JEE Main 2022 — Statistics Question with Solution

Given,

Wrong mean $={\mu }_1=30$

Wrong standard deviation $={\sigma }_1=5$

Now mean ig given by $\frac{\sum x_i}{40}=30$

$\Rightarrow \sum x_i=1200$

Variance is given by ${\sigma }_1^2=25$

$\Rightarrow \frac{\sum x_i^2}{40}-{30}^2=25$

$\Rightarrow \sum x_i^2=925\times 40=37000$

New sum of mean $=\sum x_i^{\text{'}}=1200-10-12=1178$

So, new mean $={\mu }_1^{\text{'}}=\frac{1178}{38}=31$

Now new variance $\sum x_i^2=37000-{\left(10\right)}^2-{\left(12\right)}^2=36756$

So, new S.D. ${\sigma }_1^{\text{'}}=\sqrt{\frac{36756}{38}-{\left(31\right)}^2}=\sigma$

$36756-{\left(31\right)}^2\times 38=38{\sigma }^2$

$\Rightarrow 38{\sigma }^2=238$