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

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

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.

For sums the theorem can also be used, with mean
`n*\mu` and standard deviation
`s = √(n)*\sigma`

In this problem, we have that:

`n = 47, \mu = 47*167 = 7849, s = √(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

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

`Z = (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.