document.write( "Question 858772: According to records of commonwealth Edison, the mean electric consumption
\n" ); document.write( "during January is 1,650 kilowatt hours, with a standard deviation of 320
\n" ); document.write( "kilowatt hours. The monthly electric consumption for January follows a normal
\n" ); document.write( "distribution.\r
\n" ); document.write( "\n" ); document.write( "a. Selecting a customer at random, find the probability that the customer uses
\n" ); document.write( "less than 1,000 kilowatt hours in January.\r
\n" ); document.write( "\n" ); document.write( "b. Selecting a customer at random, find the probability than the customer uses
\n" ); document.write( "between 1,500 and 2,500 kilowatt hours in January.\r
\n" ); document.write( "\n" ); document.write( "c. Commonwealth Edison wants to identify its heaviest users, which are those
\n" ); document.write( "that are in the top 6% of electric consumption. How many kilowatt hours in
\n" ); document.write( "January would qualify for the heavy user classification?
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Algebra.Com's Answer #517390 by psbhowmick(878)\"\" \"About 
You can put this solution on YOUR website!
a) Let the electric consumption be the random variable X and normally distributed with mean \"mu\" and standard deviation \"sigma\".\r
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\n" ); document.write( "\n" ); document.write( "At first calculate the Z-score corresponding to 1000 kilowatt hours.
\n" ); document.write( "\"Z=%28X-mu%29%2Fsigma=%281000-1650%29%2F320=-650%2F320=-2.03125\"\r
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\n" ); document.write( "\n" ); document.write( "To find the probability that a house uses less than 1000 kilowatt hours of electricity:
\n" ); document.write( "\"p%28X%3C1000%29+=+p%28Z%3C2.03125%29+=+0.02\"\r
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\n" ); document.write( "\n" ); document.write( "Try the other parts yourself. If you still need help then contact me. \r
\n" ); document.write( "\n" ); document.write( "Cheers!!!
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