Alpine Tire Company has determined from road tests that the

Alpine Tire Company has determined from road tests that the mean mileage of its main product is 50,000 miles with a standard deviation of 5,000 miles, and that the collected data are normally distributed. Alpine wishes to offer a warrantee providing free replacement for any new tire that fails before the guarantee mileage.

If Alpine wishes to replace no more than 10% of the tires, what should the guarantee mileage be?

In addition, the data collected indicate that the normal distribution is a reasonable assumption. What is the probability that the tire mileage will exceed 50,500miles?

Please show all calculations!!!

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.1      
          
Then, using table or technology,          
          
z =    -1.281551566      
          
As x = u + z * s,          
          
where          
          
u = mean =    50000      
z = the critical z score =    -1.281551566      
s = standard deviation =    5000      
          
Then          
          
x = critical value =    43592.24217   [ANSWER, MAXIMUM GUARANTEE]

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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 =    50500      
u = mean =    50000      
          
s = standard deviation =    5000      
          
Thus,          
          
z = (x - u) / s =    0.1      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   0.1   ) =    0.460172163 [ANSWER, MORE THAN 50500]