# STA3015Solution

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```					STA301 Solution
E(X) = np = mean of a binomial distribution = 40 Var(X) = npq = variance of a binomial distribution =? S.D(X) = √npq = standard deviation of a binomial distribution = 6 So, Var(X) = [S.D(X)] 2 Var(X) = [√npq] 2 Var(X) = [6] 2 Var(X) = 36 Var(X) = npq 36 = (40)q q = 36/40 q = 0.9 p=1–q p = 1 – 0.9 p = 0.1 E(X) = np 40 = n(0.1) n = 40/0.1

n = 400 OR Var(X) = npq 36 = n(0.1)(0.9) n(0.09) = 36 n = 36/0.09 n = 400

Answer (b): We know that: P(X=x) = (μx e-μ)/x! for x = 0, 1, 2, …, ∞ Here; e = 2.71828 n = 500 p = 0.1% or 0.001 μ = np μ = 500 * 0.001 μ = 0.5 Our Poisson formula becomes: p(x; 0.5) = (0.5x e-0.5)/x!

i. No defective p(x; 0.5) = (0.5x e-50)/x! p(0; 0.5) = (0.50 (2.71828)-0.5)/0! p(0; 0.5) = (1 * 0.606531) p(0; 0.5) = 0.606531 or 60.65% chances that there will be no defective. ii. At least two defective P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + …. P(X ≥ 2) = 1 – P(X=1) – P(X=0) So, first find P(X=1): p(x; 0.5) = (0.5x e-0.5)/x! p(1; 0.5) = (0.51 (2.71828)-0.5)/1! p(1; 0.5) = (0.5 * 0.606531)/1 p(1; 0.5) = 0.303265 or 30.33% chances that there will be exactly one defective P(X ≥ 2) = 1 – P(X=1) – P(X=0) P(X ≥ 2) = 1 – 303265 – 606531 P(X ≥ 2) = 0.090204 or 9.02% chances that at least 2 will be defective

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