain the core elements of thinking.
Answer:
Thinking is the mental activity that makes use of ideas or symbols instead of overt activity. The types of thinking are (i) Perceptual (concrete) thinking, (ii) Conceptual (abstract) thinking (iii) Reflective thinking, (iv) Creative thinking.
The core elements in thinking are-
(i) Mental representation – It is a coded internal sensation acquired by direct experiences through the sense organs or by indirect experiences such as narrations, pictures, etc. Mental representation is the mental imagery with the help of which the brain codes and stores the information. It is like a hypothetical internal cognitive symbol used by the brain to represent external reality. Example: a child is asked to imagine a peacock. The visual image of a peacock with its colouful plumage immediately comes to mind. This refers to “mental imagery” of an object.

(ii) Concepts – They form the basis for all cognitive processes, acting as building blocks by connecting with each other to form more complex concepts. Concepts involve extraction of some ‘idea’ on the basis of similarities and differences among the sensations. A concept is an idea which represents a class of objects, situations, etc which differentiates it from other classes of objects, situations, etc., for e.g. a child forms a mental image of a ‘dog’. He derives an idea of something similar in all examples of ‘dog’ that he has seen i.e. tail, fur, barking, etc. Thus, he has formed concept of dog. If he mistakenly identifies as a cat as a dog. he is corrected by others. So he tries to compare the mental images of dogs and cats. He has now learned and refined two concepts viz. dog and cat.

(iii) Schema – It is an internal representation that organizes knowledge about related concepts and relationships among them. Schema involves arranging many concepts in a particular system or organisation. It describes a pattern of thought or behavior that organises categories of information and the relationships among them. A child tries to organise the concepts that he has learned in a systematic way to generate a higher order understanding of patterns about information collected, for e.g. when a child visits a ‘zoo’ for the first time, images and concepts such as of different animals, caves/cages, etc., are activated in the brain simultaneously. Thus, he is forming the schema of ‘zoo’.

Conclusion:- Mental representation is sensory experiences in the form of mental images in the brain. Many similar mental representations denote concepts while many concepts put in a particular relationship to each other is schema.

Question 3.
With the help of your own examples explain the difference between distraction of attention and fluctuation of attention.
Answer:
Attention is the selective process by which we focus on only a few stimuli from among the various stimuli that are present in our environment. According to Norman Munn, “Attention is the mental process of bringing few stimuli into the centre of awareness out of the many stimuli present”. Attention is influenced by objective factors such as intensity, size, movement, etc., of stimuli as well as subjective factors like interest, mind-set, etc.

Distraction of attention refers to the drifting of attention from a specific stimulus to another stimulus. This occurs due to external factors such as intensity, novelty, movement, etc., of stimuli or internal factors like physical state, lack of interest, mental set, etc.

It is difficult to attend to a single stimulus for a long period of time. Our attention shifts from the original stimulus to another stimulus for a fraction of time and then comes back to the original stimulus. This is called fluctuation of attention. It may be due to factors like fatigue, decreased interest, etc.

In most cases, fluctuation of attention is due to the limitations of human attention whereas distraction of attention is due to an external powerful stimulus that has drawn our attention. When attention fluctuates, it is for a very short period of time and then it returns to the original stimulus, i.e., it is a temporary shift in attention, e.g., look at the below figure. Attention fluctuates on observing the figure as a closed book or open book.

Distraction of attention arises when you are studying and the doorbell suddenly rings, you get up to answer the door. You may or may not be able to return to the original stimulus. Fluctuation of attention cannot be controlled totally but distraction of attention should be avoided.

Class 12 Psychology Chapter 4 Cognitive Processes Intext Questions and Answers

ACTIVITIES (Textbook Page. No. 33)

Activity 1
Read the following examples and try to name the cognitive process described in each of them:

1. Reena immediately sensed the burning smell of pizza put in the oven.
2. Mahesh always pays attention to what the psychology teacher teaches in the class.
3. Professor Mr. Patil met one of his former students all of a sudden. He tried to recall the name of his student.
4. Saif was reading a story book and so he forgot to complete his homework.
5. After considering all pros and cons, Dinesh took a decision of changing his business.

Answer:

1. Sensation
2. Attention
3. Perception, Memory
4. Forgetting
5. Reasoning, Decision making.

Balbharti Maharashtra State Board Class 12 Sociology Solutions Passages Answers.

Maharashtra State Board Class 12 Sociology Passages Answers

Passage 1
Read the passages and answer the questions given below.

This is a real-life story of Rukmini Devi who lives in a small hut in Gaigotha Village in Wada Taluka of Palghar District in Maharashtra State. She belongs to the Warli tribe. Her husband is a marginal farmer who cultivates on two acres of land. They have two children, one daughter aged 10 years and an I son aged 6 years. Both the children walk to school and back daily (located about 3 miles away).

When cultivation season is over, (or if the rice crop is damaged due to heavy rains or pests) they face many hardships. A section of the crop is kept for their personal use, for the year. Tur Dal (lentil) is also grown in one small area, again for personal use. Ina small backyard, they grow vegetables like chilies, cucumber, and bitter gourd (karela).

During the off-season, both husband and wife go to the brick kilns (about 7 miles away) to do piece-rate work (That is, they get paid for each brick that they make.) While the men earn Rs. 300 per day, the women earn Rs. 150-200. Rukmini Devi stated that they prefer to walk the 7 miles both ways because the bus fare is Rs. 35/- per head one way. They cannot afford it.

Question 1.
Identify any three problems that the family of Rukmini Devi has to face.
Answer:
The main problem of the family of Rukmini Devi is poverty. The productivity of their economic activities is very less as they cannot use modern production techniques like fertilizers, pesticides, etc., their agriculture depends upon monsoons and there is a lack of adequate irrigation facilities. Due to the seasonal nature of agriculture, they have to face hardships and exploitation by non-tribal people. They are also being exploited by their employers who take maximum work from them and pay them minimum wages.

Question 2.
Point out and discuss briefly, gender discrimination in this setting.
Answer:
We find Rukmini Devi facing wage discrimination i.e., discrimination on the basis of sex in the payment of wages, where Rukmini Devi and her husband perform work of similar skill, effort, and responsibility for the same employer under similar working conditions but they don’t earn the same amount of money. This implies discriminative employers save on the cost by employing the tribal females. Rukmini Devi is working in, informal labour market where there is an absence of policies to safeguard gender rights.

Question 3.
Discuss the nature of the economy of the Warli tribes.
Answer:
The economy practiced by Warli Tribe is subsistence economy and simple. They use out model techniques therefore their production is insufficient. They cannot fulfill their basic needs. They try to fulfill their needs by collective efforts. Thus, the simple and collective economic life is an important characteristic of the tribal economy. The main occupation of the Warli tribe is agriculture which is in a state of backwardness. They live below the poverty line.

Passage 2
Read the passages and answer the questions given below.

Education, since the coming of the British to India, has been secular in content. By this we mean, the content of education did not include the study of sacred texts. Schools were open for all – to learn and climb the ladder of vertical mobility. The study of the English language, as well as the opportunity to study in the English medium, was available.

It is true that several Indians from certain social and economic strata were the first to access an English education. Many of them later constituted the intelligentsia of our society. We refer to many of them as social reformers, such as Raja Rammohan Roy, Ishwar Chandra Vidyasagar, Pandita Ramabai, Maharshi Dhondo Keshav Karve. They worked for religious, social, and educational reform in Indian society.

Such visionaries of society continue even in the post-Independence era, to the present time.

Educational opportunities have grown by leaps and bounds in the last 73 years since Independence. One questions if the educated have merely acquired education or if the education has helped citizens become gainfully employed and more importantly, enlightened enough to transform society at the micro-level.

It is necessary for the government to consider the interests of all sections of society.

Each citizen can play a dynamic role in the development of all people in our society.

Question 1.
Explain the impact of the introduction of a new education system by the British on Indian society.
Answer:
According to the new liberal education policy introduced by the British, education was not restricted to special sections of society. The spread of secular-based education widened the minds of the people living in India. Well-educated Indians were influenced by western values. They recognized that various customs and traditions were unjust and unfair. Therefore, they started various religious and social movements to reform Indian society.

Question 2.
Explain the role of education in the transformation of society at the micro-level.
Answer:
The role of education is effective to bring change at the individual level i.e., micro-level. The role of education as an agent or instrument of social change and social development is widely recognized today. Education can initiate social changes by bringing about a change in the outlook and attitude of man. It can bring about a change in the pattern of social relationships and thereby it may cause social changes Education has brought about phenomenal changes in every aspect of men’s life. Education is a process that brings about changes in the behaviour of society. It is a process that enables every individual to effectively participate in the activities of society and to make a positive contribution to the progress of society.

Question 3.
Discuss education as an instrument of social change.
Answer:
Education changes the outlook and traditional approach towards social and economic problems. It sharpens the skills, and knowledge of the children. Technical education helps in the process of industrialization which helps to bring a vast change in society. Education not only preserves the cultural traditions of the society but helps to transmit them from one generation to the next. Education fulfills needs of the society and propagates ideas to promote social change in all fields of life.

Passage 3
Read the passages and answer the questions given below.

Indian society is a melting pot of cultures. The history of Indian society gives enough evidence of the process of accommodation. From early times migrants integrated into Indian society and influenced its culture. Our historical past is testimony to this fact of cultural diffusion.

Today, we describe our society as a composite whole that includes tribal, rural, and urban communities. The way of life in these segments have their unique characteristics. However, is it also an observation that no one segment or community can be seen in its “pure” state. On the one hand, there is interdependence between communities and on the other, this would imply a certain extent of loss of cultural elements such as language, beliefs, customary practices, etc. Have we not seen how, for example, Warli or Madhubani Art has made it to T-shirts and wall hangings in many urban households? Also, how technology has reached the remotest corners of our country?

A question that may cross your mind may be, ‘Is there anything such as ‘pure culture? What constitutes “Indian culture”? ‘Can cultural extremism be valuable in the present world? These questions are valid
as they set us thinking. Perhaps there is no single “answer”?

Question 1.
What constitutes “Indian culture”, discuss with respect to cultural diffusion in Indian society.
Answer:
Over the years, India has changed a lot in terms of living standards and lifestyles, but even then the values and traditions are still intact and remain unchanged. Another aspect of India’s culture can be seen when someone is facing deep trouble. Irrespective of the class, tribe, or religion, everyone will step forward to provide help and support. Culture in India is a dimension that has been composed by its long history and its unique way of accepting customs and traditions, right since the Indus valley civilization took birth. India is a melting pot of various religions and cultures and it is the very nature of the unity in diversity, which has largely shaped the growth of Indian culture as a whole. The property of togetherness among people of various cultures and traditions has made India, a unique country.

Question 2.
Discuss tribal art and its role in cultural identity.
Answer:
Tribal art has progressed considerably due to the constant developmental efforts of the Indian government and other organizations. Tribal art generally reflects the creative energy found in the tribal areas. Tribal art ranges through a wide range of art forms, such as wall paintings, tribal dances, tribal music, and so on. Folk art in India apparently has great potential in the international market because of its traditional aesthetic sensibility. Some of the most famous folk paintings of India are the Madhubani paintings of Bihar, Warli folk paintings of Maharashtra

Question 3.
How interdependence between communities has resulted to a certain extent loss of cultural elements?
Answer:
Though the interdependence of communities connects all the cultures of the country it has also weakened cultural bonds of tribal and rural communities and also lead to the loss of cultural identity. It also makes one forget their own values, customs, and traditions. Although it has played an immense role in the unification of our country, a great amount of cultural identity and traditional values have also been lost.

Passage 4
Read the passages and answer the questions given below.

The causes of disharmony and strife are several-fold. Resistance to social change is one among many. Problems of contemporary Indian society include domestic violence, sexual abuse, child rights, problems of senior citizens, migrants, ethnocentrism, religious fundamentalism, linguistic fanaticism, environmental degradation, substance abuse and addiction to devices, mob lynching, and so on.

Given the varied types of social problems and their changing nature, there emerges a need to examine them in a scientific manner. The applicability of Sociology in its widest sense includes the exploration of various themes that cut across fields such as Masculinity Studies, Minority Studies, Film and Media Studies, Sociology of Sports, Environmental Sociology, Forensic Sociology, Gerontology, Sociology of Music, Medical Sociology, Marketing Sociology and so on.

Various government departments and voluntary organizations include sociologists on their panels to help steer policies and programmes. As Sociology is a people-centered discipline, it tends to create awareness and dialogue regarding human relationships. This is a valuable asset in governance and conflict resolution.

Question 1.
How does sociology perceive social problems in a scientific manner?
Answer:
Sociology views social problems as problems which arise out of the functioning of systems and structures in a society, or which are the result of group influences. They are also concerned with social relationships which emerge and are sustained because of the social problems. Thus, in analyzing alcoholism, a sociologist will be concerned with its effects on social relations and roles, that is, the relations with family members, with colleagues in the office, and with neighbours and friends as well as its effect on work efficiency, status, and so on. The study of social problems in sociology aspires toward a body of valid and logically related principles to get solutions for the social problems.

Question 2.
Discuss how resistance to social change leads to disharmony in society.
Answer:
Certain resistance to change is there everywhere. In no society, all the changes are welcomed by the people without questioning and resistance. To some extent, the removal of evil practices such as child marriage, human sacrifice, animal sacrifice, untouchability, taboos on inter-caste marriages, etc., could be achieved after a long struggle in India. Due to ignorance people often oppose new changes. Habit is another obstacle to social change. Individuals are very much influenced by habits and customs. People dislike or fear the unfamiliar. They are not ready to give up a practice to which they have been habituated and adopt a new one. Hence, the new practice is looked down upon or rejected which leads to social harmony.

Passage 5
Read the passages and answer the questions given below.

Given below is a make-believe scenario.
Yogini and Yogita are twins of the Patkar family who live in a small room measuring 225 sq. ft. in a
small town. Yogini is brilliant in studies and Kabbadi. Yogita is an outstanding cricketer who represents the Western India region; she also was a topper in the State-level Marathi language Competition.

Their parents come from a small village in Marathwada; they were farmers. For the sake of their daughters, they shifted to a small town to facilitate their children’s further education and sports training.

Their relatives and others in their village have heard of the Patkar girl’s success and are also encouraged to send their children to big cities with the hope that they too will become successful and famous one day.

