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 34 farm-raised trout is selected. The mean fat content for the sample is 30.5 grams per pound. Find the probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.

Answer 1

0.121 = 12.1% probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

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.

This means that
`\mu = 32, \sigma = 7.5`

Sample of 34:

This means that
`n = 34, s = (7.5)/(√(34)) = 1.2862`

Find the probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.

This is the pvalue of Z when X = 30.5. So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (30.5 - 32)/(1.2862)`

`Z = -1.17`

`Z = -1.17` has a pvalue of 0.121

0.121 = 12.1% probability of observing a sample mean of 30.5 grams of fat per pound or less in a random sample of 34 farm-raised trout.