Question

Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor.

(a) Calculate P(X = 4) and P(X ≤ 4).

(b) Determine the probability that X exceeds its mean value by more than 1 standard deviation.

(c) Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 10 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X ≤ 2) than to use the hypergeometric pmf. We can approximate the hypergeometric distribution with the (choose one: negative binomal, geometricm or binomal) distribution if the population size and the number of successes are large. Here n = (blank) and p = M/N = (blank). Approximate P(X ≤ 2) using that method.

Answer 1

(a)

Let

N is the population size,
K is the number of defective compressors in the population,
n is the number of examined refrigerators
k is the number of observed defective compressor

Given, N = 12, K = 7, n = 6

Then X ~ HyperGeometric(N = 12, K = 7, n = 6) with PMF

(b)

Mean = nK / N = 6 * 7 / 12 = 3.5

Variance = (nK / N) * ((N-K) /N) * ((N - n) / (N -1))

= (6 * 7 / 12) * ((12-7) /12) * ((12 - 6) / (12 -1))

= 0.7954

Standard deviation = = 0.8919

Probability that X exceeds its mean value by more than 1 standard deviation = P(X > 3.5 + 0.8919)

= P(X > 4.39) = P(X 5)

= P(X = 5) + P(X = 6)

(c)

Here,

N = 400, K = 40, n = 10

p = K / N = 40 / 400 = 0.1

If the population size and the number of successes are large, then we can approximate the hypergeometric distribution with the binomial distribution and X ~ Binomial(n = 10, p = 0.1)

P(X 2) = P(X = 0) + P(X = 1) + P(X = 2)

=  0.3486784 + 0.3874205 + 0.1937102

= 0.9298