Compute Probabilities of a Sample Proportion
-Smith is a weld inspector at a shipyard. He knows from keeping track of good and substandard welds that for the afternoon shift $5 \%$ of all welds done will be substandard. If Smith checks 300 of the 7500 welds completed that 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 is a binomial problem that can be approximated using the normal distribution because the sample size is large. The mean ($\mu$) of substandard welds in 300 checks is $300 \times 0.05 = 15$, and the standard deviation ($\sigma$) is $\sqrt{300 \times 0.05 \times 0.95} \approx 3.536$. Using the normal distribution, we find the Z-score for 19.5 welds (to account for continuity correction) as $Z = \frac{19.5 - 15}{3.536} \approx 1.27$. Looking up this Z-score in a standard normal distribution table or using a calculator gives a probability of approximately 0.9066 that Smith will find less than 20 substandard welds.

Compute Probabilities of a Sample Proportion -Smith Is a Weld 5%5 \%5%

Compute Probabilities of a Sample Proportion
-Smith Is a Weld $5 \%$