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Module 8 · L6 of 12 ~25 min MS12-9 ⚡ +50 XP available

Normal Distribution

Heights, weights, test scores, blood pressure — countless natural phenomena follow a pattern so consistent it has been called the most important distribution in statistics. The normal distribution is the famous bell curve: symmetric, unimodal, and governed by just two numbers — the mean and standard deviation. The 68-95-99.7 rule lets you predict proportions and identify unusual values without complex calculations.

Today's hook — Adult male heights in Australia are approximately normally distributed with mean 178 cm and standard deviation 7 cm. Without calculating, roughly what percentage of men are between 171 cm and 185 cm?
0/5QUESTS
Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

Adult male heights in Australia are approximately normally distributed with mean 178 cm and standard deviation 7 cm. Without calculating, roughly what percentage of men would you expect to be between 171 cm and 185 cm?

Before reading on — write your gut feeling. We will revisit this at the end of the lesson.

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02
Key ideas for this lesson
reference

The normal distribution is the bell-shaped curve. One rule underpins almost every normal distribution question in the HSC.

Normal distribution: A symmetric, bell-shaped curve defined entirely by its mean $\mu$ and standard deviation $\sigma$. Mean = median = mode.

68-95-99.7 rule (empirical rule): 68% of data lies within $1\sigma$ of $\mu$; 95% within $2\sigma$; 99.7% within $3\sigma$.

EMPIRICAL RULE 68% within ± 1σ 95% within ± 2σ 99.7% within ± 3σ Total area = 1 (100%)
By symmetry: 50% above mean and 50% below. Half of each band on each side.
Symmetry halves bands
Since the curve is symmetric, each percentage band splits evenly — e.g., half of the 32% outside ±1σ (= 16%) lies above μ + σ.
SD controls width
Larger $\sigma$ = wider, flatter bell. Smaller $\sigma$ = narrower, taller bell. Same mean, very different distributions.
Unusual = beyond 2SD
Values beyond $\mu \pm 2\sigma$ are unusual (only 5% of data). Values beyond $\mu \pm 3\sigma$ are very unusual (only 0.3%).
03
What you will master
Know

Key facts

  • Properties of the normal distribution
  • 68-95-99.7 (empirical) rule
  • Mean = median = mode for normal data
Understand

Concepts

  • Why the bell curve appears so often in nature
  • How standard deviation controls the spread
  • Symmetry and equal-area properties
Can do

Skills

  • Apply the empirical rule to estimate percentages
  • Find cutoff values for given percentages
  • Identify whether values are unusual
04
Key terms
Normal distributionA symmetric, bell-shaped probability distribution defined by mean $\mu$ and standard deviation $\sigma$.
Mean ($\mu$)The centre of the normal distribution. Also equals the median and mode.
Standard deviation ($\sigma$)Controls the width of the bell curve. Larger $\sigma$ = wider, flatter curve.
68-95-99.7 ruleThe empirical rule: 68%, 95%, and 99.7% of data fall within 1, 2, and 3 standard deviations of the mean respectively.
SymmetricThe left and right sides of the normal curve are mirror images. 50% of data lies above the mean, 50% below.
AsymptoticThe tails of the normal curve approach but never touch the horizontal axis — extreme values are always possible.
05
Properties of the normal distribution
core concept

The normal distribution (bell curve) has these key properties:

  • Symmetric: The left side is a mirror image of the right side
  • Unimodal: One peak at the centre
  • Mean = Median = Mode: All three measures of centre are equal
  • Asymptotic: Tails approach but never touch the horizontal axis
  • Total area under the curve = 1 (represents 100% of the data)

The normal distribution is completely defined by two parameters:

  • Mean $\mu$: Locates the centre of the distribution
  • Standard deviation $\sigma$: Controls the width and spread
Why does the bell curve appear everywhere? When many independent random factors each contribute a small amount to an outcome, the total tends to be normally distributed. This is the Central Limit Theorem in action — heights are influenced by dozens of genes and environmental factors, so the result is approximately bell-shaped.
What to write in your book
  • Normal distribution: symmetric, unimodal, bell-shaped. Mean = median = mode.
  • Defined by $\mu$ (centre) and $\sigma$ (spread). Total area under curve = 1.
  • Tails are asymptotic — they never touch the axis.

