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Module 10 · L09 of 20 ~40 min ⚡ +90 XP available

Variance and Standard Deviation

A factory tests 100 batteries from a production line where each has a 5% chance of being defective. The mean number of defective batteries is easy: $np = 5$. But how much will that count vary from one batch to the next? Variance and standard deviation answer that question, they measure spread. For the binomial, both have famously clean formulas: $\text{Var}(X) = npq$ and $\text{SD}(X) = \sqrt{npq}$. This lesson trains the calculation and the interpretation.

Today's hook, Suppose $X \sim B(100, 0.05)$. Before reading on, predict the variance and the standard deviation. Which of these is the typical "give or take" range you'd expect for the number of defectives, about 2, about 5, or about 10? Compare your guess after card 05.

0/5QUESTS

Recall, your gut answer first

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You already know that for $X \sim B(n, p)$ the mean is $E(X) = np$. Without using any formulawhat do you think happens to the spread of $X$ as $p$ gets close to $0$ or close to $1$? Is the distribution more or less spread out compared to when $p = 0.5$? Write your reasoning below.

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The two moves for spread questions

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Every variance/SD calculation rewards two habits: identify $n$, $p$, and $q = 1-p$ from the wording, then substitute into $npq$ before doing arithmetic. Skipping the $q$ step or confusing $p$ with $q$ is the single biggest cause of wrong answers.

The identify-then-substitute strategy: (1) extract $n$ (number of trials) and $p$ (success probability) from the question, (2) compute $q = 1 - p$, (3) substitute into $\text{Var}(X) = npq$ and $\text{SD}(X) = \sqrt{npq}$ before simplifying.

Mean: $\mu = np$ · Variance: $\sigma^2 = npq$ · SD: $\sigma = \sqrt{npq}$

$\sigma^2 = npq, \quad \sigma = \sqrt{npq}$

Always work out $q$ explicitly

Write $q = 1 - p$ as a separate line, never substitute $p$ where $q$ belongs. This is the #1 algebra slip in HSC binomial questions.

Variance ≠ Standard deviation

Variance has units$^2$ (e.g., "people$^2$") so we square-root to get SD, which has the same units as $X$. Always state which one you've found.

$pq$ is maximised at $p = 0.5$

For fixed $n$, the binomial spread is largest when $p = 0.5$ (since $pq = 0.25$) and shrinks toward $0$ as $p \to 0$ or $p \to 1$.

What you'll master

Know

Key facts

Variance: $\text{Var}(X) = npq$ where $q = 1-p$
Standard deviation: $\text{SD}(X) = \sqrt{npq}$
SD has the same units as $X$; variance has units$^2$

Understand

Concepts

Why spread depends on both $n$ and the product $pq$
Why $pq$ is largest at $p = 0.5$ and shrinks at the extremes
How SD describes the typical deviation from the mean

Can do

Skills

Calculate $\text{Var}(X)$ and $\text{SD}(X)$ given $n$ and $p$
Interpret SD as the typical spread around the mean $np$
Compare spreads of binomial distributions with different parameters

Key terms

Variance, $\text{Var}(X)$The expected squared deviation from the mean: $E\bigl[(X - \mu)^2\bigr]$. For binomial: $npq$.

Standard deviation, $\sigma$The positive square root of the variance: $\sigma = \sqrt{\text{Var}(X)}$. Same units as $X$.

$q$ (failure probability)The probability of failure on a single trial, $q = 1 - p$. Always compute this explicitly before substituting.

SpreadA general term for how widely the distribution's values scatter around the mean. SD is the most common measure.

Maximum spreadFor a binomial with fixed $n$, variance is maximised at $p = 0.5$, giving $\text{Var}(X) = 0.25n$.

ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including binomial distribution mean, variance and standard deviation.

The variance and SD formulas

core concept

If $X \sim B(n, p)$ counts successes in $n$ independent trials, then

Mean: $\mu = E(X) = np$, already familiar from earlier lessons.
Variance: $\sigma^2 = \text{Var}(X) = npq$ where $q = 1 - p$.
Standard deviation: $\sigma = \text{SD}(X) = \sqrt{npq}$.
Interpretation: roughly speaking, most outcomes lie within $1$–$2$ standard deviations of the mean.

Worked through the hook: For $X \sim B(100, 0.05)$:

$n = 100$, $p = 0.05$, so $q = 1 - 0.05 = 0.95$.
Mean: $\mu = np = 100 \times 0.05 = 5$ defectives.
Variance: $\sigma^2 = npq = 100 \times 0.05 \times 0.95 = 4.75$.
Standard deviation: $\sigma = \sqrt{4.75} \approx 2.179$.
So the typical count of defectives is about $5 \pm 2$, closer to "give or take 2", not 5 or 10.

