Unit 4 · Topic 3.1 · Continuous random variables

Probability density functions

In Unit 3 your random variables took separate values, and you added probabilities from a table. Now the variable can take any value in an interval, like a time or a mass, so you integrate instead of adding. If you are comfortable with definite integrals from Topic 1, most of this topic is applying them carefully. Tick off each skill once you can do it without help.

Skills in this topic

General continuous random variables (about 6 hours). Only tick a skill when you could do it in an exam.

Part 1

Probability is area

Start with data. When you measure a continuous variable, such as how long people wait in a queue, you group the results into class intervals and draw a histogram. Scale it so each bar's area is the relative frequency of that class. The total area is then 1, and the area over an interval estimates the probability of landing in it. Collect more data and narrow the classes, and the tops of the bars settle into a smooth curve. That curve is a probability density function, or PDF.

For a PDF $f$, the probability that $X$ lands between $a$ and $b$ is the area under the curve between them:

$$P(a\le X\le b)=\int_{a}^{b}f(x)\,dx$$

A function has to pass two checks to be a PDF:

1It is never negative, $f(x)\ge0$, because probability cannot be negative.
2The total area under it is 1, because all the probabilities together add to 1.

QCAA write PDFs in two pieces, the rule on its interval and zero everywhere else:

$$f(x)=\begin{cases}\dfrac{1}{8}x, & 0\le x\le4\\[4pt] 0, & \text{otherwise}\end{cases}$$

For this PDF, $P(1\le X\le2)=\int_{1}^{2}\frac{1}{8}x\,dx=\Big[\frac{x^{2}}{16}\Big]_{1}^{2}=\frac{4}{16}-\frac{1}{16}=\frac{3}{16}$.

Two things that follow from probability being area. The probability of any single exact value is zero, $P(X=a)=0$, so $P(X<a)$ and $P(X\le a)$ are the same. And the area from the bottom of the interval up to $x$ has its own name, the cumulative distribution function, $F(x)=P(X\le x)$. You will use that idea in Part 4.

Try it yourself

Move the bounds, watch the area

This is the PDF $f(x)=\frac{1}{8}x$ on $0\le x\le4$. Move $a$ and $b$, and the shaded area is $P(a\le X\le b)$. An interval of the same width holds more probability on the right, where the curve is higher. Stretch it across the whole domain and you get 1.

Lower bound Upper bound

Practise

Exam questions: finding a probability

2023 Paper 1, Question 10

Multiple choice · Technology-free · [1 mark]

The continuous random variable $Y$ has the probability density function

$$f(y)=\begin{cases}1+y, & 0\le y\le\sqrt{3}-1\\ 0, & \text{otherwise}\end{cases}$$

Determine $P\left(0\le y\le\frac{1}{2}\right)$.

$$\int_{0}^{1/2}(1+y)\,dy=\Big[y+\frac{y^{2}}{2}\Big]_{0}^{1/2}=\frac{1}{2}+\frac{1}{8}=\frac{5}{8}$$

The odd-looking upper limit, $\sqrt{3}-1$, is only there to make the total area 1. It does not affect this integral. (B) comes from subtracting $\frac{1}{8}$ instead of adding it.

Answer

2020 Paper 1, Question 7

Multiple choice · Technology-free · [1 mark]

The life expectancy (in years) of an electronic component can be represented by the probability density function

$$p(x)=\begin{cases}\dfrac{1}{x^{2}}, & x\ge1\\[4pt] 0, & \text{otherwise}\end{cases}$$

The probability that the component lasts between 1 and 10 years is

Rewrite $\frac{1}{x^{2}}$ as $x^{-2}$ and integrate from 1 to 10:

$$\int_{1}^{10}x^{-2}\,dx=\Big[-\frac{1}{x}\Big]_{1}^{10}=-\frac{1}{10}-(-1)=0.9$$

(B) is the probability that it lasts more than 10 years, the area that is left over.

Answer

Part 2

Finding an unknown constant

Often the PDF you are given has a constant in it that you do not know yet:

$$f(x)=\begin{cases}c\left(1-x^{2}\right), & 0\le x\le1\\ 0, & \text{otherwise}\end{cases}$$

You know the total area has to be 1, so integrate over the whole interval, set it equal to 1, and solve:

$$\int_{0}^{1}c\left(1-x^{2}\right)dx=c\Big[x-\frac{x^{3}}{3}\Big]_{0}^{1}=\frac{2c}{3}=1\ \Rightarrow\ c=\frac{3}{2}$$

On Paper 2 I would usually let the calculator's solver do this. On Paper 1, or when the question says "show that", you do it by hand like this.

