The personnel office at a large electronics firm regularly schedules job intervi

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 = 33 minutes and standard deviation = 8 minutes.

(a) What is the probability that a first interview will last 40 minutes or longer? (Use 3 decimal places.)


(b) Two first interviews are usually scheduled per day. What is the probability that the average length of time for the two interviews will be 40 minutes or longer? (Use 3 decimal places.)
  

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    40      
u = mean =    33      
          
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) / s =    0.875      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.875   ) =    0.190786953 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    40      
u = mean =    33      
n = sample size =    2      
s = standard deviation =    8      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    1.237436867      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.237436867   ) =    0.107962469 [ANSWER]

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