Question

The personnel office at a large electronics firm regularly schedules job interviews and maintains records of the interviews. From the past records, they have found that the length of a first interview is normally distributed, with mean μ = 36 minutes and standard deviation σ = 5 minutes. (Round your answers to four decimal places.)

(a) What is the probability that a first interview will last 40 minutes or longer?


(b) Three first interviews are usually scheduled per day. What is the probability that the average length of time for the three interviews will be 40 minutes or longer?

Answer 1

Solution :

Given that,

mean = = 36

standard deviation = = 5

(A)P(x >40 ) = 1 - P(x<40 )

= 1 - P[(x -) / < (40-36) /5 ]

= 1 - P(z <0.8 )

Using z table

= 1 -  0.7881

probability=0.2119

(B)

n=3

= =36

= / n = 5 / 3 = 2.8868

P( >40 ) = 1 - P( < 40)

= 1 - P[( - ) / < (40-36) / 2.8868]

= 1 - P(z < 1.39)

Using z table

= 1 - 0.9177

= 0.0823

probability=0.0823