Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.4 Textbook Exercise Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4

Question 1.

The marks obtained by 9 students in a test of 100 marks in Mathematics are given below:

64, 63, 72, 65, 68, 69, 66, 67, 69

Find the standard deviation of marks obtained by the students.

Answer:

Here n = 9

Mean:

x̄ = \(\frac{\Sigma x}{n}=\frac{603}{9}\) = 67 marks

Standard deviation of marks:

s = \(\sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}}\)

= \(\sqrt{\frac{64}{9}}\)

= \(\sqrt{7.111}\)

= 2.67 marks

Question 2.

The numbers of cars coming for service in five service stations of a company on a particular day are 7, 3, 11, 8, 9. Calculate the standard deviation of number of cars coming at the service station.

Answer:

Here, n = 5

Mean:

x̄ = \(\frac{\Sigma x}{n}\)

= \(\frac{38}{5}\)

= 7.6 Cars

Standard deviation of number of cars for service:

S = \(\sqrt{\frac{\Sigma(x-\bar{x})^{2}}{n}}\)

= \(\sqrt{\frac{35.20}{5}}\)

= \(\sqrt{7.04}\)

= 2.65 cars

Question 3.

The following frequency distribution represents the amounts of deposits and the number of depositors in a bank. Find the coefficient of standard deviation of the deposits.

Answer:

Mean:

x̄ = A + \(\frac{\Sigma f d}{n}\)

= 20 – \(\frac{50}{50}\)

= 20 – 1

= ₹ 19 thousand

Standard deviation of amount of deposites:

s = \(\sqrt{\frac{\Sigma f d^{2}}{n}-\left(\frac{\Sigma f d}{n}\right)^{2}}\)

= \(\sqrt{\frac{2300}{50}-\left(\frac{-50}{50}\right)^{2}}\)

= \(\sqrt{46-(-1)^{2}}\)

= \(\sqrt{46-1}=\sqrt{45}\) = ₹ 6.71 thousand

Coefficient of standard deviation

= \(\frac{s}{\bar{x}}=\frac{6.71}{19}\) = 0.35

Question 4.

The information of profits (in lakh ?) of 50 firms in the last year is given below. Find the standard deviation of the profit of the firms.

Answer:

Mean:

x̄ = A + \(\frac{\Sigma f d}{n}\) × c

= 25 + \(\frac{12}{50}\) × 10

= 25 + \(\frac{12}{5}\)

= 25 + 2.4

= ₹ 27.4 lakh

Standard deviation of profit:

Question 5.

Find the standard deviation of age of the persons from the following distribution of 125 persons living in a society. Also find the coefficient of standard deviation.

Answer:

Mean:

x̄ = A + \(\frac{\Sigma f d}{n}\) × c

= 35 + \(\frac{2}{125}\) × 100

= 35 + \(\frac{20}{125}\)

= 35 + 0.16

= 35.16

Standard deviation of age:

Coefficient of standard deviation of age

= \(\frac{s}{\bar{x}}=\frac{19.76}{35.16}\) = 0.56