Question

The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a Normal distribution with mean value 10 min and standard deviation 2 min. What is the probability that the amount of time taken is at most 12 min?

Answer 1

Normal distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = 10

Standard deviation = 2

P(at most 12) = P(X $\leq$ 12)

= P(Z < (12 - 10)/2)

= P(Z < 1)

= 0.8413

Answer 2

.44=10 o=2 p(x+12)P (25 12-10 X-M Ze (2 [P-value from 2-table - P (132 0.8413

Answer 3

X = Time taken by applicant for a montage to in out form xNC10,2) we want to find P ( X512) P z s 12-16 -10) 2 P ( 2 ) from cummulative normal table 0.8413