Let \(X\) be a random variable, and let \(P(X=x)\) denote the probability that \(X\) takes the value \(x\). Suppose that the points \((x, P(X=x)), x=0,1,2,3,4\), lie on a fixed straight line in the \(x y\)-plane, and \(P(X=x)=0\) for all \(x \in \mathbb{R}-\{0,1,2,3,4\}\). If the mean of \(X\) is \(\frac{5}{2}\), and the variance of \(X\) is \(\alpha\), then the value of \(24 \alpha\) is ___

Let equation of line is y = mx + c
\(\begin{array}{|c|c|c|c|c|c|c|}
\hline \mathrm{x} & 0 & \mathrm{l} & 2 & 3 & 4 & \mathrm{R}-\{0,1,2,3,4\} \\
\hline \mathrm{P}(\mathrm{x}) & \mathrm{c} & \mathrm{m}+\mathrm{c} & 2 \mathrm{~m}+\mathrm{c} & 3 \mathrm{~m}+\mathrm{c} & 4 \mathrm{~m}+\mathrm{c} & 0 \\
\hline
\end{array}\)
\(\sum_{x=0}^4(m x+c)=1 \Rightarrow 10 m+5 c=1 \Rightarrow 2 m+c=\frac{1}{5}...(1)\)
\(\begin{aligned} & \text { mean }=\sum x_i P_i=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \cdot \mathrm{x}_{\mathrm{i}}=30 \mathrm{~m}+10 \mathrm{c}=\frac{5}{2} \\ & \therefore 3 \mathrm{~m}+\mathrm{c}=\frac{1}{4} \ldots(2)\end{aligned}\)
\(\begin{aligned} & \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \mathrm{x}_1^2 \\ & =\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}^3+\mathrm{cx}_{\mathrm{i}}^2\right) \Rightarrow 100 \mathrm{~m}+30 \mathrm{c} \text { (Now putting } \mathrm{m} \text { and } \mathrm{c} \text { ) } \\ & \Rightarrow \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=5+3=8 \\ & \text { Variance }=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4} \\ & \therefore 24 \alpha=42\end{aligned}\)