Mathematical Methods · Units 3 & 4

The Normal Distribution

Understand the normal distribution the easy way, with plain English intuition, the 68 95 99.7 rule, z scores, worked examples and an auto marked practice test. VCE Maths Methods Units 3 and 4.

Imagine measuring the height of every adult in a city and stacking them into a giant bar chart. Almost nobody is extremely short, almost nobody is extremely tall, and a huge crowd bunches up near the middle. Smooth off the bars and you get a single, graceful hill shape called the bell curve. The normal distribution is the maths of that hill, and once you can read it you can answer almost any question about heights, test scores, weights or timings without listing every value.

The standard normal bell curve. The shaded band, from one standard deviation below the mean to one above, holds about 68 per cent of all the values.

Why one curve keeps showing up

When a measurement is pushed around by lots of small, independent effects, the results pile up into the same symmetric hill almost every time. Height depends on genes, diet, sleep and dozens of other tiny nudges, and they average out into a bell shape. That is why the normal distribution is the default model for so much real data.

Two numbers describe the whole curve. The mean $\mu$ says where the peak sits, and the standard deviation $\sigma$ says how wide the hill spreads. A small $\sigma$ gives a tall, narrow curve where values hug the mean. A large $\sigma$ gives a short, wide curve where values spread far. The total area under the curve is always $1$ , because some value must occur, and area means probability.

The 68 95 99.7 rule

You do not always need a calculator to estimate normal probabilities. Because every bell curve has the same shape, the area splits into fixed chunks measured in standard deviations from the mean.

About $68\%$ of values land within one $\sigma$ of the mean, between $\mu - \sigma$ and $\mu + \sigma$ .
About $95\%$ land within two $\sigma$ , between $\mu - 2\sigma$ and $\mu + 2\sigma$ .
About $99.7\%$ land within three $\sigma$ , between $\mu - 3\sigma$ and $\mu + 3\sigma$ .

The curve is perfectly symmetric, so each of these bands splits evenly across the mean. Within one $\sigma$ above the mean holds half of $68\%$ , which is $34\%$ , and the same below. This is the trick that turns the rule into precise answers. Notice too that the leftover tails are tiny. Outside two $\sigma$ there is only $5\%$ , split as $2.5\%$ in each tail.

Standardising with z scores

Different normal distributions have different means and spreads, so comparing a raw value from one to a raw value from another is like comparing prices in different currencies. The fix is to convert every value into the same universal unit by counting how many standard deviations it sits from its own mean. That count is the z score.

z = \frac{x - \mu}{\sigma}

A z score of $0$ is right at the mean. A z score of $2$ means two standard deviations above the mean. A negative z score means below the mean. Once a value is standardised it lives on the standard normal distribution, which always has mean $0$ and standard deviation $1$ , so a single set of tables or one calculator command handles every normal problem at once.

To go the other way, from a z score back to a real value, just rearrange the same formula:

x = \mu + z\sigma

Learn it, then fade the help

1 · I do — fully worked

Compute the z-score of $x = 78$ when $\mu = 60$ and $\sigma = 12$ . $z = \dfrac{x - \mu}{\sigma} = \dfrac{78 - 60}{12} = \dfrac{18}{12} = 1.5$ , so the score sits $1.5$ standard deviations above the mean.

2 · We do — your turn, guided

Same steps, your turn. Find the z-score of $x = 84$ when $\mu = 60$ , $\sigma = 12$ . Begin with $z = \dfrac{84 - 60}{12} = \dfrac{\;?\;}{12}$ .

Check your step

$z = \dfrac{24}{12} = 2.0$ — two standard deviations above the mean.

3 · You do — solo

No scaffolding now: find the z-score of $x = 45$ when $\mu = 60$ , $\sigma = 12$ , and say what it means.

Reveal solution

$z = \dfrac{45 - 60}{12} = \dfrac{-15}{12} = -1.25$ — the value sits $1.25$ standard deviations below the mean (negative, because it is less than $\mu$ ).

Finding probabilities and quantiles

Two kinds of question come up again and again, and they run in opposite directions.

In the first kind you are given a value and asked for a probability, such as the chance a result falls below some cutoff. You standardise the value and read the area to its left. In the second kind, called a quantile question, you are given a probability and asked for the value, such as the score that the top $10\%$ beat. Here you start from the area, find the matching z score with the inverse normal, then turn it back into a real value using $x = \mu + z\sigma$ .

Try one: test scores are normal with $\mu = 70$ and $\sigma = 8$ . What mark must a student beat to land in the top $10\%$ ?

A few habits keep marks safe in exams. Always sketch the curve and shade the region you actually want, because the symmetry makes it easy to read the wrong tail. Watch the wording carefully, since “more than” and “at least” point to the upper tail while “less than” points to the lower one. Keep your full unrounded z score until the very last line, then round once. And do not mix up the spread measures. The standard deviation $\sigma$ is the square root of the variance $\sigma^2$ , and a normal probability calculation always uses $\sigma$ , so reaching for the variance by mistake is a classic and costly error.

