document.write( "Question 1206705: The maintenance department at the main campus of a large state university receives daily requests to replace fluorecent lightbulbs. The distribution of the number of daily requests is bell-shaped and has a mean of 60 and a standard deviation of 7. Using the 68-95-99.7 rule, what is the approximate percentage of lightbulb replacement requests numbering between 60 and 81?\r
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Algebra.Com's Answer #844307 by ikleyn(52806)![]() ![]() You can put this solution on YOUR website! . \n" ); document.write( "The maintenance department at the main campus of a large state university receives daily requests \n" ); document.write( "to replace fluorescent lightbulbs. The distribution of the number of daily requests is bell-shaped \n" ); document.write( "and has a mean of 60 and a standard deviation of 7. Using the 68-95-99.7 rule, \n" ); document.write( "what is the approximate percentage of lightbulb replacement requests numbering between 60 and 81? \n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \r\n" ); document.write( "Notice that in this problem 81 is three standard deviations from the mean of 60.\r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "The Empirical Rule states that 99.7% of data observed following a normal distribution \r\n" ); document.write( "lies within 3 standard deviations of the mean. \r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Since the normal distribution is symmetric, we may conclude that the pecentage of lightbulb \r\n" ); document.write( "replacement requests numbering between 60 and 81 is half of 99.7%, i.e. 49.85%. ANSWER\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Solved.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |