Variance of Collinear Discrete Random Variable Distribution

To find the value of $24\alpha$, we begin by using the properties of discrete probability distributions and the geometric condition given in the problem.

Let $P(X = x) = p_x$ for $x \in \{0, 1, 2, 3, 4\}$. Since the points $(x, p_x)$ lie on a straight line in the $xy$-plane, we can express $p_x$ as a linear function of $x$: $p_x = mx + c$ for some constants $m$ and $c$.

Step 1: Utilize the total probability condition

The sum of all probabilities must equal $1$: $\sum_{x=0}^{4} p_x = 1$ $\sum_{x=0}^{4} (mx + c) = 1$ $m \sum_{x=0}^{4} x + c \sum_{x=0}^{4} 1 = 1$

Calculating the sums: $\sum_{x=0}^{4} x = 0 + 1 + 2 + 3 + 4 = 10$ $\sum_{x=0}^{4} 1 = 5$

Substituting these into the equation gives: $10m + 5c = 1 \quad \text{--- (Equation 1)}$

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Step 2: Utilize the given mean of $X$

The expectation (mean) of $X$ is given as $E[X] = \frac{5}{2}$: $E[X] = \sum_{x=0}^{4} x \cdot p_x = \frac{5}{2}$ $\sum_{x=0}^{4} x (mx + c) = \frac{5}{2}$ $m \sum_{x=0}^{4} x^2 + c \sum_{x=0}^{4} x = \frac{5}{2}$

Calculating the sum of squares: $\sum_{x=0}^{4} x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 = 0 + 1 + 4 + 9 + 16 = 30$

Substituting the values: $30m + 10c = \frac{5}{2} \quad \text{--- (Equation 2)}$

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Step 3: Solve for $m$ and $c$

Multiply Equation 1 by $2$: $20m + 10c = 2$

Subtract this from Equation 2: $(30m + 10c) - (20m + 10c) = \frac{5}{2} - 2$ $10m = \frac{1}{2} \implies m = \frac{1}{20}$

Substitute $m = \frac{1}{20}$ back into Equation 1: $10\left(\frac{1}{20}\right) + 5c = 1$ $\frac{1}{2} + 5c = 1 \implies 5c = \frac{1}{2} \implies c = \frac{1}{10}$

Thus, the probability distribution is: $p_x = \frac{x}{20} + \frac{1}{10} = \frac{x + 2}{20} \quad \text{for } x \in \{0, 1, 2, 3, 4\}$

Since $p_x > 0$ for all $x \in \{0, 1, 2, 3, 4\}$, this is a valid probability distribution.

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Step 4: Calculate $E[X^2]$ and the Variance $\alpha$

The second moment $E[X^2]$ is given by: $E[X^2] = \sum_{x=0}^{4} x^2 p_x = m \sum_{x=0}^{4} x^3 + c \sum_{x=0}^{4} x^2$

Calculating the sum of cubes: $\sum_{x=0}^{4} x^3 = 0^3 + 1^3 + 2^3 + 3^3 + 4^3 = 0 + 1 + 8 + 27 + 64 = 100$

Substitute the known values: $E[X^2] = \frac{1}{20}(100) + \frac{1}{10}(30) = 5 + 3 = 8$

Now, calculate the variance $\alpha$: $\alpha = \text{Var}(X) = E[X^2] - (E[X])^2$ $\alpha = 8 - \left(\frac{5}{2}\right)^2 = 8 - \frac{25}{4} = \frac{7}{4}$

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Step 5: Compute $24\alpha$

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

42