Truck tire life is normally distributed with a mean of 60000

Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. What is the probability that a tire will last 72,000 miles or more?

Solution

7.


We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    72000      
u = mean =    60000      
          
s = standard deviation =    4000      
          
Thus,          
          
z = (x - u) / s =    3      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   3   ) =    0.001349898 [ANSWER]

 Truck tire life is normally distributed with a mean of 60,000 miles and a standard deviation of 4,000 miles. What is the probability that a tire will last 72,0

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