The J.R. Simplot Company produces frozen French fries that are then sold to customers such as McDonald's. The "prime" line of fries has an average length of 6.00 inches with a standard deviation of 0.50 inch. To make sure that Simplot continues to meet the quality standard for "prime" fries, they plan to select a random sample of n = 100 fries each day. Yesterday, the sample mean was 6.05 inches. What is the probability that the mean would be 6.05 inches or more if they are meeting the quality standards?
A) 0.2350
B) 0.3413
C) 0.9413
D) 0.1587

Question

Quizplus · Accepted Answer

To find the probability that the sample mean is 6.05 inches or more, we use the Z-score formula for a sample mean: Z = (X̄ - μ) / (σ/√n), where X̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the values: Z = (6.05 - 6.00) / (0.50/√100) = 0.05 / 0.05 = 1. The Z-score of 1 corresponds to a probability of 0.8413 to the left of it. Therefore, the probability of having a mean of 6.05 inches or more is 1 - 0.8413 = 0.1587.

The JR Simplot Company Produces Frozen French Fries That Are Then