I'm looking to find what's the relationship between the total height of the normal distribution curve and the height at the point where x=standard deviation. At the plot below the height seems to be around 0.63*total height, where total height is given by ~0.4/sigma, but how to get it mathematically?


  • 3
    $\begingroup$ 1. Is this an exercise for a class? 2. Do you know the formula for the standard normal density function? To get the height at 1, substitute x=1 into that. To get the height at the center, substitute x=0 into that same density. Cancel out the constants that are in common. You'll be left with a simple term which you can do on a calculator $\endgroup$
    – Glen_b
    Commented Oct 10, 2017 at 11:51
  • $\begingroup$ 1. No, an exercise for my curiosity ;) 2. So can I substitute x=sigma to obtain the height at standard deviation point for any distribution (which would vary depending on the value of sigma)? Many thanks $\endgroup$
    – Jed
    Commented Oct 11, 2017 at 7:31

1 Answer 1


a) Standard normal case:

Height at 1/height at 0

$$=\frac{\frac{1}{\sqrt{2\pi}}e^{-\frac12 1^2}}{\frac{1}{\sqrt{2\pi}}e^{-\frac12 0^2}} = e^{-\frac12}$$

b) general normal case is in the same ratio (the shape doesn't change):

Height at 1 sd above mean/height at mean

$$=\frac{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2} \sigma^2}}{\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{1}{2\sigma^2} 0^2}} = {e^{-\frac{1}{2}}}$$

e.g. try $\sigma=2$:

> dnorm(2,0,2)/dnorm(0,0,2)
[1] 0.6065307
> exp(-1/2)
[1] 0.6065307

--- it's always $60.65\%$.


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