Question

The mean and standard and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.

Answer 1

Here n = 100, $\overline{x}-20and\sigma =3$

$\therefore \overline{x}=\frac{1}{n}\sum x_i$

$\Rightarrow \sum x_i=n\times \overline{x}=100\times 20=2000$

$\therefore$ Incorrect $\sum x_i$ = 2000

Now $\frac{1}{n}\sum x_i^2-(\overline{x}{)}^2=9$

$\Rightarrow \frac{1}{100}\sum x_i^2-(20{)}^2=9\Rightarrow \sum x_i^2=40900$

When wrong items 21, 21 and 18 are omitted from the data then we have 97 observations.

Correct $\sum x_i$ = Incorrect $\sum x_i$ - 21 - 21 -18 = 1940

$\therefore$ Correct mean = $\frac{1940}{97}$ = 20

Also

Correct $\sum x_i^2$ = Incorrect $\sum x_i^2-(21{)}^2-(21{)}^2-(18{)}^2$

= 40900 - 441 -441 324 = 39694

$\therefore$ Correct variance = $\frac{1}{97}(correct\sum x_i^2$ - $(correctmean{)}^2$

= $\frac{1}{97}\times 39694-(20{)}^2$

= 409.22 - 400 = 9.22

Correct S.D. = $\sqrt{9.22}$ = 3.036