JEE Main 2023 — Statistics Question with Solution

Given that the mean for the frequency distribution is $28$.

We know that Mean $\left(\overline{x}\right)=\frac{\sum f_i{x}_i}{\sum f_i}$

Now for the class interval $0-10$ we get $x_1=\frac{0+10}{2}=5$. 

Similarly we can calculate $x_i$ for other class intervals.

Now we know that mean $=28$

$\Rightarrow \frac{2\times 5+3\times 15+x\times 25+5\times 35+4\times 45}{14+x}=28$

$\Rightarrow x=6$

Also we know that variance $\left({\sigma }^2\right)=\frac{\sum f_i{\left(\overline{x}-x_i\right)}^2}{\sum f_i}$

${\sigma }^2=\frac{2\times {\left(28-5\right)}^2+3\times {\left(28-15\right)}^2+6\times {\left(28-25\right)}^2+5\times {\left(28-35\right)}^2+4\times {\left(28-45\right)}^2}{20}$

${\sigma }^2=\frac{2\times {\left(23\right)}^2+3\times {\left(13\right)}^2+6\times {\left(3\right)}^2+5\times {\left(7\right)}^2+4\times {\left(17\right)}^2}{20}$

${\sigma }^2=151$

Hence this is the required answer.