GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4

   

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
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 1
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

GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4

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
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 2
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.
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 3
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 4
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

GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4

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.
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 5
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 6
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:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 7

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.
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 8
Answer:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 9
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:
GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.4 10
Coefficient of standard deviation of age
= \(\frac{s}{\bar{x}}=\frac{19.76}{35.16}\) = 0.56

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