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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) 4.8I got this question in an interview for job.Origin of the question is Discrete Probability topic in section Discrete Probability of Discrete Mathematics

Answer»

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.



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