Question

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. 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?

Answer 1

Answer:

(5.871 `; 6.127)

Step-by-step explanation:

Given :

Mean = 6

Standard deviation. σ = 0.5

Samplr size, n = 100

Zcritical at 99% confidence = 2.58

The confidence interval :

Mean ± margin of error

Margin of Error = Zcritical * σ / sqrt(n)

Margin of Error = 2.58 * 0.5/sqrt(100) = 0.129

Confidence interval :

Lower boundary = 6.00 - 0.129 = 5.871

Upper boundary = 6.00 + 0.129 = 6.129

(5.871 `; 6.127)