A zipper manufacturer has found that 1.5% of its zippers are defective.The company averages 4982 zippers per day.The probability of finding between 90 and 100 defects in a given day is:
A) 0.4981.
B) 0.0310.
C) 0.0360.
D) 0.5310.
E) 0.4671.

Question

Quizplus · Accepted Answer

This problem can be approached using the binomial or normal approximation to the binomial distribution. Given the large number of trials (zippers produced per day) and the small probability of success (defect rate), the normal approximation is suitable. The mean (μ) of defects per day is 4982 * 0.015 = 74.73, and the standard deviation (σ) is sqrt(4982 * 0.015 * (1-0.015)) ≈ 8.62. Using the normal distribution, we find the Z-scores for 90 and 100 defects, then find the probabilities corresponding to these Z-scores and subtract to find the probability of having between 90 and 100 defects. The exact calculation might vary slightly based on how the normal distribution is used, but the choice closest to a reasonable calculation is B, indicating a relatively low probability, which aligns with the expectation given the mean and standard deviation.

A Zipper Manufacturer Has Found That 1