Today, if one visits the village you will notice that in many homes, there are only the elderly folk. The youth seem to have migrated to better their prospects. Can you imagine the effect of such migration on the local village community?

Question 1.
What are the challenges faced by rural people while sending their children for higher education?
Answer:
Rural people migrate to cities for a better standard of living and better future prospects. They face lot of hardships, face all sorts of exploitations for survival and to shape the lives of their children. They work hard to meet the needs of their children and try to give them better education so that they have a decent life. Patkar’s family come from a small village in Marathwada, live in a small. For the sake of their daughters, they shifted to a small town to facilitate their children’s further education and sports training.

Question 2.
What makes rural people migrate to cities?
Answer:
Rural people are plagued with various problems of agriculture, the ownership of land, lack of cottage industries, lack of educational facilities like schools and colleges, lack of health care centers, unemployment, traditionalism, and conservatism all these factors force rural people to migrate to cities. Cites attract rural people with better job opportunities, education, and a better lifestyle. Cities are centers of opportunities for the rural people so they migrate in hope of having a better standard of livings.

Question 3.
What are the effects of migration on rural communities?
Answer:
When rural people migrate to urban areas for better prospects leaving behind everything. The negative impact of migration on rural communities are there is labour shortage in farms, only senior citizens, women and children are left behind, increase in child labour, children’s are forced, to work in fields, increased workload for women’s decreased population, disorganization of family, customs and in this way rural culture slowly fades away.

Passage 6
Read the passages and answer the questions given below.

Social movements arise generally from needs felt by one or more members of any given society. Through social interactions, these needs and concerns are communicated to many more persons. A network of people who share these concerns becomes the driving force for change in that particular society. Movements are usually guided by some underlying philosophies and goals. Indeed, several movements are associated with a founder or a core group. It can take several years, or even decades for a social movement to become very wide and expansive, across vast geographical territories.

Social movements such as the Social Reform Movement, Trade Union Movement, Tribal Movement, Dalit Movement, Women’s Movement, Chipko Movement, LGBT Movement, Civil Rights Movement, Rationalist Movement and so many more have emerged and grown.

As a social movement gains momentum, greater awareness is created in society. In fact, the study of several movements has found its way into the academic curriculum as well as research. For example, courses on Labour Studies, Gender Studies, minority’ Studies, and Environmental Studies. Social movements can stimulate critical thinking about social issues in the wider society of which we are apart. Some of these concerns lead to the passing of legislation. Every era or generation has its share of concerns from which may emerge new social movements.

Question 1.
What do you understand by social movement and discuss how it functions?
Answer:
Social Movement is a collection of a large group of people, who come with the desired objective to create a change or resist change. Through social interactions, individuals communicate and show their concern on various issues where they feel it necessary to change. Social movements arise generally from needs felt by one or more members of any given society. A network of people who share these concerns becomes the driving force for change in that particular society. Movements are usually guided by some underlying philosophies and goals. Indeed, several movements are associated with a founder. It can take several years, or even decades for a social movement to become very wide and expansive, across vast geographical territories.

Question 2.
How does the social movement arise in Society?
Answer:
Social movements arise in the society when certain issues bring unrest and discontent like unwanted social order and outdated norms like early child marriage, women emancipation, human rights, LGBT rights, etc., in the society. At this junction groups of people organize themselves, raise their voices and feelings and opinions set to influence the opinion and emotions of others, and prepare for reform. The need of society to bring changes in the existing system leads to a social movement.

Question 3.
Discuss any three social movements in the given passage.
Answer:
The three types of social movements are Social Reform Movements, Trade Union Movement, and Chipko Movement.
The social Reform Movement started in the 19th century. The movement promoted to change the traditional and conservative Indian society. Issues of main concern were religion, untouchability, early child marriage, sati, widowhood, exploitation of poor, etc. With the help of various social reformers and British legislative systems, changes took place.

Trade Union Movement was organized to fight against the exploitation of workers like they worked for long hours and were paid less, poor working conditions, lack of promotions, management disputes, strikes, etc.

Chipko movement took place in Uttarakhand, where Sunderlal Bahuguna and villagers came together and hugged the trees protected them from being chopped by the contractors. This was a protest to save forests and preserve the environment. The government set up a committee to look into the matter eventually ruled in favour of the villagers.

Question 4.
In what ways do you think the social movement is beneficial for society?
Answer:
Through social movements, various issues have been raised, which has brought changes. It has changed the mindset, attitudes and, behaviour patterns for instance women’s education, acceptance of transgender, etc. As a social movement gains momentum, greater awareness is created in society. The study of several movements has found its way into the academic curriculum as well as research.

For example, courses on Labour Studies, Gender Studies, minority’ Studies, and Environmental Studies. Social movements can stimulate critical thinking about social issues in the wider society of which we are apart. Some of these concerns lead to the passing of legislation like the untouchability removal act 1955, the sati act of 1829, the marriage act of 1954, the factory act of 1948, the child labour act 1986, and many more to go.

Passage 7
Read the passages and answer the questions given below.

Can human societies be flawless? What is considered acceptable, desirable, valuable varies from time to time, place to place, and in different contexts.

There is sometimes a tendency to encourage excessive ethnocentric attitudes about one’s culture or group to which one belongs. Ethnocentrism in its extreme form is an obstacle to social harmony. For the sake of social solidarity, respect of other cultures, self-criticism, critical appraisal, reflection, and introspection is necessary. This may help to develop a pluralist way of appreciating the diversities within which we live. The life stories of people are a useful means to understand underlying feelings, beliefs, threats, and so on.

Civil society can play a part in this process to eliminate or minimize factors that hinder progress, or those which divide us.

Question 1.
Explain the term Ethnocentrism. How it is an obstacle to social harmony?
Answer:
Ethnocentrism makes one feel that one’s own culture and way of life are superior to all others. Ethnocentrism can lead to a biased understanding of other cultures. The ethnocentric group feels their culture is superior, this creates a negative outlook which can lead to arrogance and hatred for others. Ethnocentrism in its extreme form is an obstacle to social harmony as there is sometimes a tendency to encourage excessive ethnocentric attitudes about one’s culture or group to which one belongs leading to antagonism and hatred among various religions and cultures.

Question 2.
What can one do to bring social solidarity to society?
Answer:
The term social solidarity means various social groups bind together as one in society.

We need to throw away prejudices, self-interest, self-criticism, learn to respect other’s cultures, reducing inequality and injustice in society. Solidarity can be cultivated through education. Promote new policies or initiatives to eradicate poverty, volunteering and practicing in charity events, donating money, food, clothes, etc. This brings empathy towards others encourages people to bring equality, justice, and peace.

For the sake of social solidarity, respect for other cultures, self-criticism, critical appraisal, reflection, and introspection is necessary. This may help to develop a pluralist way of appreciating the diversities within which we live. The life stories of people are a useful means to understand underlying feelings, beliefs, and threats, and so on.

Question 3.
What are the divisive forces that hinder the progress of Indian society?
Answer:
India is a secular, multilingual, and multicultural country. These diversities become at times challenges that hinder the progress of society. Various divisive forces like regional disputes, language problems, discrimination on the basis of caste, communal conflicts, terrorism, unequal distribution of wealth, poverty, etc. This divisive force obstructs progress and disturbs the peace and harmony in society.

Passage 8
Read the passages and answer the questions given below.

All of you have been studying Sociology for over a year. Sociology is the scientific study of human social behaviour. However, it is not the only discipline that studies human behaviour. The study of human behaviour is of interest to historians, psychologists, sociologists, economists, political scientists, etc. Surely, this question may have crossed your mind, or your family or friends might have asked you – ‘What is the use of studying Sociology? What work will you do with a degree in Sociology? Perhaps you too have wondered about the same.

As a start, you could surely consider a career in teaching. However, you must be prepared to read extensively, be creative and develop a learner-centric personality. Indeed, you would have to be passionate about teaching and interacting with learners. For some, a career in research is another possibility, though that route is a long journey to attain the status of ‘sociologist’. Many sociology students and others too, choose to offer Sociology as their subject of special study for Civil Service Examinations like UPSC (Central Services) and MPSC (in Maharashtra). To clear these highly ‘ competitive examinations, it is necessary to read widely and be well aware of the totality of Indian society – its past, present; goals, and plans for the future.

Then, of course, there are many allied occupations where a degree in Sociology can provide insights that are useful to take on other people-oriented professions such as Policy and Programme Development, Social Work with specialization in Family and Child Welfare, Community Development, Medical and Psychiatric Social Work, School Social Work, etc.

The fact remains: it is not merely an obtaining degree in sociology that matters today, but the skill sets, sensitivity, and personality that you develop; your ability to modify and adapt to new needs and challenging situations of even daily living. Also, your ability to have a humanistic perspective whether dealing with research or creating empowerment programmes, or programmes for social change.

Question 1.
What is the scope of sociology?
Answer:
Sociology is the scientific study of human social behaviour. The scope of sociology is wide they are: It studies social relationships, social institution patterns of human behaviour in society. Sociology deals with social changes, development, and analysis of various social problems like poverty, crime, suicide. Gender inequality population etc., and suggest various measures to solve them. One can surely consider a career in teaching; however, one must be passionate about teaching and interacting with learners.

A career in research is another possibility, one may choose to offer Sociology as their subject of special study for Civil Service Examinations like UPSC (Central Services) and MPSC (in Maharashtra). There are many allied occupations where a degree in Sociology can provide insights that are useful to take on other people-oriented professions such as Policy and Programme Development, Social Work with specialization in Family and Child Welfare, Community Development, Medical and Psychiatric Social Work, School Social work, etc.

Question 2.
Discuss the uses of Sociology in present society?
Answer:
In today’s changing world the importance of sociology is growing day by day.
It makes a scientific study of society detects and solves various social problems.
Helps in planning and development. The knowledge of sociology, its application is increasing in the field of industry, social work, law, competitive examinations like UPSC and MPSC, management studies public relations, journalism, etc.
Present time sociology has become useful in framing policies and programme for development like family and child welfare schemes, community development, etc.

Question 3.
Discuss how studying Sociology is beyond obtaining a degree.
Answer:
Today, it is not merely obtaining a degree in sociology that matters, but the skill sets, sensitivity, and personality that you develop; your ability to modify and adapt to new needs and challenging situations of even daily living. Also, your ability to have a humanistic perspective whether dealing with research or creating empowerment programmes or programmes for social change plays an important role.

Passage 9
Read the passages and answer the questions given below.

Read the make-believe speech made by a representative of the Governing Body to its Executive Committee meeting, in a well-known international firm located in Pune.

“Good morning. The Board of Directors has asked me to communicate with you all a policy decision that has been taken by the higher management. Two policies have been taken by our company. One, there shall be a confidential, two-way appraisal of all employees from the coming financial year. Every employee will be assessed by one’s immediate senior, one’s team members, and by oneself through self-appraisal. Juniors will also assess the seniors to whom they report. There are specific criteria on which assessment will take place. A second policy decision is for the company to make every effort to Go Green’s keeping with the international commitment towards a cleaner and greener environment. You may please share this decision with members of your respective departments today, through our e-portal systems. Feedback from all employees is welcome but they must be made within a week from today to the Human Resource Department, via the e-portal. ”

Question 1.
Explain the 1st policy decision that has been taken by the higher management.
Answer:
Two policies have been taken by the higher management. One, there shall be a confidential, two-way appraisal of all employees from the coming financial year. Every employee will be assessed by one’s immediate senior, one’s team members, and by oneself through self-appraisal. Juniors will also assess the seniors to whom they report. There are specific criteria on which assessment will take place.

Question 2.
Discuss the action to be taken by the employee with respect to ‘Go Green’.
Answer:
A second policy decision is for the company to make every effort to ‘Go Green’ in keeping with the international commitment towards a cleaner and greener environment which they can share with members of their respective departments, through the company’s e-portal systems.

Question 3.
Explain the advantages of appraisal.
Answer:
It is said that performance appraisal is an investment for the company. Performance appraisal helps the supervisors to chalk out the promotion programmes for efficient employees.

Passage 10
Read the passages and answer the questions given below.

Indian films have a history of their emergence, growth, and development. There were the days of silent films where viewers interpreted visuals on screen and constructed their own understanding of what the films may have tried to communicate. Then came the days of audio-visual films, black and white films, and later, colour films.

People who can afford to watch films at theatres and those who can do so on their television screens at home are entertained by the stories that films tell us. There are all kinds of ideas, ideologies, tragedies, themes, and values that films communicate. Today one can watch films on the internet on one’s mobile phone. Sometimes the explicit and implicit messages are received by viewers, but they can also be lost on them.

Besides actors’ abilities to ‘play varied roles or characters, there are a whole lot of persons involved with the production process as well as its marketing. This may include the film director, screenplay writers, designers, sound engineers, make-up artists and stylists, casting experts, musicians and so on.

Fields like Visual Sociology, Sociology of Mass Communication, and Marketing Sociology have a role to play in the study of these varied dimensions. Films as a source of knowledge play multiple roles even today. The storylines and types of films are ever-increasing. Films are not limited to nor bound by standard themes, love stories, or gender stereotyping. Films can cause much upheaval on the one hand and generate much interest on the other. Regional films and international films have been added to the list of viewing possibilities and multiple interests.

Question 1.
Write an account of the popularity of Indian cinema.
Answer:
Indian cinema was always enjoyed, whether it was the days of silent films where viewers interpreted visuals on screen and constructed their own understanding of what the films may have tried to communicate. Then came the days of audio-visual films, black and white films, and later, colour films. People entertained themselves by the stories that films conveyed either by watching films at theatres or on their television screens at home. The Hindi language film industry of Mumbai also known as Bollywood, it is the largest and most popular branch of Indian cinema. Hindi cinema initially explored issues of caste and culture in films such as Achhut Kanya (1936) and Sujata (1959). The audience’s reaction towards Hindi cinema is distinctive with involvement in the films by the audience’s clapping, singing, reciting familiar dialogue with the actors.

Question 2.
What do you understand by explicit and implicit messages of films?
Answer:
The film’s main message is loud and clear through the majority of films is known as an explicit message. It also has underlying morals for its audience known as implicit messages which are not so obvious. For example, morals such as, it’s not what’s on the outside, it’s on the inside that counts.

