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 = 38 minutes and standard deviation σ = 6 minutes. (Round your answers to four decimal places.)

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


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

Answer 1

Solution :

Given that ,

mean = = 38

standard deviation = = 6

a)

P(x 40 ) = 1 - P(x 40 )

= 1 - P((x - ) / (40 - 38) / 6)

= 1 - P(z 0.33)

= 1 - 0.6293 Using standard normal table.

= 0.3707

The probability that a first interview will last 40 minutes or longer is 0.3707.

b)

n = 17

=   = 38 and

= / n = 6 / 17= 1.4552

P( 40) = 1 - P( 40)

= 1 - P(( - ) / (40 - 38) / 1.4552)

= 1 - P(z 1.37)

= 1 - 0.9147 Using standard normal table.

= 0.0853

The probability that the average length of time for the seventeen interviews will be 40 minutes or longer is 0.0853.