The normal distribution is the bell curve. Lots of measurements, like heights and fill volumes, tend to follow it, which is why it turns up on every external exam. On Paper 2 your calculator does the heavy lifting, so the skill is setting the question up properly. On Paper 1 you need the empirical rule. Tick off each skill once you can do it without help.
Skills in this topic
Normal distributions. Only tick a skill when you could do it in an exam.
Part 1
The bell curve and the standard normal
A normal distribution is a probability density function, so everything from Topic 3.1 still holds. The total area under it is 1, and it never goes below the axis. It is symmetric about the mean, $\mu$, and its spread is set by the standard deviation, $\sigma$. A bigger $\sigma$ gives a lower, wider bell.
You will almost never need it. It has appeared once on a QCAA exam (2024 Paper 2, Question 19), and the equation was given. The thing to take from it is where $\mu$ and $\sigma$ sit, in case you are asked to read them off.
You will see the notation $X\sim N\left(\mu,\sigma^{2}\right)$. Watch the square: $X\sim N\left(24,2.5^{2}\right)$ is a normal distribution with mean 24 and standard deviation 2.5. It is not 6.25.
The standard normal distribution, $Z$
The standard normal has mean 0 and standard deviation 1, and its variable is always called $Z$. If you forget, open the normal menus on your calculator: they default to $\mu=0$ and $\sigma=1$. A statement like $P(Z<2.5)$ just means the probability that a standard normal value is below 2.5.
Try it yourself
Areas under the standard normal curve
Move the bounds, or pick one of the bands, and the shaded area is the probability that $Z$ lands between them. The three bands are the empirical rule you will meet in Part 3.
Part 2
Standardising with z-scores
A z-score tells you how many standard deviations a value is from the mean. Positive means above the mean, negative means below. It turns any normal distribution into the standard normal, which is what lets you compare values from different distributions fairly.
$$z=\frac{x-\mu}{\sigma}$$
This formula is not on the formula sheet, so learn it. You will need it again in Part 5.
Student A, Mathematical Methods
Scored 60. The class had $\mu=50$ and $\sigma=5$.
$$z=\frac{60-50}{5}=2$$
Student B, General Mathematics
Scored 82. The class had $\mu=70$ and $\sigma=15$.
$$z=\frac{82-70}{15}=0.8$$
Student B's raw score is higher, but Student A did better relative to their class. A's result is 2 standard deviations above the mean, and B's is only 0.8.
Watch a value become a z-score
Part 3
Finding a probability
You are given the mean and standard deviation, and asked for the probability of being below, above or between some values. On Paper 2, use your calculator's normal cumulative distribution function (normal CD). On Paper 1, you use the empirical rule, so you must know it by heart:
68%of values lie within 1 standard deviation of the mean.
95%lie within 2 standard deviations of the mean.
99.7%lie within 3 standard deviations of the mean.
Halve each band, using the symmetry, to get the pieces in the diagram. For example, 34% lie between $\mu$ and $\mu+\sigma$, and 16% lie above $\mu+\sigma$.
Game plan: Finding a probability
1Write down the mean and standard deviation.
2Draw a quick bell curve, mark the values you have been given and shade the area you want.
3Use the picture to decide the lower and upper bounds. "More than 113" has no upper bound, so use a very large number.
4Technology-free: convert to z-scores and use the empirical rule or the probabilities you are given. Technology-active: use normal CD.
Practise
Exam questions: finding a probability
2024 Paper 2, Question 5
Multiple choice · Technology-active · [1 mark]
The mass (g) of adult kookaburras in a certain region is normally distributed with a mean of 300 g and a standard deviation of 13 g. Select the correct statement about the mass of adult kookaburras.
Work out each band from $\mu=300$ and $\sigma=13$:
$$\mu\pm\sigma=287\text{ to }313\qquad\mu\pm2\sigma=274\text{ to }326\qquad\mu\pm3\sigma=261\text{ to }339$$
Now check each option against the empirical rule. (A) is out because 68% lies within one standard deviation. It is not 34%. (B) puts 68% on the two-standard-deviation band, which should be 95%. (C) mixes a $-3\sigma$ bound with a $+2\sigma$ bound. Only (D) matches.
Answer
(D) 99.7% are between 261 g and 339 g
2021 Paper 1, Question 10
Multiple choice · Technology-free · [1 mark]
Handspans of teenagers are approximately normally distributed, with a mean of 15 cm and a standard deviation of 2 cm.
Which of the following groups is expected to be the largest?
Turn each interval into standard deviations from the mean, then use the empirical rule:
(A) 7 to 11 cm is $-4\sigma$ to $-2\sigma$, about 2.5%. (B) 11 to 15 cm is $-2\sigma$ to the mean, about 47.5%. (C) 13 to 17 cm is $\mu\pm\sigma$, about 68%. (D) 17 to 21 cm is $+\sigma$ to $+3\sigma$, about 16%.
