Mean, Variance and Standard Deviation of a Binomial Variable

For a binomial variable $X \sim B(n, p)$ with $q = 1 - p$:

• Mean (expected value): $\mu = E(X) = np$.
• This is the number of trials times the single-trial success probability — it needs no summation, no $P(X=r)$ table.
• The mean is the balance point of the distribution: for $B(n, \tfrac12)$ it sits at the centre $\tfrac{n}{2}$.

For $X \sim B(n, p)$ with $q = 1 - p$:

• Variance: $\sigma^2 = \operatorname{Var}(X) = npq$.
• Standard deviation: $\sigma = \sqrt{npq}$.
• Key inequality: since $0 < q < 1$, we always have $npq < np$, i.e. variance is always less than the mean for a binomial variable.

You can also get the variance the long way via $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$, but for a genuine binomial $npq$ is far faster.

Given a binomial's mean and variance, recover the parameters:

• Divide variance by mean: $\dfrac{npq}{np} = q$, so $q = \dfrac{\text{variance}}{\text{mean}}$.
• Then $p = 1 - q$ and $n = \dfrac{\text{mean}}{p}$.
• With $n$ and $p$ known you can compute any probability, e.g. $P(X = 0) = q^n$, $P(X \ge 1) = 1 - q^n$, or a lower tail $P(X \le r) = \sum_{k=0}^{r} {}^{n}C_{k}\,p^{k}q^{n-k}$.

When the data is a combination of mean and variance for a known n:

• Write mean $= np$ and variance $= npq = np(1-p)$, with the given $n$ substituted in.
• Sum condition: $np + npq = np(1 + q)$; replace $q = 1 - p$ and solve.
• The result is an equation in $p$ alone (frequently a quadratic). Reject any root with $p > 1$ or $p < 0$ — a probability must lie in $[0, 1]$.
• Once $p$ is fixed, back-substitute to report whichever quantity is asked (the variance, $p$, etc.).