Question 1116468: 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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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?
z(15750) = (15750-15000)/(2000/sqrt(30)) = 750*sqrt(30)/2000 = 375*sqrt(30)/1000 = 0.375*sqrt(30) = 2.0540
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P(x-bar > 15,750) = P(z > 2.0540) = normalcdf(2.0540,100) = 0.02
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Cheers,
Stan H.
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