Question 3.
Discuss types of movie genres.
Answer:
Movies consist of many genres and categories like drama, comedy, action, thriller, horror, romance, experimental, documentaries, etc. The producers, directors try to create new genres experimenting with their creativity. The storylines and types of films are ever-increasing. There is no limitation to the subject matter of the films.

Question 4.
Discuss the impact of Indian cinema on society.
Answer:
Indian cinema is no longer restricted to India and is now being well appreciated by international audiences. The contribution of the overseas market to Bollywood box office collections is quite remarkable. Indian cinema has become a part and parcel of our daily life whether it is a regional or a Bollywood movie. It has a major role to play in our society. Though entertainment is the keyword of Indian cinema it has far more responsibility as it impacts the mind of the audiences.

Passage 11
Read the passages and answer the questions given below.

How does one tackle social problems? How do societies deal with the social problems that they have to confront? Why do social problems arise? These are some questions that learners of Sociology need to address.

Societies have culture; both of these are created by people, cumulatively, through the network of relationships over thousands of years. Every society has its normative system – customs, folkways, fashions, mores, taboos, fads, laws. Social norms are guidelines for human behaviour. They tell us what is expected of us and at the same time, what to expect from others.

Are these expectations permanent and unchanging? When can they change? Who changes them? Why must they change? Again, these are questions that one might ask. Social problems can arise when the expectations are not communicated effectively, or when individuals or groups choose to disagree with the expectation. This can lead to situations of conflict – not just ideological but also a conflict that leads to hurting others’ sentiments, abuse, violence, injustice, upheavals, normlessness, and even war.

Question 1.
What do you understand by normative aspects of culture, are these expectations permanent? When do they change?
Answer:
The normative aspects of culture consist of customs, folkways, fashions, mores, taboos, fads, laws. Social norms are guidelines for human behaviour. They tell us what is expected of us and at the same time, what to expect from others. These expectations are not permanent as appropriate and inappropriate behaviour often changes dramatically from one generation to the next. Norms can and do change over time. Karl Marx believed that norms are used to promote the creation of roles in society which allows people of different levels of social class structure to be able to function properly, hence any change in social structure may lead to change in the normative aspect of culture.

Question 2.
How does the social problems arise in society?
Answer:
Although not considered to be formal laws within society, norms still work to promote a great deal of control. Norms are more specific and they are rules of conduct that guide people’s behaviour. Therefore, when an individual or a group of people behave and act in a certain way that is in contradiction to society’s values or norms, it can create a social problem. Social problems can arise when the expectations are not communicated effectively, or when individuals or groups choose to disagree with the expectation. This can lead to situations of conflict – not just ideological but also a conflict that leads to hurting others’ sentiments, abuse, violence, injustice, upheavals, normlessness, and even war.

Passage 12
Read the passages and answer the questions given below.

Sarva Shiksha Abhiyan (SSA) is a Government of India programme that makes education for children between the ages 6-14 free and compulsory. This programme was pioneered by the former Indian Prime Minister Shri Atal Behari Vajpayee in 1993-94. It became totally operational from 2000-2001. This programme made education a Fundamental Right.

Along with this, the Government of India also launched the National Programme of Nutritional Support to Primary Education (NP-NSPE) on 15th August 1995. From here emerged the concept of free ‘Midday Meal’for for children going to schools which were managed by local bodies like Gram Panchayats and Municipal Corporations. The Midday Meal is mandatory. It is taken for granted that the children should be given good, nutritious food on a daily basis. A lot of organisation goes into the cooking and delivering of these meals to the schools on time.

Universal Education goes hand in hand with Nutrition. Children of the village and municipal schools look forward to this meal. For several of them, it is perhaps the main meal of the day.

Question 1.
Discuss the various child welfare programmes launched by the government of India.
Answer:
Sarva Shiksha Abhiyan (SSA) is a Government of India programme that makes education for children between the ages 6-14 free and compulsory. This programme was pioneered by the former Indian Prime Minister Shri. Atal Behari Vajpayee in 1993-94. It became totally operational from 2000-2001. This programme made education a Fundamental Right.

The Government of India also launched the National Programme of Nutritional Support to Primary Education (NP-NSPE) on 15th August 1995.

Free ‘Midday Meal’ for children going to schools which were managed by local bodies like Gram Panchayats and Municipal Corporations was also launched later. The Midday Meal is mandatory. It is based on the fact that the children should be given good, nutritious food on a daily basis.

Question 2.
What is the objective of the Midday Meal Scheme? Where does the responsibility of implementation of midday meal scheme lie?
Answer:
The Midday Meal Scheme is a school meal programme of the government of India designed to improve the nutritional status of school children nation wise. The objective of the Midday Meal Scheme is to provide a cooked meal to the children as should be given good, nutritious on a daily basis. The meal is mandatory.

A lot of organisation goes into the cooking and delivering of these meals to the schools on time. Universal Education goes hand in hand with nutrition. Children of the village and municipal schools look forward to this meal. For several of them, it is perhaps the main meal of the day.

The responsibility of implementation of the Midday Meal Scheme lies with local bodies like Gram Panchayats and Municipal Corporations.

Question 3.
How effective are the children’s welfare programmes in India?
Answer:
Keeping in view the problems and challenges faced by children various programmes and policies are implemented for the welfare of children in India. Sarva Shiksha Abhiyan (SSA) is a Government of India programme that makes education for children between the ages 6-14 free and compulsory. As a result, the enrolment percentage of school children has gone up. Similarly, Mid-day Meal is mandatory. A lot of organisations goes into the cooking and delivering of these meals to the schools on time. Universal Education goes hand in hand with Nutrition. Children of the village and municipal schools look forward to this meal. For several of them, it is perhaps the main meal of the day.

Balbharti Maharashtra State Board Class 12 Geography Solutions Chapter 7 Region and Regional Development Textbook Exercise Questions and Answers.

Maharashtra State Board Class 12 Geography Solutions Chapter 7 Region and Regional Development

1. Identify the correct group.

Question 1.

Answer:
D

Question 2.

Answer:
C

2. Differentiate between.

Question 1.
Functional region and Formal region
Answer:

Question 2.
Physical and Political region
Answer:

3. Write short notes on.

Question 1.
Factors affecting regional development.
Answer:
The physical factors like relief, climate, location, population and land use of the region affects the regional development.

Regional development occurs near areas with favourable location, plain relief, sufficient rainfall, developed agriculture, transportation, industries and trade. For example, Asian countries have developed agriculture and trade of agricultural goods. USA has developed industries and trade.

There is poor development of economic activities in areas with unfavourable relief like mountains or hills and extreme climate. For example, in the northern part of Russia, economic development is poor due to extreme cold climate.

Location of the region is an important factor. If there is a long and broken coastline, then there is development of ports and thus development of trade and economy. For example, Japan has economically developed in spite of limited natural resources, due to development of trade.

On the other hand, landlocked countries remain isolated. They cannot develop trade due to absence of coastline. Such countries have poor economic development. For example, Afghanistan, Nepal.

The economic development in any region depends upon the quantity and quality of its people.

Where more suitable land is available, different economic activities are developed. For example, cities use land for various purposes such as industries, education, entertainment, transport, commerce, etc. Therefore, cities have economic development.

Hence, physical factors are said to affect the economic development of the region.

Question 2.
Measures to reduce regional imbalance.
Answer:

• The following measures along with numerous policies are considered by the government to reduce regional imbalance.
• The underdeveloped regions are identified and its economic, physical and social reasons for the non-development are studied.
• Regions which require monetary support in a particular sector or field are funded through sector-wise investments to improve the conditions.
• Subsidies are given to certain sectors. Investment is made for improvements of roads, schools, agriculture, irrigation, industries, housing, medical and health facilities, etc.
• Special attention is given to areas facing problems such as frequent droughts, deserts or hilly and tribal-dominated areas.
• The very important strategy to reduce regional imbalances is decentralisation of industries.

4. Give geographical reasons.

Question 1.
Regional development is dependent on physical setting.
Answer:

• The physical factors like relief, climate and location of the region affect the regional development.
• Regional development occurs where agriculture, agro-based industries and trade is developed in areas where there is plain relief and sufficient rainfall. For example, Asian countries have developed agriculture and trade of agricultural goods.
• Plain relief and sufficient rainfall encourage development of transportation and industries.
• There is poor development of economic activities in areas with unfavourable relief like mountains or hills and extreme climate. For example, in the northern part of Russia, economic development is poor due to extreme cold climate.
• Location of the country is an important factor. If there is a long and broken coastline, then there is development of trade and economy. For example, Japan has economically developed in spite of limited natural resources, due to the development of trade.
• On the other hand, landlocked countries remain isolated. They cannot develop trade due to absence of coastline. Such countries have poor economic development. For example, Afghanistan, Nepal.
• Thus, there is relationship between regional development and physical factors and therefore, we can say the regional development depends on physical factors.

Question 2.
Factors like literacy, poverty affect the regional development.
Answer:

• Literacy and poverty are the important factors affecting the regional development.
• High literacy rate determines the quality of human population. This increases the availability of employment opportunities and development of resources, which contributes to the economic growth and regional development.
• Due to high literacy rate, the population growth is controlled and resources are shared among less people in the country.
• However, poverty decreases the purchasing powers resulting in a decline in the demand for goods, which reduces the production and drops the economic development and GDP of a country.
• Hence, poverty and literacy affect regional development.

Question 3.
Development is not seen in the Himalayan region.
Answer:

• Relief and climate are the most important factors affecting the development of any region.
• The Himalayan region has high mountains with a steep slope. The climate is cold and the mountains are covered with snow. There is heavy soil erosion and hardly plain land, so agriculture is not developed.
• In this mountainous region, it is difficult and expensive to develop transportation facilities. Hence, there is poor development of industries.
• Since all above factors are neither suitable for agricultural, industrial or trade development is not seen in the Himalayan region.

5. Answer the following questions in detail.

Question 1.
What is a region?
Answer:

• A region is an area of a country or any part of the world having common features. Thus, there is a difference between two regions on the basis of its features.
• The common features may be physical (relief, climate, soil, natural vegetation, etc.), socio¬cultural factors (language, religion, etc.), man-made like political features (administrative or political boundaries).
• Regions may be small or large. The large regions include sub-regions. For example, the plateau region of India has various sub-regions such as Chota Nagpur plateau, Malwa plateau, Marwar plateau, Deccan plateau, etc.

The main characteristics of regions are :

• Regions have its geographical location mentioned in latitudes and longitude.
• Every region has a specific boundary, the area of another region starts beyond that boundary.
• The extent of a region depends upon the homogeneity of a region.
• There are sub-regions in one region and therefore regions can be arranged into different orders on the basis of same features.

There are two types of regions – formal and functional regions.
1. A formal region is also known as uniform region or homogeneous region. It is an area in which everyone shares in common one or more distinctive characteristics. It can be common physical characteristic such as relief or climate of the region. It can be socio-cultural or economic activity such as common language. For example, rice producing regions.

2. In some formal regions, there are boundaries. Thus, district, state or country may be formal region. For example, USA and Canada are the examples of formal regions.

3. In functional regions characteristics are not important but functions are more important.

4. In one functional region, there may be many political or physical regions but due to one common function it is said to be functional. For example, Mumbai metropolitan or Pune metropolitan regions.

5. Functional regions act as a focal point that connects surrounding areas by transportation and communication. For example, Mumbai metropolitan city is well connected with its surrounding areas through transportation.

Question 2.
On what factors are the regions differentiated? Give examples.
Answer:
A region is an area of a country or any part of the world with same common features. Thus, regions are differentiated on the basis of physical or man-made or socio-cultural factors.

For example, Chota Nagpur plateau region has plateau relief. Himalayan mountain region has mountainous relief. The desert region of Rajasthan has shortage of rainfall and very high temperature. Northern plain region has plain relief, fertile soil and moderate climate.

The region may be differentiated on the basis of man-made factors like political factors such as administrative boundaries or political boundaries. For example, Uttar Pradesh and Punjab are two different regions because they have different political boundaries and government bodies.

The region may be differentiated on the basis of socio-cultural factors such as language, religion or ethnicity. For example, in Mumbai there is an area, where people speaking Gujarati language live in one region. The Middle East countries form one region where people of Muslim religion stay together.

Thus, regions are differentiated on the basis of various factors.

Question 3.
Per capita income is not the real indicator of regional development. Explain.
Answer:

• The per capita income is the income of per person in the country or region. It is the ratio between the country’s national income and total population.
• Per capital income is an index of development because more the per capita income, higher the standard of living of people. But it is not always true.
• The per capita income gives an idea about the average income of people in the country but it does not explain how income is distributed among the people. Some people may be rich and some may be below the poverty line.
• The per capita income is related to only economic aspect of the country but it does not take into account literacy rate, health, sex ratio, age structure, life expectancy etc.
• Sometimes rise in per capita income is due to increase in prices of commodities.
• If the population is low the per capita income will be high, but the regional development will be slow, because such regions have shortage of skilled labour supply for further development. Therefore, we can say that the per capita income is not the real indication of regional development.

6. Find the correlation between land under permanent crops and GDP given in the table 7.5 using Spearman’s Rank Correlation. Write the conclusion in your own words.

Answer:

X₁ = Land under permanent crops
Y₁ = GDP
R₁ = Rank – 1, R₂ = Rank – 2, n = 12 (No. of pairs)
Rank correlation R = 1 – \(\frac {6Σ(R_1-R_2)^2}{n(n^2-1)}\)
= 1 – \(\frac {6×118}{12(144-1)}\)
= 1 – \(\frac {708}{1716}\)
= 1 – 0.412
= 0.588
= 0.59

The answer of correction between (1) Land under permanent crops (2) GDP is 0.59 or +0.59 which means there is a positive correlation between these two variables.

It is not perfect positive correlation (+1.0). It is slightly less than this i.e. (+0.59).

It means when the amount of land under permanent crops is more, GDP is also more and when the amount of land under permanent crops is less, GDP is also less.

This is because the production of crops contributes to the growth of GDP.

The answer 0.59 suggests that this is not true for all countries as there are other factors, activities, products which are responsible for the growth of GDP other than the production of crops.

