Find and apply the mean np and variance np(1-p) of a binomial distribution and describe the effect of n and p on its shape

What this dot point is asking

SCSA Unit 3 asks you to find the summary measures of the binomial distribution and to describe how its two parameters shape it. This dot point is examined in both sections of the WACE written examination, including questions that solve for $n$ or $p$ from a given mean or variance.

Mean and variance formulas

A binomial variable is the sum of $n$ independent Bernoulli trials, each with mean $p$ and variance $p(1-p)$. Summing $n$ of them multiplies both by $n$.

Solving for a parameter

Because the mean and variance depend on $n$ and $p$, you can recover a parameter from a given summary measure. If the mean is known you can find $p$ from $np$, or combine the mean and variance: dividing $\dfrac{\mathrm{Var}(X)}{E(X)}=\dfrac{np(1-p)}{np}=1-p$ isolates $p$ directly.

Where the variance is largest

Because $\mathrm{Var}(X)=np(1-p)$, for a fixed $n$ the spread is greatest when $p(1-p)$ is greatest, which happens at $p=0.5$. A fair process therefore carries the most uncertainty, while a process with $p$ near $0$ or $1$ is nearly certain and has small variance. This explains why political polls (with proportions near $0.5$) need larger samples to achieve a given precision than surveys about rare events.

Effect of $n$ and $p$ on shape

The parameter $p$ controls symmetry: at $p=0.5$ the distribution is symmetric; for $p<0.5$ it is skewed right, and for $p>0.5$ it is skewed left. Increasing $n$ raises the mean $np$ and the spread, and the distribution becomes more bell-shaped, which is why a binomial with large $n$ is well approximated by a normal distribution. A common rule of thumb is that the normal approximation is reasonable when both $np$ and $n(1-p)$ exceed about $5$, because then the distribution is far enough from either boundary at $0$ or $n$ to be roughly symmetric. This connection prepares the way for the sample-proportion work in Unit 4, where a binomial count is rescaled and treated as approximately normal.