Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.

Answer 1

Given n = 10, $\overline{x}=45$ and ${\sigma }^2=16$

$\therefore \overline{x}=45\Rightarrow \frac{\sum x_i}{n}=45$

$\therefore \frac{\sum x_i}{10}=45\Rightarrow \sum x_i=450$

Corrected $\sum x_i=450-52+25=423$

$\therefore \overline{x}=\frac{423}{10}=42.3$

$\Rightarrow {\sigma }^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2\Rightarrow 16=\frac{\sum x_i^2}{10}-(45{)}^2\Rightarrow \sum x_i^2=10(2025+16)$

$\Rightarrow \sum x_i^2=20410$

$\therefore$ Corrected $\sum x_i^2=20410-(52{)}^2+(25{)}^2=18331$

and corrected ${\sigma }^2=\frac{18331}{10}-(42.3{)}^2=43.81$