This question regards the basic statistics of a normal distribution, but I can't figure it out. I have been given the mean and 95% confidence intervals for a distribution, but would like to know the standard deviation. In my example:

$$ \mu=53.4\quad 95\%\ c.i.=(52.3, 54.3) $$

I had thought that the solution for $\sigma$ would be something like:

$$54.3=53.4+(SE*1.96)$$ $$SE=(54.3-53.4)/1.96=0.46$$ and then, $$SE=\frac{\sigma}{\sqrt{n}}$$ $$\sigma=0.46*\sqrt{n}$$

So, if I don't know the $n$, is this possible?


1 Answer 1


I would suggest it's not possible. There are two unknowns in one equation, implying that there can be more than one unique answer. The following graph shows a curve that lies on some combinations that can give rise to an SE of 0.46, and you can see that without assuming what the sample size is, there will be more than one possible SD.

enter image description here

In very rare cases, one may look at what the variable actually is and come up with a guess of the SD. Is it safe to assume a normal distribution? Does the variable go into the negative domain? etc. From these assumption, you can sometimes guess what the SD may be. I need to use this very occasionally when calculating sample size and power. In those situations, being able to name some conservative SD is a lot more important that nailing down the exact SD.

  • $\begingroup$ This response got me wondering if SD might not be completely related to sample size: stats.stackexchange.com/a/32385/10675 - Although the comment by @whuber seems to call this notion into question. $\endgroup$ Commented Feb 24, 2014 at 14:25
  • 1
    $\begingroup$ Marc, your present question is purely algebraic: namely, whether one thing (an SD) can be computed from the values of two other quantities (a mean and a CI). On the other hand, the comment to which you refer concerns sampling; specifically, predicting results of a second sample based on a first. There is no useful connection between the two sets of ideas. $\endgroup$
    – whuber
    Commented Feb 24, 2014 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.