document.write( "Question 1086323: Last year I ran in a 10K where the average completion time was 48.1 min with a standard deviation of 6.2 min. This year, runners completed the race in an average of 45.7 min with a 7.8 min standard deviation. The distribution of running times each year was approximately normal.
\n" ); document.write( "a. If I completed the race in 46.0 min last year and in 43.2 minutes this year, in which year did I run a better race as compared to the rest of the field? \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "b. If 6578 runners were in this year’s race, how many runners beat my time? \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "c. What time would have gotten me into the top 25%
\n" ); document.write( "

Algebra.Com's Answer #700495 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
a) Calculate the z scores for both races.
\n" ); document.write( "\"z%5B1%5D=+%2846-48.1%29%2F6.2+\"
\n" ); document.write( "\"z%5B2%5D=+%2843.2-45.7%29%2F7.8+\"
\n" ); document.write( "The lower the z score, the better the placement.
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "b) Find the probability using \"z%5B2%5D\" then multiply that value by the number of runners. Run up or down appropriately.
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "c) Find z such that \"P%28z%5B25%5D%29=0.25\"
\n" ); document.write( "\"P%28z%5B25%5D%29=0.25\"
\n" ); document.write( "\"z%5B25%5D+=-0.67449+\"
\n" ); document.write( "So,
\n" ); document.write( "\"%28X%5B25%5D-45.7%29%2F7.8+=-0.67449+\"
\n" ); document.write( "Solve for \"X%5B25%5D\".
\n" ); document.write( "
\n" );