Question

2. There are 30 calculator batteries in a box; 22 are new and the rest are used. I randomly select 4 batteries and put them into my calculator. Let X count the number of new batteries in the group of 4 selected from the box a) What is the expected number of new batteries in the group of 4 selected? b) What is the standard deviation of the number of new batteries in the group of 4 selected? c) What values of the random variable are within 2 standard deviations of the mean? d) What is the probability that none of the four selected is a new battery? e) In order for my calculator to work, at least two of the batteries must be new. What is the probability that my calculator works?

Answer 1

(a)

x has a hyper geometric distribution with N = 30, M = 22, n = 4

P(x: N, n, M) = C(M, x) * C(N - M, n - x) / C(N, n)

P(x)-C(22, x) * C(30-22, 4-х) / C(30, 4) 0 2 3 4 0.0026 0.0450 0.2360 0.4496 0.2669

P(x) x P(x) xA2 P(x) 0 0.0026 1 0.0450 0.044955 0.044955 2 0.2360 0.472031 0.944061 3 0.4496 1.348659 4.045977 4 0.2669 1.067688 4.270754 0 1.0000 2.9333 9.3057

Expected number of new batteries μ = ∑ x P(x) = 2.93

(b) Variance of the number of new batteries = ∑ x^2 P(x) – μ^2 = 9.3057 – 2.9333^2 = 0.7015

Standard deviation σ = √variance = 0.8375

(c) μ – 2 σ = 2.9333 – 2 * 0.8375 = 1.258 and μ + 2 σ = 2.9333 + 2 * 0.8375 = 4.608

Number of values in the range [1.258, 4.608] = 3 (2, 3, 4 are the values)

(d) P(0) = 0.0026

(e) P(x ≥ 2) = P(2) + P(3) + P(4) = 0.9525

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