Question

The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Answer 1

We have, n= 15, and ${\sigma }^2$ = 4

Now each observation is increased by 9

Suppose X = x+ 9 be the new data,

$\therefore \overline{(}X)=\frac{1}{15}\sum (x_i+9)=(\frac{1}{15}\times \sum x_i)+9=\overline{X}+9$

$\Rightarrow \sum x_i^2=\sum (x_i+9{)}^2=\sum x_i^2+\sum 18x_i+\sum 9^2$

Since $,{\sigma }^2=5$

$\Rightarrow \frac{1}{n}\sum x_i^2-(\overline{X}{)}^2=4$

Now, for the new data :

${\sigma }^2=\frac{1}{n}\sum x_i^2-(\overline{X}{)}^2=\frac{1}{15}(\sum x_i^2+\sum 18x_i+\sum 9^2)-(\overline{X}+9{)}^2$

$=\frac{1}{15}\sum x_i^2+\frac{1}{15}\sum 18x_i+\frac{1}{15}\sum 9^2-(9{)}^2-(18\overline{X})-(\overline{X}{)}^2$

$=[\frac{1}{15}\sum x_i^2-(\overline{X}{)}^2][\frac{1}{15}\sum 18x_i^2-(18\overline{X}{)}^2]+[\frac{1}{15}\sum 9^2-(9{)}^2]$

$=[\frac{1}{15}\sum x_i^2-(\overline{X}{)}^2][18\times \frac{1}{15}\sum x_i-(18\overline{X}\left)\right]+[\frac{1}{15}\times 15\times (9{)}^2-\left(9{)}^2\right]$

$=\frac{1}{15}\sum x_i^2-(\overline{X}{)}^2=4$