Lock it in with active recall

Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.

Active recall

Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).

What does the 68–95–99.7 rule say?

Write the z-score formula and what it measures.

How do you turn a z-score back into a real value?

What fraction lies beyond

2\sigma

in one tail?

Recall · Discrete random variables

How is standard deviation related to variance?

Recall · The binomial distribution

A binomial with large

n

is approximated by which distribution?

Work through the recipe in the Worked Examples tab, then put it to the test in Try It.

Worked Example 1A conditional probability, from a real exam

The diameter of tennis balls is a normally distributed random variable $D$ with mean $6.7$ cm and standard deviation $0.1$ cm. A ball is grade A if its diameter is between $6.54$ cm and $6.86$ cm. Given that a ball can fit through the container opening (diameter $< 6.95$ cm), find the probability that it is classed as grade A, correct to four decimal places.

1
This is conditional. Given the ball fits, the probability it is grade A is the probability of being grade A and fitting, divided by the probability of fitting.
$\Pr(\text{A} \mid \text{fits}) = \frac{\Pr(6.54 < D < 6.86)}{\Pr(D < 6.95)}$
2
Since grade A ( $D < 6.86$ ) already implies fitting ( $D < 6.95$ ), the numerator is just the grade A interval probability.
$\Pr(6.54 < D < 6.86) \approx 0.8904, \qquad \Pr(D < 6.95) \approx 0.9938$
3
Divide the two probabilities. The examiner report flags treating this as a plain intersection probability as the common error.
$\frac{0.8904}{0.9938} \approx 0.8960$

Answer

\Pr(\text{A} \mid \text{fits}) \approx 0.8960

VCAA 2023 Mathematical Methods Exam 2, Section B Q4e

Worked Example 2Working backwards to a percentile, from a real exam

For $D \sim N(6.7, 0.1^2)$ , find the minimum diameter of a tennis ball that is larger than $90\%$ of all tennis balls produced. Give your answer in centimetres, correct to two decimal places.

1
A ball larger than $90\%$ of all balls sits at the $90$ th percentile, so the area to its left is $0.90$ . Find $d$ with $\Pr(D < d) = 0.90$ .
$\Pr(D < d) = 0.90$
2
Use the inverse normal with mean $6.7$ and standard deviation $0.1$ to recover the value $d$ .
$d \approx 6.83 \text{ cm}$
3
The common error flagged in the report is solving $\Pr(D < d) = 0.10$ instead, which gives the bottom $10\%$ cut-off rather than the top.
$d \approx 6.83 \text{ cm}$

Answer

d \approx 6.83 \text{ cm}

VCAA 2023 Mathematical Methods Exam 2, Section B Q4b

Worked Example 3Reading the 68 95 99.7 rule

Adult resting heart rates are normally distributed with mean $\mu = 72$ beats per minute and standard deviation $\sigma = 8$ . Find the probability that a randomly chosen adult has a resting heart rate between $64$ and $88$ beats per minute.

1
Mark where the two values sit relative to the mean. $64$ is one $\sigma$ below and $88$ is two $\sigma$ above.
$64 = 72 - 8 = \mu - \sigma, \qquad 88 = 72 + 16 = \mu + 2\sigma$
2
The $68\%$ rule covers $\mu - \sigma$ to $\mu + \sigma$ , so each side of the mean within one $\sigma$ holds half of that.
$\Pr(\mu - \sigma \le X \le \mu) = \tfrac{1}{2}(0.68) = 0.34$
3
The $95\%$ rule covers $\mu - 2\sigma$ to $\mu + 2\sigma$ , so within two $\sigma$ above the mean holds half of that.
$\Pr(\mu \le X \le \mu + 2\sigma) = \tfrac{1}{2}(0.95) = 0.475$
4
Add the two pieces that sit on either side of the mean.
$\Pr(64 \le X \le 88) = 0.34 + 0.475 = 0.815$

Answer

\Pr(64 \le X \le 88) = 0.815

Worked Example 4Turning a value into a z score

A test score $X$ is normally distributed with mean $\mu = 60$ and standard deviation $\sigma = 12$ . Find the standardised value (z score) of a score of $X = 78$ , and say what it means.

1
The z score counts how many standard deviations a value is above or below the mean.
$z = \frac{x - \mu}{\sigma}$
2
Substitute $x = 78$ , $\mu = 60$ and $\sigma = 12$ .
$z = \frac{78 - 60}{12} = \frac{18}{12}$
3
Simplify to an exact value.
$z = 1.5$

Answer

z = 1.5, \text{ so the score sits } 1.5 \text{ standard deviations above the mean}

Worked Example 5Finding a quantile (working backwards)

The mass $X$ kg of a bag of flour is normally distributed with mean $\mu = 1.02$ kg and standard deviation $\sigma = 0.015$ kg. The heaviest $10\%$ of bags are set aside. Find, correct to three decimal places, the mass $a$ such that $\Pr(X > a) = 0.10$ .