Why $npq$? Variance is additive for independent trials. Each Bernoulli trial has variance $pq$, and adding $n$ independent trials gives total variance $npq$. This same logic gives the mean as $np$.

Binomial formulas: $E(X)=np$; $\text{Var}(X)=npq$; $\text{SD}(X)=\sqrt{npq}$ where $q=1-p$.

Pause, copy the formulas $\text{Var}(X)=npq$ and $\text{SD}(X)=\sqrt{npq}$ with a worked example for $B(100,0.05)$ into your book.

Quick check: If $X \sim B(80, 0.25)$, what is $\text{Var}(X)$?

Interpreting standard deviation

core concept

We just saw that for $X\sim B(n,p)$: $\text{Var}(X)=npq$ and $\text{SD}(X)=\sqrt{npq}$ where $q=1-p$. That raises a question: what does $\sigma=\sqrt{npq}$ actually tell you about the spread of outcomes, and how do you use it to judge whether an observed frequency is typical or unusual? This card answers it → most outcomes fall within $[\mu-\sigma,\mu+\sigma]$; a gap of more than $2\sigma$ from $\mu$ is noteworthy.

SD tells you the typical deviation from the mean. If $X \sim B(n, p)$ has mean $\mu = np$ and standard deviation $\sigma = \sqrt{npq}$, then most observed values of $X$ fall in the interval $[\mu - \sigma, \mu + \sigma]$, and almost all fall in $[\mu - 2\sigma, \mu + 2\sigma]$.

Compare two scenarios:

$X_1 \sim B(100, 0.5)$: $\mu = 50$, $\sigma = \sqrt{100 \times 0.5 \times 0.5} = \sqrt{25} = 5$. Typical range $\approx [45, 55]$.
$X_2 \sim B(100, 0.05)$: $\mu = 5$, $\sigma = \sqrt{100 \times 0.05 \times 0.95} = \sqrt{4.75} \approx 2.18$. Typical range $\approx [2.8, 7.2]$.

Even though $X_1$ has a much bigger SD ($5$ vs $2.18$), $X_2$ has a larger SD relative to its mean ($2.18/5 \approx 44\%$ versus $5/50 = 10\%$). For comparing relative variability across different distributions, scale matters.

$$\text{Var}(X) = npq \quad \text{and} \quad \text{SD}(X) = \sqrt{npq}$$

Common mistake. Students sometimes report variance when the question asks for SD (or vice versa). Always re-read the question and state your final answer with the correct label.

SD tells you the typical deviation from the mean. If $X \sim B(n, p)$ has mean $\mu = np$ and standard deviation $\sigma = \sqrt{npq}$, then most observed values of $X$ fall in the interval $[\mu - \sigma, \mu +...

Pause, copy the interpretation rule: typical observed values fall within $\pm 2\sigma$ of the mean $\mu=np$; gaps larger than this warrant comment into your book.

Did you get this? True or false: for $X \sim B(n, p)$ with $n$ fixed, the variance $npq$ is maximised when $p = 0.5$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · STRAIGHT SUBSTITUTION

A fair coin is tossed $60$ times. Let $X$ be the number of heads. Find (a) the variance of $X$ and (b) the standard deviation of $X$.

Identify parameters: $n = 60$, $p = 0.5$, $q = 1 - 0.5 = 0.5$.

A "fair coin" gives $p = 0.5$ for heads. Always state $q$ separately.

PROBLEM 2 · QUALITY CONTROL CONTEXT

A machine produces light bulbs, $4\%$ of which are defective. A quality inspector tests a random sample of $200$ bulbs. Let $X$ be the number of defective bulbs found. Find $\text{Var}(X)$ and $\text{SD}(X)$, and interpret the SD.

$X \sim B(200, 0.04)$: $n = 200$, $p = 0.04$, $q = 1 - 0.04 = 0.96$.

Each bulb is an independent trial, defective or not, with constant probability $p$.

PROBLEM 3 · COMPARING TWO DISTRIBUTIONS

Two binomial distributions are $X \sim B(50, 0.2)$ and $Y \sim B(50, 0.8)$. Compare their variances and standard deviations. Comment on what you notice.

For $X$: $n=50$, $p=0.2$, $q=0.8$. $\text{Var}(X) = 50 \times 0.2 \times 0.8 = 8$, $\text{SD}(X) = \sqrt{8} \approx 2.828$.

Straight substitution. Mean of $X$ is $np = 10$.

Fill the gap: If $X \sim B(64, 0.25)$, then $\text{Var}(X) = 64 \times 0.25 \times 0.75 = $, so $\text{SD}(X) = \sqrt{12} \approx 3.46$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $p$ where $q$ belongs (or vice versa)

Variance is $npq$, not $np^2$ or $nq^2$. Always write $q = 1 - p$ as a separate line and substitute carefully. A 5-mark question can lose all method marks from this single slip.

Trap 02

Confusing variance with standard deviation

If the question says "find the variance", do not give the SD, and vice versa. Variance has units$^2$; SD has the same units as $X$. Re-read the question, then label your final answer clearly.

Trap 03

Forgetting that $\text{SD} = \sqrt{\text{Var}}$ is positive

The standard deviation is always the positive square root. Never give a negative SD or write $\pm\sqrt{npq}$. Likewise, variance is never negative since $n, p, q \geq 0$.

Did you get this? True or false: for $X \sim B(100, 0.3)$, the standard deviation is $\sqrt{30}$.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

$X \sim B(50, 0.4)$. Find the variance and standard deviation.

A die is rolled $36$ times. Let $X$ be the number of sixes. Find $\text{Var}(X)$ and $\text{SD}(X)$.

$15\%$ of customers at a café order a flat white. In a sample of $80$ customers, find the standard deviation of the number who order a flat white.

For what value of $p$ is the variance of $B(40, p)$ greatest? What is that maximum variance?

$X \sim B(n, 0.25)$ has variance $\text{Var}(X) = 30$. Find $n$.

Odd one out: Three of these statements about $X \sim B(100, 0.5)$ are correct. Which one is NOT?

Revisit your thinking

Earlier you predicted whether $X \sim B(100, 0.05)$ would have a typical spread of about $2$, $5$, or $10$.

The answer: variance $= npq = 100 \times 0.05 \times 0.95 = 4.75$, so SD $= \sqrt{4.75} \approx 2.18$. The typical spread is about $2$. Because $p$ is so close to $0$, the distribution is tight near its mean of $5$, most batches will contain $3$–$7$ defectives. SD shrinks at the extremes of $p$ and is largest at $p = 0.5$.

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Multiple choice

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Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. $X \sim B(120, 0.25)$. Find the variance and standard deviation of $X$. (2 marks)

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ApplyBand 43 marks

Q2. In a factory, $6\%$ of phones produced are defective. A sample of $250$ phones is tested. Find the standard deviation of the number of defective phones, and interpret what it means. (3 marks)

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AnalyseBand 53 marks

Q3. A binomial random variable $X \sim B(n, p)$ has mean $24$ and variance $19.2$. Find $n$ and $p$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $q = 0.6$. $\text{Var}(X) = 50 \times 0.4 \times 0.6 = 12$. $\text{SD}(X) = \sqrt{12} \approx 3.464$.

2. $p = 1/6$, $q = 5/6$. $\text{Var}(X) = 36 \times \frac{1}{6} \times \frac{5}{6} = 5$. $\text{SD}(X) = \sqrt{5} \approx 2.236$.

3. $\text{Var}(X) = 80 \times 0.15 \times 0.85 = 10.2$. $\text{SD}(X) = \sqrt{10.2} \approx 3.194$.

4. $pq$ maximised at $p = 0.5$, giving max variance $= 40 \times 0.5 \times 0.5 = 10$.

5. $n \times 0.25 \times 0.75 = 30 \Rightarrow 0.1875 n = 30 \Rightarrow n = 160$.

Q1 (2 marks): $q = 0.75$; $\text{Var}(X) = 120 \times 0.25 \times 0.75 = 22.5$ [1]. $\text{SD}(X) = \sqrt{22.5} \approx 4.743$ [1].

Q2 (3 marks): $X \sim B(250, 0.06)$, $q = 0.94$ [1]. $\text{Var}(X) = 250 \times 0.06 \times 0.94 = 14.1$; $\text{SD}(X) = \sqrt{14.1} \approx 3.755$ phones [1]. Interpretation: mean defectives $= 15$, so most samples contain roughly $15 \pm 3.76$ defectives, i.e., between $11$ and $19$ [1].

Q3 (3 marks): $np = 24$ and $npq = 19.2$; dividing: $q = 19.2/24 = 0.8$, so $p = 0.2$ [1]. Then $n \times 0.2 = 24 \Rightarrow n = 120$ [1]. Verify: $\text{Var}(X) = 120 \times 0.2 \times 0.8 = 19.2$ ✓ [1].

Boss battle · The Spread Sentinel

earn bronze · silver · gold

Five timed questions on variance and standard deviation of binomial distributions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering variance and SD questions. Lighter alternative to the boss.

Mark lesson as complete

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Module overview · Extension 1 · Checkpoint 2