Class 12 Geography Chapter 7 Region and Regional Development Intext Questions and Answers

Try this

Question 1.
Given below are some geographical areas. Complete the table thinking about similarities between them and their types. Answer the questions that follow: (Textbook Page No. 66)
(i) While completing the above table how did you decide the common factor between these areas?
(ii) How did you differentiate one area from another?
(iii) Is the common factor the only basis of differentiation?
(iv) Make a list of characteristics which can be the basis of different area.
Answer:

(i) We decide on the basis of the function of each geographical area.
(ii) We differentiate one area from another on the basis of the type of physical, socio- cultural and political nature of the geographical areas.
(iii) Yes, the common factor is the only basis of differentiation.
(iv) Common basis is physical characteristics like river, desert, mountains, climate, soil, natural vegetation etc., and political characteristics like states, districts and cities and cultural characteristics like language, education, etc.

Question 2.
Here is the list of regions. Classify them into formal and functional regions. (Textbook Page No. 68)
Ujani Dam catchment area, area served by Pune Municipal Transport (PMT), area served by Citi Cable Service, Uttar Pradesh, District Kolhapur, Taluka Haveli, rice producing region of Konkan, Black cotton soil region of India, areas served by local trains of Mumbai, area served by a Primary Health Centre
Answer:

Question 3.
In the following table, regions are given. At the same time, some geographical factors and effects are also given. Complete the table 7.3 accordingly, with the help of solved examples. (Textbook Page No. 69 & 70)
Answer:

Question 4.
Look at the land use of the following regions in Table 7.5 and answer the questions that follow: (Textbook Page No. 72)
(i) What does the table show?
(ii) Which region has the highest land use under agriculture?
(iii) Which region has the highest GDP?
(iv) Which country (region) has more land use under forest?
(v) Which region has the highest land use under category ‘other land’? What’s its GDP?
(vi) Write your conclusion in few sentences.
Answer:
(i) The table shows land use in different regions.
(ii) The region C has the highest land use under agriculture.
(iii) The region I has the highest GDP.
(iv) The country (region) K has more land use under forest.
(iv) The region G has the highest land use under category ‘other land’. Its GDP is 1391.5 billion dollars.
(v) The region I is having highest GDP in which there is 50% arable land, it means agriculture must be major activity.

The region A is having lowest GDP where land under permanent meadows and pastures is highest. It means there is limited development of agricultural activities.

It is surprising, region F where arable land is highest in percentage even then GDP is comparatively low, because the land under permanent crop is very less.

In region K, GDP is 5747.49 and land under forest is 66%. It must be because the development of lumbering activities and forest-based industries is greater.

The region G has the highest area under other land use (95.6%) and GDP 1391.25. This region must have developed activities other than agriculture and forestry.

Use your brain power!

Question 1.
Can you tell what the identifying characteristic in the following regions?
(i) Pune Metropolitan Region
(ii) Nagpur hub
(iii) Dal Lake in Srinagar
(iv) Alleppey tourism centre
Answer:
(i) Pune Metropolitan Region – Functional region – Administration
(ii) Nagpur hub – Functional region – Industrial complex
(iii) Dal Lake in Srinagar – Functional region – Tourism
(iv) Alleppey tourism centre – Functional region – Tourism

Question 2.
(i) Identify your formal region.
(ii) Can you identify and demarcate your own functional region?
(iii) Which one is larger? (Textbook Page No. 68)
Answer:
(i) Maharashtra State
(ii) Yes, I can identify my functional region as government milk centre or D-mart departmental store.
(iii) Area served by D-mart departmental store is larger than the government milk centre.

Can you tell?

Question 1.
Read the table and answer the questions that follow. (Textbook Page No. 69)

(i) In which region do you think there are more resources available?
(ii) In which region do you think people are richer?
(iii) In which region do you think people may be happier? Why?
(iv) Do you think the given indicators or factors are enough to decide the standard of living of the people in the region?
(v) In which region would you like to stay? Why?
Answer:
(i) I think in region A more resources are available.
(ii) I think in region A people are richer.
(iii) People may be happier in region A because of higher per capita income, less population below poverty line, more urbanisation, more net sown area under irrigation.
(iv) I do not think the given indicator or factors are enough to decide the standard of living because other important factors like literacy rate, enrolment ratio / dropout ratio, sex ratio, crime rate etc., are not taken into consideration
(v) I will like to stay in region A because in this region per capita income is high, population below poverty line is low, percentage of urbanisation is highest, net sown and irrigated area is highest.

Question 2.
(i) Looking at the five stages of demographic transition theory, countries in which stage do you think, will be called developed?
(ii) Having looked at the various shapes of pyramids, which pyramid will represent a developed country? (Fig 2.1, Textbook Page No. 12)
(iii) After having answered both the questions above, what can you comment upon the role of population in regional development? (Textbook Page No. 70)
Answer:
(i) Countries in stage 4 and 5 will be called developed for e.g., USA, Sweden, Finland.
(ii) The pyramid (C) will represent a developed country.
(iii) The role of population in the regional development of any country is very important as man is a human resource. Quality of human resources is very important, which depends upon birth rate, death rate, expectancy of life, literacy ratio, male female ratio, mortality rate, etc.

A country having to qualitatively better population will help in the regional development of that country. For example, qualitatively population of the USA is better than population of India and hence, in USA there is better regional development than India.

Think about it

Question 1.
(i) Does development mean maximum utilisation of resources?
(ii) Does development consider environment as an indicator?
(iii) Can a region be called developed if it has sparse population or no population? (Textbook Page No. 69)
Answer:
(i) Development does not mean maximum utilisation of resources. Development of a region is the function of its resources and its population. There are many factors which affect the development of a region. For the holistic development of regions, skilled human resources and optimum utilisation of natural resources is necessary.

(ii) We must consider the environment as an indicator of development. Development at the cost of the environment is not real development. Since we ignore the environment while developing economy in the past 50 years, we are now suffering by many environment problems like global warming, etc.

(iii) Population is one of the most important factors in development. If there is sparse population or no population, then there will be shortage of man-power and resources will remain unutilized and therefore, quantity and quality of human resources is very important in the process of development.

Give it try

Question 1.
Draw a conclusion on the basis of the data given in table 7.4. Use Spearman’s Rank Correlation, find the correlation between the two variables. What can you comment about the development in this region? (Textbook Page No. 71)

Answer:

X₁ Percentage of malnourished children about to enter school.
Y₁ Government expenditure on education (% of total budget)
R₁ = Rank – 1, R₂ = Rank – 2, n = 8
Rank correlation R = 1 – \(\frac {6Σ(R_1-R_2)^2}{n(n^2-1)}\)
= 1 – \(\frac {6×104}{8(64-1)}\)
= 1 – \(\frac {624}{8(63)}\)
= 1 – \(\frac {624}{504}\)
= 1 – 1.24
= -0.24

The correlation between the percentage of malnourished children about to enter school and government expenditure on education is -0.24.
This value is very near to zero, which means no correlation. Hence, it is very slightly a negative correlation.

It means where the percentage of malnourished children about to enter school is more, the government expenditure on education is less e.g., in E region percentage of children is 31.7% but the expenditure is 4.09% and in C region percentage of children is only 3.4% but the expenditure is 7.10% (highest).

Considering the fact, we cannot consider this as well developed. There is no logical sequence in distribution of government expenditure on education. It is necessary that government authorities should take into consideration actual status of the region, need of the region before releasing grant to that region. This will surely help in proper allocation of the funds and overall development of the region.

Question 2.
Go through the fig. 7.2 and write two sentences on each factor. Tell how they will help in reducing regional imbalance. Also add some more to the list, if necessary. (Textbook Page No. 73)
Answer:
No country, in the world is having uniform regional development and thus, there is regional imbalance in development. Hence, a country must have a regional imbalance reduction strategy. Regional imbalance can be reverted through provision of public/services to backward areas by providing drinking water, education facilities and electricity to all.

It can also be reverted by making provision of infrastructure stimulus by allotment of special funds, development of special skills, good governance, support to infrastructural project and policy review and reform.

Question 1.
Look at the following data and answer the questions that follow: (Textbook Page No. 72)

(i) In which region is contribution of primary activities the least?
(ii) In which region is the contribution of primary activities the most?
(iii) In which region the contribution of tertiary activities is the most?
(iv) Which region has the highest HDI?
(v) Can you draw a conclusion on basis of the answers from Q (i) to Q (iv)?
Answer:
(i) In region A, the contribution of primary activities is the least.
(ii) In region E, the contribution of primary activities is the most.
(iii) In region B, the contribution of tertiary activities is the most.
(iv) In region A, there is highest HDI.
(v) We can conclude by saying that Human Development Index is high in countries where contribution of secondary and tertiary sector to GDP is high compared to the contribution of primary sector to GDP.

Region A, B, C and D have more than 30% contribution of secondary sector to GDP and more than 60% contribution of tertiary sector to GDP, it means these regions are developed regions.

The contribution of primary sector to GDP in region E is more but the contribution from secondary and tertiary sector is less compared to other regions, so region E is in the developing stage.

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Miscellaneous Exercise 8 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Miscellaneous Exercise 8

(I) Choose the correct alternative.

Question 1.
F(x) is c.d.f. of discreter r.v. X whose p.m.f. is given by P(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\), for x = 0, 1, 2, 3, 4 & P(x) = 0 otherwise then F(5) = __________
(a) \(\frac{1}{16}\)
(b) \(\frac{1}{8}\)
(c) \(\frac{1}{4}\)
(d) 1
Answer:
(d) 1

Question 2.
F(x) is c.d.f. of discrete r.v. X whose distribution is

then F(-3) = __________
(a) 0
(b) 1
(c) 0.2
(d) 0.15
Answer:
(a) 0

Question 3.
X : number obtained on uppermost face when a fair die is thrown then E(X) = __________
(a) 3.0
(b) 3.5
(c) 4.0
(d) 4.5
Answer:
(b) 3.5

Question 4.
If p.m.f. of r.v. X is given below.

then Var(X) = __________
(a) p²
(b) q²
(c) pq
(d) 2pq
Answer:
(d) 2pq

Question 5.
The expected value of the sum of two numbers obtained when two fair dice are rolled is __________
(a) 5
(b) 6
(c) 7
(d) 8
Answer:
(c) 7

Question 6.
Given p.d.f. of a continuous r.v. X as
f(x) = \(\frac{x^{2}}{3}\) for -1 < x < 2
= 0 otherwise then F(1) =
(a) \(\frac{1}{9}\)
(b) \(\frac{2}{9}\)
(c) \(\frac{3}{9}\)
(d) \(\frac{4}{9}\)
Answer:
(b) \(\frac{2}{9}\)

Question 7.
X is r.v. with p.d.f.
f(x) = \(\frac{k}{\sqrt{x}}\), 0 < x < 4
= 0 otherwise then E(X) = __________
(a) \(\frac{1}{3}\)
(b) \(\frac{4}{3}\)
(c) \(\frac{2}{3}\)
(d) 1
Answer:
(b) \(\frac{4}{3}\)

Question 8.
If X follows B(20, \(\frac{1}{10}\)) then E(X) = __________
(a) 2
(b) 5
(c) 4
(d) 3
Answer:
(a) 2

Question 9.
If E(X) = m and Var(X) = m then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(b) Possion distribution

Question 10.
If E(X) > Var(X) then X follows __________
(a) Binomial distribution
(b) Possion distribution
(c) Normal distribution
(d) none of the above
Answer:
(a) Binomial distribution

(II) Fill in the blanks.

Question 1.
The values of discrete r.v. are generally obtained by __________
Answer:
counting

Question 2.
The values of continuous r.v. are generally obtained by __________
Answer:
measurement

Question 3.
If X is dicrete random variable takes the values x₁, x₂, x₃, …… x_n then \(\sum_{i=1}^{n} p\left(x_{i}\right)\) = __________
Answer:
1

Question 4.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(\frac{x-1}{3}\) for x = 1, 2, 3, and p(x) = 0 otherwise then F(4) = __________
Answer:
1

Question 5.
If f(x) is distribution function of discrete r.v. X with p.m.f. p(x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, and p(x) = 0 otherwise then F(-1) = __________
Answer:
0

Question 6.
E(X) is considered to be __________ of the probability distribution of X.
Answer:
centre of gravity

Question 7.
If X is continuous r.v. and f(x_i) = P(X ≤ x_i) = \(\int_{-\infty}^{x_{i}} f(x) d x\) then f(x) is called __________
Answer:
Cumulative Distribution Function

Question 8.
In Binomial distribution probability of success ________ from trial to trial.
Answer:
remains constant/independent

Question 9.
In Binomial distribution, if n is very large and probability success of p is very small such that np = m (constant) then ________ distribution is applied.
Answer:
Possion

(III) State whether each of the following is True or False.

Question 1.
If P(X = x) = \(k\left(\begin{array}{l}
4 \\
x
\end{array}\right)\) for x = 0, 1, 2, 3, 4, then F(5) = \(\frac{1}{4}\) when f(x) is c.d.f.
Answer:
False

Question 2.

If F(x) is c.d.f. of discrete r.v. X then F(-3) = 0.
Answer:
True

Question 3.
X is the number obtained on the uppermost face when a die is thrown the E(X) = 3.5.
Answer:
True

Question 4.
If p.m.f. of discrete r.v.X is

then E(X) = 2p.
Answer:
True

Question 5.
The p.m.f. of a r.v. X is p(x) = \(\frac{2 x}{n(n+1)}\), x = 1, 2,……n
= 0 otherwise,
Then E(X) = \(\frac{2 n+1}{3}\)
Answer:
True

Question 6.
If f(x) = kx (1 – x) for 0 < x < 1
= 0 otherwise then k = 12
Answer:
False

Question 7.
If X ~ B(n, p) and n = 6 and P(X = 4) = P(X = 2) then p = \(\frac{1}{2}\).
Answer:
True

Question 8.
If r.v. X assumes values 1, 2, 3,………, n with equal probabilities then E(X) = \(\frac{(n+1)}{2}\)
Answer:
True

Question 9.
If r.v. X assumes the values 1, 2, 3,………, 9 with equal probabilities, E(X) = 5.
Answer:
True

(IV) Solve the following problems.

