2025 Paper 2, Q19
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.
A hundred beetles, and the fifteen short-horned ones the scientist could have caught
Both species can have short horns, so the short-horned group is a mixture. The question is what share of that group is species A. It is not asking what share of the whole population is species A.
One normal calculation, then a reversal
18 mm is exactly one standard deviation below 20 mm, so $P(\text{short}\mid A)=P(Z<-1)=0.1587$. Species B’s distribution is never given. And never needed, because its 14.6% is the conditional probability.
$P(\text{short})=0.7\times 0.1587+0.3\times 0.146=0.1549$. It is a weighted sum. The two species contribute in proportion to how common they are.
You are given $P(\text{short}\mid\text{species})$ and asked for $P(\text{species}\mid\text{short})$. That swap is what you are being tested on.
Where 0.1587 comes from, and why the mix matters
Drag the species mix. The orange slices are the short-horned beetles. The answer is the A slice as a share of the orange, and it only equals 71.7% at the 30% the question gives you.
the answer is the dark slice as a share of both orange slices
What is the 14.6%?
QCAA marking guide · 6 marks
On the GDC, use the normal distribution with lower $=0$, upper $=18$, $\mu=20$ and $\sigma=2$.
$$P(A<18)=0.1587$$
Marker: correctly determines the proportion of species A beetles with horn length shorter than 18 mm.
$$P({<}18)=P(A\ \text{and}\ {<}18)+P(B\ \text{and}\ {<}18)$$
Marker: correctly determines a method to find the proportion of the total population with horn length shorter than 18 mm. Equivalent statements accepted, e.g. $P(A\cap{<}18)+P(B\cap{<}18)$.
$$P({<}18)=0.7\times 0.1587+0.3\times 0.146=0.15489$$
Marker: determines the proportion of all beetles with horn length shorter than 18 mm. FT marks allowed for earlier errors.
$$P(A\mid{<}18)=\frac{P(A\cap{<}18)}{P({<}18)}$$
Marker: uses conditional probability to solve the problem. This mark may be implied by subsequent working.
$$\frac{0.7\times 0.1587}{0.15489}=0.717=71.7\%$$
About a 71.7% chance it is species A
Marker: determines the probability the captured beetle is from species A. Any number of decimal places is accepted, rounded or truncated.
Name the events before you use them. “Let $A$ be species A and $S$ be a horn shorter than 18 mm” turns four unlabelled decimals into a readable argument. And a tree diagram does the same job.
Marker: shows logical organisation, clear flow of the solution using appropriate mathematical terminology, symbols, conventions and representations.
Why 71.7% and not 70%
15.87% of A beetles are short, compared with 14.6% of B beetles, so the two are almost the same. That is why catching a short-horned beetle barely changes the starting 70%. It only moves to 71.7%.
0.1587 is the chance a species A beetle is short, not the chance a short beetle is A. And 0.1111 is $P(A\cap{<}18)$. That is the correct numerator, but the denominator has been forgotten.
Think of 1000 beetles. 700 are A, and 111 of those are short. 300 are B, and 44 of those are short. So 111 out of the 155 short-horned beetles are A, which is 71.7%. Under exam pressure, I find counting like this is much safer than a formula.
The word “conditional” never appears. What does appear is a sentence in the past tense, “the scientist captures a beetle with a horn length shorter than 18 mm”, and that sentence is the condition. The rest of the question is full of things that look the same but are not. The 14.6% looks like a share of the population, but it is already a conditional probability. You only need the normal distribution for one of the two species. The 30/70 split does nothing until you use it as weights. This is a fusion. The normal distribution feeds into conditional probability from Year 11, and the QCAA never tell you that is where you are going.
Question wording and marking-guide steps are from the 2025 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.