SOLUTION: A local medical research association proposes to sponsor a footrace. The average time it takes to run the course is 45.8 minutes with a standard deviation of 3.6 minutes. If the as

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Question 360241: A local medical research association proposes to sponsor a footrace. The average time it takes to run the course is 45.8 minutes with a standard deviation of 3.6 minutes. If the association decides to include only the top 25% of the racers, what should be the cut off time in the tryout run? Assume the variable is normally distributed. Would a person who runs the course in 40 minutes qualify?

Answer by neatmath(302) (Show Source):
You can put this solution on YOUR website!
In this case, we only need to find the running time that corresponds with the LOWER 25% z-score (since the lower times are the better times)! Tricky, tricky!

Remember, the z-score is simply (x-xbar)/s where xbar is the mean, x is our value of interest, and s is the standard deviation.

So first we need to find the z-score that corresponds with the lower 25% value. We can do this in several ways, by consulting z-score tables or z-score calculators.

My z-score calculator tells me we are looking for the z-score of about -0.67 to capture 25% of values to the left and 75% of values to the right.

Then we just need to convert this z-score back into a running time, like this:

Therefore our cutoff time should be about 43.388 minutes.

Notice that even though 43.388 is less than the mean, it still reflects the point where the upper 25% of the values end. This is kind of tricky, because usually the higher numbers are associated with the "upper" values.

It should be immediately obvious that a runner who runs the course in 40 minutes (which is LESS than the 43.388 cutoff time) on the tryout run should qualify for the race!

I hope this helps!