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

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.

Step-by-step 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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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:

First day

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

By the Central Limit Theorem

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

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

By the Central Limit Theorem

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.