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Specialist Mathematics · Past QCAA questions All 24 questions
2025 · Paper 2 · Technology-active

Question 17

[5 marks] Technology-active
Unit 4 · Topic 5Statistical inference Tool Kit 9.4Constructing a confidence interval Tool Kit 9.5The inversion drill (step 2) Tool Kit 9.3Probabilities for a sample mean (step 5)
The question

A variable, $X$, is assumed to be normally distributed with $\mu=24.311$ and $\sigma=5.102$.

Two 90% confidence intervals for $\mu$ were calculated from two different random samples from $X$, with the second sample being smaller than the first sample by 60.

Both confidence intervals were calculated using the population standard deviation rather than their respective sample standard deviations.

The confidence interval produced from the first sample was $(23.560,\ 25.498)$.

Determine the probability that the confidence interval produced from the second sample overlaps the confidence interval from the first sample.

Watch the situation first

The second interval only has to touch the first one

Watch the smaller sample slide along. You will see a band of centres where the two intervals share at least one point.

The first interval runs from 23.560 to 25.498 Its margin of error tells you n = 75 A sample of 15 gives a much wider interval It overlaps while its centre stays in this band The chance its centre lands in the band is 0.981 20 22 24 26 28 30 0.981 21.393 27.665 first, n = 75 23.560 25.498 second, n = 15 centre = 24.529 E = 25.498 − 24.529 = 0.969 1.645 × 5.102√n = 0.969 √n ≈ 8.661, so n = 75 n = 75 − 60 = 15 E = 1.645 × 5.102√15 = 2.167 low: 23.560 − 2.167 = 21.393 high: 25.498 + 2.167 = 27.665 mean of 15 ~ N(24.311, 1.317²) P = 0.981

The white flash shows you where the two intervals stop sharing any points.

Now for the mathematics

Take samples of 15 and count how often they overlap

Each press takes a new sample of 15 and draws its 90% interval against the first one. You can run it hundreds of times and watch the overlap rate settle.

first interval, n = 75 your latest interval, n = 15 20 22 24 26 28 30 21.393 27.665 sample means so far
Samples taken
Intervals that overlap
The exact answer
0.981
P(21.393 ≤ mean ≤ 27.665)

QCAA marking guide · 5 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Small differences in rounding are fine, and the marking guide allows for them.

Step 11 mark

You start with the interval you were given. Its centre is the first sample mean, and its half-width is the margin of error.

$$\bar{x}_{1}=\frac{23.560+25.498}{2}=24.529$$

$$E_{1}=25.498-24.529=0.969$$

You could also use the full width of 1.938. This mark can be implied by your later working, and your answer can vary a little depending on how you round.

Marker: correctly determines the margin of error for the CI from the first sample

Step 21 mark

Now you run the margin of error formula backwards to find $n_{1}$. For a 90% interval you use $z=1.645$, and you use $\sigma$ because the question tells you to.

$$1.645\left(\frac{5.102}{\sqrt{n_{1}}}\right)=0.969$$

$$n_{1}\approx75$$

Be careful here. Your calculator will give you about 75.02. When you are planning a sample you round up, but this sample already exists, so you round to the nearest whole number. Follow-through marks are allowed.

Marker: determines first sample size

Step 31 mark

The second sample is 60 smaller, so you now have its size. Put it back into the same formula to get its margin of error.

$$n_{2}=75-60=15$$

$$E_{2}=1.645\left(\frac{5.102}{\sqrt{15}}\right)\approx2.167$$

You could also give the width of 4.333. This mark can be implied by your later working.

Marker: determines the margin of error for the CI from the second sample

Step 4 · insight mark ★1 mark

This is where you turn a question about intervals into a question about one number. The second interval reaches $2.167$ either side of its centre $\bar{x}_{2}$. It still touches the first interval as long as $\bar{x}_{2}$ is within $2.167$ of either end.

$$23.560-2.167=21.393\qquad25.498+2.167=27.665$$

$$21.393\le\bar{x}_{2}\le27.665$$

Rounding to $21.4\le\bar{x}_{2}\le27.7$ is accepted.

Marker: determines range of values of second sample mean that allows the second CI to overlap the first CI

Step 51 mark

Now you need the distribution of $\bar{X}$ for samples of 15. You know the population exactly, so you can write it down straight away.

$$\bar{X}\sim N\left(24.311,\ \left(\frac{5.102}{\sqrt{15}}\right)^{2}\right)$$

$$P\left(21.393\le\bar{X}\le27.665\right)\approx0.981$$

$P\approx0.981$

Giving the answer as 98% is also accepted. Notice that you centre this distribution on $\mu=24.311$, not on the first sample mean of 24.529, because the second sample is drawn from the population.

Marker: determines probability that the CI from the second sample overlaps the CI from the first sample

Putting it all together

Each phrase of the question gave you something

“normally distributed with $\mu=24.311$ and $\sigma=5.102$”
You know the distribution of the second sample mean exactly, which you need in the last step.
“Two 90% confidence intervals”
You use $z=1.645$ in both margins of error.
“smaller than the first sample by 60”
You get $n_{2}=n_{1}-60$, so you have to find $n_{1}$ first.
“using the population standard deviation”
You put $\sigma=5.102$ into both intervals, so you never need a sample standard deviation.
“$(23.560,\ 25.498)$”
You read off the first centre and margin of error, and from those you get the first sample size.
“overlaps”
You need the band of centres where the two intervals share at least one point.
What makes this complex unfamiliar

Nothing in this question hands you a sample size. You have to run the confidence interval formula backwards to find $n_{1}$, then forwards again for $n_{2}$. Then you have to change a question about two intervals into a question about where one sample mean lands.

I expect a lot of you will find $n_{1}$ and round it up to 76, because that is what you do when you are planning a sample. The QCAA like to check whether you know the difference. Here the sample already exists, so 75.02 is just 75 with a little rounding error.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics 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.