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.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q2
Solution:
Here, largest value (L) = 40, smallest value (S) = 10
∴ Range = L – S
= 40 – 10
= 30

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

Question 3.
Find the range of the following data.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q3
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:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q4
Find Q.D.
Solution:
The given data can be arranged in ascending order as follows:
24, 36, 40, 58, 62, 80
Here, n = 6
Q1 = 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
∴ Q1 = 33
Q3 = 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

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

Question 5.
Calculate Q.D. from the following data.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q5
Solution:
We construct the less than cumulative frequency table as follows:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q5.1
Here, N = 35
Q1 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.
∴ Q1 lies in the class 20-30.
∴ L = 20, c.f. = 8, f = 7, h = 10
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q5.2

Question 6.
Calculate the appropriate measure of dispersion for the following data.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q6
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:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q6.1
Here N = 250
Q1 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.
∴ Q1 lies in the class 35-40.
∴ L = 35, c.f. = 15, f = 50, h = 5
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q6.2
The cumulative frequency which is just greater than (or equal to) 187.5 is 190.
∴ Q3 lies in the class 45-50.
∴ L = 45, c.f. = 150, f = 40, h = 5
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q6.3

Question 7.
Calculate Q.D. of the following data.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q7
Solution:
We construct the less than cumulative frequency table as follows:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q7.1
Here, N = 120
Q1 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.
∴ Q1 lies in the class 4-6.
∴ L = 4, c.f. = 15, f = 20, h = 2
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q7.2
Cumulative frequency which is just greater than (or equal to) 90 is 90.
∴ Q3 lies in the class 10-12.
∴ L = 10, c.f. = 72, f = 18, h = 2
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q7.3

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.:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q8

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

Question 9.
Following data gives no. of goals scored by a team in 90 matches.
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q9
Compute the variance and standard deviation for the above data.
Solution:
We prepare the following table for the calculation of variance and S.D:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q9.1

Question 10.
Compute the arithmetic mean and S.D. and C.V. (Given √296 = 17.20)
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q10
Solution:
We prepare the following table for calculation of arithmetic mean and S.D.:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q10.1

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
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q11
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 = 202
∴ Variance = 400
∴ \(\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} x_{\mathrm{i}}^{2}-(\bar{x})^{2}=400\)
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q11.1
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 – (32 + 672) + (132 + 172)
= 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.

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

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:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q12

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 n1 and n2 be the number of boys and girls respectively.
Let n = 200, \(\bar{x}_{\mathrm{c}}\) = 65, \(\bar{x}_{1}\) = 70, \(\bar{x}_{1}\) = 62, σ1 = 8, σ2 = 10
Here, n1 + n2 = n
∴ n1 + n2 = 200 …….(i)
Combined mean is given by
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q13
∴ 70n1 + 62n2 = 13000
∴ 35n1 + 31n2 = 6500 ……..(ii)
Solving (i) and (ii), we get
n1 = 75, n2 = 125
Combined standard deviation is given by,
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q13.1

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:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q14
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q14.1

Question 15.
The mean and standard deviations of two brands of watches are given below:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q15
Calculate the coefficient of variation of the two brands and interpret the results.
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q15.1
Since C.V. (I) > C.V. (II)
∴ the brand I is more variable.

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

Question 16.
Calculate the coefficient of variation for the data given below. [Given √3.3 = 1.8166]
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q16
Solution:
Maharashtra Board 11th Commerce Maths Solutions Chapter 2 Measures of Dispersion Miscellaneous Exercise 2 Q16.1