Quick check: In a perfectly normal distribution, which statement is always true?

06
The 68-95-99.7 rule (empirical rule)
core concept

For any normal distribution with mean $\mu$ and standard deviation $\sigma$:

  • 68% of data falls within $\mu \pm \sigma$ (within 1 standard deviation)
  • 95% of data falls within $\mu \pm 2\sigma$ (within 2 standard deviations)
  • 99.7% of data falls within $\mu \pm 3\sigma$ (within 3 standard deviations)

Example: IQ scores: $\mu = 100$, $\sigma = 15$.

  • 68% of people have IQ between $100 - 15 = 85$ and $100 + 15 = 115$
  • 95% of people have IQ between $100 - 30 = 70$ and $100 + 30 = 130$
  • 99.7% of people have IQ between $100 - 45 = 55$ and $100 + 45 = 145$

Only 0.3% of people have IQ below 55 or above 145.

Symmetry shortcut: Since the curve is symmetric, the percentage outside a band splits equally. For example: 32% falls outside $\mu \pm 1\sigma$, so 16% is above $\mu + \sigma$ and 16% is below $\mu - \sigma$.
What to write in your book
  • 68% within $\pm 1\sigma$, 95% within $\pm 2\sigma$, 99.7% within $\pm 3\sigma$.
  • By symmetry: 16% above $\mu + \sigma$, 2.5% above $\mu + 2\sigma$, 0.15% above $\mu + 3\sigma$.
  • Values beyond $\pm 2\sigma$ are unusual. Values beyond $\pm 3\sigma$ are very unusual.

True or false: For a normal distribution with mean 50 and SD 5, exactly 2.5% of values lie above 60.

PROBLEM 1 · APPLYING THE EMPIRICAL RULE

Exam marks: $\mu = 72$, $\sigma = 10$, normally distributed. (a) What percentage score between 62 and 82? (b) What percentage score above 92? (c) What is the minimum mark for the top 2.5%?

1
(a) $62 = 72 - 10 = \mu - \sigma$ and $82 = 72 + 10 = \mu + \sigma$
Identify how many SDs each boundary is from the mean
PROBLEM 2 · REVERSE APPLICATION

Weights of apples: $\mu = 150$ g, $\sigma = 15$ g, normally distributed. (a) What percentage weigh between 135 g and 165 g? (b) What percentage weigh less than 120 g? (c) What weight separates the heaviest 16%?

1
(a) $135 = 150 - 15 = \mu - \sigma$ and $165 = 150 + 15 = \mu + \sigma$
Check: both boundaries are exactly 1 SD from the mean
08
Applications — finding percentages and unusual values
core concept

Finding percentages in a range:

Example: Year 12 student heights — $\mu = 170$ cm, $\sigma = 8$ cm. What percentage are taller than 178 cm?

  • $178 = 170 + 8 = \mu + \sigma$
  • 68% are within $\mu \pm \sigma$, so 32% are outside
  • By symmetry, 16% are above $\mu + \sigma = 178$ cm

Identifying unusual values:

  • Values more than 2 SD from the mean are unusual (only 5% of data)
  • Values more than 3 SD from the mean are very unusual (only 0.3% of data)
HSC strategy: Always start by expressing the given value in terms of SDs from the mean. Ask: "Is this $\mu + \sigma$, $\mu + 2\sigma$, or $\mu + 3\sigma$?" Once you know the SD multiple, the percentage follows directly from the 68-95-99.7 rule.
What to write in your book
  • Always convert to "how many SDs from the mean" first.
  • 16% above $\mu + \sigma$. 2.5% above $\mu + 2\sigma$. 0.15% above $\mu + 3\sigma$.
  • Unusual = beyond $\pm 2\sigma$. Very unusual = beyond $\pm 3\sigma$.

Fill the gap: A normal distribution has $\mu = 60$ and $\sigma = 8$. The percentage of values between 44 and 76 is %.

Trap 01
Forgetting to halve tail percentages
The empirical rule gives the total percentage within ±nσ. To find one tail (e.g., above μ + 2σ), you must halve the outside percentage. Above μ + 2σ = (100 − 95)/2 = 2.5%, not 5%.
Trap 02
Applying the rule to non-normal data
The 68-95-99.7 rule only applies to data that is approximately normally distributed. Skewed data (e.g., house prices, incomes) will not follow these percentages.
Trap 03
Confusing mean and SD in boundary calculations
To find μ + 2σ, you add two full standard deviations to the mean — not one. Always write $\mu + n\sigma$ explicitly before substituting numbers.
09
Comparing two normal distributions
core concept

Two normal distributions with the same mean but different standard deviations look very different:

  • Same mean: Both centred at the same point on the horizontal axis
  • Larger $\sigma$: Wider, flatter bell — data more spread out
  • Smaller $\sigma$: Narrower, taller bell — data more concentrated around the mean

Example: Factory A: $\mu = 100$ mm, $\sigma = 2$ mm. Factory B: $\mu = 100$ mm, $\sigma = 5$ mm. Customer requires parts within $100 \pm 4$ mm.

  • Factory A: $100 \pm 4 = \mu \pm 2\sigma$. By the 95% rule, 95% of parts meet spec.
  • Factory B: $100 \pm 4 = \mu \pm 0.8\sigma$. This is less than 1 SD — roughly 58% meet spec.
Key lesson: Same mean does not mean same quality. The standard deviation is critical for quality control, manufacturing tolerances, and comparing distributions. A manager who looks only at the mean misses this crucial difference.
What to write in your book
  • Two normal distributions with same mean but different SD look very different.
  • Larger $\sigma$ = wider bell = more variability = more values outside specifications.
  • Always compare both mean (centre) and SD (spread) when comparing normal distributions.

Match each percentage to the correct empirical rule band:

1

Test scores: $\mu = 65$, $\sigma = 12$, normally distributed. Find: (a) the range containing 95% of scores, (b) the percentage of students scoring above 89, (c) whether a score of 40 is unusual.

2

Factory A: $\mu = 50$ mm, $\sigma = 2$ mm. Factory B: $\mu = 50$ mm, $\sigma = 5$ mm. Customer spec: 46 mm to 54 mm. Calculate the percentage meeting spec from each factory and explain which is better.

Top 3 list: Name THREE real-world examples of data that is approximately normally distributed. For each, identify what the mean and standard deviation might represent.

10
Revisit your thinking

For Australian male heights: $171\text{ cm} = 178 - 7 = \mu - \sigma$ and $185\text{ cm} = 178 + 7 = \mu + \sigma$. By the 68-95-99.7 rule, approximately 68% of Australian men are between 171 cm and 185 cm tall. This is exactly one standard deviation either side of the mean. The rule is powerful because it works for any normal distribution — you only need to know $\mu$ and $\sigma$ to estimate proportions for entire populations.

What has changed in your understanding? What did you get right? What surprised you?

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01
Multiple choice
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Pick your answer, then rate your confidence — that tells the system what to drill next.

Q1. A data set is normally distributed with $\mu = 50$ and $\sigma = 6$. What percentage of values fall between 44 and 56?

Q2. Heights: $\mu = 168$ cm, $\sigma = 6$ cm. What percentage of people are taller than 180 cm?

Q3. Battery life: $\mu = 500$ hours, $\sigma = 50$ hours. A battery lasts 650 hours. This value is:

Q4. Which property is always true for a normal distribution?

Q5. Exam marks: $\mu = 72$, $\sigma = 10$. What is the minimum mark for the top 16% of students?

02
Short answer
ApplyBand 42 marks

SA 1. A factory produces bolts with lengths normally distributed: $\mu = 50$ mm, $\sigma = 2$ mm. (a) What percentage of bolts are between 48 mm and 52 mm? (b) What percentage are shorter than 46 mm? (c) Specifications require bolts between 46 mm and 54 mm. What percentage meet specifications? (2 marks)

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ApplyBand 42 marks

SA 2. IQ scores: $\mu = 100$, $\sigma = 15$. (a) What IQ score separates the top 16% from the rest? (b) Between what two scores do the middle 95% of IQs fall? (c) A person has IQ 130. What percentage of people have higher IQs? (2 marks)

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AnalyseBand 53 marks

SA 3. Two factories produce the same part. Factory A: $\mu = 100$ mm, $\sigma = 2$ mm. Factory B: $\mu = 100$ mm, $\sigma = 5$ mm. Both normally distributed. (a) Describe and compare the two distributions. Which is more consistent? (b) A quality engineer rejects parts more than 3 SD from the mean. What percentage does each factory reject? (c) Customers require parts within $100 \pm 4$ mm. What percentage from each factory meets this specification? (d) A manager argues that since both factories have the same mean, they are equivalent. Refute this claim with quantitative evidence. (3 marks)

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Comprehensive answers (click to reveal)

MC 1 — B: $44 = 50 - 6 = \mu - \sigma$ and $56 = 50 + 6 = \mu + \sigma$. The range $\mu \pm 1\sigma$ always contains 68%.

MC 2 — C: $180 = 168 + 12 = \mu + 2\sigma$. 95% within $\pm 2\sigma$, so 5% outside. By symmetry, 2.5% above 180 cm.

MC 3 — D: $650 = 500 + 3 \times 50 = \mu + 3\sigma$. Exactly 3 SD above mean — very unusual (only 0.15% above this).

MC 4 — A: For a perfectly normal distribution, mean = median = mode by the symmetry property.

MC 5 — C: Top 16% means above $\mu + \sigma = 72 + 10 = 82$. (32% outside $\pm 1\sigma$; by symmetry 16% in each tail.)

SA 1 (2 marks): (a) 68% [0.5]. (b) 46 = $\mu - 2\sigma$; 2.5% below [0.5]. (c) $\mu \pm 2\sigma$; 95% [1].

SA 2 (2 marks): (a) $\mu + \sigma = 115$ [0.5]. (b) $\mu \pm 2\sigma = 70$ to $130$ [0.5]. (c) $130 = \mu + 2\sigma$; 2.5% above [1].

SA 3 (3 marks): (a) Both centred at 100 mm. Factory A: narrow, tall bell ($\sigma = 2$). Factory B: wide, flat bell ($\sigma = 5$). Factory A more consistent. [0.5] (b) Both reject 0.3% — the 3 SD rule gives the same rejection rate for any normal distribution. [0.5] (c) Factory A: $100 \pm 4 = \mu \pm 2\sigma$ → 95% meet spec. Factory B: $100 \pm 4 = \mu \pm 0.8\sigma$ → approximately 58% meet spec. [1] (d) Same mean conceals dramatically different quality: 95% vs 58% of parts meet customer spec. Standard deviation is the critical measure of quality, not the mean alone. [1]

Drill 1: (a) $65 \pm 24 = 41$ to $89$. (b) $89 = \mu + 2\sigma$; 2.5% above. (c) $40 = 65 - 25$ is about 2.08 SD below mean — beyond 2 SD, so unusual.

Drill 2: Factory A: $50 \pm 4 = \mu \pm 2\sigma$ → 95%. Factory B: $50 \pm 4 = \mu \pm 0.8\sigma$ → ~58%. Factory A is better — much higher proportion of parts meets specifications despite the same mean.

01
Boss battle · The Bell Curve Master
earn bronze · silver · gold

Five timed questions on the normal distribution, the 68-95-99.7 rule, and applying the empirical rule to real contexts. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering normal distribution questions. Pool: lesson 6.

Mark lesson as complete

Tick when you've finished the practice and review.