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zysi [14]
3 years ago
11

Suppose the discrete random variable X has the probability distribution below:

Mathematics
1 answer:
Talja [164]3 years ago
4 0

(a) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.11 + 0.52 + 0.19 = 0.82

(b) P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.52 + 0.19 + 0.12 + 0.06 = 0.89

(c) µ = 0×0.11 + 1×0.52 + 2×0.19 + 3×0.12 + 4×0.06 = 1.5

(d) σ² = (0²×0.11 + 1²×0.52 + 2²×0.19 + 3²×0.12 + 4²×0.06) - µ² = 1.07

σ = √(σ²) ≈ 1.03

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