Question

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was late found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and S. D.

Answer 1

We have, n = 100, $\overline{x}=40$ and $\sigma =5.1$

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

$\Rightarrow \sum x_i=n\overline{x}=100\times 40=4000$

$\therefore$ Incorrect $\sum x_i=4000$

and,

$\sigma =5.1$

$\Rightarrow {\sigma }^2=26.01$

$\Rightarrow \frac{1}{n}\sum x_i^2-(Mean{)}^2=26.01$

$\Rightarrow \frac{1}{100}\sum x_i^2-1600=26.01$

$\Rightarrow \sum x_i^2=1626.01\times 100$

$\therefore$ Incorrect $\sum x_i^2=162601$

When the incorrect observation 50 is replaced by 40 :

We have , Incorrect $\sum x_i=4000$

$\therefore$ Corrected $\sum x_i=4000-50+40=3990$

and,

Incorrect $\sum x_i^2=162601$

$\therefore$ Corrected $\sum x_i^2=162601-{50}^2+40=161701$

Now, Corrected mean = $\frac{3990}{100}=30.90$

$\text{Corrected variance}=\frac{1}{100}(\text{Corrected}\sum x_i^2)-(\text{Corrected mean}{)}^2$

$\Rightarrow$ Corrected variance = $\frac{161701}{100}-{\left(\frac{3990}{100}\right)}^2$

$\Rightarrow$Corrected variance = $\frac{161701\times -(3990{)}^2}{(100{)}^2}$

$\Rightarrow$ Corrected standard deviation = $\sqrt{25}=5$