I'm not a math or statistics expert and only have a self-taught basic understanding of these things. I'm working on a problem where I know the mean of the population and I want to estimate the standard deviation. This assumes a normal distribution of the population. Is this possible?

For example, if the mean if 32, with possible values between 0 and 100, can I calculate what the standard deviation is with just this information?

Thank you for your help!

  • $\begingroup$ Is the mean just a given? or was it calculated from data? If you have access to the data, estimating the shape of the distribution is just a few steps of R codes away. $\endgroup$ – horaceT Jun 16 '16 at 16:57
  • $\begingroup$ It is calculated from data, but I don't have immediate access to the data. So the real world example of this is if an organization of 100 sales reps has a win rate of 32%. My client can tell me that their win rate is 32%. What I want to be able to do is estimate the percentage of their reps that have win rates in different ranges using standard deviations (e.g. so we can focus on who needs help, who are our model sales reps, etc.). Eventually, I'll have access to the individual data and can do the calculations with all of the data, but it would be great if I could estimate it first. $\endgroup$ – Zigrivers Jun 16 '16 at 17:00
  • $\begingroup$ No, you can't get the standard deviation from the mean unless it's a special case such as Poisson distribution. $\endgroup$ – Aksakal Jun 16 '16 at 20:48

Normal distributions with very different standard deviation can have the same mean, so knowing the mean doesn't tell you which standard deviation you had. Indeed for samples from the normal distribution, the sample mean and sample standard deviation are independent, so the mean doesn't tell you anything about the standard deviation.

if the mean if 32, with possible values between 0 and 100

Then you cannot have a normal distribution (normal distributions are necessarily unbounded). On the other hand, the mean and the two bounds together do impose an upper limit on the standard deviation, but it's pretty weak.

  • $\begingroup$ That's a great point and should have been obvious to me. It is indeed a Skewed Distribution curve. With it being a skewed distribution, is there any way to estimate the standard deviation? Or do I need the individual values to be able to calculate the standard deviation? $\endgroup$ – Zigrivers Jun 16 '16 at 16:27
  • $\begingroup$ 1. While skewness would also indicate non-normality, it could well have been perfectly symmetric yet we would still know - for certain - that the distribution from which your sample was drawn could not actually be a normal distribution. 2. Saying that it's skewed doesn't really help us pin down the standard deviation. ... but the fact that you were able to go back and check that it was skewed clearly indicates you have an additional source of information than the mean. ... ctd $\endgroup$ – Glen_b -Reinstate Monica Jun 16 '16 at 23:09
  • $\begingroup$ ctd... Please include all available information about the numbers and what they measure, including any graphical displays if possible, in the text of your question (via an edit) and if you would be so kind, also comment here (and under the other answer) to indicate any substantive changes. You should also include what you're ultimately trying to achieve (why do you need to know the standard deviation?) ... @Zig $\endgroup$ – Glen_b -Reinstate Monica Jun 16 '16 at 23:09

The later clarification, in a comment to the question, that these are "win rates" ranging from 0% to 100% makes this a bit more feasible than the earlier answer from @Glen_b posted before that information was available.

You might think of this problem like flipping a biased coin, with a probability p of showing "heads" (a "win"; p = 0.32 for your example). Flipping this coin N times (e.g., N sales attempts) is a binomial sampling problem in terms of the number of "wins." This can look like a normal distribution, but as values can only be non-negative integers it can't strictly be normal.

As documented on the linked Wikipedia page and on this Cross Validated page, the variance of the estimate of p among repeated trials with N flips per trial is: $p(1-p)/N$. So to determine whether a particular sales rep is a "model" or "needs help," you also have to take into account the number of sales attempts.

  • $\begingroup$ OP's case seem to be different from binomial distribution, which you suggest. Each sales rep has a different winning rate. And he wants the distribution of salesmen by winning rate. You're suggesting that all salesmen have the same winning rate, but they play and either win or lose, then your distribution is for the overall winning rate, not idnividual $\endgroup$ – Aksakal Jun 16 '16 at 20:47
  • $\begingroup$ My sense is that the OP wants to determine whether individual sales reps have significantly different win rates from an overall company average of 32%, for which a binomial model would be OK. I await further clarification and follow-up from the OP, should he desire. $\endgroup$ – EdM Jun 16 '16 at 20:52

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