Question

According to R.L. Polk & Co., the average age of cars and light trucks in the U.S. is 10.8 years. Assume that the standard deviation for this population is 3.7 years. A sample of 35 randomly selected vehicles was selected. What is the probability that the sample mean will be greater than 10 years?

Answer 1

10.03% probability that the sample mean will be greater than 10 years

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.

In this question, we have that:

`\mu = 10.8, \sigma = 3.7, n = 35, s = (3.7)/(√(35)) = 0.6254`

What is the probability that the sample mean will be greater than 10 years?

This is 1 subtracted by the pvalue of Z when X = 10. So

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

By the Central Limit Theorem

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

`Z = (10 - 10.8)/(0.6254)`

`Z = -1.28`

`Z = -1.28` has a pvalue of 0.1003

10.03% probability that the sample mean will be greater than 10 years