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 less than 20 substandard welds?
A) 0.9066
B) 0.4066
C) 0.0934
D) 0.5934

Question

Quizplus · Accepted Answer

This problem can be solved 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 in 300 checks is n*p = 300*0.05 = 15, and the standard deviation (σ) is sqrt(n*p*(1-p)) = sqrt(300*0.05*0.95) ≈ 3.63. To find the probability of finding fewer than 20 substandard welds, we calculate the Z-score for 19.5 (using 19.5 for continuity correction) and then find the corresponding cumulative probability from the Z-table. The calculation shows a high probability, which matches closest to choice A, indicating it's the correct answer.

Smith Is a Weld Inspector at a Shipyard