I understand the concept of standard deviation as the square root of the square of the mean of each sample value - the mean of the sample values.
Here is the mathematical representation (I've solved out the proof independently) :

1.) $\sigma = \sqrt{\{x^2\} - \{x\}^2}$
where $\{\,\}$ is the average and $x$ is a sample value.

2.) There is an alternate mathematical representation using summation sigma (for discrete random variable also) that more people are probably acquainted with. Or does this one have a slightly different meaning, I'm not sure?

My question is, can someone explicitly show me the derivation for the standard deviation of a binomial distribution.
Here is the information I know:

1.) Final formula: $\sigma = \sqrt{pqN}$

2.) $p =$ probability of event A occurring AKA $p = n(A)/N$

where $A$ is an event OR the first binomially distributed random variable, $n(A)$ is the amount of times event $A$ happens, and $N$ is the total number of events

3.) $q =$ probability of event $B$ occurring AKA $p = n(B)/N$ where $B$ is an event OR the second binomially distributed random variable, $n(B)$ is the amount of times event $B$ happens, and $N$ is the total number of events. Also, $q = 1-p$ because there are only two events, $A$ and $B$.


1 Answer 1


First prove that if $n=1$ then the variance is $p(1-p)$.

Then prove that if $X_1,\ldots,X_n$ are independent, then $\operatorname{var}(X_1+\cdots+X_n)$ $=\operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n)$.

Then show that if each of $X_1,\ldots,X_n$ has a Bernoulli distribution with parameter $p$ and they are independent, i.e. each has a binomial distribution with parameters, $1,p$ and they are independent, then $X_1+\cdots+X_n$ has a binomial distribution with parameters $n,p$.

(But "first" and "then" need not be construed literally; you can do the steps above in any order.)

Then put it all together.

  • $\begingroup$ Hi Michael! Thanks so much for all the edits and this answer! I think your explanation may have lit a light bulb in my head. Is the binomial distribution a special case of the normal distribution (where you only have TWO events instead of MANY)? $\endgroup$
    – Nova
    Feb 10, 2015 at 2:59
  • $\begingroup$ Also what is Xn in this case? Any random event value? For example, when rolling two dice, Xn is any number between and including 2 and 12? $\endgroup$
    – Nova
    Feb 10, 2015 at 3:02
  • $\begingroup$ That $X$ has a Bernoulli distribution means $\Pr(X=1)=p$ and $\Pr(X=0)=1-p$. The case of two dice doesn't fit here. The binomial distribution is not a special case of the normal distribution; that would mean that every binomial distribution is a normal distribution. $\endgroup$ Feb 10, 2015 at 3:27
  • $\begingroup$ I'm a bit confused now. If X can only have two values (1 and 0) what is the point of talking about Xn in your explanation? $\endgroup$
    – Nova
    Feb 10, 2015 at 3:30
  • $\begingroup$ You have a sequence of Bernoulli-distributed random variables: $X_1,X_2,X_3,\ldots$. Each of them can assume either of the two values. Suppose, for example, that $n=3$. Then $(X_1,X_2,X_3)$ can be any of eight triples: $(0,0,0)$, $(0,0,1)$, $(0,1,0)$, $(1,0,0)$, $(0,1,1)$, $(1,0,1)$, $(1,1,0)$, $(1,1,1)$. The probabilities assigned to these are $(1-p)^3$ for the first one, $p(1-p)^2$ for each of the next three, $p^2(1-p)$ for each of the three after that, and $p^3$ for the last one. The sum $X_1+X_2+X_3$ has a binomial distribution, so it is equal to $0$ with probability $(1-p)^3$,$\ldots$ $\endgroup$ Feb 10, 2015 at 3:35

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