Each of 11 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 8 of these refrigerators have a defective compressor and the other 3 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 7 examined that have a defective compressor
(a) Calculate P(X -5) and P(X S 5). (Round your answers to four decimal places.) P(X = 5) =
(b) Determine the probability that χ exceeds its mean value by more than 1 standard deviation. Round your answer to four decimal places.
(c) Consider a large shipment of 500 refrigerators, of which 50 have defective compressors. If X is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X S 2) than to use the hypergeometric pmf distribution if the population size and the number of successes are large. Here We can approximate the hypergeometric distribution with the binomial n= and p = M/N = Approximate P(X S 2) using that method. (Round your answer to three decimal places.) P(X 2) 0.8183 X

Question

contestada

Respuesta :

fichoh fichoh · Answer 1 · 2021-03-16T06:29:43+01:00

Answer:

Kindly check explanation

Step-by-step explanation:

1.)

p = 7/8 = 0.875 ; 1 - p = 1 - 0.875 = 0.125

Using the relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

P(x = 5) = 8C5 * 0.875^5 * 0.125^3 = 0.0561

P(x ≤ 5) = p(x=0)+p(x=1)+p(x=2)+p(x=3)+p(x=4)+p(x=5) = 0.0673 (binomial distribution calculator)

2.) mean value, m = np = 8 * 0.875 = 7

P(x > 7) = 8C7 * 0.875^7 * 0.125^1 = 0.344

C.)

p = 15 / 50 = 0.3 ; 1-p = 1 - 0.3 = 0.7

P(x =x) = nCx * p^x * (1 - p)^(n - r)

P(x ≤2) = p(x=0)+p(x=1)+p(x=2)

P(x=0) = 15C0 * 0.3^0 * 0.7^15 = 0.0047

P(x=1) = 15C1 * 0.3^1 * 0.7^14 = 0.0305

P(x=2) = 15C2 * 0.3^2 * 0.7^13 = 0.0916

(0.0047 + 0.0305 + 0.0916) = 0.1268