Question

. The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal probability distribution with average time 10 minutes and a standard deviation of 2 minutes. If five individuals fill out the form on Day 1 and six individuals fill out the form on Day 2, what is the probability that the sample average time taken is less than 11 minutes for BOTH days?

Answer 1

Answer:

Probability that the sample average time taken is less than 11 minutes for Day 1 is 0.86864.

Probability that the sample average time taken is less than 11 minutes for Day 2 is 0.88877.

Step-by-step explanation:

We are given that the time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal probability distribution with average time 10 minutes and a standard deviation of 2 minutes.

Also, five individuals fill out the form on Day 1 and six individuals fill out the form on Day 2.

(a) Let $\bar X$ = sample average time taken

The z score probability distribution for sample mean is given by;

                                Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 10 minutes

            $\sigma$ = standard deviation = 2 minutes

            n = sample of individuals fill out form on Day 1 = 5

Now, the probability that the sample average time taken is less than 11 minutes for Day 1 is given by = P($\bar X$ < 11 minutes)

         P($\bar X$ < 11 minutes) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{11-10}{\frac{2}{\sqrt{5} } }$ ) = P(Z < 1.12) = 0.86864

The above probability is calculated by looking at the value of x = 1.12 in the z table which has an area of 0.86864.

(b) Let $\bar X$ = sample average time taken

The z score probability distribution for sample mean is given by;

                                Z  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 10 minutes

            $\sigma$ = standard deviation = 2 minutes

            n = sample of individuals fill out form on Day 2 = 6

Now, the probability that the sample average time taken is less than 11 minutes for Day 2 is given by = P($\bar X$ < 11 minutes)

         P($\bar X$ < 11 minutes) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{11-10}{\frac{2}{\sqrt{6} } }$ ) = P(Z < 1.22) = 0.88877

The above probability is calculated by looking at the value of x = 1.22 in the z table which has an area of 0.88877.