JEE Main 2024 — Statistics Question with Solution

$X=$ denotes number of defective $\begin{array}{|c|c|c|c|c|} \hline \mathrm{x} & 0 & 1 & 2 & 3 \\ \hline \mathrm{P}(\mathrm{x}) & \frac{7}{15} & \frac{5}{12} & \frac{5}{12} & \frac{1}{12} \\ \hline \mathrm{x}_1{ }^2 & 0 & 1 & 4 & 9 \\ \hline \mathrm{P}_{\mathrm{i}} \mathrm{x}_1{ }^2 & 0 & \frac{5}{12} & \frac{20}{12} & \frac{9}{12} \\ \hline \mathrm{p}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}} & 0 & \frac{5}{12} & \frac{10}{12} & \frac{3}{12} \\ \hline \end{array}$ $\mu =\Sigma {\mathrm{p}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}=\frac{18}{12}$ $\begin{aligned} & \Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_1^2=\frac{34}{12} \\ & \sigma^2=\Sigma \mathrm{p}_{\mathrm{i}} \mathrm{x}_1^2-(\mu)^2 \end{aligned}$ $\begin{array}{rl}& =\frac{34}{12}-{\left(\frac{18}{12}\right)}^2=\frac{17}{6}-\frac{9}{4}\\ & \frac{34-27}{12}=\frac{7}{12}\\ & 96{\sigma }^2=96\times \frac{7}{12}=56\end{array}$