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 inches.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 determine the probability, we need to calculate the z-score:

z = (sample mean - population mean)/(standard deviation/sqrt(sample size))

z = (6.05 - 6)/[0.5/sqrt(100)] = 1

Using a standard normal table or calculator, the probability of getting a z-score of 1 or greater is 0.1587. Therefore, the probability that the mean would be 6.05 inches or more if they are meeting the quality standards is 0.1587 or 15.87%. The correct answer is D.

The JRSimplot Company Produces Frozen French Fries That Are Then Sold