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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 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. Correct Answer 3.7

Expected Value: μ = E(X) = ∑x * P(x) = 4 × 0.32 + 5 × 0.47 = 3.63. Next find ∑x2 * P(x): ∑x2 * P(x) = 16 × 0.32 + 25 × 0.47 = 16.87. Therefore, Var(X) = ∑x2P(x) − μ2 = 16.87 − 13.17 = 3.7.

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