Question

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 23,383 miles, with a variance of 21,436,900. What is the probability that the sample mean would be less than 22,909 miles in a sample of 286 tires if the manager is correct? Round your answer to four decimal places.

Answer 1

Solution:

Given that,

mean =  = 23383

variance = ² = 21436900

standard deviation =  = 4630

n = 286

So,

= 23383

=  ( /n) = (4630286) = 273.7778

p ( x < 22909 )

= p ( x -  / ) < (22909 - 23383 / 273.7778 )

= p ( z < - 474 / 273.7778)

= p ( z < - 1.73 )

Using z table

= 0.0418

Probability = 0.0418