Question 722361
follows a normal distribution, but
now with a mean of 30 minutes per car, and a standard deviation
of 4.0 minutes. 
b) if you select a random sample of 16 oil & lube jobs, what is 
the probability the sample mean “service time” from that group 
will be shorter than 28 minutes ?
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x-bar = 28 min
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z(28) = (28-30)/[4/sqrt(16)) = -2
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P(x-bar < 28) = P(z < -2) = 0.0228
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Cheers,
Stan H.
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