What is the normal distribution?

The normal distribution is the bell-shaped model written X N(μ, σ^2). It is symmetric about the mean μ, and σ controls its spread.

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How do we describe continuous random variables and work with the normal distribution?

Use probability density functions and the normal distribution, including standardisation to z-scores, to find probabilities.

A continuous random variable is described by a probability density function whose area gives probability; the normal distribution is the key model, and standardising to z-scores lets you compute its probabilities.

Generated by Claude Opus 4.89 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

A continuous random variable can take any value in an interval, so individual values have probability zero and we work with intervals instead. Its distribution is given by a probability density function $f(x)$ .

Because area is what matters, $P(X = c) = 0$ for any single value, and therefore $P(a \le X \le b) = P(a < X < b)$ ; the endpoints do not change a continuous probability.

The mean and variance are the continuous analogues of the discrete formulas:

\mu = \int_{-\infty}^{\infty} x\,f(x)\,dx, \qquad \sigma^2 = \int_{-\infty}^{\infty} (x-\mu)^2 f(x)\,dx.

The normal distribution

The normal distribution is the bell-shaped model written $X \sim N(\mu, \sigma^2)$ . It is symmetric about the mean $\mu$ , and $\sigma$ controls its spread.

Standardising to z-scores

Every normal distribution can be converted to the standard normal $Z \sim N(0,1)$ by standardising.

Worked example: a normal probability

The heights of a group of students are normally distributed with mean $\mu = 170$ cm and standard deviation $\sigma = 8$ cm. Find the probability that a randomly chosen student is taller than $182$ cm.

Step 1: Standardise the value to a $z$ -score

Convert $x = 182$ into a $z$ -score, which measures how many standard deviations $182$ cm is above the mean. This converts the problem to the standard normal distribution $Z \sim N(0, 1)$ , which we can look up directly:

z = \frac{x - \mu}{\sigma} = \frac{182 - 170}{8} = \frac{12}{8} = 1.5.

Step 2: Identify the probability we need

The question asks for the probability of being taller than $182$ cm, which corresponds to the area to the right of $z = 1.5$ under the standard normal curve:

P(X > 182) = P(Z > 1.5).

Step 3: Use the standard normal table and convert to the upper tail

Standard normal tables and most calculators give the lower-tail probability $P(Z \le z)$ . Subtract from $1$ to get the upper-tail area:

P(Z \le 1.5) \approx 0.9332 \quad\Rightarrow\quad P(Z > 1.5) = 1 - 0.9332 = 0.0668.

Final answer: $P(X > 182) \approx 0.0668$ , so about $6.7\%$ of students are taller than $182$ cm.

Inverse normal problems

Many questions run the standardisation backwards: you are told a probability and must find the value of $x$ that achieves it. First find the $z$ -value with that lower-tail probability (from a table or the inverse normal function on a calculator), then unstandardise with $x = \mu + z\sigma$ . For example, to find the mark that the top $10\%$ of a normally distributed cohort exceed, with $\mu = 60$ and $\sigma = 12$ , you need $P(Z > z) = 0.10$ , so $z = 1.2816$ , giving $x = 60 + 1.2816\times 12 \approx 75.4$ . Watch the direction: "top $10\%$ " means $10\%$ above, so the $z$ -value is positive.

An inverse normal calculation

Problem

Battery lifetimes are normally distributed with mean $\mu = 50$ hours and standard deviation $\sigma = 6$ hours. The manufacturer wants to quote a guaranteed lifetime that $95\%$ of batteries will exceed. Find it.

Set up the probability

" $95\%$ exceed" means $P(X > x) = 0.95$ , so $P(X < x) = 0.05$ . The corresponding $z$ -value has $5\%$ below it: $z = -1.6449$ .

Unstandardise

x = \mu + z\sigma = 50 + (-1.6449)(6) = 50 - 9.87 = 40.13 \text{ hours}.

The manufacturer should guarantee about

40

hours. The negative

z

correctly places the threshold below the mean, since most batteries must beat it.

