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I am asked to put together some KPIs. I'm breaking down what the general lifespan of our subscribers is by industry.

There are some industries I feel outliers are impacting what the average would be, so I figure I should probably use median. But there are others where I think average might be a better course of action. What would be the best way to tell if I'm better off using average or median? Or if there's something else you think would be better, feel free to share.

In my Excel workbook, I actually did both average and median for each category. The issue is, for some of them, there is a WIDE gap in the difference.

For example, I have one category with a sample size of nearly 6000 where the average is 129, but the median is 162. I should also note that the sample size by industry can vary wildly. Some industries the same size is in the thousands while others might be as low as 100.

Here is a snapshot of the workbook:

Industry|SampleSize|Median Tenure|Average Tenure

Industry 1|5973|62|129

Industry 2|2090|90|155

Industry 3|1862|62|135

Industry 4|2491|161|111

Industry 5|2740|62|121

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    $\begingroup$ Picture's worth a thousand words. Plot the distributions by industry. $\endgroup$ – jlemaitre Nov 4 '15 at 22:35
  • $\begingroup$ Using mean (and standard deviation) implies that the distribution is Normal. It might be that your data is far from being distributed according to the normal distribution. A plot, as suggested by @jlemaitre, would shed a light. $\endgroup$ – Vladislavs Dovgalecs Nov 4 '15 at 22:47
  • $\begingroup$ Right, but the issue is, its a lot of different industries. Hundreds of them. So that would be rather tedious. $\endgroup$ – wizkids121 Nov 4 '15 at 22:57
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    $\begingroup$ Sorry, but @xeon's statement is quite incorrect. Mean and SD can be calculated for non-normal distributions and (some mathematical subtleties apart) are almost always well defined. It's a different point that mean and SD may not be the best way to summarize particular samples in practice, but much depends on purpose. Using the mean to summarize exponential and Poisson distributions is for example utterly standard. $\endgroup$ – Nick Cox Nov 4 '15 at 23:10
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    $\begingroup$ Look through any text on distributions and you will see mean and SD discussed routinely for non-normal distributions. The exponential is conventionally always parameterised in terms of either the mean or its reciprocal. The Poisson is always parameterised in terms of the mean and in fact its median is very awkward to work with. I agree naturally that with many skewed or long-tailed samples, indeed virtually all data, it is prudent to work with median as well as mean, and perhaps other summaries, but your statement that using mean and SD implies normality is false. $\endgroup$ – Nick Cox Nov 4 '15 at 23:23
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Honestly, the best case scenario is to present both mean and median. Doing so provides an accurate picture of your data, even if you don't include a picture.

If you can only provide one, remember that these are both measures of center. Which number better measures the center of your data? Generally, I would compute both and if there are many cases where the mean is far from the median, then report the median for every group. However, if there are not many cases where the two are far apart, then I would report the mean for every group.

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When the mean and the median are different, data distribution is said to be skewed. In the example you mentioned, the median (162) is larger than the mean (129), which means there are more observations towards larger values than towards smaller ones. In your context, this means that the lifespan of the subscribers in that industry tend to be larger, with a few low-valued observations pulling down the value of the mean.

This may help you.

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  • $\begingroup$ Thank you for that insight, but I still have a problem: How do I know which one I should be using? $\endgroup$ – wizkids121 Nov 4 '15 at 23:11
  • $\begingroup$ When the gap between them is very high, this means there are really few values influencing the mean. In that case, use the median. You could measure a percentage gap from the median by abs(median - mean) / median, for example. $\endgroup$ – Samuel Barbosa Nov 4 '15 at 23:12
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    $\begingroup$ You can get skewed distributions even if the mean and median are the same. $\endgroup$ – Nick Cox Nov 4 '15 at 23:13
  • $\begingroup$ Shoot I messed up my initial post. The median is actually 62, not 162. None the less, the percentage gap from the median is 1.079. What would that generally infer? $\endgroup$ – wizkids121 Nov 4 '15 at 23:16
  • $\begingroup$ @NickCox I know. Notice that I argued the other way around. $\endgroup$ – Samuel Barbosa Nov 4 '15 at 23:17

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