In order to be able to create a histogram for the probabilities of an event (e.g. scoring 0,1,2,3...n heads in a row) that matches a normal curve the distribution must be symmetric.

Something like:
enter image description here

My question is: is there in general a relation between how big the n is with whether we can have a histogram such as the one in the image or is it irrelevant?
What would it depend on then?

Update after the comments:
Yes my example with the coin is indeed a binomial distribution but it was only an example of a case that can be represented by a histogram.
My question was about the cases that the histogram matches a normal curve.

  • 2
    $\begingroup$ Sounds like you want to know about the Central Limit Theorem of a Law of Large Numbers. $\endgroup$ Commented Jul 24, 2017 at 17:42
  • $\begingroup$ What? Why must a histogram be symmetric? Yes—in order to fit a normal curve to a histogram it needs to be symmetrical—but the body reads "in order to be able to create a histogram for the probabilities of an event (...) the distribution must be symmetric". Where does this claim come from? $\endgroup$ Commented Jul 24, 2017 at 20:16
  • $\begingroup$ @LoveTooNap29: You are right. I corrected the post $\endgroup$
    – Jim
    Commented Jul 24, 2017 at 22:04
  • 1
    $\begingroup$ @Sean: You could have included a link... Central limit theorem $\endgroup$
    – user856
    Commented Jul 24, 2017 at 22:18
  • $\begingroup$ A "head/tail" random variable doesn't follow a normal curve but a binomial one. It is symmetrical for a fair coin. $\endgroup$
    – user65203
    Commented Jul 25, 2017 at 6:52

2 Answers 2


If you have a binomial random variable with $n$ trials and Success probability $p$ on each trial, then the number $X$ of heads observed will become very nearly normal for large $n.$ The convergence to normal is faster of $p = 1/2.$

Often one looks at $Z_n = \frac{X - np}{\sqrt{np(1-p)}},$ so that $E(Z_n) = 0$ and $Var(Z_n) = 1.$ Then $Z_n$ 'converges in distribution' to standard normal. Roughly speaking, that means $\lim_{n \rightarrow \infty}P(Z_n \le c) = \Phi(c),$ where $\Phi$ denotes the CDF of the standard normal distribution.

For example, suppose $n = 100$ and $p = 1/2$ and let $X$ be the number of successes. Then the exact value of $P(X \le 40) = 0.0284,$ to four places. This can be found by summing the terms of the binomial distribution from $P(X = 0)$ up through $P(X = 40).$ In R statistical software, the computation is as follows:

pbinom(40, 100, .5)
## 0.02844397

Because this binomial distribution is nearly normal we can compute this as $$P(X \le 40) = P(X \le 40.5) = P\left(\frac{X - np}{\sqrt{np(1-p)}} \le \frac{40.5 - 50}{5} = -1.9 \right) \approx \Phi(-1.9),$$

which can be evaluated as 0.0287, using printed normal tables or software.

## 0.02871656

[Using 40.5 instead of 40 allows a better match between the discrete binomial distribution and the continuous normal distribution. This is sometimes called a 'continuity correction'.]

The plot below shows values of $\mathsf{Binom}(100,.25)$ with substantial probability, along with the density function of $\mathsf{Norm}(\mu=50,\,\sigma=5).$ The relevant probability is to the left of the vertical red line.

enter image description here

Also, if $n = 1000$ and $p = 1/4$ then $E(X) = 250$ and $SD(X) = 13.6931.$ If we seek $P(X \le 275)$ the exact value to four places is 0.9677, and the value from a normal approximation is 0.9687. This illustrates that the convergence of binomial to normal is usefu even if $p \ne 1/2,$ provided $n$ is sufficiently large. [Usually, the normal approximation gives about two-place accuracy.]

pbinom(275, 1000, .25)
## 0.9677063
pnorm(275.5, 250, 13.6931)
## 0.9687162

enter image description here


The binomial $\text{pdf}$ is

$$B_k=\binom nkp^kq^{n-k}$$

while the corresponding normal approximation would be

$$\frac1{\sqrt{2\pi npq}}e^{-(k-np)^2/2npq}.$$

More precisely, we obtain the probability of the $k^{th}$ bin by integrating between half integers,

$$N_k=\int_{k-1/2}^{k+1/2}\frac1{\sqrt{2\pi npq}}e^{-(k-np)^2/2npq}dt=\Phi\left(\frac{k+\frac12-np}{\sqrt{npq}}\right)-\Phi\left(\frac{k-\frac12-np}{\sqrt{npq}}\right).$$

In the case of a fair coin,

$$B_k=\binom nk\frac1{2^n},$$

$$N_k=\Phi\left(\frac{2k-n+1}{\sqrt{n}}\right)-\Phi\left(\frac{2k-n-1}{\sqrt{n}}\right)\approx\sqrt{\frac2{\pi n}}e^{-(2k-n)^2/2n}=\tilde N_k.$$

Comparing these curves seems barely tractable by analytical methods.

The plot below shows the discrepancies between the Binomial and Normal laws, either exact or approximated, for $n=100$.

enter image description here

Fortunately, the largest error seems to occur in the middle, where

$$B_{n/2}=\binom n{n/2}\frac1{2^n},$$ $$N_{n/2}=2\Phi\left(\frac1n\right).$$

Note that approximating the Binomial term by Stirling makes the two values equal, so that a better approximation is required.

From Wikipedia,

$$\binom n{n/2}\approx2^n\sqrt{{\frac 2{\pi n}}}\left(1-\frac{2c_n}n\right)$$ with $\frac19<c_n<\frac18$. Hence the maximum error decreases like $$\frac1n.$$


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