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I've been given the following which shows new membership sign-ups per campaign:

Average: 291
25% Percentile: 219
50% Percentile: 283
75% Percentile: 347

And need to calculate the percentage for a given value.

I Googled for a solution but from what I see, it seems as though I need the full range of data to calculate percentage for a value.

Is it possible to do with just these three percentiles? For example, if a campaign were to have 300 new membership sign-ups, can I calculate percentage?

Ian Lunn
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2 Answers2

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You can't do it exactly without knowing the distribution. I could create a number of distributions that match your percentiles. Almost anything sensible that you do will be close. I think I would just take your points $(130,0),(219,25),(283,50),(347,75),(390,100)$ and do a linear interpolation between them. So to look up a percentile for $300$, we note it is between $283$ and $347$, so the percentile is $50+\frac{300-283}{347-283}(75-50)\approx 56.6$ As the distribution is probably peaked around the middle, this probably underestimates the percentile a bit.

Ross Millikan
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It largely depends on what specific distribution the data follows, but you might be able to get some sort of idea if you make assumptions such as that it might be normally distributed (wiki - normal distribution). This is often a reasonable assumption as many naturally occurring sets of data are normally distributed (iq, height, size of trees, number of leaves on trees, etc...).

If it so happens that it is normally distributed, (which looks reasonable), then we can center our focus on the average calculated: 291.

From there, we want to guess as to what the standard deviation might be. If it is in fact normally distributed, the $25^{th}$ percentile mark should be at about the same distance away from the average as the $75^{th}$ percentile mark at about $0.6745$ times the standard deviation away from the average.

With that, we guess then $291-219 = 72\approx 0.6745\sigma$ and $347-291=56\approx 0.6745\sigma$. Well... they aren't quite the same distance apart, so that implies that if it were normally distributed that there would be some outliers messing up the average.

Let us guess instead that the average should naturally have been closer to the $50^{th}$ percentile mark at 283. $283-219 = 64\approx 0.6745\sigma$ and $347-283=64 \approx 0.6745\sigma$. These we find are perfectly evenly spaced, which helps support our hypothesis that it were normally distributed. Calculating for $\sigma$ then gives us $\sigma = 64/0.6745\approx 95$.

We suppose then that our data is normally distributed with average 283 and standard deviation 95. Using a conversion to work with the standard normal distribution (where average was 0 and standard deviation is 1), by setting $z = \frac{283 - x}{95}$ we can then use a chart of values to find out in what percentile a specific value of $x$ is.

This gives way to the following chart:

$\begin{array}{l|l} 10^{th}\text{percentile} & 161\\ 20^{th}\text{percentile} & 203\\ 30^{th}\text{percentile} & 233\\ 40^{th}\text{percentile} & 259\\ 50^{th}\text{percentile} & 283\\ 60^{th}\text{percentile} & 307\\ 70^{th}\text{percentile} & 333\\ 80^{th}\text{percentile} & 363\\ 90^{th}\text{percentile} & 404\\ 99^{th}\text{percentile} & 504 \end{array}$

This will not be entirely accurate for a few reasons. First: we already know that the average was calculated and was in fact different than the $50^{th}$ percentile marker (which if it was normally distributed should not have happened). Second, it is possible that the data is not well approximated by a normal distribution. Still, without more information, the assumptions made seem reasonable.

JMoravitz
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  • you cannot do it for this question. – Seyhmus Güngören Mar 27 '15 at 19:11
  • @SeyhmusGüngören I very clearly stated that I was making the assumption that it was normally distributed (as many sets of data are). Of course, there is absolutely no way of knowing the exact percentiles without knowing the exact distribution, however given the lack of information, it is very reasonable to assume that it might fall under this type of distribution and try anyways. – JMoravitz Mar 27 '15 at 19:13
  • you cannot make a wrong assumption and continue solving the question based on a wrong assumption. For all Gaussian distributions mean and median, $50%$ percentile, are the same. – Seyhmus Güngören Mar 27 '15 at 19:17
  • The further information that there were no observations greater than $390$ also throws the assumption into serious doubt. It seems not credible that the $90$th percentile could be anywhere near $404$. – David K Mar 27 '15 at 22:45
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    @DavidK when you take into account that the number of observations is quite obviously finite (and even possibly small), the "no observations greater than 390" does not imply that the maximum is 390. Indeed, with any normal approximation to the curve, there are some absurdities that will arise, such as for average male height. There is supposedly a positive probability that someone could be not only 23foot tall, but also positive probability that someone could be negative 2feet tall. Even so, normal distributions are often a good approximation to data, though only an approximation. – JMoravitz Mar 27 '15 at 22:50
  • Your examples of absurd heights are well outside the six-sigma range. This example is quite different. $390$ is barely more than one s.d. away from the mean; if the distribution were as you assume, we could expect $13$% of the observations to exceed $390$. You're asserting that it's credible that we never observed an event that had prior probability $1-(0.87)^N$, where $N$ is the sample size. OK for small enough $N$, but if $N$ is so small is there any point to computing percentiles this way? – David K Mar 27 '15 at 23:14