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 $xy$-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 _______.

$\text{Given: } (x, P(X=x)) \text{ lie on a straight line for } x = 0,1,2,3,4$ $P(X=x) = 0 \ \text{ for } \ x \in \mathbb{R} - \{0,1,2,3,4\}$ $\text{Mean}(X) = \dfrac{5}{2}, \qquad \text{Variance}(X) = \alpha$ $\text{To find: } 24\alpha$ $P(X=k) = a + bk, \qquad k = 0,1,2,3,4$ $P(X=k) = a + bk \text{ is the line through these five points}$ $\sum_{k=0}^{4} P(X=k) = 1, \qquad \sum_{k=0}^{4} k\,P(X=k) = \dfrac{5}{2}$ $\sum_{k=0}^{4}(a+bk) = 5a + 10b = 1$ $\sum_{k=0}^{4} k(a+bk) = 10a + 30b = \dfrac{5}{2}$ $2 \times (5a + 10b = 1): \quad 10a + 20b = 2$ $10b = \dfrac{5}{2} - 2 = \dfrac{1}{2} \quad \Rightarrow \quad b = \dfrac{1}{20}$ $5a + \dfrac{1}{2} = 1 \quad \Rightarrow \quad a = \dfrac{1}{10}$ $P(X=k) = \dfrac{1}{10} + \dfrac{k}{20} = \dfrac{2+k}{20}$ $P = \left(\dfrac{2}{20}, \dfrac{3}{20}, \dfrac{4}{20}, \dfrac{5}{20}, \dfrac{6}{20}\right)$ $E[X^2] = \sum_{k=0}^{4} k^2 P(X=k) = \dfrac{0 + 3 + 16 + 45 + 96}{20} = \dfrac{160}{20} = 8$ $\alpha = E[X^2] - (E[X])^2 = 8 - \left(\dfrac{5}{2}\right)^2$ $= 8 - \dfrac{25}{4} = \dfrac{32 - 25}{4} = \dfrac{7}{4}$ $24\alpha = 24 \times \dfrac{7}{4} = 6 \times 7 = 42$