Continuous random variables and the normal distribution - TCE Mathematics Methods (Tasmania)

Syllabus dot point

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

TCEMath MethodsUnit 49 min readUpdated 2 Jun 2026

Quick answer

A continuous random variable is described by a probability density function (pdf); probability equals the area under the pdf, found by integration. The normal distribution is the central model, and standardising any value to a z-score, $z=\dfrac{x-\mu}{\sigma}$ , lets you read probabilities from one standard table or calculator.

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 $f(x)$ .

Properties of a probability density function

A pdf satisfies $f(x) \ge 0$ everywhere, and the total area underneath is $1$ : $\displaystyle\int_{-\infty}^{\infty} f(x)\,dx = 1$ . The probability that $X$ lies between $a$ and $b$ is the area $\displaystyle\int_a^b f(x)\,dx$ .

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.

The empirical (68-95-99.7) rule

For a normal distribution, about $68\%$ of values lie within $1$ standard deviation of the mean, about $95\%$ within $2$ standard deviations, and about $99.7\%$ within $3$ standard deviations.

Standardising to z-scores

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

The z-score

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

This measures how many standard deviations $x$ is from the mean. You then find $P(Z \le z)$ from a standard normal table or calculator.

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 $P(X = k) = \binom{n}{k}p^{k}(1-p)^{n-k}$ .

Forgetting to convert "greater than" correctly

Tables and many calculators give the lower-tail area $P(Z \le z)$ . If the question asks for $P(Z > z)$ , you must subtract from $1$ . A common slip is reporting $0.9332$ as the answer to "taller than 182 cm" when the correct value is $1 - 0.9332 = 0.0668$ . Always sketch the bell curve and shade the region you actually want.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice5 marks

Calculator-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 $x = \mu + 3\sigma = 10 + 6 = 16$ .

Practice6 marks

Calculator-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.

Practice4 marks

Calculator-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.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC

What this dot point is asking

The normal distribution

Standardising to z-scores

Step 1: Standardise the value to a zzz-score

Step 2: Identify the probability we need

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

Inverse normal problems

Problem

Set up the probability

Unstandardise

Combining the normal and binomial

Summary

Exam-style questions

Practise this

Sources & how we know this

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