All four intervals are 4 cm wide, but the one centred on the mean holds the most, because that is where the curve is highest.
Answer
(C) between 13 cm and 17 cm
2020 Paper 1, Question 4
Multiple choice · Technology-free · [1 mark]
Pulse rates of adult men are approximately normally distributed with a mean of 70 and a standard deviation of 8.
Which of the following choices correctly describes how to determine the proportion of men that have a pulse rate greater than 78?
$$z=\frac{78-70}{8}=1$$
"Greater than" means the area to the right. (A) is the proportion below 78.
Answer
(B) the area to the right of $z=1$
2022 Paper 1, Question 4
Multiple choice · Technology-free · [1 mark]
The weekly amount of money a company spends on repairs is normally distributed, with a mean of $1200 and a standard deviation of $100.
Given that $P(Z\le-2.5)=0.0062$ and $P(Z>1)=0.1587$, where $Z$ is a standard normal random variable, determine the probability that the weekly repair costs will be between $950 and $1300.
Convert both bounds to z-scores so you can use the probabilities you have been given:
The area between them is everything, minus the left tail, minus the right tail:
$$P(-2.5<Z<1)=1-0.0062-0.1587=0.8351$$
(D) is $1-0.1587$, which forgets to remove the left tail.
Answer
(C) 0.8351
2021 Paper 2, Question 1
Multiple choice · Technology-active · [1 mark]
The scores obtained on a test can be assumed to be normally distributed with a mean of 102 and a standard deviation of 19.
What proportion of scores are over 113?
Normal CD with lower bound 113, a very large upper bound, $\sigma=19$ and $\mu=102$:
$$P(X>113)\approx0.2813$$
A sense check: 113 is above the mean, so fewer than half the scores are over it. That rules out (B), (C) and (D) straight away. (B) is the z-score, $\frac{11}{19}$, and (C) is the proportion below 113.
Answer
(A) 0.2813
Part 4
Finding a quantile
This is Part 3 backwards. You are given the probability and asked for the value, which is called a quantile. "60% of vehicles travelled less than $x$ km" gives you the area and asks for the bound. Use your calculator's inverse normal function.
The one thing to get right is the tail. Inverse normal usually wants the area to the left. If you are given an area to the right, such as "the top 2%" or "80% contain more than", either choose the right tail in the menu or convert first: the top 2% starts at the 98th percentile.
Game plan: Finding a quantile
1Write down the mean and standard deviation.
2Draw a bell curve and shade the area you have been given.
3Use the picture to decide whether the shaded area is a left tail or a right tail.
4Technology-free: use the empirical rule. Technology-active: use inverse normal with the right tail setting. Check the answer is on the side of the mean your picture says it should be.
Practise
Exam questions: probabilities and quantiles
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 6
Multiple choice · Technology-active · [1 mark]
A stall at the school fete sells cups of lemonade. Assuming the amount of lemonade in a cup is normally distributed with a mean of 60 mL and a standard deviation of 3 mL, 80% of the cups contain more than
If 80% contain more than the value, then 20% contain less. That is a left-tail area of 0.2:
Most cups are above this value, so it has to be below the mean of 60. That rules out (C) and (D). (D) is the answer to the wrong tail.
Answer
(B) 57.5 mL
2022 Paper 2, Question 12
Simple familiar · Technology-active · [4 marks]
Suppose that the distance travelled by vehicles in a year can be modelled by a normal distribution. In 2021, vehicles travelled a mean of 13 700 km with a standard deviation of 3400 km.
(a) Determine the probability that a vehicle chosen at random travelled less than 12 000 km in 2021.[2 marks]
(b) Determine the value of $x$ where 60% of vehicles travelled less than $x$ km in 2021.[2 marks]
(a) This is a probability, so use normal CD with a very small lower bound and an upper bound of 12 000:
$$P(X<12\,000)\approx0.3085$$
Write down the distribution and the bounds you entered as well as the number. That is where the second mark is.
1 markShowing your approach: the distribution and the bounds you entered.
1 markThe probability.
(b) This time you know the area, 0.6 to the left, so it is a quantile:
1 markShowing your approach: invNorm with the area and the parameters.
1 markThe value of $x$.
Answer
(a) about 0.3085 (b) about 14 561 km
2021 Paper 2, Question 14
Simple familiar · Technology-active · [7 marks]
The heights of students at School A are normally distributed with a mean of 165 cm and a standard deviation of 15 cm.