Game plan: An unknown constant

1Write the integral over the whole interval where the PDF is not zero, and set it equal to 1.
2Integrate, keeping the constant as a letter, then solve for it. On Paper 2, a solver is fine.
3Put the value back into the PDF before you answer any later parts.

Watch the area become 1

Practise

Exam question: an unknown constant

For the written-answer questions, the solutions show where each mark goes, based on the QCAA marking guides. Many marks allow follow-through: if an earlier step is wrong but you use your wrong answer correctly, you can still get the later marks. That is another reason to show every step.

2022 Paper 2, Question 16

Complex familiar · Technology-active · [4 marks]

The time spent waiting in a queue at a certain supermarket is given by $(X+11)$ minutes, where $X$ is a random variable with the probability density function

$$f(x)=\begin{cases}\dfrac{a\left(4-x^{2}\right)}{32}, & -2\le x\le2\\[4pt] 0, & \text{otherwise}\end{cases}$$

Determine the probability of waiting between 10 and 12 minutes in a queue at this supermarket.

First find $a$. The total area must be 1:

$$\int_{-2}^{2}\frac{a\left(4-x^{2}\right)}{32}\,dx=\frac{a}{32}\Big[4x-\frac{x^{3}}{3}\Big]_{-2}^{2}=\frac{a}{32}\times\frac{32}{3}=\frac{a}{3}=1\ \Rightarrow\ a=3$$

1 markThe integral equation, total area equal to 1.

1 markThe value of $a$.

Next, translate the question into $X$. The wait is $X+11$, so a wait between 10 and 12 minutes means $10\le X+11\le12$, which is $-1\le X\le1$. This is the step most people miss.

1 markThe interval for $X$, from $-1$ to $1$.

$$P(-1\le X\le1)=\int_{-1}^{1}\frac{3\left(4-x^{2}\right)}{32}\,dx=\frac{11}{16}=0.6875$$

1 markThe probability.

Answer

$\frac{11}{16}\approx0.688$

The full walkthrough of 2022 Paper 2, Question 16 goes through it mark by mark.

Part 3

Mean, variance and standard deviation

These work exactly like the discrete versions from Unit 3, with integrals in place of sums. For the mean, multiply the PDF by $x$ and integrate over the interval:

$$E(X)=\mu=\int x\,f(x)\,dx$$

The formula sheet gives the variance as $\int(x-\mu)^{2}f(x)\,dx$. I find this version quicker. It is not on the formula sheet, so learn it:

$$\mathrm{Var}(X)=E\left(X^{2}\right)-\big[E(X)\big]^{2}\qquad\text{where}\qquad E\left(X^{2}\right)=\int x^{2}f(x)\,dx$$

The standard deviation is the square root of the variance, $\sigma=\sqrt{\mathrm{Var}(X)}$.

Worked example: $f(x)=\frac{1}{8}x$ on $0\le x\le4$

$$E(X)=\int_{0}^{4}x\times\frac{1}{8}x\,dx=\Big[\frac{x^{3}}{24}\Big]_{0}^{4}=\frac{8}{3}$$

$$E\left(X^{2}\right)=\int_{0}^{4}x^{2}\times\frac{1}{8}x\,dx=\Big[\frac{x^{4}}{32}\Big]_{0}^{4}=8$$

$$\mathrm{Var}(X)=8-\left(\frac{8}{3}\right)^{2}=\frac{8}{9}\qquad\sigma=\sqrt{\frac{8}{9}}=\frac{2\sqrt{2}}{3}\approx0.943$$

Game plan: Mean, variance and standard deviation

1If the PDF has an unknown constant, find it first (Part 2).
2Mean: integrate $x\,f(x)$ over the interval.
3Variance: integrate $x^{2}f(x)$ to get $E\left(X^{2}\right)$, then subtract the square of the mean.
4Standard deviation: square root the variance. Read the question again to check which one it wants, because both are always in the options.

Practise

Exam questions: mean, variance and standard deviation

2023 Paper 1, Question 7

Multiple choice · Technology-free · [1 mark]

Determine the mean of the continuous random variable $X$ with the probability density function

$$f(x)=\begin{cases}\dfrac{1}{8}x, & 0\le x\le4\\[4pt] 0, & \text{otherwise}\end{cases}$$

$$E(X)=\int_{0}^{4}\frac{1}{8}x^{2}\,dx=\Big[\frac{x^{3}}{24}\Big]_{0}^{4}=\frac{64}{24}=\frac{8}{3}$$

A quick check: the density rises to the right, so the mean should sit above the middle of the interval, 2, and it does.

Answer

(D) $\frac{8}{3}$

2021 Paper 1, Question 8

Multiple choice · Technology-free · [1 mark]

The continuous random variable $X$ has the probability density function

$$f(x)=\begin{cases}\dfrac{3}{x^{2}}, & 1\le x\le\dfrac{3}{2}\\[4pt] 0, & \text{otherwise}\end{cases}$$

The mean of $X$ is

Multiplying by $x$ cancels one power of $x$, which leaves a log integral:

$$E(X)=\int_{1}^{3/2}x\times\frac{3}{x^{2}}\,dx=\int_{1}^{3/2}\frac{3}{x}\,dx=3\Big[\ln x\Big]_{1}^{3/2}=3\ln\frac{3}{2}=\ln\frac{27}{8}$$

The last step uses the log law $k\ln m=\ln m^{k}$. (A) forgets the 3.

Answer

(B) $\ln\left(\frac{27}{8}\right)$

2024 Paper 2, Question 2

Multiple choice · Technology-active · [1 mark]

Calculate the expected value of a continuous random variable $X$ with the probability density function

$$p(x)=\begin{cases}\dfrac{1}{4}x^{2}, & 0\le x\le\sqrt[3]{12}\\[4pt] 0, & \text{otherwise}\end{cases}$$

$$E(X)=\int_{0}^{\sqrt[3]{12}}x\times\frac{1}{4}x^{2}\,dx=\Big[\frac{x^{4}}{16}\Big]_{0}^{\sqrt[3]{12}}=\frac{12^{4/3}}{16}\approx1.72$$

On your calculator, type the cube root as $12^{1/3}$ in the upper bound. The answer has to sit inside the interval, which runs from 0 to about 2.29, and 1.72 does.

Answer

(A) 1.72

2022 Paper 2, Question 5

Multiple choice · Technology-active · [1 mark]

The continuous random variable $X$ has the probability density function

$$f(x)=\begin{cases}\dfrac{\cos(x)}{2}, & -\dfrac{\pi}{2}\le x\le\dfrac{\pi}{2}\\[4pt] 0, & \text{otherwise}\end{cases}$$

The standard deviation of $X$ is

The PDF is symmetric about $x=0$, so $E(X)=0$ without any working. Then use your calculator for $E\left(X^{2}\right)$ (radian mode):

$$E\left(X^{2}\right)=\int_{-\pi/2}^{\pi/2}x^{2}\times\frac{\cos x}{2}\,dx\approx0.4674$$

$$\mathrm{Var}(X)=0.4674-0^{2}=0.4674\qquad\sigma=\sqrt{0.4674}\approx0.684$$

(A) is the variance. The question asks for the standard deviation.

Answer

(B) 0.684

2021 Paper 2, Question 13

Simple familiar · Technology-active · [7 marks]

The amount of gravel (in tonnes) sold by a construction company in a given week is a continuous random variable $X$ and has a probability density function defined by:

$$f(x)=\begin{cases}c\left(1-x^{2}\right), & 0\le x\le1\\ 0, & \text{otherwise}\end{cases}$$

(a) Show that $c=\frac{3}{2}$.[1 mark]

(b) Determine $P(X<0.25)$.[2 marks]

(a) For a "show that", set the total area equal to 1 and write every step:

$$\int_{0}^{1}c\left(1-x^{2}\right)dx=c\Big[x-\frac{x^{3}}{3}\Big]_{0}^{1}=\frac{2c}{3}=1\ \Rightarrow\ c=\frac{3}{2}$$

1 markShowing the total area is 1 when $c=\frac{3}{2}$.

(b) Integrate from the bottom of the interval, 0, up to 0.25:

$$P(X<0.25)=\int_{0}^{0.25}\frac{3}{2}\left(1-x^{2}\right)dx=\frac{3}{2}\left(0.25-\frac{0.25^{3}}{3}\right)\approx0.367$$

1 markThe definite integral from 0 to 0.25.

1 markThe probability.

$$E(X)=\int_{0}^{1}\frac{3}{2}\left(x-x^{3}\right)dx=\frac{3}{2}\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{3}{8}$$

1 markThe definite integral for the mean.

1 markThe mean.

$$E\left(X^{2}\right)=\int_{0}^{1}\frac{3}{2}\left(x^{2}-x^{4}\right)dx=\frac{3}{2}\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{1}{5}$$

$$\mathrm{Var}(X)=\frac{1}{5}-\left(\frac{3}{8}\right)^{2}=\frac{19}{320}\approx0.0594$$

1 markA definite integral for the variance. Using $E\left(X^{2}\right)$ like this or $\int(x-\mu)^{2}f(x)\,dx$ both work.

1 markThe variance.

Answer

(b) about 0.367 (c) $\frac{19}{320}\approx0.0594$

2025 Paper 2, Question 17

Complex familiar · Technology-active · [4 marks]

The probability density function approximates the time (s) that people spend viewing a particular photo on a social media page.

$$p(t)=\begin{cases}ke^{-\frac{7t}{50}}, & 0\le t\le20\\ 0, & \text{otherwise}\end{cases}$$

Determine the mean time people spend viewing the photo.

The question does not mention $k$, but you cannot find the mean without it. That is what makes this one unfamiliar. Find $k$ first, from the total area:

$$\int_{0}^{20}ke^{-\frac{7t}{50}}\,dt=1\ \Rightarrow\ k\times\frac{50}{7}\left(1-e^{-2.8}\right)=1\ \Rightarrow\ k\approx0.1491$$

1 markSetting the integral over the domain equal to 1.

1 markThe value of $k$, to any number of decimal places.

Store $k$ in your calculator rather than typing the rounded value. Then find the mean:

$$E(T)=\int_{0}^{20}t\times ke^{-\frac{7t}{50}}\,dt\approx5.85$$

1 markSubstituting the domain and the density into the mean integral.

1 markThe mean time.

Check it is reasonable: the density is highest at $t=0$ and falls away, so the mean should be well below the middle of the interval, 10.

Answer

The mean viewing time is about 5.85 seconds.

The full walkthrough of 2025 Paper 2, Question 17 goes through it mark by mark.

Part 4

Medians and percentiles

The $n$th percentile is the value $a$ with $n\%$ of the distribution below it. The median is the 50th percentile. You are looking for a bound this time. You know the area, so you solve for the bound. So you solve $F(a)=\frac{n}{100}$.

For the PDF $f(x)=\frac{3}{2}\left(1-x^{2}\right)$ on $0\le x\le1$, the 47th percentile is the value of $a$ that solves

$$\int_{0}^{a}\frac{3}{2}\left(1-x^{2}\right)dx=0.47$$

On Paper 2, use your calculator's solver. Be warned that the Casio fx-CG50 can take one to three minutes to solve an equation with an integral in it. Set it going, then start reading the next question while you wait.

Game plan: Percentiles

1Write the integral from the bottom of the interval up to $a$, and set it equal to the percentage as a decimal.
2Integrate and solve for $a$, by hand or with a solver.
3Reject any solution that is outside the interval where the PDF is defined.

Watch the area split

Practise

Exam question: a percentile

2022 Paper 1, Question 18

Complex unfamiliar · Technology-free · [4 marks]

A percentile is a measure in statistics showing the value below which a given percentage of observations occur.

The continuous random variable $X$ has the probability density function

$$f(x)=\begin{cases}2x-2, & 1\le x\le2\\ 0, & \text{otherwise}\end{cases}$$

Determine the 36th percentile of $X$.

Call the percentile $a$. The area from the bottom of the interval, 1, up to $a$ is 0.36:

$$\int_{1}^{a}(2x-2)\,dx=\Big[x^{2}-2x\Big]_{1}^{a}=a^{2}-2a+1=0.36$$

1 markThe definite integral from 1 to $a$.

1 markThe quadratic equation.

