The mileage in thousands of miles that car owners get with a

The mileage (in thousands of miles) that car owners get with a certain kind of radial tire is normally distributed with a mean of 45,000 and a standard deviation of 3,000 miles, respectively.

A.) To 4 decimal places what is the probability that a tire will last for more than 50,000 miles?

B.) To 4 decimal places what is the probability that a tire will last for no more than 35,000 miles?

C.) To the nearest mile, 90% of the tires will last for how many miles?

Solution

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    50000      
u = mean =    45000      
          
s = standard deviation =    3000      
          
Thus,          
          
z = (x - u) / s =    1.666666667      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.666666667   ) =    0.047790352 [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 =    35000      
u = mean =    45000      
          
s = standard deviation =    3000      
          
Thus,          
          
z = (x - u) / s =    -3.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -3.333333333   ) =    0.00042906 [ANSWER]

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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,          
          
where          
          
u = mean =    45000      
z = the critical z score =    1.281551566      
s = standard deviation =    3000      
          
Then          
          
x = critical value =    48844.6547   [ANSWER]