(a) Determine the probability that a student chosen at random from School A is shorter than 180 cm.[1 mark]
(b) Determine the minimum integer value of the height of a student who is in the top 2% of this distribution.[3 marks]
The heights of students at School B are also normally distributed. A student at School B has the same height as the height determined in Question 14b) but their corresponding $z$-score is 3.
(c) Determine which student's height ranks higher in terms of percentile for their school.[3 marks]
(a) 180 cm is exactly one standard deviation above the mean, so $P(X<180)=P(Z<1)\approx0.8413$.
1 markThe probability.
(b) The top 2% is a right tail, which starts at the 98th percentile:
The School B student has $z=3$, and $P(Z<3)\approx0.9987$. So the School A student is at about the 98th percentile and the School B student at about the 99.9th.
1 markA statement that justifies the decision, comparing the z-scores or percentiles.
1 markDeciding which student ranks higher.
Answer
(a) about 0.8413 (b) 196 cm (c) the School B student ranks higher
Part 5
An unknown mean or standard deviation
As soon as you see a normal distribution, look for the mean and the standard deviation. If one of them is missing, z-scores will be involved. The question gives you another piece of information instead, something like this:
It is known that 30% of donuts weigh less than 45 g.
That one sentence gives you two things. The value is $x=45$. And because $P(X<45)=0.3$, you can get the z-score from inverse normal on the standard normal, with a left-tail area of 0.3: $z=\text{invNorm}(0.3,\ 0,\ 1)\approx-0.5244$. Then substitute into the z-score formula. If the mean were 50 g, for example:
The empirical rule's "2 standard deviations" is an approximation. It gives 15.84, which is not an option, so you need the exact z-score here. (C) is the distance from the mean to each end. It is not the standard deviation.
Answer
(A) 16.16
2020 Paper 2, Question 16
Complex familiar · Technology-active · [4 marks]
Bottles of soft drink should contain a volume with a mean of 591 mL, but some variation is expected.
Any bottle at or below the 20th percentile of the volume distribution is rejected. A percentile is a measure in statistics that shows the values below which a given percentage of observations occur.
Thirty-five per cent of the bottles contain 593 mL or more of soft drink.
Assuming the volumes are normally distributed, determine the smallest volume (in mL) that will be accepted.
The standard deviation is missing, and the extra information is "35% contain 593 mL or more". That is a right tail of 0.35, so a left tail of 0.65:
The term extremely tall is used to describe any person whose height is three standard deviations or more above the mean height of the population.
A person who just qualifies as extremely tall in a country where heights are normally distributed with a mean height of 180 cm and a standard deviation of 10 cm travels to another country. The person discovers they are taller than exactly 90% of the destination country's population.
Assuming that the standard deviation of both countries is the same, determine the minimum height required to be considered extremely tall in the destination country.
The person's height: just qualifying at home means exactly three standard deviations above the mean, $180+3(10)=210$ cm.
1 markThe cut-off for extremely tall in the original country, 210 cm.
In the destination country, 210 cm has 90% of people below it, so its z-score is $\text{invNorm}(0.9,\ 0,\ 1)\approx1.2816$. The mean there is the missing parameter:
Extremely tall in the destination country starts three standard deviations above that mean: $197.18+3(10)\approx227.2$ cm.
1 markThe minimum height in the destination country.
Answer
About 227.2 cm
2023 Paper 2, Question 17
Complex familiar · Technology-active · [5 marks]
Model bridges were constructed for a competition. The models that could support the heaviest loads before collapsing were given awards.
The load results of the competition were normally distributed, with a mean of 1.36 kg and a standard deviation of 0.12 kg.
Three award categories were used: honours for the top 15% of load results; distinction for the next 15%; and commendation for the next 15%.
The model bridge constructed by Finley only just missed out on a commendation. Kirby's model bridge only just qualified for honours. Determine the difference, to the nearest gram, between the loads supported by Finley and Kirby's models.
Draw the bell curve and mark the awards from the right. Honours is the top 15%, so it starts at the 85th percentile. Distinction runs from the 70th to the 85th, and commendation from the 55th to the 70th.
Kirby only just made honours, so Kirby's load is the 85th percentile. Finley only just missed commendation, so Finley's load is the 55th percentile.
1 markIdentifying both areas you need, the 85th and 55th percentiles.
The hardest normal distribution questions join it to something else. Often you find a probability from a normal distribution and then use it somewhere new, such as a binomial distribution or a conditional probability. Sometimes there are two distributions, and the link between them is what lets you find a missing mean or standard deviation.
Game plan: Complex unfamiliar questions
1Work out what the question is actually asking for.
2Connect it to the topics you know, such as normal, binomial or conditional probability.
3Pick out the key information.
4Ignore the information you do not need.
5Set out your response so each step leads to the next.