Part – I

Question 1.
Identify the random variable as discrete or continuous in each of the following. Identify its range if it is discrete.
(i) An economist is interested in knowing the number of unemployed graduates in the town with a population of 1 lakh.
Solution:
X = No. of unemployed graduates in a town.
∵ The population of the town is 1 lakh
∴ X takes finite values
∴ X is a Discrete Random Variable
∴ Range of = {0, 1, 2, 4, …. 1,00,000}

(ii) Amount of syrup prescribed by a physician.
Solution:
X : Amount of syrup prescribed.
∴ X Takes infinite values
∴ X is a Continuous Random Variable.

(iii) A person on a high protein diet is interested in the weight gained in a week.
Solution:
X : Gain in weight in a week.
X takes infinite values
∴ X is a Continuous Random Variable.

(iv) Twelve of 20 white rats available for an experiment are male. A scientist randomly selects 5 rats and counts the number of female rats among them.
Solution:
X : No. of female rats selected
X takes finite values.
∴ X is a Discrete Random Variable.
Range of X = {0, 1, 2, 3, 4, 5}

(v) A highway safety group is interested in the speed (km/hrs) of a car at a checkpoint.
Solution:
X : Speed of car in km/hr
X takes infinite values
∴ X is a Continuous Random Variable.

Question 2.
The probability distribution of a discrete r.v. X is as follows.

(i) Determine the value of k.
(ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).
Solution:
(i) Assuming that the given distribution is a p.m.f. of X
∴ Each P(X = x) ≥ 0 for x = 1, 2, 3, 4, 5, 6
k ≥ 0
ΣP(X = x) = 1 and
k + 2k + 3k + 4k + 5k + 6k = 1
∴ 21k = 1 ∴ k = \(\frac{1}{21}\)

(ii) P(X ≤ 4) = 1 – P(X > 4)
= 1 – [P(X = 5) + P(X = 6)]
= 1 – [latex]\frac{5}{21}+\frac{6}{21}[/latex]
= 1 – \(\frac{11}{21}\)
= \(\frac{10}{21}\)
P(2 < X < 6) = p(3) + p(4) + p(5)
= 3k + 4k + 5k
= \(\frac{3}{21}+\frac{4}{21}+\frac{5}{21}\)
= \(\frac{12}{21}\)
= \(\frac{4}{7}\)

(iii) P(X ≥ 3) = p(3) + p(4) + p(5) + p(6)
= 3k + 4k + 5k + 6k

Question 3.
Following is the probability distribution of an r.v. X.

Find the probability that
(i) X is positive.
(ii) X is non-negative.
(iii) X is odd.
(iv) X is even.
Solution:
(i) P(X is positive)
P(X = 0) = p(1) + p(2) + p(3)
= 0.25 + 0.15 + 0.10
= 0.50

(ii) P(X is non-negative)
P(X ≥ 0) = p(0) + p(1) + p(2) + p(3)
= 0.20 + 0.25 + 0.15 + 0.10
= 0.70

(iii) P(X is odd)
P(X = -3, -1, 1, 3)
= p(- 3) +p(-1) + p(1) + p(3)
= 0.05 + 0.15 + 0.25 + 0.10
= 0.55

(iv) P(X is even)
= 1 – P(X is odd)
= 1 – 0.55
= 0.45

Question 4.
The p.m.f of a r.v. X is given by
\(P(X=x)= \begin{cases}\left(\begin{array}{l}
5 \\
x
\end{array}\right) \frac{1}{2^{5}}, & x=0,1,2,3,4,5 . \\
0 & \text { otherwise }\end{cases}\)
Show that P(X ≤ 2) = P(X ≥ 3).
Solution:
For x = 0, 1, 2, 3, 4, 5

Question 5.
In the following probability distribution of an r.v. X

Find a and obtain the c.d.f. of X.
Solution:
Given distribution is p.m.f. of r.v. X
ΣP(X = x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) = 1

Question 6.
A fair coin is tossed 4 times. Let X denote the number of heads obtained. Identify the probability distribution of X and state the formula for p.m.f. of X.
Solution:
A fair coin is tossed 4 times
∴ Sample space contains 16 outcomes
Let X = Number of heads obtained
∴ X takes the values x = 0, 1, 2, 3, 4.
∴ The number of heads obtained in a toss is an even

Question 7.
Find the probability of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appearing in at least one toss.
Solution:
S : A die is tossed two times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
(i) X : No. is greater than 4
Range of X = {0, 1, 2}

(ii) X : Six appears on aleast one die.
Range of X = {0, 1, 2}

Question 8.
A random variable X has the following probability distribution.

Determine (i) k, (ii) P(X < 3), (iii) P(X > 6), (iv) P(0 < X < 3).
Solution:
(i) It is a p.m.f. of r.v. X
Σp(x) = 1
p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
9k + 10k² = 1
10k² + 9k – 1 = 0
10k² +10k – k – 1 = 0
∴ 10k(k + 1) – 1(k + 1) = 0
∴ (10k – 1) (k + 1) = 0
∴ 10k – 1 = 0r k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
k = -1 is not accepted, p(x) ≥ 0, ∀ x ∈ R
∴ k = \(\frac{1}{10}\)

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iii) P(X > 6) = p(7)
= 7k² + k
= \(7\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{7}{100}+\frac{1}{10}\)
= \(\frac{17}{100}\)

(iv) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

Question 9.
The following is the c.d.f. of a r.v. X.

Find the probability distribution of X and P(-1 ≤ X ≤ 2).
Solution:

P(-1 ≤ X ≤ 2) = p(-1) + p(0) + p(1) + p(2)
= 0.2 + 0.15 + 0.10 + 0.10
= 0.55

Question 10.
Find the expected value and variance of the r.v. X if its probability distribution is as follows.
(i)

Solution:

(ii)

Solution:
E(X) = Σx . p(x)

(iii)

Solution:

(iv)

Solution:

= 1.25
S.D. of X = σ_x = √Var(X)
= √1.25
= 1.118

Question 11.
A player tosses two coins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears, and ₹ 2 if no head appears. Find the expected value and variance of the winning amount.
Solution:
S : Two fair coin are tossed
S = {HH, HT, TT, TH}
n(S) = 4
∴ Range of X = {0, 1, 2}
∴ Let Y = amount received corresponds to values of X

Expected winning amount
E(Y) = Σpy = \(\frac{22}{4}\) = ₹ 5.5
V(Y) = Σpy² – (Σpy)²
= \(\frac{154}{4}\) – (5.5)²
= 38.5 – 30.25
= ₹ 8.25

Question 12.
Let the p.m.f. of the r.v. X be
\(p(x)= \begin{cases}\frac{3-x}{10} & \text { for } x=-1,0,1,2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate E(X) and Var(X).
Solution:

Question 13.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
We know that

Question 14.
The p.d.f. of the r.v. X is given by
\(f(x)= \begin{cases}\frac{1}{2 a} & \text { for } 0<x<2 a \\ 0 & \text { otherwise }\end{cases}\)
Show that P(X < \(\frac{a}{2}\)) = P(X > \(\frac{3a}{2}\))
Solution:

Question 15.
Determine k if
\(f(x)= \begin{cases}k e^{-\theta x} & \text { for } 0 \leq x<\infty, \theta>0 \\ 0 & \text { otherwise }\end{cases}\)
is the p.d.f. of the r.v. X. Also find P(X > \(\frac{1}{\theta}\)). Find M if P(0 < X < M) = \(\frac{1}{2}\)
Solution:
We know that

Question 16.
The p.d.f. of the r.v. X is given by
\(f_{x}(x)=\left\{\begin{array}{l}
\frac{k}{\sqrt{x}}, 0<x<4 \\
0, \text { otherwise }
\end{array}\right.\)
Determine k, c.d.f. of X and hence find P(X ≤ 2) and P(X ≥ 1).
Solution:
We know that

Question 17.
Let X denote the reaction temperature (in °C) of a certain chemical process. Let X be a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{1}{10}, & -5 \leq x \leq 5 \\ 0, & \text { otherwise }\end{cases}\)
Compute P(X < 0).
Solution:
Given p.d.f. is f(x) = \(\frac{1}{10}\), for -5 ≤ x ≤ 5
Let its c.d.f. F(x) be given by

Part – II

Question 1.
Let X ~ B(10, 0.2). Find (i) P(X = 1) (ii) P(X ≥ 1) (iii) P(X ≤ 8)
Solution:
X ~ B(10, 0.2)
n = 10, p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
(i) P(X = 1) = ¹⁰C₁ (0.2)¹ (0.8)⁹ = 0.2684

(ii) P(X ≥ 1) = 1 – P(X < 1)
= 1 – P(X = 0)
= 1 – ¹⁰C₀ (0.2)⁰ (0.8)¹⁰
= 1 – 0.1074
= 0.8926

(iii) P(X ≤ 8) = 1 – P(x > 1)
= 1 – [p(9) + p(10)]
= 1 – [¹⁰C₉ (0.2)⁹ (0.8)¹ + ¹⁰C₁₀ (0.2)¹⁰]
= 1 – 0.00000041984
= 0.9999

Question 2.
Let X ~ B(n, p) (i) If n = 10 and E(X) = 5, find p and Var(X), (ii) If E(X) = 5 and Var(X) = 2.5, find n and p.
Solution:
X ~ B(n, p)
(i) n = 10, E(X) = 5
∴ np = 5
∴ 10p = 5
∴ p = \(\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
V(X) = npq
= 10 × \(\frac{1}{2}\) × \(\frac{1}{2}\)
= 2.5

(ii) E(X) = 5, V(X) = 2.5
∴ np = 5, ∴ npq = 2.5
∴ 5q = 2.5
∴ q = \(\frac{2.5}{5}\) = 0.5, p = 1 – 0.5 = 0.5
But np = 5
∴ n(0.5) = 5
∴ n = 10

Question 3.
If a fair coin is tossed 4 times, find the probability that it shows (i) 3 heads, (ii) head in the first 2 tosses, and tail in the last 2 tosses.
Solution:
n : No. of times a coin is tossed
∴ n = 4
X : No. of heads
P : Probability of getting heads

Question 4.
The probability that a bomb will hit the target is 0.8. Find the probability that, out of 5 bombs, exactly 2 will miss the target.
Solution:
X : No. of bombs miss the target
p : Probability that bomb miss the target
∴ q = 0.8
∴ p = 1 – q = 1 – 0.8 = 0.2
n = No. of bombs = 5
∴ X ~ B(5, 0.2)
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁵C₂ (0.2)² (0.8)^5-2
= 10 × 0.04 × (0.8)³
= 10 × 0.04 × 0.512
= 0.4 × 0.512
= 0.2048

Question 5.
The probability that a lamp in the classroom will burn is 0.3. 3 lamps are fitted in the classroom. The classroom is unusable if the number of lamps burning in it is less than 2. Find the probability that the classroom can not be used on a random occasion.
Solution:
X : No. of lamps not burning
p : Probability that the lamp is not burning
∴ q = 0.3
∴ p = 1 – q = 1 – 0.3 = 0.7
n = No. of lamps fitted = 3
∴ X ~ B(3, 0.7)
∴ p(x) = ⁿC_x p^x q^n-x
P(classroom cannot be used)
P(X < 2) = p(0) + p(1)
= ³C₀ (0.7)⁰ (0.3)^3-0 + ³C₁ (0.7)¹ (0.3)^3-1
= 1 × 1 × (0.3)³ + 3 × 0.7 × (0.3)²
= (0.3)2 [0.3 + 3 × 0.7]
= 0.09 [0.3 + 2.1]
= 0.09 [2.4]
= 0.216

Question 6.
A large chain retailer purchases an electric device from the manufacturer. The manufacturer indicates that the defective rate of the device is 10%. The inspector of the retailer randomly selects 4 items from a shipment. Find the probability that the inspector finds at most one defective item in the 4 selected items.
Solution:
X : No. of defective items
n : No. of items selected = 4
p : Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
P(At most one defective item)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)^4-0 + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)3 [0.9 + 4 × 0.1]
= (0.9)3 × [0.9 + 0.4]
= 0.729 × 1.3
= 0.9477

Question 7.
The probability that a component will survive a check test is 0.6. Find the probability that exactly 2 of the next 4 components tested survive.
Solution:
p = 0.6, q = 1 – 0.6 = 0.4, n = 4
x = 2
∴ p(x) = ⁿC_x p^x q^n-x
P(X = 2) = ⁴C₂ (0.6)² (0.4)² = 0.3456

Question 8.
An examination consists of 5 multiple choice questions, in each of which the candidate has to decide which one of 4 suggested answers is correct. A completely unprepared student guesses each answer randomly. Find the probability that this student gets 4 or more correct answers.
Solution:
n : No. of multiple-choice questions
∴ n = 5
X : No. of correct answers
p : Probability of getting correct answer
∵ There are 4 options out of which one is correct
∴ p = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∵ X ~ B(5, \(\frac{1}{4}\))
∴ p(x) = ⁿC_x p^x q^n-x
P(Four or more correct answers)
P(X ≥ 4) = p(4) + p(5)

Question 9.
The probability that a machine will produce all bolts in a production run with in the specification is 0.9. A sample of 3 machines is taken at random. Calculate the probability that all machines will produce all bolts in a production run within the specification.
Solution:
n : No. of samples selected
∴ n = 3
X : No. of bolts produce by machines
p : Probability of getting bolts
∴ p = 0.9
∴ q = 1 – p = 1 – 0.9 = 0.1
∴ X ~ B(3, 0.9)
∴ p(x) = ⁿC_x p^x q^n-x
P(Machine will produce all bolts)
P(X = 3) = ³C₃ (0.9)³ (0.1)^3-3
= 1 × (0.9)³ × (0.1)⁰
= 1 × (0.9)³ × 1
= (0.9)³
= 0.729

Question 10.
A computer installation has 3 terminals. The probability that anyone terminal requires attention during a week is 0.1, independent of other terminals. Find the probabilities that (i) 0 (ii) 1 terminal requires attention during a week.
Solution:
n : No. of terminals
∴ n = 3
X : No. of terminals need attention
p : Probability of getting terminals need attention
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
∵ X ~ B(3, 0.1)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(No attention)
∴ P(X = 0) = ³C₀ × (0.1)⁰ (0.9)^3-1
= 1 × 1 × (0.9)³
= 0.729

(ii) P(One terminal need attention)
∴ P(X = 1) = ³C₁ (0.1)¹ (0.9)^3-1
= 3 × 0.1 × (0.9)²
= 0.3 × 0.81
= 0.243

