Question

Despite its nutritional value, seafood is only a tiny part of the American diet, with the average American eating just 16 pounds of seafood per year with a standard deviation of 3.2 pounds. Janice takes a random sample of 42 Americans. Find the probability that the sample mean seafood consumption of the 42 Americans is more than 17.5 pounds?

Answer 1

0.12% probability that the sample mean seafood consumption of the 42 Americans is more than 17.5 pounds

Explanation:

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

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore 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 pvalue, 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.

In this problem, we have that:

`\mu = 16, \sigma = 3.2, n = 42, s = (3.2)/(√(42)) = 0.4938`

Find the probability that the sample mean seafood consumption of the 42 Americans is more than 17.5 pounds?

This is 1 subtracted by the pvalue of Z when X = 17.5. So

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

By the Central Limit Theorem

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

`Z = (17.5 - 16)/(0.4938)`

`Z = 3.04`

`Z = 3.04` has a pvalue of 0.9988

1 - 0.9988 = 0.0012

0.12% probability that the sample mean seafood consumption of the 42 Americans is more than 17.5 pounds