1
If $\Pr(X > a) = 0.10$ then the area to the left is $0.90$ . Find the matching z score where $\Pr(Z < z) = 0.90$ .
$z = 1.2816 \ (\text{from the inverse normal})$
2
Undo the standardisation. Rearranging $z = \dfrac{a - \mu}{\sigma}$ gives a formula for the value $a$ .
$a = \mu + z\sigma$
3
Substitute $\mu = 1.02$ , $z = 1.2816$ and $\sigma = 0.015$ .
$a = 1.02 + 1.2816 \times 0.015 = 1.03922\ldots$

Answer

a \approx 1.039 \text{ kg}

Practice test

Try it yourself

9 questions, 11 marks

Choose your answers, then submit to see your score and the full worked solutions. Multiple choice is marked for you, just like Exam 2 Section A.

Q1.Consider the normal random variable $X$ that satisfies $\Pr(X < 10) = 0.2$ and $\Pr(X > 18) = 0.2$ . The value of $\Pr(X < 12)$ is closest to:

1mark

VCAA 2024 Mathematical Methods Exam 2, Section A Q19

Q2.For a normal random variable $X$ , it is known that $\Pr(X > 200) = 0.325$ and $\Pr(180 < X < 200) = 0.589$ . The mean and standard deviation of $X$ are closest to:

1mark

VCAA 2025 Mathematical Methods Exam 2, Section A Q12

Q3.The diameter of tennis balls is a normally distributed random variable $D$ with mean $6.7$ cm and standard deviation $0.1$ cm. Find $\Pr(D > 6.8)$ , correct to four decimal places.

1mark

Work this on paper. The worked solution appears once you submit.

VCAA 2023 Mathematical Methods Exam 2, Section B Q4a

Q4.A random variable $X$ is normally distributed with mean $\mu = 50$ and standard deviation $\sigma = 5$ . Using the $68$ $95$ $99.7$ rule, $\Pr(45 \le X \le 55)$ is closest to:

1mark

Need a hint?

Check how many

\sigma

each endpoint is from the mean:

45 = \mu - \sigma

and

55 = \mu + \sigma

, so the interval is one

\sigma

either side.

Q5.The variable $X$ is normally distributed with mean $\mu = 30$ and standard deviation $\sigma = 4$ . The standardised value (z score) of $x = 23$ is:

1mark

Need a hint?

Use

z = \dfrac{x - \mu}{\sigma}

and keep the sign: since

23

is below the mean, expect a negative answer.

Q6.Heights of a large group of trees are normally distributed with mean $\mu = 12$ m and standard deviation $\sigma = 2$ m. Using the $68$ $95$ $99.7$ rule, the proportion of trees taller than $16$ m is closest to:

1mark

Need a hint?

Note

16 = \mu + 2\sigma

, so you want just the upper tail beyond two

\sigma

: take half of the

5\%

left outside

\mu \pm 2\sigma

Q7. $X$ is normally distributed with $\mu = 100$ and $\sigma = 15$ . A value has standardised score $z = -2$ . The original value $x$ is:

1mark

Need a hint?

Work backwards with

x = \mu + z\sigma

, and remember

z = -2

is negative so you subtract.

Q8.Two students sit different tests, each scaled to a normal distribution. Anna scores $z = 1.2$ and Ben scores $z = 0.9$ . Which statement is correct?

1mark

Need a hint?

A z score already standardises each test, so no raw marks are needed: a larger positive z means further above the mean.

Q9.The time $X$ minutes a student takes to finish a quiz is normally distributed with mean $\mu = 24$ and standard deviation $\sigma = 3$ . Using the $68$ $95$ $99.7$ rule, find $\Pr(18 \le X \le 27)$ . Give an exact value and show your reasoning.

3marks

Work this on paper. The worked solution appears once you submit.

Frequently asked questions

What is the 68 95 99.7 rule in the normal distribution?

It is a quick way to estimate normal probabilities without a calculator. About 68 percent of values fall within one standard deviation of the mean, about 95 percent within two, and about 99.7 percent within three. Because the curve is symmetric, you can halve each band at the mean to find one sided pieces.

How do I calculate a z score?

Subtract the mean from your value and divide by the standard deviation. The z score tells you how many standard deviations your value sits above or below the mean. A positive z score is above the mean, a negative one is below, and a z score of zero is exactly at the mean.

What is the difference between standard deviation and variance in normal problems?

The variance is the square of the standard deviation, so the standard deviation is the square root of the variance. Every normal probability and z score calculation uses the standard deviation, not the variance, so if a question gives you the variance you must take its square root first.

The Normal Distribution

Learn

Why one curve keeps showing up

The 68 95 99.7 rule

Standardising with z scores

Finding probabilities and quantiles

Lock it in with active recall

Worked examples

Practice questions

Try it yourself

Frequently asked questions