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 = 9 minutes. (Round your answers to four decimal places.)

1.What is the probability that a first interview will last 40 minutes or longer?

2. Twelve first interviews are usually scheduled per day. What is the probability that the average length of time for the twelve interviews will be 40 minutes or longer?

Answer 1

Answer is in detail below. If you've not understood any part of it, please write back. Have a good weekend!

We have given that X follows normal distribution

We have also been given the parameters of normal distribution, mean = 36, standard deviation = 9

1.

probability that a first interview will last 40 minutes or longer = P(X>=40)

P(X>=40)

Lets convert the raw score into Z score by standardizing it.

To standardize it we use the formula , Z = (X-Mu)/SD = (X-36)/9

= P(Z>= (40-36)/9 )

= P(Z>=.4444)

= 1- P(Z<=.4444)

We can either use the formula NORMSDIST(Z) to find the cumulative probability , or the Z table to convert Z score to a probability

= 1- .6716

= .3284

Answer is 0.3284

2.

The sample size is 12. We need to now find P(Xbar >= 40) = ?

To solve this we have to first standardize score for the sample, we do this by :

Z = (Xbar - Mean)/(standard deviation/sqrt(n)) = (40-36)/(9/sqrt(12)) = 1.540

So, P(Xbar > 40) = P(Z> (1.540) = .0618 [Used 1 - NORMSDIST(1.540) on excel]

Answer is 0.0618