Question

ASK YOUR TEACHER Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 8 min and standard deviation 3 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min

Answer 1

On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.

On the second day, 99.29% probability that the sample average amount of time taken is at most 11 min.

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 problem, we have that:

`\mu = 8, \sigma = 3`

First day

`n = 5, s = (3)/(√(5)) = 1.34`

The probability is the pvalue of Z when X = 11. So

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

By the Central Limit Theorem

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

`Z = (11 - 8)/(1.34)`

`Z = 2.24`

`Z = 2.24` has a pvalue of 0.9875

On the first day, 98.75% probability that the sample average amount of time taken is at most 11 min.

Second day

`n = 6, s = (3)/(√(5)) = 1.2247`

The probability is the pvalue of Z when X = 11. So

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

By the Central Limit Theorem

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

`Z = (11 - 8)/(1.2247)`

`Z = 2.45`

`Z = 2.45` has a pvalue of 0.9929

On the second day, 99.29% probability that the sample average amount of time taken is at most 11 min.