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 $B$.
Best Answer
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.