document.write( "Question 1171133: A student scores 60 on a Mathematics test that has a mean of 54 and a standard deviation of 3, and she scores 80 on a history test with a mean of 75 and a standard deviation of 2. On which test did she perform better?
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Algebra.Com's Answer #796038 by Theo(13342) You can put this solution on YOUR website! use the z-score formula to determine the answer to this. \n" ); document.write( "z-score formula is: \n" ); document.write( "z = (x - m) / s \n" ); document.write( "z is the z-score \n" ); document.write( "x is the raw score \n" ); document.write( "m is the mean \n" ); document.write( "s is the standard deviation\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "math test has mean of 54 with standard deviation of 3. \n" ); document.write( "z = (60 - 54) / 3 = 6/3 = 2. \n" ); document.write( "area to the left of that z-score = .977249938. \n" ); document.write( "round to 3 decimal places to get .977 \n" ); document.write( "multiply by 100 to get z-score of 2 is greater than 97.7% of all possible z-scores.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "history test has mean of 75 with standard deviation of 2. \n" ); document.write( "z = (80 - 75) / 2 = 5/2 = 2.5 \n" ); document.write( "area to the left of that z-score = .9937903201. \n" ); document.write( "round to 3 decimal places to get .994 \n" ); document.write( "multiply by 100 to get z-score of 2.5 is greater than 99.4% of all possible z-scores.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "she performed better on the history exam.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i used the t-84 plus to find the probabilities. \n" ); document.write( "if you used the z-score table, you would do the following. \n" ); document.write( "look up the z-score (rounded to 2 decimal places) in the table to find the area under the normal distribution curve to the left of that z-score.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "with a z-score of 2.00, the area to the left of the z-score is shown as .97725 = .977 rounded to 3 decimal places * 100 = 97.7%. \n" ); document.write( "with a z-score of 2.50, the area to the left of the z-score is shown as .99379 = .994 rounded to 3 decimal places * 100 = 99.4%.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the table i used can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |