document.write( "Question 139732: Having a lot of trouble with this. Please help and show me how so I can actually learn from all this. Thanks!\r
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document.write( "PS- I am allowed to use Megastat on this so any direction there would be welcomed as well.\r
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document.write( "The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of $2,625 and a standard deviation of $350. (shos all your work please)\r
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document.write( "A. What is the z-value for a salary of $2,200?\r
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document.write( "B. What is the approximate percent of statisticians making between $2,625 & $2,975?\r
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document.write( "C. What is the approximate percent of statisticians making between $2,275 & $2,625? \n" );
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Algebra.Com's Answer #102054 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The bi-monthly starting salaries of recent statistician graduates follow the normal distribution with a mean of $2,625 and a standard deviation of $350. (shos all your work please)\r \n" ); document.write( "\n" ); document.write( "A. What is the z-value for a salary of $2,200? \n" ); document.write( "z(2200) = (2200-2625)/350 = -1.21428... \n" ); document.write( "-----------------------------------------------------\r \n" ); document.write( "\n" ); document.write( "B. What is the approximate percent of statisticians making between $2,625 & $2,975? \n" ); document.write( "Since the mean is 2625 the z-value of 2625 is zero \n" ); document.write( "z(2975) = (2975-2625)/350 = 1 \n" ); document.write( "P(2625 < x < 2975) = P(0 < z < 1) = 0.3413.. \n" ); document.write( "--------------------------------\r \n" ); document.write( "\n" ); document.write( "C. What is the approximate percent of statisticians making between $2,275 & $2,625? \n" ); document.write( "P( 2275 < x < 2625) = P(-1 < z < 0) = 0.3413 \n" ); document.write( "================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |