Maybe a stupid question but I was trying to approximate the binomial distribution with a normal distribution and I can't understand where the problem is.

Online I read that a binomial distribution can be approximated like this using the central limit theorem:

If we have i.i.d. binomial distributed random variables $X_{i}$ than we can write $P(\sum_{0}^{n}X_{i}\leqslant z)=Φ(\frac{z-np}{\sqrt{np(1-p)}})$

I have tried to come to this solution but applying the central limit theorem that is formulated: "$\lim_{n \to \infty} P(\frac{S_{n}-nμ}{\sqrt{nσ2}}\leqslant z)\Rightarrow Φ(z)$" I come to the solution that the normal approximation is $Φ(\frac{z-n^2p}{\sqrt{n^2p(1-p)}})$. This is because of the $n$ in the numerator and the $\sqrt{n}$ in the denominator.

Where do these two go in the formula I found online?

  • $\begingroup$ It looks like you take $\mu$ and $\sigma^2$ to be the mean and variance of $S_n$. It should be the mean and variance of $X_1$. $\endgroup$
    – jakobdt
    Jul 12, 2022 at 12:23
  • $\begingroup$ What do you mean of $X_{1}$? If $X_{1}$ has binomial distribution than $μ=np$ and $σ^2=np(1-p)$, right? $\endgroup$
    – MarcoDJ01
    Jul 13, 2022 at 9:52
  • $\begingroup$ You need to be a bit careful as you cannot use $n$ for both a parameter in the distribution of $X_1$ and for the number of binomial random variables. You might want to use e.g. $N$ for the latter. If $X_1,X_2,\dotsc$ are i.i.d. random variables with $X_1\sim\text{bin}(n,p)$, then $\mu:=\mathbb{E}[X_1]=np$ and $\sigma^2:=\text{Var}(X_1)=np(1-p)$. The CLT gives the convergence $(S_N-N\mu)/\sqrt{N\sigma^2}\overset{d}{\to}N(0,1)$. That is, $\mathbb{P}((S_N-N\mu)/\sqrt{N\sigma^2}\leq z)\to\Phi(z)$. $\endgroup$
    – jakobdt
    Jul 13, 2022 at 13:11

1 Answer 1


If you want to approximate $X_n\sim\mathsf{Bin}(n,p)$ with a normal distribution then you must go for $X_n=\sum_{i=1}^nB_i$ where the $B_i$ are iid and have Bernoulli distribution with parameter $p$.

Then: $$\frac{X_n-\mathbb EX_n}{\mathsf{SD}(X_n)}=\frac{X_n-np}{\sqrt{np(1-p)}}\to U\text{ a.s.}$$where $U$ has standard normal distribution.

Note that here:$$\frac{X_n-np}{\sqrt{np(1-p)}}=\frac{\bar B_n-p}{\sigma/\sqrt{n}}$$for $\sigma=\mathsf{SD}(B_1)=\sqrt{p(1-p)}$, showing the connection with CLT.

Edit concerning questions in comments on this question:

If a random variable $Y$ has a second moment then it has a standardized form: $$Y^*:=\frac{Y-\mu}{\sigma}$$where $\mu:=\mathbb EY$ and $\sigma^2:=\mathsf{Var}Y$.

Characteristic for this form are: $$\mathbb EY^*=0\text{ and }\mathsf{Var}Y^*=1$$

Formulation of CLT: If $X_1,X_2,\dots$ are iid random variables that have a second moment and: $$S_n:=X_1+\cdots+X_n$$ then standard form $S_n^*$ converges to a random variable $Z$ that has standard normal distribution.

If in this context $\mathbb EX_1=\mu$ and $\mathsf{Var}(X_1)=\sigma^2$ then we find $\mathbb ES_n=n\mu$ and $\mathsf{Var}(S_n)=n\sigma^2$ so that we find:$$S_n^*:=\frac{S_n-n\mu}{\sigma\sqrt{n}}$$

Applying this on special case where $X_i$ have Bernoulli distribution with parameter $p$ we get:$$S_n^*:=\frac{S_n-np}{\sqrt{np(1-p)}}$$ In this situation $S_n$ has binomial distribution with parameters $n$ and $p$.

Applying this on special case where $X_i$ have Poisson distribution with parameter $\lambda$ we get:$$S_n^*:=\frac{S_n-n\lambda}{\sqrt{n\lambda}}$$ In this situation $S_n$ has Poisson distribution with parameter $n\lambda$.

  • $\begingroup$ Yes but the CLT says "$\lim_{n \to \infty} P(\frac{S_{n}-nμ}{\sqrt{nσ^2}}\leqslant z)\Rightarrow Φ(z)$" But in your solution is "$\lim_{n \to \infty} P(\frac{S_{n}-μ}{\sqrt{σ^2}}\leqslant z)\Rightarrow Φ(z)$". Where do these two n go in your solution? $\endgroup$
    – MarcoDJ01
    Jul 13, 2022 at 9:49
  • $\begingroup$ In your comment let $S_n:=B_1+\cdots+B_n$ where the $B_i$ are iid Bernoulli rv's with parameter $p$. Then $\mu=p$ and $\sigma^2=p(1-p)$. Substituting that in your statement about CLT we get:$$P\left(\frac{S_n-np}{\sqrt{np(1-p)}}\leq z\right)\to\Phi(z)$$which is exactly the statement in my answer with the only difference that $S_n$ has notation $X_n$ there. Dividing numerator and denominator both by $n$ does not change things. What you are saying about my solution is not correct. In the denominator you find factor $\sqrt{n}$ or - after dividing by $n$ up and down by $n$- factor $1/\sqrt{n}$. $\endgroup$
    – drhab
    Jul 13, 2022 at 10:57
  • $\begingroup$ So you are taking the $μ$ and $σ^2$ of only one Bernoulli distributed random variable? $\endgroup$
    – MarcoDJ01
    Jul 13, 2022 at 11:18
  • $\begingroup$ Yes. My $X_n$ is a summation of $n$ iid Bernoulli rv's having parameter $p$. Consequently $X_n$ has binomial distribution with parameters $n$ and $p$. $\endgroup$
    – drhab
    Jul 13, 2022 at 11:21
  • $\begingroup$ Okay to make sure that I have understood. If we take that the $X_{n}$ are poisson distributed with parameter $λ$ (so we have that$μ=λ$ and $σ^2=λ$), than we have for the convergence with the CLT $\lim_{n \to \infty} P(\frac{S_{n}-λ}{\sqrt{λ}}\leqslant z)\Rightarrow Φ(z)$, right? $\endgroup$
    – MarcoDJ01
    Jul 13, 2022 at 12:30

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