Q. The mean and the standard deviation( $s.d$.) of five observations are $9$ and $0$, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes $10$, then their $s.d$. is :
Solution:
In the first case, Mean of 5 observations $=9$
Standard Deviation $=0$
Since, standard deviation is 0, all the five observations remain the same having value of mear n 9
That is, 5 observations $=9,9,9,9,9$
In the second case,
One of the observations is changed so that mean becomes 10 , that is,
$\frac{9+9+9+9+x}{5} =10$
$36+x =50$
$x=14$
Now,
observations (x)
Mean $\bar{(x)}$
$(x-\bar{x})$
$(x-\bar{x})^{2}$
9
10
-1
1
9
10
-1
1
9
10
-1
1
9
10
-1
1
14
10
4
16
Therefore, s.d. $=\sqrt{\frac{(x-\bar{x})^{2}}{n}}$
$=\sqrt{\frac{20}{5}}$
$=\sqrt{4}=2$
| observations (x) | Mean $\bar{(x)}$ | $(x-\bar{x})$ | $(x-\bar{x})^{2}$ |
|---|---|---|---|
| 9 | 10 | -1 | 1 |
| 9 | 10 | -1 | 1 |
| 9 | 10 | -1 | 1 |
| 9 | 10 | -1 | 1 |
| 14 | 10 | 4 | 16 |