<PRE><B>Question 796332</B>
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 &lt;= x-bar &lt;= 1095) = P(0.4865&lt;= z &lt;=1.4595) 
= normalcdf(0.4865,1.4595) = 0.2411
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Cheers,
Stan H. </PRE>