document.write( "Question 1173215: In a Statistics examination, the average score was 80 and the variance was 9. Students with scores ranging from 83 to 88 received a grade of 1.50. If the scores are approximately normally distributed and 11 students received a grade of 1.50, determine the percentage of students with scores lower than 83.
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document.write( "Let X be the scores in Statistics exam. \n" );
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Algebra.Com's Answer #798396 by Theo(13342)![]() ![]() You can put this solution on YOUR website! mean = 80 \n" ); document.write( "standard deviation = square root of variance = sqrt(9) = 3\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z-score is found by the following formula: \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( "when x = 83, z = (83 - 80) / 3 = 3 / 3 = 1 \n" ); document.write( "when x = 88, z = (88 - 80) / 3 = 8 / 3 = 8/3 \n" ); document.write( "area to the left of a z-score of 1 is equal to .841345 \n" ); document.write( "area to the left of a z-score of 8/3 is equal to .996170 \n" ); document.write( "area in between 83 and 88 is equal to .996170 - .841345 = .154825\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "11 students had grades between 83 and 88. \n" ); document.write( "11/.154825 = approximately 71 students in total. \n" ); document.write( ".841345 * 71 = approximately 60 students who got grades less than 83. \n" ); document.write( "the balance of the students got grades greater than 88 = 1 - .84345 - .154825 = .00382 * 71 = less than 1. \n" ); document.write( "rounded that becomes 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the numbers don't really come out very well because they are not whole numbers and the percentages, especially of the scores above 88, don't relate very well to a whole number. \n" ); document.write( "in fact, a percentage that low would lead to a whole number of 0.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to answer the question, i think all you had to do was find the area under the normal distribution curve to the left of a score of 83. \n" ); document.write( "the rest of the information was superfluous, since it wasn't necessary to answer the problem.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to answer the problem, all you had to do was find the area under the normal distribution curve to the left of a score of 83. \n" ); document.write( "the z-score was (83 - 80) / 3 = 1 \n" ); document.write( "the area to the left of a z-score of 1 was .841345. \n" ); document.write( "that's 84.1345%. \n" ); document.write( "that would be your answer, without any further analysis. \n" ); document.write( "it was not necessary to determine the number of students, like i attempted to do. \n" ); document.write( "in fact, i'm kind of sorry i did go the extra mile.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "i would recommend going with 84.1345% rounded to whatever number of decimal points required.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "let me know how you did with that. \n" ); document.write( "theo\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |