Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that forthe afternoon shift 5% of all welds done will be substandard. If Smith checks 300 of the 7500 welds completedthat shift, what is the probability that he will find between 10 and 20 substandard welds?
A) 0.8132
B) 0.4066
C) 0.2033
D) 0.6377

Question

Quizplus · Accepted Answer

This problem can be approached using the binomial probability formula, but given the large number of trials (n = 300) and the small probability of success (p = 0.05 for a weld being substandard), it's more practical to use the normal approximation to the binomial distribution. The mean (μ) of substandard welds is np = 300 * 0.05 = 15, and the standard deviation (σ) is sqrt(np(1-p)) = sqrt(300*0.05*0.95) ≈ 3.63. Using the normal distribution, we find the probability of observing between 10 and 20 substandard welds. The Z-scores for 10 and 20 are (10-15)/3.63 ≈ -1.38 and (20-15)/3.63 ≈ 1.38, respectively. Looking these up in the Z-table or using a calculator for the cumulative probabilities and subtracting gives us approximately 0.8132, indicating the probability of finding between 10 and 20 substandard welds.

Smith Is a Weld Inspector at a Shipyard