The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 31713171 miles, with a standard deviation of 356356 miles.
If he is correct, what is the probability that the mean of a sample of 4040 cars would be less than 31023102 miles? Round your answer to four decimal places.
Solution :
Given that ,
mean = = 3171
standard deviation = = 356
n = 40
= 3171
= / n = 356 / 40 = 56.2885
P( < 3102) = P(( - ) / < (3102 - 3171) /56.2885 )
= P(z < -1.23)
= 0.1093
probability =0.1093
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