Question 11.
In a large school, 80% of the students like mathematics. A visitor asks each of 4 students, selected at random, whether they like mathematics, (i) Calculate the probabilities of obtaining an answer yes from all of the selected students, (ii) Find the probability that the visitor obtains the answer yes from at least 3 students.
Solution:
X : No. of students like mathematics
p: Probability that students like mathematics
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
n : No. of students selected
∴ n = 4
∵ X ~ B(4, 0.8)
∴ p(x) = ⁿC_x p^x q^n-x
(i) P(All students like mathematics)
∴ P(X = 4) = ⁴C₄ (0.8)⁴ (0.2)^4-4
= 1 × (0.8)⁴ × (0.2)⁰
= 1 × (0.8)⁴ × 1
= 0.4096

(ii) P(Atleast 3 students like mathematics)
∴ P(X ≥ 3) = p(3) + p(4)
= ⁴C₃ (0.8)³ (0.2)^4-3 + 0.4096
= 4 × (0.8)³ (0.2)¹ + 0.4096
= 0.8 × (0.8)³ + 0.4096
= (0.8)⁴ × 0.4096
= 0.4096 + 0.4096
= 0.8192

Question 12.
It is observed that it rains on 10 days out of 30 days. Find the probability that
(i) it rains on exactly 3 days of a week.
(ii) it rains at most 2 days a week.
Solution:
X : No. of days it rains in a week
p : Probability that it rains
∴ p = \(\frac{10}{30}=\frac{1}{3}\)
∴ q = 1 – p = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
n : No. of days in a week
∴ n = 7
∴ X ~ B(7, \(\frac{1}{3}\))
(i) P(Rains on Exactly 3 days of a week)

(ii) P(Rains on at most 2 days of a week)
∴ P(X ≤ 2) = p(0) + p(1) + p(2)

Question 13.
If X follows Poisson distribution such that P(X = 1) = 0.4 and P(X = 2) = 0.2, find variance of X.
Solution:
X : Follows Possion Distribution

∴ m = 1
∴ Mean = m = Variance of X = 1

Question 14.
If X has Poisson distribution with parameter m, such that
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
find probabilities P(X = 1) and P(X = 2), when X follows Poisson distribution with m = 2 and P(X = 0) = 0.1353.
Solution:
Given that the random variable X follows the Poisson distribution with parameter m = 2
i.e. X ~ P(2)
Its p.m.f. is satisfying the given equation.
\(\frac{P(X=x+1)}{P(X=x)}=\frac{m}{x+1}\)
When x = 0,
\(\frac{\mathrm{P}(\mathrm{X}=1)}{\mathrm{P}(\mathrm{X}=0)}=\frac{2}{0+1}\)
P(X = 1) = 2P(X = 0)
= 2(0.1353)
= 0.2706
When x = 1,
\(\frac{\mathrm{P}(\mathrm{X}=2)}{\mathrm{P}(\mathrm{X}=1)}=\frac{2}{1+1}\)
P(X = 2) = P(X = 1) = 0.2706

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.4 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.4

Question 1.
If X has Poisson distribution with m = 1, then find P(X ≤ 1) given e^-1 = 0.3678.
Solution:
∵ m = 1
∵ X follows Poisson Distribution

= e^-m × 1 + e^-m × 1
= e^-1 + e^-1
= 2 × e^-1
= 2 × 0.3678
= 0.7356

Question 2.
If X ~ P(\(\frac{1}{2}\)), then find P(X = 3) given e^-0.5 = 0.6065.
Solution:

Question 3.
If X has Poisson distribution with parameter m and P(X = 2) = P(X = 3), then find P(X ≥ 2). Use e^-3 = 0.0497
Solution:
∵ X follows Poisson Distribution

Question 4.
The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter m = 4. Find the probability that the manager receives (i) only two complaints on a given day, (ii) at most two complaints on a given day. Use e^-4 = 0.0183.
Solution:
∵ m = 1
∵ X ~ P(m = 4)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
X = No. of complaints recieved
(i) P(Only two complaints on a given day)

(ii) P(Atmost two complaints on a given day)
P(X ≤ 2) = p(0) + p(1) + p(2)
= \(\frac{e^{-4} \times 4^{0}}{0 !}+\frac{e^{-4} \times 4^{1}}{1 !}\) + 0.1464
= e^-4 + e^-4 × 4 + 0.1464
= e^-4 [1 + 4] + 0.1464
= 0.0183 × 5 + 0.1464
= 0.0915 + 0.1464
= 0.2379

Question 5.
A car firm has 2 cars, which are hired out day by day. The number of cars hired on a day follows a Poisson distribution with a mean of 1.5. Find the probability that
(i) no car is used on a given day.
(ii) some demand is refused on a given day, given e^-1.5 = 0.2231.
Solution:
Let X = No. of demands for a car on any day
∴ No. of cars hired
n = 2
m = 1.5
∵ X ~ P(m = 1.5)

Question 6.
Defects on plywood sheets occur at random with an average of one defect per 50 sq. ft. Find the probability that such a sheet has (i) no defect, (ii) at least one defect. Use e^-1 = 0.3678.
Solution:
∵ X = No. of defects on a plywood sheet
∵ m = -1
∵ X ~ P(m = -1)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(No defect)
P(X = 0) = \(\frac{e^{-1} \times 1^{0}}{0 !}\)
= e^-1
= 0.3678

(ii) P(At least one defect)
P(X ≥ 1) = 1 – P(X < 1)
= 1 – p(0)
= 1 – 0.3678
= 0.6322

Question 7.
It is known that, in a certain area of a large city, the average number of rats per bungalow is five. Assuming that the number of rats follows Poisson distribution, find the probability that a randomly selected bungalow has
(i) exactly 5 rats
(ii) more than 5 rats
(iii) between 5 and 7 rats, inclusive. Given e^-5 = 0.0067.
Solution:
X = No. of rats
∵ m = 5
∴ X ~ P(m = 5)
∴ p(x) = \(\frac{e^{-m} \cdot m^{x}}{x !}\)
(i) P(Exactly five rats)

(ii) P(More than five rats)
P(X > 5) = 1 – P(X ≤ 5)

(iii) P(between 5 and 7 rats, inclusive)
P(5 ≤ x ≤ 7) = p(5) + p(6) + p(7)

= 0.0067 × 3125 × 0.02
= 0.0067 × 62.5
= 0.42

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.3 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.3

Question 1.
A die is thrown 4 times. If ‘getting an odd number’ is a success, find the probability of (i) 2 successes (ii) at least 3 successes (iii) at most 2 successes.
Solution:
X: Getting an odd no.
p: Probability of getting an odd no.
A die is thrown 4 times
∴ n = 4
∵ p = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
∵ X ~ B(3, \(\frac{1}{2}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(Two Successes)

(ii) P(Atleast 3 Successes)

(iii) P(Atmost 2 Successes)

Question 2.
A pair of dice is thrown 3 times. If getting a doublet is considered a success, find the probability of two successes.
Solution:
n: No. of times die is thrown = 3
X: No. of doublets
p: Probability of getting doublets
Getting a doublet means, same no. is obtained on 2 throws of a die
There are 36 outcomes
No. of doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Question 3.
There are 10% defective items in a large bulk of items. What is the probability that a sample of 4 items will include not more than one defective item?
Solution:
n: No of sample items = 4
X: No of defective items
p: Probability of getting defective items
∴ p = 0.1
∴ q = 1 – p = 1 – 0.1 = 0.9
X ~ B(4, 0.1)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} \mathrm{q}^{n-x}\)
P(Not include more than 1 defective)
P(X ≤ 1) = p(0) + p(1)
= ⁴C₀ (0.1)⁰ (0.9)⁴ + ⁴C₁ (0.1)¹ (0.9)^4-1
= 1 × 1 × (0.9)⁴ + 4 × 0.1 × (0.9)³
= (0.9)³ [0.9 + 0.4]
= (0.9)³ × 1.3
= 0.977

Question 4.
Five cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability that (i) all the five cards are spades, (ii) only 3 cards are spades, (iii) none is a spade.
Solution:
X: No. of spade cards
Number of cards drawn
∴ n = 5
p: Probability of getting spade card

(i) P(All five cards are spades)

(ii) P(Only 3 cards are spades)

(iii) P(None is a spade)

Question 5.
The probability that a bulb produced by a factory will use fuse after 200 days of use is 0.2. Let X denote the number of bulbs (out of 5) that fuse after 200 days of use. Find the probability of (i) X = 0, (ii) X ≤ 1, (iii) X > 1, (iv) X ≥ 1.
Solution:
X : No. of bulbs fuse after 200 days of use
p : Probability of getting fuse bulbs
No. of bulbs in a sample
∴ n = 5
∴ p = 0.2
∴ q = 1 – p = 1 – 0.2 = 0.8
∵ X ~ B(5, 0.2)
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
(i) P(X = 0) = ⁵C₀ (0.2)⁰ (0.8)^5-0
= 1 × 1 × (0.8)⁵
= (0.8)⁵

(ii) P(X ≤ 1) = p(0) + p(1)
= ⁵C₀ (0.2)⁰ (0.8)^5-0 + ⁵C₁ (0.2)¹ (0.8)^5-1
= 1 × 1 × (0.8)⁵ + 5 × 0.2 × (0.8)⁴
= (0.8)⁴ [0.8 + 1]
= 1.8 × (0.8)⁴

(iii) P(X > 1) = 1 – [p(0) + p(1)]
= 1 – 1.8 × (0.8)⁴

(iv) P(X ≥ 1) = 1 – p(0)
= 1 – (0.8)⁵

Question 6.
10 balls are marked with digits 0 to 9. If four balls are selected with replacement. What is the probability that none is marked 0?
Solution:
X : No. of balls drawn marked with the digit 0
n : No. of balls drawn
∴ n = 4
p : Probability of balls marked with 0.
∴ p = \(\frac{1}{10}\)
∴ q = 1 – p = 1 – \(\frac{1}{10}\) = \(\frac{9}{10}\)
p(x) = \({ }^{n} C_{x} p^{x} q^{n-x}\)
P(None of the ball is marked with digit 0)

Question 7.
In a multiple-choice test with three possible answers for each of the five questions, what is the probability of a candidate getting four or more correct answers by random choice?
Solution:
n: No. of Questions
∴ n = 5
X: No. of correct answers by guessing
p: Probability of getting correct answers

Question 8.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
X : No. of sixes in 6 throws
n : No. of times dice thrown
∴ n = 6
p : Probability of getting six
∴ p = \(\frac{1}{6}\)
∴ q = 1 – p = 1 – \(\frac{1}{6}\) = \(\frac{5}{6}\)
∵ X ~ B(6, \(\frac{1}{6}\))
∴ p(x) = \({ }^{n} \mathrm{C}_{x} p^{x} q^{n-x}\)
P(At most 2 sixes)
P(X ≤ 2) = p(0) + p(1) + p(2)

Question 9.
Given that X ~ B(n, p),
(i) if n = 10 and p = 0.4, find E(X) and Var(X).
(ii) if p = 0.6 and E(X) = 6, find n and Var(X).
(iii) if n = 25, E(X) = 10, find p and Var(X).
(iv) if n = 10, E(X) = 8, find Var(X).
Solution:
∵ X ~ B (n, p), E(X) = np, V(X) = npq, q = 1 – p
(i) E(X) = np = 10 × 0.4 = 4
∵ q = 1 – p = 1 – 0.4 = 0.6
V(X) = npq = 10 × 0.4 × 0.6 = 2.4

(ii) ∵ p = 0.6
∴ q = 1 – p = 1 – 0.6 = 0.4
E(X) = np
∴ 6 = n × 0.6
∴ n = 10
∴ V(X) = npq = 10 × 0.6 × 0.4 = 2.4

(iii) E(X) = np
∴ 10 = 25 × p
∴ p = 0.4
∴ q = 1, p = 1 – 0.4 = 0.6
∴ S.D.(X) = √V(X)
= \(\sqrt{n p q}\)
= \(\sqrt{25 \times 0.4 \times 0.6}\)
= √6
= 2.4494

(iv) ∵ E(X) = np
∴ 8 = 10p
∴ p = 0.8
∴ q = 1 – p = 1 – 0.8 = 0.2
∵ V(X) = npq = 10 × 0.8 × 0.2 = 1.6

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.2 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.2

Question 1.
Check whether each of the following is p.d.f.
(i) \(f(x)= \begin{cases}x & \text { for } 0 \leq x \leq 1 \\ 2-x & \text { for } 1<x \leq 2\end{cases}\)
Solution:
Given function is
f(x) = x, 0 ≤ x ≤ 1
Each f(x) ≥ 0, as x ≥ 0.


∴ The given function is a p.d.f. of x.

(ii) f(x) = 2 for 0 < x < 1
Solution:
Given function is
f(x) = 2 for 0 < x < 1 Each f(x) > 0,

∴ The given function is not a p.d.f.

