Consider a coin-tossing experiment in which you tossed a coin 12 times and recorded the number of heads. If you performed this experiment over and over again, what would the mean number of heads be? On average, you would expect half the coin tosses to come up heads. Therefore the mean number of heads would be 6 . In general, the mean of a binomial distribution with parameters \mathrm{N} (the number of trials) and \sqcap (the probability of success on each trial) is:

\begin{align*} \mu=\mathrm{Nr} \end{align*}

where \mu is the mean of the binomial distribution. The variance of the binomial distribution is:

\begin{align*} \sigma^{2}=N \pi(1-\pi) \end{align*}

where \sigma^{2} is the variance of the binomial distribution.

Let's return to the coin-tossing experiment. The coin was tossed 12 times, so N=12 . A coin has a probability of 0.5 of coming up heads. Therefore, \Pi=0.5 The mean and variance can therefore be computed as follows:

\begin{align*} \begin{aligned} &\mu=\mathrm{Nm}=(12)(0.5)=6 \\ &\sigma^{2}=\mathrm{Nm}(1-\pi)=(12)(0.5)(1.0-0.5)=3.0 \end{aligned} \end{align*}

Naturally, the standard deviation (\sigma) is the square root of the variance \left(\sigma^{2}\right).

\begin{align*} \sigma=\sqrt{N \pi(1-\pi)} \end{align*}