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 ________.

The five probabilities are $$P(X = 0), P(X = 1), P(X = 2), P(X = 3), P(X = 4)$$. Because the points $$(x, P(X = x))$$ for $$x = 0,1,2,3,4$$ lie on a straight line, the probability is an affine (linear) function of $$x$$:

$$P(X = x) = a + bx,\qquad x = 0,1,2,3,4$$

Here $$a$$ and $$b$$ are constants to be determined. Outside these five values the probability is zero, so the total probability is

$$\sum_{x=0}^{4} P(X = x) = 1$$ $$\Rightarrow \sum_{x=0}^{4} (a + bx) = 1$$

The useful sums are $$\sum_{x=0}^{4} 1 = 5,\qquad \sum_{x=0}^{4} x = 0+1+2+3+4 = 10$$

Therefore $$5a + 10b = 1 \qquad -(1)$$

Next use the given mean $$E[X] = \dfrac{5}{2} = 2.5$$:

$$E[X] = \sum_{x=0}^{4} x\,P(X = x) = \sum_{x=0}^{4} x(a + bx)$$ Break the sum into two parts:

$$\sum_{x=0}^{4} xa = a\sum_{x=0}^{4} x = a\cdot10$$ $$\sum_{x=0}^{4} bx^{2} = b\sum_{x=0}^{4} x^{2}$$ Since $$\sum_{x=0}^{4} x^{2} = 0^{2}+1^{2}+2^{2}+3^{2}+4^{2} = 30$$,

$$E[X] = 10a + 30b = 2.5 \qquad -(2)$$

Solve the simultaneous equations. Divide $$(1)$$ by $$5$$: $$a + 2b = 0.2$$ Divide $$(2)$$ by $$10$$: $$a + 3b = 0.25$$ Subtract the first from the second:

$$b = 0.25 - 0.2 = 0.05 = \frac{1}{20}$$ Back-substitute into $$a + 2b = 0.2$$:

$$a = 0.2 - 2(0.05) = 0.2 - 0.1 = 0.1$$

Compute $$E[X^{2}]$$ to obtain the variance. First, $$E[X^{2}] = \sum_{x=0}^{4} x^{2}P(X = x) = \sum_{x=0}^{4} x^{2}(a + bx)$$

Break it again: $$\sum_{x=0}^{4} x^{2}a = a\sum_{x=0}^{4} x^{2} = a\cdot30$$ $$\sum_{x=0}^{4} x^{3}b = b\sum_{x=0}^{4} x^{3}$$ Here $$\sum_{x=0}^{4} x^{3} = 0^{3}+1^{3}+2^{3}+3^{3}+4^{3} = 100$$

Thus $$E[X^{2}] = 30a + 100b = 30(0.1) + 100(0.05) = 3 + 5 = 8$$

The variance is $$\alpha = \operatorname{Var}(X) = E[X^{2}] - (E[X])^{2} = 8 - (2.5)^{2} = 8 - 6.25 = 1.75 = \frac{7}{4}$$

Finally, $$24\alpha = 24 \times \frac{7}{4} = 6 \times 7 = 42$$

Therefore, the required value is 42.