Question 1064873: In one region, the September energy consumption levels for single -family homes are found to be normally distributed with a mean of 1060 kWh and a standard deviation of 233 kWh. If 44 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1045 kWh.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! In one region, the September energy consumption levels for single -family homes are found to be normally distributed with a mean of 1060 kWh and a standard deviation of 233 kWh. If 44 different homes are randomly selected, find the probability that their mean energy consumption level for September is greater than 1045 kWh.
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z(1045) = (1045-1060)/(233/sqrt(44)) = -15*sqrt(44)/233 = -0.4270
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P(x > 1045) = P(z > -0.4270) = 0.6653
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Cheers,
Stan H.
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