Question

The distribution of weights of adult males in a certain county is strongly right-skewed with a mean weight of 185 lbs and standard deviation 16 lbs. What is the probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs

Answer 1

0.962 = 96.2% probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs.

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
`\mu` and standard deviation
`\sigma`, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

The distribution of weights of adult males in a certain county is strongly right-skewed with a mean weight of 185 lbs and standard deviation 16 lbs.

This means that
`\mu = 185, \sigma = 16`

Sample of 100:

This means that
`n = 100, s = (16)/(√(100)) = 1.6`

What is the probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs?

This is the p-value of Z when X = 188 subtracted by the p-value of Z when X = 172. So

X = 188

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

By the Central Limit Theorem

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

`Z = (188 - 185)/(1.6)`

`Z = 1.875`

`Z = 1.875` has a p-value of 0.9620

X = 172

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

`Z = (172 - 185)/(1.6)`

`Z = -8.125`

`Z = -8.125` has a p-value of 0.

0.9620 - 0 = 0.962

0.962 = 96.2% probability that a simple random sample of 100 adult males from this county has a mean weight between 172 and 188 lbs.