Combining the normal and binomial

A frequent extended-response structure first finds a normal probability $p$ , then uses that $p$ as the success probability in a binomial calculation about a sample of items. The link is that "an individual exceeds a threshold" is a single yes/no trial, so counting how many of $n$ individuals exceed it is binomial with that $p$ . Keep the two stages clearly separated: compute the normal probability first, then feed it into $P(X = k) = \binom{n}{k}p^{k}(1-p)^{n-k}$ .

Summary

Treat continuous probability as area under the pdf, computed by integration, with total area $1$ . For normal problems, standardise with $z = \dfrac{x-\mu}{\sigma}$ , look up the lower-tail probability, then adjust for the region you need: subtract from $1$ for an upper tail, or subtract two lower-tail values for a between-values range. Use the 68-95-99.7 rule as a quick sanity check on your answer.

Exam-style practice questions

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

TCE 20245 marksCalculator-free. A random variable

X

is normally distributed with mean

10

and standard deviation

2

. Using the empirical rule, find: (a)

P(6 \le X \le 14)

; (b)

P(6 \le X \le 8)

; (c) the value of

x

so that

P(X \ge x) = 0.0015

Show worked answer →

Use the empirical ( $68$ - $95$ - $99.7$ ) rule. With $\mu = 10$ and $\sigma = 2$ : one SD spans $8$ to $12$ , two SDs span $6$ to $14$ , three SDs span $4$ to $16$ .

(a) $6$ and $14$ are $\mu - 2\sigma$ and $\mu + 2\sigma$ , so $P(6 \le X \le 14) \approx 0.95$ .

(b) $8 = \mu - \sigma$ and $6 = \mu - 2\sigma$ . The area between one and two SDs below the mean is $\dfrac{0.95 - 0.68}{2} = 0.135$ .

(c) $P(X \ge x) = 0.0015$ is the upper tail beyond three standard deviations (about $99.7\%$ lies within $3$ SDs, leaving $0.15\%$ in each tail). So $x = \mu + 3\sigma = 10 + 6 = 16$ .

TCE 20236 marksCalculator-assumed. Forearm length is normally distributed with mean

40

cm and standard deviation

10

cm. (a) Find the probability a randomly selected person has a forearm length greater than

45

cm. (b) What length puts a person in the longest

20\%

? (c) If

10

people are selected, find the probability that at least

2

have a forearm length greater than

45

cm.

Show worked answer →

(a) Standardise: $z = \dfrac{45 - 40}{10} = 0.5$ . $P(X > 45) = P(Z > 0.5) \approx 0.309$ .

(b) The longest $20\%$ means $P(X > x) = 0.20$ , so $x$ is the $80$ th percentile. The $z$ -value with $20\%$ above is $z = 0.8416$ . Then $x = 40 + 0.8416\times 10 = 48.42$ cm.

P(Y \ge 2) = 1 - P(Y = 0) - P(Y = 1).

P(Y = 0) = (0.6915)^{10} = 0.0247

and

P(Y = 1) = 10\times 0.3085\times(0.6915)^{9} = 0.1102

.
So

P(Y \ge 2) = 1 - 0.0247 - 0.1102 = 0.865

. This part links the normal probability from (a) into a binomial calculation.

TCE 20244 marksCalculator-assumed. A normal distribution has

10\%

above

30

and

25\%

below

10

. Find

\mu

and

\sigma

to two decimal places.

Show worked answer →

Convert each statement into a standardised equation. From standard normal tables: $P(Z > z) = 0.10$ gives $z = 1.2816$ , and $P(Z < z) = 0.25$ gives $z = -0.6745$ .

So $30 = \mu + 1.2816\sigma$ and $10 = \mu - 0.6745\sigma$ .

Subtract the second from the first: $20 = (1.2816 + 0.6745)\sigma = 1.9561\sigma$ , so $\sigma = \dfrac{20}{1.9561} = 10.22$ .

Substitute back: $\mu = 10 + 0.6745\times 10.22 = 16.89$ .
So $\mu = 16.89$ and $\sigma = 10.22$ (to two d.p.). The skill is setting up two simultaneous equations from two given tail areas and solving them.