Question

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 ∈ R - {0,1,2,3,4}. If the mean of X is \(\frac{5}{2}\) , and the variance of X is α, then the value of 24α is ______.

Answer 1

Correct Answer: 42

To solve the problem, use the fact that the probabilities lie on a straight line and sum to 1, then calculate mean and variance.

Step 1: Express the probabilities
Since points \((x, P(X=x))\) for \(x=0,1,2,3,4\) lie on a line, the pmf is linear in \(x\): \[ P(X=x) = a x + b, \quad x = 0,1,2,3,4 \] with \(P(X=x) = 0\) elsewhere.

Step 2: Use normalization condition
Sum of probabilities: \[ \sum_{x=0}^4 P(X=x) = \sum_{x=0}^4 (a x + b) = a (0+1+2+3+4) + 5 b = a \times 10 + 5 b = 1 \] So, \[ 10 a + 5 b = 1 \quad \Rightarrow \quad 2 a + b = \frac{1}{5} \] (Equation 1)

Step 3: Use mean condition
Mean \(\mu = E(X) = \frac{5}{2}\): \[ E(X) = \sum_{x=0}^4 x P(X=x) = \sum_{x=0}^4 x (a x + b) = a \sum_{x=0}^4 x^2 + b \sum_{x=0}^4 x \] Calculate sums: \[ \sum_{x=0}^4 x = 10, \quad \sum_{x=0}^4 x^2 = 0^2 + 1^2 + 2^2 + 3^2 + 4^2 = 30 \] So, \[ E(X) = a \times 30 + b \times 10 = 30 a + 10 b = \frac{5}{2} \] Divide both sides by 5: \[ 6 a + 2 b = \frac{1}{2} \] (Equation 2)

Step 4: Solve for \(a\) and \(b\)
From Equation 1: \[ b = \frac{1}{5} - 2 a \] Substitute into Equation 2: \[ 6 a + 2 \left( \frac{1}{5} - 2 a \right) = \frac{1}{2} \] \[ 6 a + \frac{2}{5} - 4 a = \frac{1}{2} \] \[ 2 a + \frac{2}{5} = \frac{1}{2} \] \[ 2 a = \frac{1}{2} - \frac{2}{5} = \frac{5}{10} - \frac{4}{10} = \frac{1}{10} \] \[ a = \frac{1}{20} \] Then, \[ b = \frac{1}{5} - 2 \times \frac{1}{20} = \frac{1}{5} - \frac{1}{10} = \frac{2}{10} - \frac{1}{10} = \frac{1}{10} \]

Step 5: Compute \(E(X^2)\)
\[ E(X^2) = \sum_{x=0}^4 x^2 P(X=x) = \sum_{x=0}^4 x^2 (a x + b) = a \sum_{x=0}^4 x^3 + b \sum_{x=0}^4 x^2 \] Calculate sums: \[ \sum_{x=0}^4 x^3 = 0 + 1 + 8 + 27 + 64 = 100 \] \[ \sum_{x=0}^4 x^2 = 30 \] Thus, \[ E(X^2) = a \times 100 + b \times 30 = 100 \times \frac{1}{20} + 30 \times \frac{1}{10} = 5 + 3 = 8 \]

Step 6: Compute variance \(\alpha = Var(X)\)
\[ Var(X) = E(X^2) - [E(X)]^2 = 8 - \left(\frac{5}{2}\right)^2 = 8 - \frac{25}{4} = \frac{32}{4} - \frac{25}{4} = \frac{7}{4} \]

Step 7: Compute \(24 \alpha\)
\[ 24 \alpha = 24 \times \frac{7}{4} = 6 \times 7 = 42 \]

Final Answer:
\[ \boxed{42} \]

Answer 2

Let \(P(X=x) = mx + c\) for \(x = 0, 1, 2, 3, 4\). Since \(P(X=x)\) is a probability mass function, we must have \(\sum_{x=0}^4 P(X=x) = 1\). 
So, \(\sum_{x=0}^4 (mx + c) = 1\). 
\(c + (m+c) + (2m+c) + (3m+c) + (4m+c) = 1\) 
\(10m + 5c = 1\). 
The mean of \(X\) is given by \(\mu = E[X] = \sum_{x=0}^4 xP(X=x) = \frac{5}{2}\). 
\(\sum_{x=0}^4 x(mx+c) = 0(c) + 1(m+c) + 2(2m+c) + 3(3m+c) + 4(4m+c) = \frac{5}{2}\) \(m+c + 4m+2c + 9m+3c + 16m+4c = \frac{5}{2}\) 
\(30m + 10c = \frac{5}{2}\) \(60m + 20c = 5\) \(12m + 4c = 1\)

We have two equations: \(10m + 5c = 1 \Rightarrow 2m + c = \frac{1}{5}\) 
\(12m + 4c = 1 \Rightarrow 3m + c = \frac{1}{4}\) 
Subtracting the first equation from the second equation, we have \((3m+c) - (2m+c) = \frac{1}{4} - \frac{1}{5} = \frac{5-4}{20} = \frac{1}{20}\) 
\(m = \frac{1}{20}\) 
Substituting \(m = \frac{1}{20}\) in \(2m + c = \frac{1}{5}\), 
we have \(2(\frac{1}{20}) + c = \frac{1}{5}\) \(\frac{1}{10} + c = \frac{1}{5}\) 
\(c = \frac{1}{5} - \frac{1}{10} = \frac{2-1}{10} = \frac{1}{10}\) 
Thus, \(P(X=x) = \frac{1}{20}x + \frac{1}{10}\). 
\(E[X^2] = \sum_{x=0}^4 x^2 P(X=x) = \sum_{x=0}^4 x^2(\frac{1}{20}x + \frac{1}{10}) = \sum_{x=0}^4 (\frac{x^3}{20} + \frac{x^2}{10})\) \(E[X^2] = \frac{1}{20} \sum_{x=0}^4 x^3 + \frac{1}{10} \sum_{x=0}^4 x^2 = \frac{1}{20} (0^3 + 1^3 + 2^3 + 3^3 + 4^3) + \frac{1}{10} (0^2 + 1^2 + 2^2 + 3^2 + 4^2)\) 
\(= \frac{1}{20}(0 + 1 + 8 + 27 + 64) + \frac{1}{10}(0 + 1 + 4 + 9 + 16) = \frac{100}{20} + \frac{30}{10} = 5 + 3 = 8\) 
Variance \(\alpha = Var(X) = E[X^2] - (E[X])^2 = 8 - (\frac{5}{2})^2 = 8 - \frac{25}{4} = \frac{32-25}{4} = \frac{7}{4}\). 
\(24\alpha = 24(\frac{7}{4}) = 6 \cdot 7 = 42\).

Final Answer: The final answer is \(\boxed{42}\)