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 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P(X ?5) than to use the hypergeometric pmf.

Answer: Step 1 of 3 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. Taking everything mentioned above into consideration we can see that this is a hypergeometric distribution, where h(x;n, M,N) = h(x;6, 7, 12). a. Calculate P(X = 4) and P(X 4) Compute P(X = 4), P(X 4) The Hypergeometric distribution formula is as shown below i) We have, MC x (N M)C (n x) P(X = x) = NC n 7C 4 (12 7)C(6 4) P(X = 4) = 12C 6 P(X = 4) = 924 = 0.3787 Therefore, P(X = 4) = 0.3787 ii) Now, P(X 4) = 1 P(X > 4) P(X 4) = 1 (P(X = 5) + P(X = 6)) 7C5 (12 7(6 5)7C 6 (12 7(6 6) = 1 ( 12C 6 + 12C6 ) 105 7 = 1 ( 924 + 924 ) = 1 - 0.1212 = 0.8788 Hence,P(X 4) = 0.8788. Step 2 of 3 b. Determine the probability that X exceeds its mean value by more than 1 standard deviation. Calculate the mean value and standard deviation of X. h(x;n, M,N) = h(x;6, 7, 12). We know that N = 12, M = 7, n = 6 nM E(X) = N = 42 12 3.5 SD(X) = N n× nM (1 M ) N 1 N N 12 6 42 7 = 12 1 12(1 12) = 0.5454 × 1.4583 = 0.8918 Therefore, the mean value and standard deviation of X is 3.5 and 0.8918. Hence, + = 4.39 The probability that X exceeds its mean value by more than 1 standard deviation is ( > + ) = 1 ( < + ) = 1 ( + ). We used Excel function “=HYPGEODIST(x ;n, M,N)” (4.39) = HYPGEODIST(4.39,6,7,12, TRUE) = 0.8788; ( > + ) = 1 0.8788 = 0.1212.