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.The quality analysts will compute the mean length for the sample.They want to establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99.What should these limits be?
A)Approximately ± 0.13 inches
B)Within the approximate range of 5.87 inches to 6.13 inches
C)Within the range of about 4.71 inches to 7.29 inches
D)Approximately ± 1.29 inches

Question

Quizplus · Accepted Answer

To establish limits on either side of the 6.00 inch mean so that the chance of the sample mean falling within the limits is 0.99, we need to find the values of z that satisfy the condition:

P(-z < Z < z) = 0.99

From the standard normal distribution table, the value of z for the area between -z and z being 0.99 is 2.576. We can use this to find the limits as:

Lower limit = 6.00 - (2.576 * 0.50 / sqrt(100)) = 5.87 inches
Upper limit = 6.00 + (2.576 * 0.50 / sqrt(100)) = 6.13 inches

Thus, the limits are within the approximate range of 5.87 inches to 6.13 inches, which is choice B.

The JRSimplot Company Produces Frozen French Fries That Are Then Sold