document.write( "Question 973618: Over 10000 runners competed in a marathon. The times of the runners follow a normal distribution. Using the information given below, determine the means and the standard deviation for the runners' times.
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document.write( "- only 16% of the runners were able to finish in less than 3 hr 43 min 31 sec.
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document.write( "- 97.5% of the runners needed at least 2 hr 54min 54 sec to complete the marathon. \n" );
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Algebra.Com's Answer #595750 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! Mean = t sd= s\r \n" ); document.write( "\n" ); document.write( "we know that half the runners finished in less than some mean time\r \n" ); document.write( "\n" ); document.write( "we know that 16% runners finished in fewer than 3.725 hours, 1 sd below the mean. This makes the mean greater than 3.725 hours.\r \n" ); document.write( "\n" ); document.write( "we also know that 2.5% runners finished in fewer than 2.915 hours. 1.96 sd below the mean\r \n" ); document.write( "\n" ); document.write( "Therefore, 0.810 hours is 0.96 sd; therefore 0.844 hours must be the standard deviation.\r \n" ); document.write( "\n" ); document.write( "That makes the mean 4.569 hours. ;;; That is 4 hours 34 minutes and 8 seconds.\r \n" ); document.write( "\n" ); document.write( "The standard deviation is 50 m 38 s.\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( " |