Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2

Question 1.

Find the range for the following data.

116, 124, 164, 150, 149, 114, 195, 128, 138, 203, 144

Solution:

Here, largest value (L) = 203, smallest value (S) = 114

∴ Range = L – S

= 203 – 114

= 89

Question 2.

Given below the frequency distribution of weekly w ages of 400 workers. Find the range.

Solution:

Here, largest value (L) = 40, smallest value (S) = 10

∴ Range = L – S

= 40 – 10

= 30

Question 3.

Find the range of the following data.

Solution:

Here, upper limit of the highest class (L) = 175, lower limit of the lowest class (S) = 115

∴ Range = L – S

= 175 – 115

= 60

Question 4.

The city traffic police issued challans for not observing the traffic rules:

Find Q.D.

Solution:

The given data can be arranged in ascending order as follows:

24, 36, 40, 58, 62, 80

Here, n = 6

Q_{1} = value of \(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of \(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation

= value of (1.75)th observation

= value of 1st observation + 0.75(value of 2nd observation – value of 1st observation)

= 24 + 0.75(36 – 24)

= 24 + 0.75(12)

= 24 + 9

∴ Q_{1} = 33

Q_{3} = value of 3\(\left(\frac{n+1}{4}\right)^{\text {th }}\) observation

= value of 3\(\left(\frac{6+1}{4}\right)^{\text {th }}\) observation

= value of (3 × 1.75)th observation

= value of (5.25)th observation

= value of 5th observation + 0.25(value of 6th observation – value of 5th observation)

= 62 + 0.25(80 – 62)

= 62 + 0.25(18)

= 62 + 4.5

= 66.5

∴ Q.D. = \(\frac{Q_{3}-Q_{1}}{2}=\frac{66.5-33}{2}=\frac{33.5}{2}\) = 16.75

Question 5.

Calculate Q.D. from the following data.

Solution:

We construct the less than cumulative frequency table as follows:

Here, N = 35

Q_{1} class = class containing \(\left(\frac{N}{4}\right)^{\text {th }}\) observation

∴ \(\frac{\mathrm{N}}{4}=\frac{35}{4}\) = 8.75

Cumulative frequency which is just greater than (or equal to) 8.75 is 15.

∴ Q_{1} lies in the class 20-30.

∴ L = 20, c.f. = 8, f = 7, h = 10

Question 6.

Calculate the appropriate measure of dispersion for the following data.

Solution:

Since open-ended classes are given, the appropriate measure of dispersion that we can compute is the quartile deviation.

We construct the less than cumulative frequency table as follows:

Here N = 250

Q_{1} class class containing \(\left(\frac{N}{4}\right)^{t h}\) observation

∴ \(\frac{\mathrm{N}}{4}=\frac{250}{4}\) = 62.5

Cumulative frequency which is just greater than (or equal to) 62.5 is 65.

∴ Q_{1} lies in the class 35-40.

∴ L = 35, c.f. = 15, f = 50, h = 5

The cumulative frequency which is just greater than (or equal to) 187.5 is 190.

∴ Q_{3} lies in the class 45-50.

∴ L = 45, c.f. = 150, f = 40, h = 5

Question 7.

Calculate Q.D. of the following data.

Solution:

We construct the less than cumulative frequency table as follows:

Here, N = 120

Q_{1} class = class containing \(\left(\frac{\mathrm{N}}{4}\right)^{\mathrm{th}}\) observation

∴ \(\frac{\mathrm{N}}{4}=\frac{120}{4}\) = 30

Cumulative frequency which is just greater than (or equal to) 30 is 35.

∴ Q_{1} lies in the class 4-6.

∴ L = 4, c.f. = 15, f = 20, h = 2

Cumulative frequency which is just greater than (or equal to) 90 is 90.

∴ Q_{3} lies in the class 10-12.

∴ L = 10, c.f. = 72, f = 18, h = 2

Question 8.

Find variance and S.D. for the following set of numbers.

