The operation manager at a tire manufacturing company believ

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 46,264 miles, with a variance of 17,833,730. What is the probability that the sample mean would be less than 46,539 miles in a sample of 208 tires if the manager is correct? Round your answer to four decimal places.

Solution

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    46539      
u = mean =    46264      
n = sample size =    208      
s = standard deviation =    4223.000118      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    0.939167959      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   0.939167959   ) =    0.826177742 [ANSWER]
          

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 46,264 miles, with a variance of 17,833,730. What is the proba

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