Question

The mean and standard deviation of $50$ items were found to be $75$ and $10$ respectively. At the time of checking it was found that one item $65$ was incorrect. Calculate the mean and standard deviation, if the wrong item is omitted.

Answer 1

$Mean=75$

$\Rightarrow \frac{\sum x_i}{n}=75\Rightarrow \sum x_i=75\times 50=3750$

$\Rightarrow 3750-65=3685$

New mean $=\frac{3685}{49}=75.2041$

Standard deviation $\left(\sigma \right)=10$

Variance $\left({\sigma }^2\right)=100$

$\therefore \frac{\sum x_1^2}{n}-{\left(\frac{\sum x_1}{n}\right)}^2=100$

$\Rightarrow \frac{\sum x_1^2}{50}-{\left(75\right)}^2=100$

$\Rightarrow \sum x_1^2=286250$

New $\sum x_1^2=286250-{\left(65\right)}^2$

$=282025$

New variance $=\frac{282025}{49}-{\left(75.2041\right)}^2=99.9556$

New standard deviation $=\sqrt{99.9556}=9.99778$

Hence, solved.