Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7.5 grams of fat per pound. A random sample of 40 farm-raised trout is selected. The mean fat content for the sample is 31.5 grams per pound. Find the probability of observing a sample mean of 31.5 grams of fat per pound or less in a random sample of 40 farm-raised trout.
Question
Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per pound, with a standard deviation of 7.5 grams of fat per pound. A random sample of 40 farm-raised trout is selected. The mean fat content for the sample is 31.5 grams per pound. Find the probability of observing a sample mean of 31.5 grams of fat per pound or less in a random sample of 40 farm-raised trout.
Solution
To solve this problem, we will use the concept of z-score in statistics. The z-score is a measure of how many standard deviations an element is from the mean.
Step 1: Identify the given values. The population mean (μ) = 32 grams of fat per pound The standard deviation (σ) = 7.5 grams of fat per pound The sample mean (x̄) = 31.5 grams of fat per pound The sample size (n) = 40
Step 2: Calculate the standard error of the mean (SE). The standard error of the mean is a statistical term that measures the accuracy with which a sample represents a population. In statistics, the standard error equals the standard deviation divided by the square root of the sample size. SE = σ / √n = 7.5 / √40 = 1.187
Step 3: Calculate the z-score. The z-score is the number of standard errors that x̄ is away from μ. z = (x̄ - μ) / SE = (31.5 - 32) / 1.187 = -0.42
Step 4: Look up the z-score in the z-table to find the probability. The z-table shows the probability that a z-score is less than -0.42 is 0.3372.
So, the probability of observing a sample mean of 31.5 grams of fat per pound or less in a random sample of 40 farm-raised trout is 0.3372 or 33.72%.
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