Question 795278
A car rental has cars with an average of 9000 miles and standard deviation of 1200. If 25 cars are in a lot, what is the probability there average miles will be less than 9500 miles.
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t(9500) = (9500-9000)/(1200/sqrt(25)) = 500/240 = 2.0833
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P(x-bar < 9500) = P(t < 2.0833) = tcdf(-100,2.0833,24) = 0.9760
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Cheers,
Stan H.
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