Question 951167
Suppose that the average weekly earnings for employees in general automotive repair shops is $450, and that the standard deviation for the weekly earnings for such employees is $50. A sample of 100 such employees is selected at random.
 Find probability that the mean of the sample is less than $445. 
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Note: The std for sample means is s/sqrt(n)
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z(445) = (445-450)/[50/sqrt(100) = -1
P(x-bar < 445) = P(z < -1) = 0.1587
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Cheers,
Stan H.
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