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

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.

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 =
`(\bar X-\mu)/((\sigma)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(11-10)/((2)/(√(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 =
`(\bar X-\mu)/((\sigma)/(√(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(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(11-10)/((2)/(√(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.