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 5/2 , and the variance of X is α, then the value of 24α is ______

Answer 1

Correct answer is : 42

\( \sum_{x=0}^4 x P(x)=\frac{5}{2}\)

\( \sum_{x=0}^4 x^2 P(x)=? \)

\((0, P(0)),(1, P(1)),(2, P(2)),(3, P(3)),(4, P(4))\)

\(K=P(1)-P(0)=P(2)-P(1)=P(3)-P(2)=P(4)-P(3) \)

\(P(1)=K+P(0)\)

\(P(2)=2 K+P(0)\)

\(P(3)=3 K+P(0)\)

\(P(4)=4 K+P(0)\)

\(P(0)+P(1)+P(2)+P(3)+P(4)=1\)

\(\Rightarrow 5 P(0)+10 K=1\)

\(K+P(0)+4 K+2 P(0)+9 K+3 P(0)+16 K+4 P(0)=\frac{5}{2}\)

\(30 K+10 P(0)=\frac{5}{2} \)

\(\therefore 10 K=\frac{1}{2} \)

\(K=\frac{1}{20}, P(0)=\frac{1}{10}\)

\(P(1)=\frac{3}{20}, P(2)=\frac{4}{20}\)

\(P(3)=\frac{5}{20}, P(4)=\frac{6}{20}\)

\(\sum_{x=0}^4 x^2 P(x)=8\)

\(\therefore\) Variance = \(8- \frac {25}{4} = \frac {32 -25}{4} =\frac {7}{4}\)

\(\therefore\) \(24 \alpha = \frac {24 \times 7}{4} = 42\)