I am following code given here- http://www.bigdataexaminer.com/how-to-implement-these-5-powerful-probability-distributions-in-python/

enter image description here Under "Normal Distribution" section, the graph peaks at .40 when mean is 0. Shouldn't the value of probability be 1 intuitively? In other words, probability of finding something right at mean should be very high but it is only .40

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    $\begingroup$ The probability density shown here is just that, a probability density, not a probability. It's the total area under the curve that is $1$, not the highest value on the curve, which as you report is about $0.4$. To see that this checks out, imagine rearranging the area of the curve into a rectangle with base from $-5$ to $5$. The average density, the height of the rectangle, will be of the order of $0.1$, and $0.1 \times (5 - (-5))$ is $1$. (Note further that if the probability of value equal to the mean is $1$, that leaves zero scope for any other value to be observed.) $\endgroup$
    – Nick Cox
    Commented Apr 12, 2015 at 23:10
  • $\begingroup$ @user777 looks like a near duplicate to me, but I'd really like to let some others make that assessment as well. $\endgroup$
    – Glen_b
    Commented Apr 13, 2015 at 0:26

1 Answer 1


Probability of finding something right at mean should be very high...

Theoretically speaking, if $\mu$ is the mean of the distribution then the probability density function evaluated at $\mu $ is $$p(\mu) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp{\big(-\frac{(\mu-\mu)^2}{2\sigma^2}\big)} = \frac{1}{\sqrt{2\pi\sigma^2}}$$ $$\sigma^2 = 1 \implies p(\mu) = \frac{1}{\sqrt{2\pi}} \approx 0.39894228$$ So its value is determined by $\sigma^2$, and can be any positive finite number. (Yes, even larger than 1).

Notice that since this is a continuous distribution, $$P(X = \mu) = \int^{\mu}_{\mu} p(x)dx = 0$$ So the probability of you getting exactly the mean is actually $0$. But becomes non-zero (and bounded by one) as soon as you consider an interval around $\mu$: $(\mu - \delta, \mu + \delta)$ in which case you find $$P\big(X \in (\mu - \delta, \mu + \delta)\big) = \Phi(\mu + \delta) - \Phi(\mu - \delta)$$ Where $\Phi$ is the cumulative distribution function, and $\delta > 0$.

  • $\begingroup$ Helpful, but you don't mean "any finite number" as densities can't be negative. $\endgroup$
    – Nick Cox
    Commented Apr 12, 2015 at 23:27

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