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Module 8 · L6 of 12 ~25 min MST-12-S2-10 ⚡ +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
1

Get oriented

Set up your goals, key ideas and key terms for the bell curve.

Worksheets

Practise this lesson

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

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
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 $3\sigma$.

The empirical rule for a normal distribution.
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%).
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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
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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.
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Describe the normal curve

Learn the shape, symmetry and centre of a normal distribution.

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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.

The normal distribution is bell-shaped, symmetric about the mean μ, with spread determined by standard deviation σ. Mean = median = mode. Total area under the curve = 1. Most data lies within 3σ of the mean.

Pause, copy the five properties of the normal distribution: bell-shaped, symmetric about μ, mean = median = mode, total area under curve = 1, and most data lies within 3σ of the mean into your book.

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

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Use the 68-95-99.7 rule

Learn the empirical rule and follow it through two worked problems.

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

The normal distribution is bell-shaped, symmetric about the mean μ, with spread determined by σ; mean = median = mode, and the total area under the curve equals 1. The five core properties describe the shape, but the empirical rule gives the critical percentages: 68% of data lies within 1σ of the mean, 95% within 2σ, and 99.7% within 3σ.

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 $\mu - \sigma$.

Empirical rule: 68% of data lies within 1σ, 95% within 2σ, 99.7% within 3σ of the mean. By symmetry, half these percentages lie on each side. Memorise as 68-95-99.7.

Pause, copy the empirical rule: 68% within ±1σ, 95% within ±2σ, 99.7% within ±3σ, and the symmetric split: half of each percentage lies on each side of the mean (e.g., 34% between μ and μ+1σ) into your book.

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

Worked examples · reveal each step

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%?

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(a) $62 = 72 - 10 = \mu - \sigma$ and $82 = 72 + 10 = \mu + \sigma$
Identify how many SDs each boundary is from the mean
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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%?

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

Turn the rule into percentages, and avoid the three traps that cost marks.

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Applications, finding percentages and unusual values
core concept

The empirical rule gives 68% within ±1σ, 95% within ±2σ, 99.7% within ±3σ, by symmetry, half each percentage lies on each side. For one-sided intervals or non-standard bands, use symmetry: the percentage from μ to μ+1σ is 68/2 = 34%, and from μ+1σ to μ+2σ is (95−68)/2 = 13.5%. Build any interval by adding or subtracting these half-band values.

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.

To find the percentage of data in a given interval: calculate how many standard deviations each boundary is from the mean, apply 68/95/99.7 (whole-band or half-band using symmetry), then subtract or add as needed.

Pause, copy the three-step interval process: (1) count how many standard deviations each boundary is from μ; (2) apply the matching 68/95/99.7 percentage; (3) use symmetry to split or construct one-sided/non-standard intervals, into your book.

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

Common errors · the 3 traps that cost marks

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.
5

Compare two distributions

Use mean and standard deviation to compare two normal curves.

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Comparing two normal distributions
core concept

Applying the empirical rule, using symmetry and subtraction to find percentages in any interval, gives numerical answers. When comparing two normal distributions, you need a qualitative description: the distribution with the larger mean has a higher centre (typically larger values), and the distribution with the larger standard deviation is more spread out (less consistent data).

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.

To compare two normal distributions: state which has the larger mean (higher centre) and which has the larger standard deviation (more spread). A smaller σ means data is more consistent; a larger σ means more variability.

Pause, copy the comparison framework: larger mean → higher centre (typically larger values); larger standard deviation → more spread (less consistent); always describe the comparison using the actual variable names and units in context into your book.

Match each percentage to the correct empirical rule band:

  • 68%
  • 95%
  • 99.7%
  • 16%
  • Within ± 3 standard deviations
  • Above μ + 1σ (by symmetry)
  • Within ± 1 standard deviation
  • Within ± 2 standard deviations
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Drill and consolidate

Work the quick-fire practice, then revisit your first answer.

Quick-fire practice · 2 activities

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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.

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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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Practise the normal distribution

Answer the multiple-choice questions on the bell curve and the empirical rule.

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Multiple choice
+5 XP per correct · +25 XP all-correct

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?

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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.