Asked by laura
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 of0.50 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?
Answers
Answered by
PsyDAG
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
SEdiff = √(SEmean1^2 + SEmean2^2)
SEm = SD/√n
If only one SD is provided, you can use just that to determine SEdiff.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
Answered by
oojaj
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Answered by
Sha
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