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.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 1
5.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 2
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.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 3
∴ Σ(xi – \(\bar {x} \) = 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
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 4

3. First 10 multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 5
Thus, mean = 16.5 and variance = 74.25.

4.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 6

Short-Cut Method:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 7

5.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 8

Short-Cut Method:
Let yi = xi – 98, where we have assumed A = 98. Then,
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 9

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2

Question 6.
Find the mean and standard deviation of the following, using short-cut method:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 10
Solution:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 11
∴ σ = \(\sqrt{2.86}\) = 1.69.

Find the mean and variance for the following frequency distribution in questions 7 and 8:
7.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 12
8.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 13
Solutions to questions 7 and 8:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 14
= 30[76 – 0.13]
= 30 × 75.87 = 2276.

8.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 15
= 2(68 – 2) = 2 × 66 = 132.

Question 9.
Find the mean, variance and standard deviation of the following, using short-cut method:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 16
Solution:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 17
∴ Standard deviation σ = \(\sqrt{105.58}\) = 10.27.

GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2

Question 10.
The diameter of circles (in mm) drawn in a design are given below:
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 18
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.
GSEB Solutions Class 11 Maths Chapter 15 Statistics Ex 15.2 img 19
= 5.55.

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