Question 796332: In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1,050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September would be between 1065 and 1095 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 1,050 kWh and a standard deviation of 218 kWh. If 50 different homes are randomly selected, find the probability that their mean energy consumption level for September would be between 1065 and 1095 kWh.
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z(1065) = (1065-1050)/[218/sqrt(50)] = 0.4865
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z(1095) = (1095-1050)/218 = 1.4595
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P(1065 <= x-bar <= 1095) = P(0.4865<= z <=1.4595)
= normalcdf(0.4865,1.4595) = 0.2411
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Cheers,
Stan H.
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