document.write( "Question 1170017: A teacher gave a test on which the students' marks were normally distributed, but the results were pathetic. The mean was 52%, and the standard deviation was 12%. The teacher decided that the top 10% of the students should get A's, the next 20% should get B's, the next 40% should get C's, the next 20% should get D's, and the bottom 10% should get F's. To the nearest percent, what was the minimum percent a student had to achieve in order to recieve a B? Do not include the percent sign in your answer. \n" ); document.write( "
| Algebra.Com's Answer #794909 by Theo(13342)     You can put this solution on YOUR website! mean = 52% \n" ); document.write( "standard deviation = 12% \n" ); document.write( "top 10% get A \n" ); document.write( "next 20% get B \n" ); document.write( "next 40% get C \n" ); document.write( "next 20% get D \n" ); document.write( "bottom 10% get F\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "minimum B score will be the z-score that has 10% + 20% = 30% of the area under the normal distribution curve to the right of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the table are set to calculate off of ratios, not percents. \n" ); document.write( "just divide the percent by 100 and you get an area of .30 to the right of the z-score. \n" ); document.write( "since the tables only look for area to the left of the z-score, you are looking for a z-score that has 1 - .30 = .70 of the area under the normal distribution curve to the left of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "if you look in the z-score table, you will find that a z-score of .52 has .69847 of the area under the normal distribution curve to the left of it and a z-score of .53 has .70194 of the area under the normal distribution curve to the left of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a manual interpolation leads to a z-score of .524409 rounded to 5 decimal digits.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that's pretty close to having an area of .7 to the left of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "a check with an online calculator shows that a z-score of .524409 has an area of .7 to the left of it.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the display of the calculator results are shown below.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "  \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the calculator i used can be found at https://www.calculator.net/z-score-calculator.html?c1raw=3.075&c1mean=3.2&c1sd=.1&calctype=zscore&x=81&y=18\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to find the raw score associated with that, use the z-score formula of:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "z = (x - m) / s\r \n" ); document.write( " \n" ); document.write( "\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( "the formula becomes .524409 = (x - 52) / 12.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "solve for x to get x = .524409 * 12 + 52 = 58.292908.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "that's the lowest score you can get and still get a B.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |