Binomial mean/variance | Year 12

01 / LEARN

The idea, step by step

A binomial count is a sum of indicators

Write a binomial count as $X = B_{1} + \dots + B_{n}$ , where independent indicators each have mean p and variance p(1 − p). Adding the means and independent variances gives $E (X) = n p$ and $Var (X) = n p (1 - p)$ .

The standard deviation is √(np(1 − p)). Do not use np as a spread measurement or leave the variance under a standard-deviation heading. For a success proportion X/n, the mean is p and the variance is p(1 − p)/n.

Work backwards and test feasibility

If $μ = n p > 0$ and $v = n p (1 - p)$ , then $p = 1 - v / μ$ and $n = μ / p$ , provided p is non-zero. The resulting n must be a non-negative integer and p must lie in [0,1].

For a non-degenerate binomial count, variance is smaller than the mean. A sample mean and sample variance may not satisfy the exact relationship because they are estimates; exact stated model moments must satisfy it. Edge cases p = 0 or 1 need separate reasoning.

A common trap

For a binomial variable, variance is np and mean is np(1 − p).

Instead: Mean counts the expected successes, np. Variance includes the additional factor 1 − p.

02 / WORKED EXAMPLES

See the thinking

Example 1

Summaries for a count

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

Mean np = 20 × 0.4 = 8.
Variance np(1 − p) = 20 × 0.4 × 0.6 = 4.8.
Standard deviation = √4.8 ≈ 2.1909.

Mean 8; SD about 2.1909.

Example 2

Recover the parameters

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

Divide variance by mean: 7.2/12 = 0.6 = 1 − p.
Therefore p = 0.4.
Use np = 12 to find n = 12/0.4 = 30; this is an integer.

n = 30 and p = 0.4.

Check your understanding

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

Show answer

0.2.

E(X/50) = E(X)/50 = 10/50 = p.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF 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 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
np = 12 × 0.25 = 3.
np(1 − p) = 3 × 0.75 = 2.25.
Mean 3; variance 2.25.
$X \sim Bin (100, 0.6)$ . Find SD(X).
2 marks
Variance = 100 × 0.6 × 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 = 4/8 = 0.5.
p = 0.5 and n = 8/0.5 = 16.
n = 16, p = 0.5.
Can a binomial variable have exact mean 5 and exact variance 6? Explain.
2 marks
Variance/mean would give 1 − p = 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 X/40.
3 marks
Var(X) = 40 × 0.3 × 0.7 = 8.4.
Var(X/40) = 8.4/40² = 0.00525.
0.00525.
A binomial variable has p = 1 and mean 7. State n and describe its spread.
2 marks
np = 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 ↗

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