document.write( "Question 1044667:  Use the normal distribution of SAT critical reading scores for which the mean is 512 and the standard deviation is 113. Assume the variable x is normally distributed.
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document.write( "left parenthesis a right parenthesis(a)
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document.write( "What percent of the SAT verbal scores are less than 675?
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document.write( "left parenthesis b right parenthesis(b)
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document.write( "If 1000 SAT verbal scores are randomly selected, about how many would you expect to be greater than 575? \n" );
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| Algebra.Com's Answer #660036 by rothauserc(4718)     You can put this solution on YOUR website! a) z-score = (X - mean) / std dev = (675 - 512) / 113 = 1.4424 approx 1.44 \n" ); document.write( ": \n" ); document.write( "we consult the table of Z-values for our z-score \n" ); document.write( ": \n" ); document.write( "********************************************** \n" ); document.write( "Probability(P) ( X < 675 ) = 0.9251 approx 93% \n" ); document.write( "********************************************** \n" ); document.write( ": \n" ); document.write( "b) sample is 1000 and sample mean is the same as population mean = 512 \n" ); document.write( "sample std. dev. = std. dev / square root(sample size) = 113 / square root(1000) = 3.5733 approx 3.57 \n" ); document.write( ": \n" ); document.write( "P ( X > 575 ) = 1 - P ( X < 575 ) \n" ); document.write( ": \n" ); document.write( "z-score = (575 - 512) / 3.57 = 17.6471 approx 17.65 \n" ); document.write( ": \n" ); document.write( "P ( X < 575 ) = 100% or 1.00 \n" ); document.write( ": \n" ); document.write( "*************************************************** \n" ); document.write( "We expect no SAT scores > 575 in our sample of 1000 \n" ); document.write( "*************************************************** \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( " |