Question 1079527
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 = 32 minutes and standard deviation σ = 9 minutes. (Round your answers to four decimal places.)
(a) What is the probability that a first interview will last 40 minutes or longer?
z(40) = (40-32)/9 = 8/9
P(x >= 40) = P(z >= 8/9) = normalcdf(8/9,100) = 0.1870
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(b) Nine first interviews are usually scheduled per day. What is the probability that the average length of time for the nine interviews will be 40 minutes or longer?
z(40) = (40-32)/(9/sqrt(9)) = 8/3
P(x-bar >= 40) = P(z >= 8/3) = normalcdf(8/3,100) = 0.0038
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cheers,
Stan H.
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