The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4764 miles, with a variance of 193,600. If he is correct, what is the probability that the mean of a sample of 34 cars would differ from the population mean by less than 149 miles? Round your answer to four decimal places.
Solution :
Given that,
mean = = 4764
variance = 193600
standard deviation = = 440
n = 34
= 4764
= / n = 440 / 34
= P[-149 / 440 / 34 < ( - ) / < 149 / 440 / 34)]
= P(-1.97 < Z < 1.97)
= P(Z < 1.97) - P(Z < -1.97)
= 0.9756 - 0.0244
= 0.9512
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