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The times taken by buses to travel between two stops on a route follow a normal

ID: 3131486 • Letter: T

Question

The times taken by buses to travel between two stops on a route follow a normal patern of variation with a mean of 13.9 minutes. A communter makes this journey ten times each week. One week she times the ten journeys and works out that the standard deviation of them is 1.7 minutes. Treating the ten journeys in a week as a random sample:

a) What mean journey time will be exceeded one week in twenty?

b) What mean journey time will be exceeded one week in forty?

c) What mean journey time will not be exceeded 90% of the time?

Explanation / Answer

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 1/20 =   0.95      
          
Then, using table or technology,          
          
z =    1.644853627      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    13.9      
z = the critical z score =    1.644853627      
s = standard deviation =    1.7      
n = sample size =    10      
Then          
          
x = critical value =    14.78425226   [ANSWER]

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b)

First, we get the z score from the given left tailed area. As          
          
Left tailed area = 1 - 1/40 =   0.975      
          
Then, using table or technology,          
          
z =    1.959963985      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    13.9      
z = the critical z score =    1.959963985      
s = standard deviation =    1.7      
n = sample size =    10      
Then          
          
x = critical value =    14.95365155   [ANSWER]  

**************************

C)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.9      
          
Then, using table or technology,          
          
z =    1.281551566      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    13.9      
z = the critical z score =    1.281551566      
s = standard deviation =    1.7      
n = sample size =    10      
Then          
          
x = critical value =    14.58894572   [ANSWER]  

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