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What are the properties of the normal distribution and why is it so widely used?

The normal distribution is a symmetric bell-shaped density defined by its mean and standard deviation, with predictable probabilities given by the empirical rule.

The shape and properties of the normal distribution, the role of the mean and standard deviation, and the 68-95-99.7 empirical rule for estimating probabilities.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Properties of the bell curve
The empirical rule (68-95-99.7)
Why so many quantities are normal
Comparing two normal distributions
Common errors
Why it matters

What this dot point is asking

The normal distribution is the most important continuous distribution in statistics. Many natural and measured quantities - heights, exam scores, measurement errors - follow it closely, and it is the limiting shape of the binomial distribution for large $n$ .

Properties of the bell curve

The notation $X\sim N(\mu,\sigma^2)$ means $X$ is normally distributed with mean $\mu$ and variance $\sigma^2$ (so standard deviation $\sigma$ ).

The empirical rule (68-95-99.7)

For any normal distribution, fixed proportions of the data fall within whole numbers of standard deviations of the mean:

Because the curve is symmetric, each tail beyond $\pm 1\sigma$ holds about $16\%$ , beyond $\pm 2\sigma$ about $2.5\%$ , and beyond $\pm 3\sigma$ about $0.15\%$ .

Applying the empirical rule

Heights in a population are $N(170,\,8^2)$ cm, meaning a mean of $170$ and $\sigma=8$ . This example estimates the proportion of people taller than $186$ cm.

Step 1: express 186 cm in terms of $\mu$ and $\sigma$

To use the empirical rule, we first check how many standard deviations $186$ cm is from the mean. Computing $186 - 170 = 16 = 2\times 8$ , so $186 = \mu + 2\sigma$ . The value sits exactly two standard deviations above the mean.

Step 2: apply the empirical rule to find the tail proportion

The empirical rule tells us that $95\%$ of values lie within $2\sigma$ of the mean, leaving $5\%$ in the two tails combined. Because the normal distribution is symmetric, each tail holds equal probability, so the proportion above $\mu + 2\sigma$ is half of $5\%$ :

P(X>186)\approx \frac{5\%}{2}=2.5\%.

Final answer: approximately $2.5\%$ of the population are taller than $186$ cm.

Proportion within a range

For the same heights $N(170,8^2)$ , this example estimates the proportion of people with height between $162$ cm and $178$ cm.

Step 1: identify both endpoints as $\mu \pm \sigma$

Before applying any rule, we express both endpoints in terms of the mean and standard deviation. Here $162 = 170 - 8 = \mu - \sigma$ and $178 = 170 + 8 = \mu + \sigma$ , so the interval from $162$ to $178$ is exactly the region within one standard deviation of the mean.

Step 2: apply the 68% rule

The empirical rule states directly that approximately $68\%$ of values in any normal distribution lie within one standard deviation of the mean:

P(162\le X\le 178)\approx 0.68=68\%.

Final answer: approximately $68\%$ of the population have heights between $162$ cm and $178$ cm.

Why so many quantities are normal

The normal distribution appears everywhere because of the central limit theorem, which you meet in Topic 6: when many small independent effects add together, their sum tends toward a normal shape regardless of the individual distributions. Human heights, measurement errors and manufacturing variations all arise this way, as the accumulation of many tiny influences. This is also why the binomial distribution, which counts the sum of many independent Bernoulli trials, looks more and more normal as $n$ grows. Recognising the underlying additive structure is what justifies modelling a quantity as normal in the first place.

Comparing two normal distributions

Two normal distributions with the same mean but different standard deviations have their peaks at the same place, but the one with the larger $\sigma$ is shorter and wider because the fixed total area of $1$ must spread over a greater range. Two distributions with the same $\sigma$ but different means are identical in shape, just shifted along the axis. SACE diagrams often show two bell curves and ask you to compare their means and spreads; read the centre for the mean and the width (or the height of the peak) for the standard deviation.

Common errors

Why it matters

The normal distribution underpins the rest of Topic 5 (z-scores and exact probabilities) and all of Topic 6 (sampling distributions and confidence intervals), where sample means are approximately normal. Recognising when data is normal and using the empirical rule for quick estimates is a core Stage 2 skill.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20242 marksCalculator-assumed. Let

X

be normally distributed with mean

54

and standard deviation

6

. Determine (a)

\Pr(50 < X < 52)

and (b)

\Pr(X > 58)

Show worked answer →

Use the normal distribution with $\mu = 54$ , $\sigma = 6$ (calculator normal CDF, or standardise with $z = \dfrac{x - 54}{6}$ ).

(a) $z$ ranges from $\dfrac{50 - 54}{6} = -0.667$ to $\dfrac{52 - 54}{6} = -0.333$ , so $\Pr(-0.667 < Z < -0.333) \approx 0.117$ . (1 mark)

(b) $z = \dfrac{58 - 54}{6} = 0.667$ , so $\Pr(Z > 0.667) = 1 - 0.7475 \approx 0.252$ . (1 mark)

Calculator values to three significant figures are accepted. Both regions lie within one standard deviation of the mean, so neither probability is extreme.

SACE 20231 marksCalculator-assumed. The daily milk production of a randomly chosen cow is normally distributed with mean

21.9

litres and standard deviation

3.26

litres. Determine the probability that production is less than

23

litres.

Show worked answer →

Standardise: $z = \dfrac{23 - 21.9}{3.26} = \dfrac{1.1}{3.26} \approx 0.337$ .

Then $\Pr(X < 23) = \Pr(Z < 0.337) \approx 0.632$ , from the standard normal or the calculator's normal CDF.

So the probability is $0.632$ . Because $23$ is just above the mean $21.9$ , the answer is a little more than $0.5$ , which checks out.

SACE 20222 marksCalculator-free. The masses of apples are

N(150, 20^2)

grams. Using the empirical rule, estimate (a) the proportion of apples between

130

g and

170

g, and (b) the proportion heavier than

190

Show worked answer →

(a) $130 = \mu - \sigma$ and $170 = \mu + \sigma$ , so this is within one standard deviation: about $68\%$ . (1 mark)

(b) $190 = \mu + 2\sigma$ . The empirical rule gives $95\%$ within $2\sigma$ , so $5\%$ in the two tails, and by symmetry the upper tail holds half: about $2.5\%$ . (1 mark)

Marks reward recognising the values as whole numbers of standard deviations from the mean and applying the $68$ - $95$ - $99.7$ rule with the tail halving.