Question 2.
The following is the p.d.f. of a r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
Find (i) P(X < 1.5), (ii) P(1 < X < 2), (iii) P(X > 2)
Solution:

Question 3.
It is felt that error in measurement of reaction temperature (in Celsius) in an experiment is a continuous r.v. with p.d.f.
\(f(x)= \begin{cases}\frac{x^{3}}{64} & \text { for } 0 \leq x \leq 4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Verify whether f(x) is a p.d.f.
(ii) Find P(0 < X ≤ 1).
(iii) Find the probability that X is between 1 and 3.
Solution:
(i) f(x) is p.d.f. of r.v. X if
(a) f(x) ≥ 0, ∀ x ∈ R
(b) \(\int_{0}^{4} f(x) d x\) = 1

Question 4.
Find k, if the following function represents the p.d.f. of a r.v. X.
(i) \(f(x)= \begin{cases}k x & \text { for } 0<x<2 \\ 0 & \text { otherwise }\end{cases}\)
Also find P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)]
Solution:

(ii) \(f(x)= \begin{cases}k x(1-x) & \text { for } 0<x<1 \\ 0 & \text { otherwise }\end{cases}\)
Also find (a) P[\(\frac{1}{4}\) < X < \(\frac{1}{2}\)], (b) P[X < \(\frac{1}{2}\)]
Solution:
We know that

Question 5.
Let X be the amount of time for which a book is taken out of the library by a randomly selected student and suppose that X has p.d.f.
\(f(x)= \begin{cases}0.5 x & \text { for } 0 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Calculate (i) P(X ≤ 1), (ii) P(0.5 ≤ X ≤ 1.5), (iii) P(X ≥ 1.5).
Solution:
Given p.d.f. of X is f(x) = 0.5x for 0 ≤ x ≤ 2
∴ Its c.d.f. F(x) is given by

(i) P(X < 1) = F(1)
= 0.25(1)²
= 0.25

(ii) P(0.5 < X < 1.5) = F(1.5) – F(0.5)
= 0.25(1.5)² – 0.25(0.5)²
= 0.25[2.25 – 0.25]
= 0.25(2)
= 0.5

(iii) P(X ≥ 1.5) = 1 – P(X ≤ 1.5)
= 1 – F(1.5)
= 1 – 0.25(1.5)²
= 1 – 0.25(2.25)
= 1 – 0.5625
= 0.4375

Question 6.
Suppose X is the waiting time (in minutes) for a bus and its p.d.f. is given by
\(f(x)=\left\{\begin{array}{cl}
\frac{1}{5} & \text { for } 0 \leq x \leq 5 \\
0 & \text { otherwise }
\end{array}\right.\)
Find the probability that (i) waiting time is between 1 and 3 minutes, (ii) waiting time is more than 4 minutes.
Solution:
p.d.f. of r.v. X is given by
f(x) = \(\frac{1}{5}\) for 0 ≤ x ≤ 5
This is a constant function.
(i) Probability that waiting time X is between 1 and 3 minutes

(ii) Probability that waiting time X is more than 4 minutes

Question 7.
Suppose error involved in making a certain measurement is a continuous r.v. X with p.d.f.
\(f(x)= \begin{cases}k\left(4-x^{2}\right) & \text { for }-2 \leq x \leq 2 \\ 0 & \text { otherwise }\end{cases}\)
Compute (i) P(X > 0), (ii) P(-1 < X < 1), (iii) P(X < -0.5 or X > 0.5)
Solution:
Since given f(x) is a p.d.f. of r.v. X
Since -2 ≤ x ≤ 2
∴ x² ≤ 4
∴ 4 – x² ≥ 0
∴ k(4 – x²) ≥ 0
∴ k ≥ 0 [∵ f(x) ≥ 0]

Question 8.
Following is the p.d.f. of a continuous r.v. X.
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) Find an expression for the c.d.f. of X.
(ii) Find F(x) at x = 0.5, 1.7, and 5.
Solution:
The p.d.f. of a continuous r.v. X is
\(f(x)= \begin{cases}\frac{x}{8} & \text { for } 0<x<4 \\ 0 & \text { otherwise }\end{cases}\)
(i) c.d.f. of continuous r.v. X is given by

(ii) F(0.5) = \(\frac{(0.5)^{2}}{16}=\frac{0.25}{16}=\frac{1}{64}\) = 0.015
F(1.7) = \(\frac{(1.7)^{2}}{16}=\frac{2.89}{16}\) = 0.18
For any of x greater than or equal to 4, F(x) = 1
∴ F(5) = 1

Question 9.
The p.d.f. of a continuous r.v. X is
\(f(x)=\left\{\begin{array}{cl}
\frac{3 x^{2}}{8} & \text { for } 0<x<2 \\
0 & \text { otherwise }
\end{array}\right.\)
Determine the c.d.f. of X and hence find (i) P(X < 1), (ii) P(X < -2), (iii) P(X > 0), (iv) P(1 < X < 2).
Solution:
The p.d.f. of a continuous r.v. X is

Question 10.
If a r.v. X has p.d.f.
\(f(x)= \begin{cases}\frac{c}{x} & \text { for } 1<x<3, c>0 \\ 0 & \text { otherwise }\end{cases}\)
Find c, E(X) and V(X). Also find f(x).
Solution:
The p.d.f. of r.v. X is
f(x) = \(\frac{c}{x}\), 1 < x < 3, c > 0
For p.d.f. of X, we have

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 8 Probability Distributions Ex 8.1 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 8 Probability Distributions Ex 8.1

Question 1.
Let X represent the difference between a number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X?
Solution:
∵ A coin is tossed 6 times
S = {6H and 0T, 5H and 1T, 4H and 2T, 3H and 3T, 2H and 4T, 1H and 5T, 0H and 6T}
X: Difference between no. of heads and no. of tails.
X = 6 – 0 = 6
X = 5 – 1 = 4
X = 4 – 2 = 2
X = 3 – 3 = 0
X = 2 – 4 = -2
X = 1 – 5 = -4
X = 0 – 6 = -6
X = {-6, -4, -2, 0, 2, 4, 6}

Question 2.
An urn contains 5 red and 2 black balls. Two balls are drawn at random. X denotes the number of black balls drawn. What are the possible values of X?
Solution:
S : Two bolts are drawn from the Urn
S = {RR, RB, BR, BB}
X : No. of black balls
∴ X = {0, 1, 2}

Question 3.
Determine whether each of the following is a probability distribution. Give reasons for your answer.
(i)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.4 + 0.4 + 0.2
= 1
∴ The function is a p.m.f.

(ii)

Solution:
Here, p(3) = -0.1 < 0
∴ P(X = x) ≯ 0, ∀ x
∴ The function is not a p.m.f.

(iii)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{2} \mathrm{P}(\mathrm{X}=x)\) = p(0) + p(1) + p(2)
= 0.1 + 0.6 + 0.3
= 1
∴ The function is a p.m.f.

(iv)

Solution:
Here, P(Z = z) ≥ 0, ∀ z and
\(\sum_{x=-1}^{3} \mathrm{P}(\mathrm{Z}=z)\) = p(-1) + p(0) + p(1) + p(2) + p(3)
= 0.05 + 0 + 0.4 + 0.2 + 0.3
= 0.95
≠ 1
∴ The function is not a p.m.f.

(v)

Solution:
Here, P(Y = y) ≥ 0, ∀ y and
\(\sum_{x=-1}^{2} \mathrm{P}(\mathrm{Y}=y)\) = p(-1) + p(0) + p(1)
= 0.1 + 0.6 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

(vi)

Solution:
Here, P(X = x) ≥ 0, ∀ x and
\(\sum_{x=0}^{0} \mathrm{P}(X=x)\) = p(-2) + p(-1) + p(0)
= 0.3 + 0.4 + 0.2
= 0.9
≠ 1
∴ The function is not a p.m.f.

Question 4.
Find the probability distribution of
(i) number of heads in two tosses of a coin,
(ii) number of trails in three tosses of a coin,
(iii) number of heads in four tosses of a coin.
Solution:
(i) S: Coin is tossed two times
S = {HH, HT, TH, TT}
n(S) = 4
X: No. of heads
Range of X = {0, 1, 2}
p.m.f. Table

(ii) S: 3 coin are tossed
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
X: No. of heads
Range of X = {0, 1, 2, 3}
p.m.f. Table

(iii) S: Four coin are tossed
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
n(S) = 16
X: No. of heads
Range of X = {0, 1, 2, 3, 4}
p.m.f. Table

Question 5.
Find the probability distribution of the number of successes in two tosses of a die if successes are defined as getting a number greater than 4.
Solution:
S = A die is tossed 2 times
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36
X = No. getting greater than 4
Range of X = {0, 1, 2}
p(0) = \(\frac{16}{36}=\frac{4}{9}\)
p(1) = \(\frac{16}{36}=\frac{4}{9}\)
p(2) = \(\frac{4}{36}=\frac{1}{9}\)

Question 6.
A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.
Solution:
Total no. of bulbs = 30
No. of defective bulbs = 6
A sample of 4 bulbs are drawn from 30 bulbs.
∴ n(S) = \({ }^{30} \mathrm{C}_{4}\)
∴ No. of non-defective bulbs = 24
Let X = No. of defective bulbs drawn in sample of 4 bulbs.

Question 7.
A coin is biased so that the head is 3 times as likely to occur as the tail. Find the probability distribution of a number of tails in two tosses.
Solution:
Here, the head is 3 times as likely to occur as the tail.
i.e., If 4 times coin is tossed, 3 times there will be a head and 1 time there will be the tail.
∴ p(H) = \(\frac{3}{4}\) and p(T) = \(\frac{1}{4}\)
Let X : No. of tails in two tosses.
And coin is tossed twice.
∴ X = {0, 1, 2}
For X = 0,
p(0) = p(both heads)
= p(H) × p(H)
= \(\frac{3}{4} \times \frac{3}{4}\)
= \(\frac{9}{16}\)
For X = 1,
p(1) = p(HT or TH)
= p(HT) + p(TH)
= p(H) × p(T) + p(T) × p(H)
= \(\frac{3}{4} \times \frac{1}{4}+\frac{1}{4} \times \frac{3}{4}\)
= \(\frac{6}{16}\)
For X = 2,
p(2) = p(both tails)
= p(T) × p(T)
= \(\frac{1}{4} \times \frac{1}{4}\)
= \(\frac{1}{16}\)
The probability distribution of the number of tails in two tosses is

Question 8.
A random variable X has the following probability distribution:

Determine (i) k, (ii) P(X < 3), (iii) P(0 < X < 3), (iv) P(X > 4).
Solution:
(i) It is a p.m.f. of r.v. X
∴ Σp(x) = 1
∴ p(1) + p(2) + p(3) + p(4) + p(5) + p(6) + p(7) = 1
∴ k + 2k + 2k + 3k + k² + 2k² + (7k² + k) = 1
∴ 10k² + 9k = 1
∴ 10k² + 9k – 1 = 0
∴ 10k² + 10k – k – 1 = 0
∴ 10k(k + 1) – (k + 1) = 0
∴ (10k – 1)(k + 1) = 0
∴ 10k – 1 = 0 or k + 1 = 0
∴ k = \(\frac{1}{10}\) or k = -1
but k = -1 is not accepted
∴ k = \(\frac{1}{2}\) is accepted

(ii) P(X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iii) P(0 < X < 3) = p(1) + p(2)
= k + 2k
= 3k
= 3 × \(\frac{1}{10}\)
= \(\frac{3}{10}\)

(iv) P(X > 4) = p(5) + p(6) + p(7)
= k² + 2k² + (7k² + k)
= 10k² + k
= \(10\left(\frac{1}{10}\right)^{2}+\frac{1}{10}\)
= \(\frac{2}{10}\)
= \(\frac{1}{5}\)

Question 9.
Find expected value and variance of X using the following p.m.f.

Solution:

E(X) = Σxp = -0.05
V(X) = Σx²p – (Σxp)²
= 2.25 – (-0.05)²
= 2.25 – 0.0025
= 2.2475

Question 10.
Find expected value and variance of X, the number on the uppermost face of a fair die.
Solution:
S : A fair die is thrown
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
X: No obtained on uppermost face of die
Range of X = {1, 2, 3, 4, 5, 6}

E(X) = Σxp = \(\frac{21}{6}=\frac{7}{2}\) = 3.5
V(X) = Σx²p – (Σxp)²
= \(\frac{91}{6}\) – (3.5)²
= 15.17 – 12.25
= 2.92

Question 11.
Find the mean of the number of heads in three tosses of a fair coin.
Solution:
S : A coin is tossed 3 times
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Range of X = {0, 1, 2, 3}

∴ Mean = E(X) = Σxp = \(\frac{12}{8}=\frac{3}{2}\) = 1.5

Question 12.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
S : Two dice are thrown
S = {(1, 1), (1, 2), (1, 3), ……, (6, 6)}
n(S) = 36
Range of X = {0, 1, 2}
First 6 positive integers are 1, 2, 3, 4, 5, 6
X = Larger two numbers selected
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(S) = 36

∴ E(X) = Σxp = \(\frac{12}{36}=\frac{1}{3}\)

Question 13.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers. Find E(X).
Solution:
First 6 positive integers are 1, 2, 3, 4, 5, 6
X : The larger of the selected two numbers
S = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
n(S) = 30

E(X) = Σxp = \(\frac{140}{30}=\frac{14}{3}\) = 4.67

Question 14.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance of X.
Solution:
S : Two fair dice are rolled
S = {(1, 1), (1, 2), (1, 4), ……, (6, 6)}
n(S) = 36
X : Sum of the two numbers.
Range of X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}


V(X) = Σx²p – (Σxp)²
= \(\frac{1952}{36}-\left(\frac{252}{36}\right)^{2}\)
= 54.22 – (7)²
= 5.22
SD(X) = √V(X) = √5.22 = 2.28

Question 15.
A class has 15 students whose ages are 14, 17, 15, 14, 21, 17, 19, 20, 16, 18, 20, 17, 16, 19 and 20 years. If X denotes the age of a randomly selected student, find the probability distribution of X. Find the mean and variance of X.
Solution:

Question 16.
70% of the member’s favour and 30% oppose a proposal in a meeting. The random variable X takes the value 0 if a member opposes the proposal and the value 1 if a member is in favour. Find E(X) and V(X).
Solution:

E(X) = Σxp = 0.7
V(X) = Σx²p – (Σxp)²
= 0.7 – (0.7)²
= 0.7 – 0.49
= 0.21

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7 Questions and Answers.

Maharashtra State Board 12th Commerce Maths Solutions Chapter 7 Assignment Problem and Sequencing Miscellaneous Exercise 7

(I) Choose the correct alternative.