This is Paper 1, so spot the perfect square: $a^{2}-2a+1=(a-1)^{2}$.

$$(a-1)^{2}=0.36\ \Rightarrow\ a-1=\pm0.6\ \Rightarrow\ a=1.6\ \text{or}\ a=0.4$$

1 markBoth values of $a$.

Reject $a=0.4$, because it is outside $1\le x\le2$.

1 markJudging which value is reasonable, using the domain.

Answer

The 36th percentile is 1.6.

Practice

Practice questions

Eight questions of my own, covering every skill. Each one is labelled with the skill it practises, so if one goes wrong you know which part of the guide to go back to.

Question 1 of 8

Skill 1: estimates from data · Multiple choice · Technology-free · [1 mark]

A company tested 200 phone batteries. 46 of them lasted between 20 and 25 hours. An estimate of the probability that a battery of this type lasts between 20 and 25 hours is

The relative frequency is the number in the interval divided by the total: $\frac{46}{200}=0.23$. On a histogram, this is the area of the bar for that interval.

Option (C) has divided by 100 instead of 200, and (D) is the probability of lasting outside that interval.

Answer

(B) 0.23

Question 2 of 8

Skill 2: probability as area · Technology-free · [3 marks]

The continuous random variable $X$ has the probability density function shown. Determine the value of $k$.

$$f(x)=\begin{cases}kx(3-x), & 0\le x\le3\\ 0, & \text{otherwise}\end{cases}$$

1It is a PDF, so the total area under it is 1.

$$\int_{0}^{3}k\left(3x-x^{2}\right)dx=1$$

2Integrate and substitute the bounds.

$$k\Big[\tfrac{3x^{2}}{2}-\tfrac{x^{3}}{3}\Big]_{0}^{3}=k\left(\frac{27}{2}-9\right)=\frac{9k}{2}=1$$

3Solve for $k$.

Answer

$k=\frac{2}{9}$

Question 3 of 8

Skill 2: probability as area · Technology-free · [3 marks]

The continuous random variable $X$ has the probability density function shown. Determine $P\left(\frac{\pi}{3}\le X\le\frac{\pi}{2}\right)$.

$$f(x)=\begin{cases}\tfrac{1}{2}\sin x, & 0\le x\le\pi\\ 0, & \text{otherwise}\end{cases}$$

1The probability is the area between those two values, so integrate with them as the bounds.

$$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\tfrac{1}{2}\sin x\,dx=\Big[-\tfrac{1}{2}\cos x\Big]_{\frac{\pi}{3}}^{\frac{\pi}{2}}$$

2Substitute, using the unit circle: $\cos\frac{\pi}{2}=0$ and $\cos\frac{\pi}{3}=\frac12$.

$$-\tfrac{1}{2}(0)+\tfrac{1}{2}\left(\tfrac{1}{2}\right)=\frac{1}{4}$$

Answer

$\frac{1}{4}$

Question 4 of 8

Skill 2: probability as area · Technology-free · [3 marks]

The cumulative distribution function of a continuous random variable $X$ is $F(x)=\dfrac{x^{2}}{16}$ for $0\le x\le4$. Determine $P(1\le X\le3)$, and the probability density function of $X$.

1$F(x)$ is the area up to $x$, so the area between 1 and 3 is $F(3)-F(1)$.

$$P(1\le X\le3)=\frac{9}{16}-\frac{1}{16}=\frac{8}{16}=\frac{1}{2}$$

2The PDF is the derivative of the CDF.

$$f(x)=F'(x)=\frac{2x}{16}=\frac{x}{8},\qquad 0\le x\le4$$

This is the PDF from the median animation in Part 4.

Answer

$P(1\le X\le3)=\frac{1}{2}$, and $f(x)=\frac{x}{8}$ for $0\le x\le4$ (0 otherwise)

Question 5 of 8

Skill 4: variance and standard deviation · Technology-free · [5 marks]

The continuous random variable $X$ has the probability density function shown. Determine the mean and standard deviation of $X$.

$$f(x)=\begin{cases}\dfrac{3x^{2}}{8}, & 0\le x\le2\\ 0, & \text{otherwise}\end{cases}$$

1For the mean, multiply by $x$ inside the integral.

$$E(X)=\int_{0}^{2}x\cdot\frac{3x^{2}}{8}\,dx=\Big[\tfrac{3x^{4}}{32}\Big]_{0}^{2}=\frac{48}{32}=\frac{3}{2}$$

