Question

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

X 0 1 2 3 4
P(X) 0.11 0.52 0.19 0.12 0.06
1pt a right parenthesis space F i n d space P left parenthesis X less than 3 right parenthesis

2pt b right parenthesis space F i n d space P left parenthesis X greater or equal than 1 right parenthesis

2pt c right parenthesis space F i n d space mu subscript X

2pt, 1pt d right parenthesis space F i n d space sigma subscript X squared space a n d space sigma subscript X

Answer 1

(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