Question

A Random Variable X can take only two values, 4 and 5 such that P(4) = 0.32 and P(5) = 0.47. Determine the Variance of X.

(a) 8.21

(b) 12

(c) 3.7

(d) 4.8

I got this question in an interview for job.

Origin of the question is Discrete Probability topic in section Discrete Probability of Discrete Mathematics

Answer 1

Right answer is (c) 3.7

To explain: Expected Value: μ = E(X) = ∑x * P(x) = 4 × 0.32 + 5 × 0.47 = 3.63. Next find ∑x^2 * P(x): ∑x^2 * P(x) = 16 × 0.32 + 25 × 0.47 = 16.87. Therefore, Var(X) = ∑x^2P(x) − μ^2 = 16.87 − 13.17 = 3.7.