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.
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
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 onwrite your gut feeling. We will revisit this at the end of the lesson.
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 $3\sigma$.
Key facts
- Properties of the normal distribution
- 68-95-99.7 (empirical) rule
- Mean = median = mode for normal data
Concepts
- Why the bell curve appears so often in nature
- How standard deviation controls the spread
- Symmetry and equal-area properties
Skills
- Apply the empirical rule to estimate percentages
- Find cutoff values for given percentages
- Identify whether values are unusual
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
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?
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.
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
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%?
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%?
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)
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
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.
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:
Quick-fire practice · 2 activities
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.
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.
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?
Pick your answer, then rate your confidencethat 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?
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)
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)
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)
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.
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 arenaClimb platforms by answering normal distribution questions. Pool: lesson 6.
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