Binomial mean/variance | Year 12

01 / LEARN

Understand the idea.

A binomial count is a sum of indicators

A binomial count is the sum of $n$ independent Bernoulli indicators. Each has mean $p$ and variance $p (1 - p)$ , giving $E (X) = n p$ and $Var (X) = n p (1 - p)$ . Standard deviation is $\sqrt{n p (1 - p)}$ .

Keep counts separate from proportions. Dividing $X$ by $n$ gives mean $p$ and variance $p (1 - p) / n$ . The mean describes the centre; the variance and standard deviation describe different measures of spread and need their correct labels.

Work backwards and test feasibility

For exact positive mean $μ$ and variance $v$ , divide their binomial formulas: $v / μ = 1 - p$ . Then find $p = 1 - v / μ$ and $n = μ / p$ when $p > 0$ . This avoids solving both parameters at once.

Check that the result has $0 \leq p \leq 1$ and an integer trial count. A non-degenerate binomial model has variance below its mean. Exact model moments obey these restrictions; estimated sample moments need not fit them exactly.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Summaries for a count

$X \sim Bin (20, 0.4)$ . Find its mean and standard deviation.

Worked solution

1
Compute the mean count.
$E (X) = n p = 20 (0.4) = 8$
Twenty trials each contribute an expected success indicator of 0.4.
2
Compute the variance.
$Var (X) = n p (1 - p) = 20 (0.4) (0.6) = 4.8$
The Bernoulli variances add because the trials are independent.
3
Take the square root for standard deviation.
$SD (X) = \sqrt{4.8} \approx 2.1909$
The requested spread is standard deviation, rather than its square.

Answer

Mean 8; SD about 2.1909.

EXAMPLE 02

Recover the parameters

A binomial variable has mean 12 and variance 7.2. Find n and p.

Worked solution

1
Divide variance by mean.
$\frac{7.2}{12} = \frac{n p (1 - p)}{n p} = 1 - p = 0.6$
The positive stated mean allows division by the common factor.
2
Recover the success probability.
$p = 1 - 0.6 = 0.4$
The ratio gives the failure probability.
3
Recover and validate the trial count.
$n = \frac{12}{0.4} = 30$
The trial count is an integer and $p$ lies in $[0, 1]$ .
4
Check both moments.
$30 (0.4) = 12, 30 (0.4) (0.6) = 7.2$
Both supplied summaries are reproduced by the inferred model.

Answer

n = 30 and p = 0.4.

03 / YOUR TURN

Try it yourself.

Check your understanding

If X is Bin(50, 0.2), what is E(X/50)?

Show answer & reasoning

0.2.

Linearity gives $E (X / 50) = E (X) / 50 = 50 (0.2) / 50 = 0.2$ . The expected success proportion equals $p$ .

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6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

$X \sim Bin (12, 0.25)$ . Find its mean and variance.
2 marks
$X \sim Bin (100, 0.6)$ . Find $SD (X)$ .
2 marks
A binomial variable has mean 8 and variance 4. Find $n$ and $p$ .
3 marks
Can a binomial variable have exact mean 5 and exact variance 6? Explain.
2 marks
$X \sim Bin (40, 0.3)$ . Find the variance of the proportion $\frac{X}{40}$ .
3 marks
A binomial variable has $p = 1$ and mean 7. State $n$ and describe its spread.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$X \sim Bin (12, 0.25)$ . Find its mean and variance.
2 marks
$n p = 12 \times 0.25 = 3$ .
$n p (1 - p) = 3 \times 0.75 = 2.25$ .
Mean 3; variance 2.25.
$X \sim Bin (100, 0.6)$ . Find $SD (X)$ .
2 marks
Variance $= 100 \times 0.6 \times 0.4 = 24$ .
Take its positive square root.
$\sqrt{24} \approx 4.8990$ .
A binomial variable has mean 8 and variance 4. Find $n$ and $p$ .
3 marks
$1 - p = \frac{4}{8} = 0.5$ .
$p = 0.5$ and $n = \frac{8}{0.5} = 16$ .
$n = 16$ , $p = 0.5$ .
Can a binomial variable have exact mean 5 and exact variance 6? Explain.
2 marks
$\frac{Variance}{mean}$ would give $1 - p = \frac{6}{5}$ .
This would require $p = - 0.2$ , which is not a probability.
No.
$X \sim Bin (40, 0.3)$ . Find the variance of the proportion $\frac{X}{40}$ .
3 marks
$Var (X) = 40 \times 0.3 \times 0.7 = 8.4$ .
$Var (\frac{X}{40}) = \frac{8.4}{40^{2}} = 0.00525$ .
0.00525.
A binomial variable has $p = 1$ and mean 7. State $n$ and describe its spread.
2 marks
$n p = n = 7$ .
All seven trials succeed with certainty, so there is no variation.
$n = 7$ ; variance and SD are zero.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods: discrete random variables and probability distributions, expected value and variance, transformations and independent sums, Bernoulli and binomial models. Original supporting lessons and exercises follow the approved Ascent chapter structure.

Official curriculum source ↗

In this lesson

Understand the idea.

A binomial count is a sum of indicators

Work backwards and test feasibility

See the method in action.

Summaries for a count

Compute the mean count.

Compute the variance.

Take the square root for standard deviation.

Recover the parameters

Divide variance by mean.

Recover the success probability.

Recover and validate the trial count.

Check both moments.

Try it yourself.

Check your understanding

Keep practising