What mile time is equidistant from 4:00 and 5:00, in terms of percentile

I would think much higher, perhaps 4:40 or 4:42. I think quite a few high school kids can run in the 4:50's and 4:40's but not many folks are running in the 4:00's and 4:10's.

What do you plan on doing with this magic number?

# of 4:00 of faster milers in the world - 300?

# of 5:00 milers in the world - 100,000?

So your ratio is 5.4%. or 27/500

300/100000 = sqrt(OP's magic percentage^2)

Assuming that those numbers of runners capable of those times are correct (which they are probably not), you're looking for a time that roughly 5400 people can run.

I'd say your estimate of 4:20 is probably close. It's most likely a little faster though.

Let's say for the sake of argument that there are 500,000 people with a 5:00 or faster PR

The ratio then becomes 2.4%.

So we're looking for a time that 12247 people can run. I'd still put that in the 4:25 range.

Larry83 wrote:

I would think much higher, perhaps 4:40 or 4:42. I think quite a few high school kids can run in the 4:50's and 4:40's but not many folks are running in the 4:00's and 4:10's.

Larry83 is closer to correct here. What you're asking for is the median mile time for the sample of times between 4:00 and 5:00. Naturally this distribution is going to left skewed and the median time or 50th percentile will be greater than the 4:30 midpoint.

The other interesting feature of this distribution is that there will be no tail on the right-side of the distribution since we are truncating the distribution well before times starts to trail off. In other words there are plenty of times between 4:30 and 5:00 so the distribution will have a left tail and then more or less a plateau throughout the rest.

I have data on mile times. This sample includes about 9,000 times between 4:00 and 5:00. It looks like so:

The median here is 4:34.

Try again.

Seriously, try again. Because I think that you may have some worthwhile data to answer the question (I'm too lazy to check myself) but you have provided an incorrect answer.

What is being asked by the OP is decidedly NOT the median mile time for the sample. Read the OP's question again and get back to us with the correct answer this time.

Go!

Here's the same distribution but using a smoothed line and also overlaying what normal distribution would look like.

http://tinypic.com/view.php?pic=1zgh1y&s=8

This is exactly what the OP wants. In the OP's words, "where the ratio of number of 5:00 runners to 4:xx runners is the same as 4:xx runners to 4:00," which is would in more precise language be called the 50th percentile, also know as the median.

See for example,

. "In statistics and probability theory, the median is the number separating the higher half of a data sample, a population, or a probability distribution, from the lower half."

Rob E wrote:

Here's the same distribution but using a smoothed line and also overlaying what normal distribution would look like.

http://tinypic.com/view.php?pic=1zgh1y&s=8

Dude, come on, get with the program. You come back on here but are still thinking of the wrong problem.

Here, I'll break it down for you:

Let

X = # runners who have a PR under 4:00

Y = # runners who have a PR under 4:AB

Z = # runners who have a PR under 5:00

Find AB such that Z/Y = Y/X

Thanks in advance.

Dude, I'm trying to be polite here. Because I think that you are putting in an honest effort. And also because I think that you may provide the correct answer with a bit of help here.

I already posted the correct interpretation of the question. And it definitely has nothing to do with the median.

The OP did NOT ask for the point where the number above and below would be equal. That would be the median.

Read my post just prior to this one. Let me know if you still have trouble understanding.

I see other's posted this while I was typing. I'll post it anyway since I have another illustration.

OP is not asking for the median. I'll illustrate. Let's say we have sample of 101 runners. There's two runners under 4:00 and 60 runners under 5:00. (No need to cut off either left or right tail).

The median is 4:ab meaning there are 50 runners under 4:ab (and 50 over).

The ratio of under 5:00 runners to under 4:ab runners is 60/50 = 1.2.

Is the ratio of under 4:ab runners to under 4:00 runners also 1.2?

If it is, median indeed is the answer OP was looking for.

And to make a guess before Rob E figures out the correct answer: I'll go for 4:19.

4:20 sounds really nice, because it's "one third of the way". But 4:19 is "one PIth of the way", which sounds even more likely to be how nature would work.

I already answered the question....I just don't have the data of:

# of runners at 4:00

# of runners at 5:00

Answer is:

Ratio = Sqrt[(# of runners at 4:00)/(# of runners at 5:00)]

Multiply that ratio times the number of runners who run 5:00 and you'll have the nth person that represents the time for which the OP is looking.

So if 100 people run 4:00 and 20000 people run 5:00:

Ratio = Sqrt[(100/20000)]

Ratio is 7.07%

7.07% X 20000 = 1414

You're looking for the time that the 1414th fastest person can run.

The only point of argument is: How many people can run 4:00 and how many can run 5:00. The math is finite.

Jeff Albertson wrote:

I already answered the question....I just don't have the data of:

# of runners at 4:00

# of runners at 5:00

Answer is:

Ratio = Sqrt[(# of runners at 4:00)/(# of runners at 5:00)]

Multiply that ratio times the number of runners who run 5:00 and you'll have the nth person that represents the time for which the OP is looking.

So if 100 people run 4:00 and 20000 people run 5:00:

Ratio = Sqrt[(100/20000)]

Ratio is 7.07%

7.07% X 20000 = 1414

You're looking for the time that the 1414th fastest person can run.

The only point of argument is: How many people can run 4:00 and how many can run 5:00. The math is finite.

Dude, what you have presented is self evident. And has been said much more succinctly elsewhere.

The only question lies in the numbers and you're not supplying those. So WTF?

No. One person said that we need to find the ratio, but supplied no method of getting there. The OP clearly did not know how to arrive at the ratio or else he would not have started the thread.

Then you have some Bozo trying to solve for the median. That Bozo needs to go back to 6th grade.

You are correct that I have not supplied the numbers. I doubt anyone can.

Even looking at a sample size like NCAAs, let's say current D-1 runners (estimated):

4:00 or faster = 30

5:00 or faster = 274 schools X average 15 5:00 milers = 4110

Ratio is 8.5%

You're looking for the time that would make you 351st in NCAA D-1.

I'd guess that would be somewhere in the 4:15-4:25 range.

so that 351st runner would be faster than 91.5% of runners that are sub 5.

Likewise, a 4:00 miler would be faster than 91.5% of runners out of that top 351.

To take it even further:

The 2nd fastest miler in NCAA would be faster than 91.5% of all sub-4 milers.

The original question is so convoluted that I'm going to assume that either Try Try Again or Jeff Albertsons are the OP and are answering his/her own question as some kind of dick measuring contest. Furthermore, it wouldn't surprise me if one of the two wrote this question on an exam, no one was able to comprehend it, and as some kind of validation they've come to LetsRun.