Question

5. The average age of men at the time of their first marriage is 24.8 years. Suppose the

standard deviation is 2.8 years. Forty-nine married males are selected at random and asked the
age at which they were first married. Find the probability that the sample mean will be more than 26.

Answer 1

0.0013 = 0.13% probability that the sample mean will be more than 26.

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 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.

The average age of men at the time of their first marriage is 24.8 years. Suppose the standard deviation is 2.8 years.

This means that
`\mu = 24.8, \sigma = 2.8`

Forty-nine married males are selected at random and asked the age at which they were first married.

This means that
`n = 49, s = (2.8)/(√(49)) = 0.4`

Find the probability that the sample mean will be more than 26.

This is 1 subtracted by the pvalue of Z when X = 26. So

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

By the Central Limit Theorem

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

`Z = (26 - 24.8)/(0.4)`

`Z = 3`

`Z = 3` has a pvalue of 0.9987

1 - 0.9987 = 0.0013

0.0013 = 0.13% probability that the sample mean will be more than 26.