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The help desk for a software company logs the lengths of times of each call that

ID: 3126621 • Letter: T

Question

The help desk for a software company logs the lengths of times of each call that technicians have with customers. Historical records based on the past 3 years indicate that call length (L) is a normally distributed variable with an average () of 8.00 minutes and a standard deviation () of 1.15 min. For each of the problems below, find the appropriate probability.

A) P(L 5.34)    

B) P(L > 8.08)   

C) P(7.28 L < 9.69)   

D) The call times that enclose the central 89% of the distribution are:
    Note: Round your answers to two decimal places for this problem.
    Low: to High:

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    5.34      
u = mean =    8      
          
s = standard deviation =    1.15      
          
Thus,          
          
z = (x - u) / s =    -2.313043478      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.313043478   ) =    0.010360124 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    8.08      
u = mean =    8      
          
s = standard deviation =    1.15      
          
Thus,          
          
z = (x - u) / s =    0.069565217      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.069565217   ) =    0.472269861 [ANSWER]

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c)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    7.28      
x2 = upper bound =    9.69      
u = mean =    8      
          
s = standard deviation =    1.15      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -0.626086957      
z2 = upper z score = (x2 - u) / s =    1.469565217      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.265628954      
P(z < z2) =    0.929160227      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.663531273   [ANSWER]

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d)

MIDDLE VALUES          
          
As the middle area is          
          
Middle Area = P(x1<x<x2) =    0.89      
          
Then the left tailed area of the left endpoint is          
          
P(x<x1) = (1-P(x1<x<x2))/2 =    0.055      
          
Thus, the z score corresponding to the left endpoint, by table/technology, is          
          
z1 =    -1.59819314      
By symmetry,          
z2 =    1.59819314      
          
As          
          
u = mean =    8      
s = standard deviation =    1.15      
          
Then          
          
x1 = u + z1*s =    6.162077889   [ANSWER]  
x2 = u + z2*s =    9.837922111   [ANSWER]