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?

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.

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To find the probability that the mean would be 6.05 inches or more if they are meeting the quality standards, we can use the concept of the sampling distribution of the sample mean.

The sampling distribution of the sample mean is a probability distribution that represents all possible values of the sample mean for different sample sizes that can be drawn from a population. The sampling distribution of the sample mean follows a normal distribution when the sample size is large enough, due to the Central Limit Theorem.

In this case, we are given that the population has a mean of 6.00 inches and a standard deviation of 0.50 inch. Since the sample size is large (n = 100), we can assume that the sampling distribution of the sample mean will follow a normal distribution.

To calculate the probability that the sample mean would be 6.05 inches or more, we need to calculate the z-score and then find the area under the normal distribution curve to the right of this z-score.

The z-score is calculated using the formula:

z = (x - μ) / (σ / √n)

Where:
x = sample mean
μ = population mean
σ = population standard deviation
n = sample size

Plugging in the given values:
x = 6.05 inches
μ = 6.00 inches
σ = 0.50 inch
n = 100

z = (6.05 - 6.00) / (0.50 / √100)
= 0.05 / (0.50 / 10)
= 0.05 / 0.05
= 1

Now, we need to find the area under the normal distribution curve to the right of z = 1. We can use a standard normal table or a statistical calculator to find this area.

By looking up the z-score 1 in a standard normal table, we find that the area to the right of z = 1 is approximately 0.1587. This is the probability that the mean would be 6.05 inches or more if they are meeting the quality standards.

Therefore, the probability is approximately 0.1587, or 15.87%.