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.

Question

Zane Mccormick

Mathematics

5 July, 05:09

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.

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Answers (1)

Know the Answer?

William Flynn · Answer 1

a) 0.2010, 0.799

(b) 0.2082

(c) 0.9952

Step-by-step explanation:

Since there are 7 defective refrigerator out of 12

p = 7/12 = 0.5833, q = 0.4167

a.) Given n = 6

P (X = 4) = 6C4 * (0.5833) ⁴ * (0.4167) ²

= 0.2010

P (<=4) = 1 - P (X=4)

= 0.799

(b) if X is saidvto exceed it mean by 1

We calculate the probability for

X+1, X+2

X+1 = 5

X+2 = 6

P (X = 5 or 6) = 6C5 * (0.5833) ^5 * (0.4167) + 6C6 * (0.5833) ^6 * (0.4167) ^0.

= 0.16882 + 0.039387

= 0.2082

(c) p = 40/400 = 0.1, q = 0.9

n = 15, and X< = 5

We could adopt the normal distribution.

Where mean (u) = np = 15 * 0.1 = 1.5

Sd = npq = 15 * 0.1 * 0.9 = 1.35

P (X<=5) = P (Z<=5)

= P (Z < = X-U/Sd)

= P (Z< = 2.59)

= 0.99520