The typical American changes the oil in his car every 5000 m

The typical American changes the oil in his car every 5.000 miles (population mean = 5.000. Standard deviation of the population =1.500). Cart\'s Garage schedules 40 oil changes per week. What percent of the time will the cars serviced each week at Carl\'s have not had an oil change for a mean of 5,500 miles or more?

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    5500      
u = mean =    5000      
n = sample size =    40      
s = standard deviation =    1500      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    2.108185107      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   2.108185107   ) =    0.017507491 OR 1.75% [ANSWER, OPTION 1]