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

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $s = \frac{\sigma}{\sqrt{n}}$.

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

$n = 5, s = \frac{3}{\sqrt{5}} = 1.34$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 8}{1.34}$

$Z = 2.24$

$Z = 2.24$ 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

$n = 6, s = \frac{3}{\sqrt{5}} = 1.2247$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11 - 8}{1.2247}$

$Z = 2.45$

$Z = 2.45$ 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.