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 8 of these refrigerators have a defective compressor and the other 4 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 PCX S 5). (Round your answers to four decimal places.) P(X = 5) = P(X 55) = (b) Determine the probability that X 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 20 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 ---Select-- n = and p = M/N = Approximate P(X s 2) using that method. (Round your answer to three decimal places.) PIX s 2) =

Answer 1

& Solution! - net X: No. of refrigerators having defective compressor. x =40,1,2,...} p=prob. of having defective compreses 3 = 0.67 .-91-P = 0.33 Here X NBinomial (n=7, P = 0.67) . Its pmf is P(x)= x (0.670 (0.337, n=0) 2.7 ow a) P(x+5) = 7C (0.67)5(0.33)2 - 0-3088 P(X85) = 1-P(x >5) =1-[ P(6) + P(7)] = 1-["66 (0-6736(0.33)' +7C760-67360335) -1-(0.2089 +0.0606) P(X 55) = 0.7305 b) Here mean = np=760.67) = 4.69 Std.der. = Inpg =760.6) (0.33) = 1.2441 ..mean & I std.dev = 4.69 + 1.2441=5.9340 . Poob that x enceeds ean value by more than 1 std dev .= P(x> 5.9340) = P(6) +P(7) - 0.2089 +0.0606 (from port (a) = 0.2695 c) Gives – n=20, p=prob. of defective compressor = 5000l ex No. of refrigerators with defective compressors x=40, 12 o} ,9=1-p= 0.9 70 Binomial (n=29 p=0.1) . Its poot is pla)= 20 cm (0.1) (0.9) 20- X=011---20 8(x42) = P(O)+P (1) + P (2) = 2000 (0.1) (0.g) 20 + 200 Co.1) (0.9) 19+ 20C2C0.))(0.9) - 0.1216 + 0.2702 + 0.2852 P(X82) = 0.677