6Check that you have actually answered the question.
A scientist is gathering data on two species of horned beetle, species A and B. Horn length is a method of distinguishing the species.
Species A horn lengths are normally distributed with a mean of 20 mm and a standard deviation of 2 mm.
It is known that 14.6% of species B beetles have horns shorter than 18 mm.
In the particular population the scientist is studying, 30% of the beetles are species B and 70% are species A.
The scientist captures a beetle with a horn length shorter than 18 mm.
Determine the probability that the beetle is from species A.
You already know the beetle is short, and you want the chance it is species A. That is a conditional probability, $P(A\mid\text{short})$. Connect it to a tree diagram.
For species A, 18 mm is one standard deviation below the mean:
1 markThe probability that the beetle is species A.
Check: more of the population is species A, and A beetles are short slightly more often than B beetles, so an answer above 0.7 makes sense.
Answer
About 0.717
1 markLogical organisation: define your events and use correct probability notation.
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. Sketch the bell curve and shade the area for every one.
Question 1 of 8
Skill 1: the bell curve · Multiple choice · Technology-free · [1 mark]
IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. The percentage of people with an IQ between 70 and 130 is approximately
70 and 130 are each two standard deviations from the mean, $100\pm2(15)$. The empirical rule says about 95% of values lie within two standard deviations.
Option (A) is for one standard deviation and (C) is for three. Option (D) is only one side of the mean.
Answer
(B) 95%
Question 2 of 8
Skill 2: z-scores · Technology-free · [2 marks]
Ana scored 72 on a test with a mean of 65 and a standard deviation of 5. Ben scored 80 on a different test with a mean of 74 and a standard deviation of 8. Who did better compared with the others who sat their test?
2A bigger z-score means further above the mean, measured in standard deviations. Ben's raw score is higher, but Ana is further above her mean.
Answer
Ana did better, with $z=1.4$ compared with Ben's $z=0.75$.
Question 3 of 8
Skill 3: probabilities and quantiles · Technology-free · [2 marks]
The time a bus takes to complete its route is normally distributed, with a mean of 40 minutes and a standard deviation of 4 minutes. Given that $P(Z\le-1.5)=0.0668$ and $P(Z>2)=0.0228$, determine the probability that the bus takes between 34 and 48 minutes.
1Convert each bound to a z-score, so you can use the probabilities you were given.
2Draw the curve. The shaded area is everything except the two tails.
$$P(34\le X\le48)=1-0.0668-0.0228=0.9104$$
Answer
0.9104
Question 4 of 8
Skill 3: probabilities and quantiles · Technology-active · [2 marks]
The mass of mangoes from an orchard is normally distributed, with a mean of 420 g and a standard deviation of 35 g. Determine the probability that a randomly selected mango has a mass between 400 g and 480 g.
1Sketch it. The region covers the mean, so expect an answer bigger than 0.5.
2Use the normal cumulative function with the bounds and the parameters.
Skill 3: probabilities and quantiles · Technology-active · [2 marks]
The finishing times in a fun run are normally distributed, with a mean of 62 minutes and a standard deviation of 9 minutes. The fastest 10% of runners receive a medal. Determine the slowest finishing time that receives a medal.
1The fastest runners have the smallest times, so the 10% sits in the left tail.
2You know the area, so it is a quantile. Use invNorm.
The mass of rice in a packet is normally distributed with a mean of 250 g. It is known that 8% of packets contain less than 243 g. Determine the standard deviation of the mass of rice in a packet.
1The extra information is $P(X<243)=0.08$. Find the z-score for a left tail of 0.08 on the standard normal.
$$z=\text{invNorm}(0.08,\ 0,\ 1)\approx-1.4051$$
2Substitute into the z-score formula and solve for $\sigma$.
The duration of a regular flight is normally distributed with a mean of 95 minutes and a standard deviation of 6 minutes. A flight is on time if it takes less than 100 minutes. Determine the probability that at least 18 of the next 20 flights are on time.
1Find the probability that one flight is on time. This is the normal part.
$$p=P(X<100)\approx0.7977$$
2Each of the 20 flights is on time or not, independently, with the same $p$. So the number on time is binomial.
$$Y\sim\text{Bin}(20,\ 0.7977)$$
3"At least 18" means 18, 19 or 20. Use the binomial cumulative function.
$$P(Y\ge18)\approx0.199$$
This is the same structure as the IQ question in Topic 5.2. Find $p$ with the normal distribution, then use it in a binomial.
Answer
About 0.199
Keep practising
Ready for something harder?
Two normal distribution questions have their own complex unfamiliar walkthroughs: the beetle question above, and 2024 Paper 2, Question 19, which gives you the normal equation itself. Try them from a blank page first.