2For the variance, find $E\left(X^{2}\right)$ first.

$$E\left(X^{2}\right)=\int_{0}^{2}x^{2}\cdot\frac{3x^{2}}{8}\,dx=\Big[\tfrac{3x^{5}}{40}\Big]_{0}^{2}=\frac{96}{40}=\frac{12}{5}$$

$$\text{Var}(X)=\frac{12}{5}-\left(\frac{3}{2}\right)^{2}=\frac{12}{5}-\frac{9}{4}=\frac{3}{20}$$

3Take the square root, and rationalise.

$$\text{SD}(X)=\sqrt{\frac{3}{20}}=\frac{\sqrt{15}}{10}$$

Answer

Mean $\frac{3}{2}$, standard deviation $\frac{\sqrt{15}}{10}$

Question 6 of 8

Skill 3: expected value · Multiple choice · Technology-free · [1 mark]

A continuous random variable has the probability density function $f(x)=2x$ for $0\le x\le1$, and 0 otherwise. Its expected value is

$E(X)=\int_{0}^{1}x(2x)\,dx=\Big[\tfrac{2x^{3}}{3}\Big]_{0}^{1}=\frac{2}{3}$.

Option (B) is the middle of the interval, which would only be right if the density were flat. Here the density is bigger near 1, so the mean is pulled above $\frac12$.

Answer

Question 7 of 8

Skill 2: probability as area · Technology-free · [3 marks]

The continuous random variable $X$ has the probability density function shown. Determine the median of $X$.

$$f(x)=\begin{cases}\dfrac{x}{18}, & 0\le x\le6\\ 0, & \text{otherwise}\end{cases}$$

1The median $m$ has half the area below it. Set the area from 0 up to $m$ equal to $\frac12$.

$$\int_{0}^{m}\frac{x}{18}\,dx=\Big[\tfrac{x^{2}}{36}\Big]_{0}^{m}=\frac{m^{2}}{36}=\frac{1}{2}$$

2So $m^{2}=18$ and $m=\pm3\sqrt2$.

3Reject $-3\sqrt2$, because it is outside $0\le x\le6$.

Answer

$m=3\sqrt2$

Question 8 of 8

Skill 4: variance and standard deviation · Technology-active · [4 marks]

The continuous random variable $X$ has the probability density function shown. Determine the standard deviation of $X$, and the probability that $X$ is within one standard deviation of the mean.

$$f(x)=\begin{cases}\tfrac{3}{4}\left(1-x^{2}\right), & -1\le x\le1\\ 0, & \text{otherwise}\end{cases}$$

1The density is symmetric about 0, so $E(X)=0$. You can check this on your calculator.

2With a mean of 0, the variance is just $E\left(X^{2}\right)$.

$$\text{Var}(X)=\int_{-1}^{1}x^{2}\cdot\tfrac{3}{4}\left(1-x^{2}\right)dx=\frac{1}{5},\qquad \sigma=\sqrt{0.2}\approx0.4472$$

3"Within one standard deviation of the mean" means $-0.4472\le X\le0.4472$. Store $\sigma$ and integrate between those bounds.

$$\int_{-0.4472}^{0.4472}\tfrac{3}{4}\left(1-x^{2}\right)dx\approx0.626$$

Answer

$\sigma\approx0.447$, and the probability is about 0.626

Keep practising

Ready for something harder?

The percentile question has its own complex unfamiliar walkthrough. Try it again from a blank page first. The hard part is realising you are solving for a bound rather than working out an area.

Open 2022 Paper 1, Question 18 All complex unfamiliar questions

Game plan

Your game plan for PDFs

1Unknown constant in the PDF? Set the total area equal to 1 and solve for it before anything else.
2Translate the question into a statement about $X$, especially if the variable has been shifted or scaled.
3Probability: integrate $f(x)$ between the bounds. Mean: integrate $x\,f(x)$. Variance: $E\left(X^{2}\right)-\big[E(X)\big]^{2}$. Standard deviation: square root the variance.
4Percentile or median: you know the area, so solve for the upper bound. Reject answers outside the interval.
5Sanity check: probabilities are between 0 and 1, and the mean and percentiles sit inside the interval.

Past questions and marking-guide steps are from the 2020–2025 QCAA Mathematical Methods external assessments, © State of Queensland (QCAA) 2020–2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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