25, 21, 23, 29, 27, 22, 28, 23, 27, 25 (Given √6.6 = 2.57)

Solution:

We prepare the following table for the calculation of variance and S.D.:

Question 9.

Following data gives no. of goals scored by a team in 90 matches.

Compute the variance and standard deviation for the above data.

Solution:

We prepare the following table for the calculation of variance and S.D:

Question 10.

Compute the arithmetic mean and S.D. and C.V. (Given √296 = 17.20)

Solution:

We prepare the following table for calculation of arithmetic mean and S.D.:

Question 11.

The mean and S.D. of 200 items are found to be 60 and 20 respectively. At the time of calculation, two items were wrongly taken as 3 and 67 instead of 13 and 17. Find the correct mean and variance.

Solution:

Here, n = 200, \(\bar{x}\) = Mean = 60, S.D. = 20

Wrongly taken items are 3 and 67.

Correct items are 13 and 17.

Now, \(\bar{x}\) = 60

Correct value of \(\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i}\) (sum of wrongly taken items) + (sum of correct items)

= 12000 – (3 + 67) + (13 + 17)

= 12000 – 70 + 30

= 11960

Correct value of mean = \(\frac{1}{n}\) × correct value of \(\sum_{i=1}^{n} x_{i}\)

= \(\frac{1}{200}\) × 11960

= 59.8

Now, S.D. = 20

Variance = (S.D.)^{2} = 20^{2}

∴ Variance = 400

∴ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^{2}-(\bar{x})^{2}=400\)

Correct value of \(\sum_{i=1}^{n} x_{i}^{2}\)

= \(\sum_{i=1}^{n} x_{i}^{2}\) – (Sum of squares of wrongly taken items) + (Sum of squares of correct items)

= 800000 – (3^{2} + 67^{2}) + (13^{2} + 17^{2})

= 800000 – (9 + 4489) + (169 + 289)

= 800000 – 4498 + 458

= 795960

∴ Correct value of Variance = (\(\frac{1}{n}\) × \(\sum_{i=1}^{n} x_{i}^{2}\)) – (correct value of \(\bar{x}\))^{2}

= \(\frac{1}{200}\) × 795960 – (59.8)^{2}

= 3979.8 – 3576.04

= 403.76

∴ The correct mean is 59.8 and correct variance is 403.76.

Question 12.

The mean and S.D. of a group of 48 observations are 40 and 8 respectively. If two more observations 60 and 65 are added to the set, find the mean and S.D. of 50 items.

Solution:

Question 13.

The mean height of 200 students is 65 inches. The mean heights of boys and girls are 70 inches and 62 inches respectively and the standard deviations are 8 and 10 respectively. Find the number of boys and combined S.D.

Solution:

Let n_{1} and n_{2} be the number of boys and girls respectively.

Let n = 200, \(\bar{x}_{\mathrm{c}}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{1}\) = 62, σ_{1} = 8, σ_{2} = 10

Here, n_{1} + n_{2} = n

∴ n_{1} + n_{2} = 200 …….(i)

Combined mean is given by

∴ 70n_{1} + 62n_{2} = 13000

∴ 35n_{1} + 31n_{2} = 6500 ……..(ii)

Solving (i) and (ii), we get

n_{1} = 75, n_{2} = 125

Combined standard deviation is given by,

Question 14.

From the following data available for 5 pairs of observations of two variables x and y, obtain the combined S.D. for all 10 observations,

where \(\sum_{i=1}^{n} x_{i}=30, \sum_{i=1}^{n} y_{i}=40, \sum_{i=1}^{n} x_{i}^{2}=225, \sum_{i=1}^{n} y_{i}^{2}=340\)

Solution:

Question 15.

The mean and standard deviations of two brands of watches are given below:

Calculate the coefficient of variation of the two brands and interpret the results.

Solution:

Since C.V. (I) > C.V. (II)

∴ the brand I is more variable.

Question 16.

Calculate the coefficient of variation for the data given below. [Given √3.3 = 1.8166]

Solution: