document.write( "Question 1170523: A city mayor in Michigan is planning the number of new snow plows he must purchase to remove the snow next winter. The average snowfall in the past 10 years has been normally distributed with a mean of 112 inches and a standard deviation of 14 inches. What amount, in inches, separates the lowest 20% of the distribution of yearly snowfall in the past 10 years from the highest 20%.
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document.write( "Use the z-table below: \n" );
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Algebra.Com's Answer #795516 by Theo(13342)![]() ![]() You can put this solution on YOUR website! mean = 112 \n" ); document.write( "standard deviation = 14 \n" ); document.write( "bottom 20% would have a z-score of -.8416212335 \n" ); document.write( "top 20% would have a z-score of .8416212335 \n" ); document.write( "uswe the z-score formula to find the raw score. \n" ); document.write( "the z-score formula ia: \n" ); document.write( "z = (x - m) / sd \n" ); document.write( "x = the raw score \n" ); document.write( "m = the mean \n" ); document.write( "sd = the standard deviation.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "with your numbers, ...\r \n" ); document.write( "\n" ); document.write( "the low z-score formula is: \n" ); document.write( "-.8416212335 = (x - 112) / 14 \n" ); document.write( "solve for x to get: \n" ); document.write( "x = 14 * -.8416212335 + 112 = 100.2173027\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the high z-score formula is: \n" ); document.write( ".8416212335 = (x - 112 / 14 \n" ); document.write( "solve for x to get: \n" ); document.write( "x = 14 * .8416212335z + 112 = 123.7826973\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the amount in inches that separates the lowest 20% of the normal distribution from the highest 20% of the normal distribution is equal to 123.7826973 minus 100.2173027 = 23.56539454 inches.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you have .6 of the area under the normal distribution curve between those 2 z-scores and you have .4 of the area under the normal distribution curve outside of those 2 z-score values.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "pictures of the z-score distributions and the raw score distributions are shown below.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " ![]() |