Math Detours · Grade 12 · Sharpen · 15

The binomial distribution

n trials, probability p. Mean np, variance np(1 − p), and why variances add but standard deviations do not.

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Tonight's idea

Repeat an experiment \(n\) times, independently, succeeding each time with probability \(p\). Let \(X\) count the successes. One particular pattern of \(k\) successes and \(n - k\) failures has probability \(p^k(1-p)^{n-k}\), and there are \(\binom nk\) such patterns, so

\[P(X = k) = \binom nk p^k (1-p)^{n-k}.\]

These sum to 1 by the binomial theorem: \(\big(p + (1-p)\big)^n = 1\).

0 2 4 6 8 10 12 14 mean 6, SD 2.05 k
\(P(X = k)\) for \(n = 20\), \(p = 0.3\): peak at \(np = 6\), bracket one SD either side.

For the mean and variance, leave that formula alone. Write \(X\) as a sum of indicators: \(I_j = 1\) if trial \(j\) succeeds, \(0\) if not, so \(X = I_1 + \dots + I_n\). One indicator has \(E[I] = p\), and since \(0^2 = 0\) and \(1^2 = 1\), also \(E[I^2] = p\). So \(\mathrm{Var}(I) = E[I^2] - E[I]^2 = p - p^2 = p(1-p)\).

Expectations always add. Variances add when the variables are independent. For \(U\), \(V\) with means \(\mu\), \(\nu\), expand the square in \(\mathrm{Var}(U + V) = E\big[((U-\mu) + (V-\nu))^2\big]\):

\[\mathrm{Var}(U+V) = \mathrm{Var}(U) + \mathrm{Var}(V) + 2\,E[(U-\mu)(V-\nu)].\]

Independence lets the last expectation split as \(E[U-\mu]\,E[V-\nu] = 0 \cdot 0\). Adding \(n\) indicators:

\(E[X] = np\), \(\quad \mathrm{Var}(X) = np(1-p)\), \(\quad \mathrm{SD}(X) = \sqrt{np(1-p)}\).

Standard deviations do not add: \(\sqrt{a + b} \neq \sqrt a + \sqrt b\). Double \(n\) and the variance doubles but the SD grows only by \(\sqrt2\), so the spread of a total grows like \(\sqrt n\).

Watch one: \(n = 20\), \(p = 0.3\)

The mean is \(np = 6\), the variance \(20 \times 0.3 \times 0.7 = 4.2\), the SD \(\sqrt{4.2} \approx 2.05\). The chance of exactly 6 successes:

\[P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14} = 38{,}760 \times 0.000729 \times 0.00678 \approx 0.192.\]

Adding the terms from \(k = 10\) to \(20\) gives \(P(X \geq 10) \approx 0.048\). Ten successes sit nearly two SDs above the mean, and the grade 10 bell-curve rule puts about 2.5% beyond two SDs in one tail: rough, but the right size.

Each ball in a Galton board is a binomial with \(p = \tfrac12\): mathdetours.com/demos/galton.html.

Your turn

  1. A fair die is rolled 12 times and \(X\) counts the sixes. Find the mean, variance and SD of \(X\), and \(P(X = 2)\).
  2. Thirty multiple-choice questions, four options each, all guessed. Find the mean and SD of the score. A student gets 15 right: how many SDs above the mean, and roughly how surprising?
  3. One die has mean \(3.5\) and variance \(\tfrac{35}{12}\). (a) Find the variance and SD of the total of two dice, and compare the SD with twice the SD of one die. (b) Find the variance and SD of the total of 100 dice.
  4. (a) For fixed \(n\), which \(p\) makes \(\mathrm{Var}(X)\) largest, and how large is it? (b) A poll of \(n = 1{,}000\) reports \(\hat p = X/n\). Using \(\mathrm{Var}(cX) = c^2\,\mathrm{Var}(X)\), show \(\mathrm{SD}(\hat p) = \sqrt{p(1-p)/n} \leq 1/(2\sqrt n)\). Why do polls quote a margin of about ±3 points?

Stretch: when variances refuse to add

Let \(Y = n - X\) count the failures. Find \(\mathrm{Var}(Y)\) by the indicator method. Then compute \(\mathrm{Var}(X + Y)\) directly. Why is it not \(\mathrm{Var}(X) + \mathrm{Var}(Y)\), and what must the cross term \(2\,E[(X - np)(Y - n(1-p))]\) be?

Answers

  1. Mean \(12 \times \tfrac16 = 2\), variance \(12 \times \tfrac16 \times \tfrac56 = \tfrac53\), SD \(\approx 1.29\). \(P(X = 2) = \binom{12}{2}\left(\tfrac16\right)^2\left(\tfrac56\right)^{10} = 66 \times \tfrac{5^{10}}{6^{12}} \approx 0.296\).
  2. \(n = 30\), \(p = \tfrac14\): mean \(7.5\), variance \(30 \times \tfrac14 \times \tfrac34 = 5.625\), SD \(\approx 2.37\). Fifteen right is \((15 - 7.5)/2.37 \approx 3.2\) SDs above the mean. By the 68–95–99.7 rule, beyond three SDs is well under 1%; the exact binomial tail is \(P(X \geq 15) \approx 0.0027\). Guessing almost never does this.
  3. (a) Variance \(2 \times \tfrac{35}{12} = \tfrac{35}{6}\), SD \(\approx 2.42\). Twice the SD of one die is \(2 \times 1.71 = 3.42\): bigger, because SDs do not add. (b) Variance \(100 \times \tfrac{35}{12} = \tfrac{875}{3} \approx 291.7\), SD \(\approx 17.1\), which is \(\sqrt{100} = 10\) times the SD of one die, not 100 times.
  4. (a) \(p(1-p)\) is a downward parabola with vertex at \(p = \tfrac12\), so the largest variance is \(n/4\) at \(p = \tfrac12\). (b) \(\mathrm{Var}(\hat p) = \mathrm{Var}(X)/n^2 = np(1-p)/n^2 = p(1-p)/n\); take the root. Since \(p(1-p) \leq \tfrac14\), \(\mathrm{SD}(\hat p) \leq \tfrac{1}{2\sqrt n}\); for \(n = 1{,}000\) that is \(0.0158\). Two SDs, the 95% band, is about \(0.032\): three points either way, whatever the true \(p\).

Stretch. \(Y\) is a sum of \(n\) indicators \(1 - I_j\), each with probability \(1 - p\) of being 1, so \(\mathrm{Var}(Y) = n(1-p)p\), the same as \(\mathrm{Var}(X)\). But \(X + Y = n\) always, so \(\mathrm{Var}(X + Y) = 0\), not \(2np(1-p)\). The additivity rule needs independence, and \(X\), \(Y\) are perfectly dependent: knowing one fixes the other. The cross term must cancel the two variances exactly: \(2\,E[(X - np)(Y - n(1-p))] = -2np(1-p)\).

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