document.write( "Question 1120406: 9. A normal distribution of utility bills shows the mean to be $100 and the standard deviation of $12.\r
\n" ); document.write( "\n" ); document.write( "a) What percent of the utility bills are more than $125.
\n" ); document.write( "b) If 300 utility bills are randomly selected, about how many of them would you expect to be less than $90?
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Algebra.Com's Answer #736076 by rothauserc(4718)\"\" \"About 
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a) Probability (P) ( X > 125 ) = 1 - P ( X < 125 )
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\n" ); document.write( "z-score(125) = (125 - 100)/12 = 2.0833
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\n" ); document.write( "using table of z-scores, the probability associated with 2.0833 is 0.9812
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\n" ); document.write( "(P) ( X > 125 ) = 1 - 0.9812 = 0.0188
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\n" ); document.write( "b) because we know the standard deviation of the population, we can use the normal distribution in our calculations
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\n" ); document.write( "z-score(90) = ( 90 - 100 ) / 12 = −0.8333
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\n" ); document.write( "P ( X < 90 ) = 0.2033
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\n" ); document.write( "300 * 0.2033 = 60.99 approximately 61
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