Mean and Variance of the Binomial Distribution

For a general discrete distribution, computing the mean and variance means summing $x_i P(x_i)$ and $x_i^2 P(x_i) - \mu^2$ term by term, as in section 10.3. For the Binomial distribution specifically, this sum works out to a strikingly simple closed form (the algebra behind it is left aside here, but the result is the single most useful fact about the distribution):

Theorem. If $X \sim B(n,p)$, then $\text{mean } \mu = np$ and $\text{variance } \sigma^2 = npq$.

This matches intuition well: if a coin with probability $p$ of heads is tossed $n$ times, the expected number of heads is simply $n$ times the chance of heads on each toss, i.e. $np$. The variance $npq$ shows that the spread is largest when $p = q = 0.5$ (a fair, unbiased trial) and shrinks toward $0$ as $p$ moves toward either extreme $0$ or $1$ (an almost-certain outcome has almost no variability left in it).

Worked Example (finding a probability). A coin is tossed $20$ times. Treating heads as success with $p = \frac12$, the number of heads $X \sim B(20, 0.5)$ has mean $\mu = np = 20 \times 0.5 = 10$ and variance $\sigma^2 = npq = 20 \times 0.5 \times 0.5 = 5$, so the standard deviation is $\sigma = \sqrt5 \approx 2.24$. On average we expect $10$ heads, and the typical deviation from that average is a little over $2$ heads either way — which matches the everyday sense that getting exactly $10$ heads out of $20$ tosses is common, while getting, say, $19$ heads would be extraordinarily rare. …