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

Answer:

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

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 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:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 16, \sigma = 3.2, n = 42, s = \frac{3.2}{\sqrt{42}} = 0.4938$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{17.5 - 16}{0.4938}$

$Z = 3.04$

$Z = 3.04$ 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