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 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 1

a) 0.2010, 0.799

(b) 0.2082

(c) 0.9952

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