Time taken by a randomly selected applicant for a mortgage t

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 2 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.)

Solution

For the day with 5 individuals:

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    11      
u = mean =    9      
n = sample size =    5      
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.236067977      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.236067977   ) =    0.987326341

For the day with 6 individuals:

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    11      
u = mean =    9      
n = sample size =    6      
s = standard deviation =    2      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.449489743      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   2.449489743   ) =    0.992847061

Thus, for this to happen together,

P(at most 11 min for both days) = 0.987326341*0.992847061 = 0.980264056 [ANSWER]