<PRE><B>Question 1116849</B>
The distribution of annual earnings of all bank tellers with five years of experience is skewed negatively. This distribution has a mean of Birr 15,000 and a standard deviation of Birr 2000. If we draw a random sample of 30 tellers, what is the probability that their earnings will average more than Birr 15,750 annually?
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z(15750) = (15750-15000)/[2000/sqrt(30)] = 750*sqrt(30)/2000 = 2.0540
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P(x-bar &lt; 15750) = P(z &gt; 2.0540) = normalcdf(2.0540,100) = 0.02
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Cheers,
Stan H.
P.S. This problem was answered a couple of days ago.
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