Question

Recently, the average amount of time to foreclose on a house in the U.S. was reported to be 359 days. Assume that the standard deviation for this population is 90.4 days. A random sample of 42 homes that have completed the foreclosure process was selected. What is the probability that the sample average was less than 375 days?

Answer 1

87.49% probability that the sample average was less than 375 days

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 = 359, \sigma = 90.4, n = 42, s = (90.4)/(√(42)) = 13.95`

What is the probability that the sample average was less than 375 days?

This is the pvalue of Z when X = 375. So

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

By the Central Limit Theorem

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

`Z = (375 - 359)/(13.95)`

`Z = 1.15`

`Z = 1.15` has a pvalue of 0.8749.

87.49% probability that the sample average was less than 375 days