JEE Main 2024 — Statistics Question with Solution

Let the incorrect mean be $${\mu }^{\prime }$$ and standard deviation be $${\sigma }^{\prime }$$

We have

$${\mu }^{\prime }=\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}}{15}=12\Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}=180$$

As per given information correct $$\Sigma {\mathrm{x}}_{\mathrm{i}}=180-10+12$$

$$\Rightarrow \mu (\text{ correct mean})=\frac{182}{15}$$

Also

$${\sigma }^{\prime }=\sqrt{\frac{\Sigma {\mathrm{x}}_{\mathrm{i}}^2}{15}-144}=3\Rightarrow \Sigma {\mathrm{x}}_{\mathrm{i}}^2=2295$$

Correct $$\Sigma {\mathrm{x}}_{\mathrm{i}}{}^2=2295-100+144=2339$$

$${\sigma }^2(\text{ correct variance })=\frac{2339}{15}-\frac{182\times 182}{15\times 15}$$

Required value

$$\begin{array}{rl}& =15\left(\mu +{\mu }^2+{\sigma }^2\right)\\ & =15\left(\frac{182}{15}+\frac{182\times 182}{15\times 15}+\frac{2339}{15}-\frac{182\times 182}{15\times 15}\right)\\ & =15\left(\frac{182}{15}+\frac{2339}{15}\right)\\ & =2521\end{array}$$