Question 1.
In sequencing, an optimal path is one that minimizes ___________
(a) Elapsed time
(b) Idle time
(c) Both (a) and (b)
(d) Ready time
Answer:
(c) Both (a) and (b)

Question 2.
If job A to D have processing times as 5, 6, 8, 4 on first machine and 4, 7, 9, 10 on second machine then the optimal sequence is:
(a) CDAB
(b) DBCA
(c) BCDA
(d) ABCD
Answer:
(b) DBCA

Question 3.
The objective of sequence problem is
(a) to find the order in which jobs are to be made
(b) to find the time required for the completing all the job on hand
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs
(d) to maximize the cost
Answer:
(c) to find the sequence in which jobs on hand are to be processed to minimize the total time required for processing the jobs

Question 4.
If there are n jobs and m machines, then there will be ___________ sequences of doing the jobs.
(a) mn
(b) m(n!)
(c) n^m
(d) (n!)^m
Answer:
(d) (n!)^m

Question 5.
The Assignment Problem is solved by
(a) Simple method
(b) Hungarian method
(c) Vector method
(d) Graphical method
Answer:
(b) Hungarian method

Question 6.
In solving 2 machine and n jobs sequencing problem, the following assumption is wrong
(a) No passing is allowed
(b) Processing times are known
(c) Handling times is negligible
(d) The time of passing depends on the order of machining
Answer:
(d) The time of passing depends on the order of machining

Question 7.
To use the Hungarian method, a profit maximization assignments problem requires
(a) Converting all profit to opportunity losses
(b) A dummy person or job
(c) Matrix expansion
(d) Finding the maximum number of lines to cover all the zeros in the reduced matrix
Answer:
(a) Converting all profits to opportunity losses

Question 8.
Using the Hungarian method the optimal assignment obtained for the following assignment problem to minimize the total cost is:

(a) 1 – C, 2 – B, 3 – D, 4 – A
(b) 1 – B, 2 – C, 3 – A, 4 – D
(c) 1 – A, 2 – B, 3 – C, 4 – D
(d) 1 – D, 2 – A, 3 – B, 4 – C
Answer:
(a) 1 – C, 2 – B, 3 – D, 4 – A

Question 9.
The assignment problem is said to be unbalanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 10.
The assignment problem is said to be balanced if
(a) Number of rows is greater than the number of columns
(b) Number of rows is lesser than number of columns
(c) Number of rows is equal to the number of columns
(d) If the entry of rows is zero
Answer:
(c) Number of rows is equal to number of columns

Question 11.
The assignment problem is said to be balanced if it is a
(a) Square matrix
(b) Rectangular matrix
(c) Unit matrix
(d) Triangular matrix
Answer:
(a) Square matrix

Question 12.
In an assignment problem if the number of rows is greater than the number of columns then
(a) Dummy column is added
(b) Dummy row is added
(c) Row with cost 1 is added
(d) Column with cost 1 is added
Answer:
(a) Dummy column is added

Question 13.
In a 3 machine and 5 jobs problem, the least of processing times on machines A, B, and C are 5, 1 and 3 hours and the highest processing times are 9, 5 and 7 respectively, then it can be converted to a 2 machine problem if the order of the machines is:
(a) B – A – C
(b) A – B – C
(c) C – B – A
(d) Any order
Answer:
(b) A – B – C

Question 14.
The objective of an assignment problem is to assign
(a) Number of jobs to equal number of persons at maximum cost
(b) Number of jobs to equal number of persons at minimum cost
(c) Only the maximize cost
(d) Only to minimize cost
Answer:
(b) Number of jobs to equal number of persons at minimum cost

(II) Fill in the blanks.

Question 1.
An assignment problem is said to be unbalanced when ___________
Answer:
the number of rows is not equal to the number of columns

Question 2.
When the number of rows is equal to the Number of columns then the problem is said to be ___________ assignment problem.
Answer:
balanced

Question 3.
For solving assignment problem the matrix should be a ___________
Answer:
square matrix

Question 4.
If the given matrix is not a ___________ matrix, the assignment problem is called an unbalanced problem.
Answer:
square

Question 5.
A dummy row(s) or column(s) with the cost elements as ___________ the matrix of an unbalanced assignment problem as a square matrix.
Answer:
zero

Question 6.
The time interval between starting the first job and completing the last, job including the idle time (if any) in a particular order by the given set of machines is called ___________
Answer:
Total elapsed time

Question 7.
The time for which a machine j does not have a job to process to the start of job i is called ___________
Answer:
Idle time

Question 8.
The maximization assignment problem is transformed to minimization problem by subtracting each entry in the table from the ___________ value in the table.
Answer:
maximum

Question 9.
When the assignment problem has more than one solution, then it is ___________ optimal solution.
Answer:
multiple

Question 10.
The time required for printing four books A, B, C, and D is 5, 8, 10, and 7 hours. While its data entry requires 7, 4, 3, and 6 hrs respectively. The sequence that minimizes total elapsed time is ___________
Answer:
A – D – B – C

(III) State whether each of the following is True or False.

Question 1.
One machine – one job is not an assumption in solving sequencing problems.
Answer:
False

Question 2.
If there are two least processing times for machine A and machine B, priority is given for the processing time which has the lowest time of the adjacent machine.
Answer:
True

Question 3.
To convert the assignment problem into a maximization problem, the smallest element in the matrix is deducted from all other elements.
Answer:
False

Question 4.
The Hungarian method operates on the principle of matrix reduction, whereby the cost table is reduced to a set of opportunity costs.
Answer:
True

Question 5.
In a sequencing problem, the processing times are dependent on the order of processing the jobs on machines.
Answer:
False

Question 6.
The optimal assignment is made in the Hungarian method to cells in the reduced matrix that contain a Zero.
Answer:
True

Question 7.
Using the Hungarian method, the optimal solution to an assignment problem is fund when the minimum number of lines required to cover the zero cells in the reduced matrix equals the number of people.
Answer:
True

Question 8.
In an assignment problem, if a number of columns are greater than the number of rows, then a dummy column is added.
Answer:
False

Question 9.
The purpose of a dummy row or column in an assignment problem is to obtain a balance between a total number of activities and a total number of resources.
Answer:
True

Question 10.
One of the assumptions made while sequencing n jobs on 2 machines is: two jobs must be loaded at a time on any machine.
Answer:
False

(IV) Solve the following problems.

Part – I

Question 1.
A plant manager has four subordinates, and four tasks to be performed. The subordinates differ in efficiency and the tasks differ in their intrinsic difficulty. This estimate of the times each man would take to perform each task is given in the effectiveness matrix below.

How should the tasks be allocated, one to a man, as to minimize the total man-hours?
Solution:
The hr matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum no. of lines covering ail the zeros (4) is equal to the order of the matrix (4)
∴ The assignment is possible.

The assignment is
A → I, B → III, C → II, D → IV
For the minimum hrs. take the corresponding value from the hr matrix.
Minimum hrs = 7 + 3 + 18 + 9 = 37 hrs

Question 2.
A dairy plant has five milk tankers, I, II, III, IV & V. These milk tankers are to be used on five delivery routes A, B, C, D & E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.

How should the milk tankers be assigned to the chilling centre so as to minimize the distance travelled?
Solution:
The distance matrix is given by

Subtracting row minimum from all values in that row we get

Subtracting column minimum from each value in that column we get

The number of lines covering all the zeros (3) is less than the order of the matrix (5) so the assignment is not possible. The modification is required.
The minimum uncovered value (15) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same. We get

The minimum lines covering all the zeros (4) are less than the order of the matrix (5) so the assignment is not possible. The modification is required the minimum uncovered value (5) is subtracted from uncovered values and added to the values at the intersection. The numbers on the lines remain the same we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) So assignment is possible.
The assignment is
A → II, B → III, C → V, D → I, E → IV
Total minimum distance is = 120 + 120 + 175 + 40 + 70 = 525 kms.

Question 3.
Solve the following assignment problem to maximize sales:

Solution:
As it is a maximization problem so we need to convert it into a minimization problem.
Subtracting all the values from the maximum value (19) we get

Also, it is an unbalanced problem so we need to add a dummy row (E) with all values zero, we get

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get the same matrix

The minimum number of lines covering all the zero (4) is less than the order of the matrix (5) So assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered value and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

The assignment is
A → V, B → II, C → IV, D → III, E → I
No salesman goes to I as E is a dummy row.
For the maximum value take the corresponding values from the original matrix.
We get Maximum value = 15 + 19 + 14 + 17 + 0 = 65 units

Question 4.
The estimated sales (tons) per month in four different cities by five different managers are given below:

Find out the assignment of managers to cities in order to maximize sales.
Solution:
This is a maximizing problem. To convert it into minimizing problem subtract all the values of the matrix from the maximum (largest) value (39) we get

Also as it is an unbalanced problem so we have to add a dummy column (T) with all the values as zero. We get

Subtracting row minimum from all values in that row we get the same matrix
Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so assignments are not possible. The modification is required. The minimum uncovered value (1) is subtracted from the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So I → S, II → T, III → Q, IV → P, V → R.
As T is dummy manager II is not given any city.
To find the maximum sales we take the corresponding value from the original matrix
Total maximum sales = 35 + 39 + 36 + 35 = 145 tons

Question 5.
Consider the problem of assigning five operators to five machines. The assignment costs are given in the following table.

Operator A cannot be assigned to machine 3 and operator C cannot be assigned to machine 4. Find the optimal assignment schedule.
Solution:
This is a restricted assignment problem, so we assign a very high cost (oo) to the prohibited cells we get

Subtracting row minimum from all values in that row we get.

Subtracting column minimum from all values in that column we get

As the minimum number of lines covering all the zeros (4) is equal to the order of the matrix (5) so the assignment is not possible. The modification is required. The minimum uncovered value (2) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same. We get

As the minimum number of lines covering all the zeros (5) is equal to the order of the matrix, assignment is the possible

So A → 4, B → 3, C → 2, D → 1, E → 5
For the minimum cost take the corresponding values from the cost matrix we get
Total minimum cost = 3 + 3 + 4 + 3 + 7 = 20 units

Question 6.
A chartered accountant’s firm has accepted five new cases. The estimated number of days required by each of their five employees for each case are given below, where-means that the particular employee can not be assigned the particular case. Determine the optimal assignment of cases of the employees so that the total number of days required to complete these five cases will be minimum. Also, find the minimum number of days.

Solution:
This is a restricted assignment problem so we assign a very high cost (∞) to all the prohibited cells. The day matrix becomes

Subtracting row minimum from all values in that row we get

Subtracting column minimum from all values in that column we get

The minimum number of lines covering all the zeros (4) is less than the order of the matrix (5) so the assignment is not possible, The modification is required. The minimum uncovered value (1) is subtracted from all the uncovered values and added to the values at the intersection. The values on the lines remain the same, we get

The minimum number of lines covering all the zeros (5) is equal to the order of the matrix (5) so the assignment is possible.

So E₁ → I, E₂ → IV, E₃ → II, E₄ → V, E₅ → III
To find the minimum number of days we take the corresponding values from the day matrix.
Total minimum number of days = 6 + 6 + 6 + 6 + 3 = 27 days

Part – II

Question 1.
A readymade garments manufacture has to process 7 items through two stages of production, namely cutting and sewing. The time taken in hours for each of these items in different stages are given below:

Find the sequence in which these items are to be processed through these stages so as to minimize the total processing time. Also, find the idle time of each machine.
Solution:
Let A = cutting and B = sewing. So we have

Observe min {A, B} = 2 for item 1 for B.

The problem reduces to

Now min {A, B} = 3 for item 3 for A

The problem reduces to

New min {A , B} = 4 for item 4 for A.

The problem reduces to

Now min(A, B} = 5 for item 6 for B

The problem reduces to

Now min {A, B} = 6 for item 5 for A and item 2 for B

Now only 7 is left
∴ The optimal sequence is

Worktable

Total elapsed time = 46 hrs
Idle time for A (cutting) = 46 – 44 = 2 hrs
Idle time for B (Sewing) = 4 hrs

Question 2.
Five jobs must pass through a lathe and a surface grinder, in that order. The processing times in hours are shown below. Determine the optimal sequence of the jobs. Also, find the idle time of each machine.

Solution:
Let A = lathe and B = surface grinder. We have

Observe min {A, B} = 1 for job II for A

The problem reduces to

Now min {A, B} = 2 for job IV for A

The problem reduces to

Now min {A, B} = 3 for job I for B

The problem reduces to

Now min {A, B} = 5 for jobs III and V for A
∴ We have two options

or

We take the first one.
Worktable

Total elapsed time = 21 hrs
Idle time for A (lathe) = 21 – 17 = 4 hrs
Idle time for B (surface grinder) = 3 hrs

Question 3.
Find the sequence that minimizes the total elapsed time to complete the following jobs. Each job is processed in order AB.

Determine the sequence for the jobs so as to minimize the processing time. Find the total elapsed time and the idle time for both machines.
Solution:
Observe min {A, B} = 3 for job VII on B.

The problem reduces to

Now min {A, B} = 4 for job IV on B.

The problem reduces to

Now min {A, B} = 5 for job III & V on A. we have two options

or

We take the first one
The problem reduces to

Now min {A, B} = 5 for job II on A

The problem reduces to

Now min {A, B} = 7 for a job I on B and for job VI on A
∴ The optional sequence is

Worktable

Total elapsed time = 55 units
Idle time for A = 55 – 52 = 3 units
Idle time for B = 9 units.

Question 4.
A toy manufacturing company has five types of toys. Each toy has to go through three machines A, B, C in the order ABC. The time required in hours for each process is given in the following table.

Solve the problem for minimizing the total elapsed time.
Solution:
Min A = 12, Max B = 12
As min A ≥ max B.
The problem can be converted into two machine problems.
Let G and H be two fictitious machines such that G = A + B and H = B + C, We get

Now min {G, H} = 16 for type 3 on G

The problem reduces to

Min (G, H} = 18 for type 1, 4 & 5 on H
We have more than one option, we take

Now only type 2 is left.
∴ The optional sequence is

Worktable

Total elapsed time = 102 hours
Idle time for A = 102 – 84 = 18 hours
Idle time for B = 54 + (102 – 94) = 62 hours
Idle time for C = 38 hours

Question 5.
A foreman wants to process 4 different jobs on three machines: a shaping machine, a drilling machine, and a tapping, the sequence of operations being shaping-drilling-tapping. Decide the optimal sequence for the four jobs to minimize the total elapsed time. Also, find the total elapsed time and the idle time for every machine.

Solution:
The time matrix is

Min A = 8, Max B = 8, as min A ≥ max B.
The problem can be converted into a two-machine problem.
Let G and H be two fictitious machines such that
G = A + B and H = B + C we get

Observe min (G, H} = 12 for job 2 on H

The problem reduces to

Now min {G, H} = 14 for job 3 on G and job 4 on H

Now only job 1 is left.
∴ The optimal sequence is

Worktable

Total elapsed time = 74 min
Idle time for A (shapping) = 74 – 62 = 12 min
Idle time for B (Drilling) = 47 + (74 – 70) = 51 min
Idle time for C (trapping) = 31 min