Question

Find mean and variance of the set of observations 3, 5, 7, 9, 11.​

Answer 1

Answer:

Step-by-step explanation:

Mean =3+5+7+9+11/5=7

1st observation=7-3=4

2nd. =7-5=2

....all 5th on is 4,2,0,-2,-4

Square of all ob =16+4+0+(-4)+(-16)=0

Hence variance =0

Answer 2

Given:

A set of observations 3, 5, 7, 9, 11.​

To find:

Mean and variance.

Solution:

Set of observations = 3, 5, 7, 9, 11.​

$Mean for Ungrouped Data: \bar{x}=\frac{x_{1}+x_{2}+\cdots x_{n}}{n} = \frac{3+ 5 + 7 + 9 + 11}{5}= 7.$

$Variance for ungrouped data = \frac{\left(x_{i}-x\right)^{2}}{n} \\\\ = \frac{\begin{array}{l}(3-5)^{2}+(5-5)^{2}+(7-5)^{2}+(9-5)^{2} +(11-5)^{2}\end{array}}{5} \\\\ = \frac{4+0+4+16+36}{5} = 12.$

Therefore, the mean and variance of the set of observations 3, 5, 7, 9, 11 are 7 and 12 respectively.