SOLUTION: 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
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Question 1086243: 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.
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?
b. If 6578 runners were in this year’s race, how many runners beat my time?
c. What time would have gotten me into the top 25%
You can put this solution on YOUR website! z-value1=(46-48.1)/6.2=-0.339
z-value2=(43.2-45.7)/7.8=-0.321 Better overall the first year, the z is more negative so further from the mean in terms of lower time.
0.3743 or 37th percentile or 2462 runners.
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25th percentile is z=-0.675
(x-45.7)/7.8=-0.675
x-45.7=-5.265
=40.435 or 40.4 minutes.
