Question

Time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 9 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? (Round your answer to four decimal places.)

Answer 1

Answer:

a) The probability that the sample average amount of time taken on each day is at most 11 min.

P(X>11) = 0.2546

Step-by-step explanation:

Step1:-

Given data in a  normal distribution with mean value 9 min and standard deviation 3 min.

Let 'X' be the amount of time taken on each day

Given x = 11 , μ=9min and σ=3min

by using normal distribution

$Z= \frac{x-mean}{S.D} = \frac{11-9}{3}$

Z = 0.66>0

Step2:-

The probability that the sample average amount of time taken on each day is at most 11 min.

P(X>11) = P(z>z₁) = 0.5-A(z₁) ( check diagram)

                          = 0.5 - A(0.66)

                          = 0.5 - 0.2454 ( check normal table)

                          = 0.2546

Conclusion:-

The probability that the sample average amount of time taken on each day is at most 11 min.

P(X>11) = 0.2546