GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2

Find the mean and variance for the following data in questions 1 to 5:
1. 6, 7, 10, 12, 13, 4, 8, 12
2. First n natural numbers.
3. First 10 multiples of 3.
4.

5.

Solutions to questions 1 to 5:
1. Mean \(\bar {x} \) = \(\frac{\Sigma x_{i}}{n}\) = \(\frac{6+7+10+12+13+4+8+12}{8}\)
= \(\frac{72}{8}\) = 9.

∴ Σ(x_i – \(\bar {x} \) = 9 + 4 + 1 + 9 + 16 + 25 + 1 + 9 = 74.
∴ Variance = \(\frac{\Sigma\left(x_{i}-\bar{x}\right)^{2}}{\Sigma f_{i}}\) = \(\frac{74}{8}\) = 9.25.

2. The first n natural numbers are 1, 2, 3, …, n

3. First 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.

Thus, mean = 16.5 and variance = 74.25.

4.

Short-Cut Method:

5.

Short-Cut Method:
Let y_i = x_i – 98, where we have assumed A = 98. Then,

Question 6.
Find the mean and standard deviation of the following, using short-cut method:

Solution:

∴ σ = \(\sqrt{2.86}\) = 1.69.

Find the mean and variance for the following frequency distribution in questions 7 and 8:
7.

8.

Solutions to questions 7 and 8:

= 30[76 – 0.13]
= 30 × 75.87 = 2276.

8.

= 2(68 – 2) = 2 × 66 = 132.

Question 9.
Find the mean, variance and standard deviation of the following, using short-cut method:

Solution:

∴ Standard deviation σ = \(\sqrt{105.58}\) = 10.27.

Question 10.
The diameter of circles (in mm) drawn in a design are given below:

Calculate the mean diameter and standard deviation of the circles.
Solution:
Firstly, the data is to be made continous by making classes as 32.5 – 36.5, 36.5 – 40.5, 40.5 – 44.5, 44.5 – 48.5, 48.5 – 52.5.

= 5.55.