document.write( "Question 1117263: For a sample with a mean of M = 50 and a standard deviation of s = 10, a z-score of z = +2.00 corresponds to X = 70.
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document.write( "True
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document.write( "False \n" );
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Algebra.Com's Answer #732297 by Theo(13342)![]() ![]() You can put this solution on YOUR website! formula for z-score is:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "x is the score you are comparing to the mean. \n" ); document.write( "m is the mean. \n" ); document.write( "s is the standard deviation / standard error\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "when z = 2 and m = 50 and s = 10, you get:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "2 = (x - 50) / 10\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for x to get x = 20 + 50 = 70\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "therefore true.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score tells you the number of standard deviations / standard errors you are above or below the mean.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the z-score of 2 tells you you are 2 standard deviations / standard errors you are above the mean.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "2 * 10 = 20 + 50 = 70\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "standard error is involved with sample means which you will learn about later if you haven't already.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "so, until then, the formula is z = (x - m) / s, where s is the standard deviation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |