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:

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.

Step-by-step 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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

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 [tex]\mu = 32, \sigma = 7.5[/tex]

Sample of 34:

This means that [tex]n = 34, s = \frac{7.5}{\sqrt{34}} = 1.2862[/tex]

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{30.5 - 32}{1.2862}[/tex]

[tex]Z = -1.17[/tex]

[tex]Z = -1.17[/tex] 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.

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