Question

Data from the Centers for Disease Control and Prevention indicate that weights of American adults in 2005 had a mean of 167 pounds and a standard deviation of 35 pounds. Use this information to estimate the probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005. Explain your reasoning and justify your calculations throughout.

Answer 1

Answer:

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

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.

For sums the theorem can also be used, with mean $n*\mu$ and standard deviation $s = \sqrt{n}*\sigma$

In this problem, we have that:

$n = 47, \mu = 47*167 = 7849, s = \sqrt{47}*35 = 240$

Use this information to estimate the probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.

This is 1 subtracted by the pvalue of Z when X = 7500. So

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

By the Central Limit Theorem

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

$Z = \frac{7500 - 7849}{240}$

$Z = -1.45$

$Z = -1.45$ has a pvalue of 0.0735

1 - 0.0735 = 0.9265

92.65% probability that the total weight in a random sample of 47 American adults exceeds 7500 pounds in 2005.