Question 13

Since the points $(x, P(X=x))$ lie on a straight line, let $P(X=x) = mx + c$ for $x \in \{0, 1, 2, 3, 4\}$. Sum of probabilities must be 1: $\sum_{x=0}^4 (mx + c) = 1 \implies c + (m+c) + (2m+c) + (3m+c) + (4m+c) = 1$ $10m + 5c = 1 \quad \dots (1)$ Mean of $X$ is $E[X] = \frac{5}{2}$: $E[X] = \sum_{x=0}^4 x P(X=x) = 0(c) + 1(m+c) + 2(2m+c) + 3(3m+c) + 4(4m+c) = \frac{5}{2}$ $m(1 + 4 + 9 + 16) + c(1 + 2 + 3 + 4) = \frac{5}{2}$ $30m + 10c = 2.5 \quad \dots (2)$ From (1), $20m + 10c = 2$. Subtracting this from (2): $10m = 0.5 \implies m = 0.05 = \frac{1}{20}$ Substitute $m$ into (1): $10(0.05) + 5c = 1 \implies 0.5 + 5c = 1 \implies 5c = 0.5 \implies c = 0.1 = \frac{1}{10}$. Now calculate $E[X^2]$: $E[X^2] = \sum_{x=0}^4 x^2 P(X=x) = m \sum x^3 + c \sum x^2$ $\sum_{x=0}^4 x^3 = 0 + 1 + 8 + 27 + 64 = 100$ $\sum_{x=0}^4 x^2 = 0 + 1 + 4 + 9 + 16 = 30$ $E[X^2] = (0.05)(100) + (0.1)(30) = 5 + 3 = 8$ Variance $\alpha = E[X^2] - (E[X])^2 = 8 - (2.5)^2 = 8 - 6.25 = 1.75$. The value of $24\alpha = 24 \times 1.75 = 24 \times \frac{7}{4} = 6 \times 7 = 42.$