Specialist Mathematics · Past QCAA questions All simple familiar questions
2023 · Paper 2 · Simple familiar

Question 15

[7 marks] Technology-active
Topic guidePremiumUnit 4 Topic 5.1 · The distribution of sample meansProbabilities and quantiles
Game planPremium9.3 Probabilities and quantiles for a sample mean
The question

The travel time for students attending a certain university is assumed to be normally distributed, with a population mean of 25.2 minutes and standard deviation of 4.7 minutes.

Travel times are collected from a random sample of 120 of these students and used to calculate a sample mean, $\bar{X}_1$, in minutes.

a)

Determine $P\left(\bar{X}_1\le25\right)$.

[2 marks] b)

Given $P\left(\bar{X}_1>k\right)=0.9$, determine the value of $k$.

[1 mark]

Travel times are collected from a second random sample of the university’s students and used to calculate a second sample mean, $\bar{X}_2$, in minutes.

c)

Given $P\left(\bar{X}_2\le25\right)\approx0.4$, determine the number of students in the second sample.

[4 marks]
Watch the situation first

The sample size sets the width of the curve

Means of samples vary less than single travel times. Watch the curve narrow for 120 students, then widen again for a smaller sample.

Single travel times vary a lot: σ = 4.7 minutes Means of 120 students vary much less: 4.7 ÷ √120 ≈ 0.429 The area to the left of 25 is about 0.32 90% of the area lies to the right of k ≈ 24.65 A smaller sample gives a wider curve The area left of 25 reaches 0.4 when n ≈ 35 Mean travel time (minutes) k ≈ 24.65 0.9 0.32 0.40 one student n = 120 n ≈ 35

Every curve has the same mean, 25.2 minutes. Only the spread, $\frac{\sigma}{\sqrt n}$, changes with the sample size.

Now for the mathematics

Find the sample size that gives 0.4

Change the sample size and watch the area to the left of 25 minutes. Small samples give wide curves, so more of the area spills below 25.

25

QCAA marking guide · 7 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. The steps follow the QCAA sample response.

Step 1 · part a)1 mark

The sample mean has the same mean as the population, and a standard deviation of $\frac{\sigma}{\sqrt n}$.

$$\mu_{\bar{X}}=25.2\qquad\sigma_{\bar{X}}=\frac{4.7}{\sqrt{120}}\approx0.429\ \text{minutes}$$

Leaving it as $\frac{4.7}{\sqrt{120}}$ is also accepted.

Marker: correctly calculates $\sigma_{\bar{X}}$ for the first sample

Step 2 · part a)1 mark

Use the normal cumulative distribution on your GDC with mean $25.2$ and standard deviation $0.429$. As a check, the $z$-score of $25$ is about $-0.47$.

$P\left(\bar{X}_1\le25\right)\approx0.32$

An answer of $0.3$ is also accepted.

Marker: calculates required probability

Step 3 · part b)1 mark

90% of the area is to the right of $k$, so 10% is to the left. Use inverse normal with an area of $0.1$ to the left, or set your GDC’s tail to the right with an area of $0.9$.

$$P\left(\bar{X}_1>k\right)=0.9\iff P\left(\bar{X}_1\le k\right)=0.1$$

$k\approx24.65$ minutes

$24.6$ is also accepted. $k$ is below the mean, which makes sense because most of the area is to its right.

Marker: calculates $k$

Step 4 · part c)1 mark

You don’t know $n$ yet, so you can’t use the GDC with $\bar{X}_2$ directly. Work with the standard normal instead. Inverse normal with mean $0$ and standard deviation $1$ gives the $z$-value with $0.4$ to its left.

$$P(Z\le z)=0.4\ \Rightarrow\ z\approx-0.253$$

Marker: correctly calculates the $z$-value based on given probability

Step 5 · part c)1 mark

Write the $z$-score of $25$ for the second sample. The standard deviation is $\frac{4.7}{\sqrt n}$.

$$z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt n}}\ \Rightarrow\ -0.253=\frac{25-25.2}{\frac{4.7}{\sqrt n}}$$

Marker: determines an equation in terms of the sample size $n$

Step 6 · part c)1 mark

Solve on your GDC, or rearrange by hand.

$$\sqrt n=\frac{0.253\times4.7}{0.2}\approx5.95\ \Rightarrow\ n\approx35.3$$

Using the unrounded $z$-value gives about $35.4$. Values like $34.5$ from early rounding are also accepted.

Marker: determines an approximate value of $n$

Step 7 · part c)1 mark

A number of students must be a whole number. The probability was only approximately $0.4$, so round to the nearest integer.

There are 35 students in the second sample.

The QCAA also accept $36$, and $34$ if it came from early rounding.

Marker: evaluates the reasonableness of the solution by rounding $n$ to an integer value

Putting it all together

Each phrase of the question gave you something

“assumed to be normally distributed”
The sample mean is then normal for any sample size, which is why part c) works with a small sample.
“population mean of 25.2 … standard deviation of 4.7”
These are $\mu$ and $\sigma$. The sample mean keeps $\mu$ but has spread $\frac{\sigma}{\sqrt n}$.
“$P\left(\bar{X}_1>k\right)=0.9$”
This is a right-hand area. Change it to $0.1$ on the left before using inverse normal, unless your GDC lets you choose the tail.
“$\approx0.4$”
The probability is rounded, so $n$ won’t come out as a whole number.
“the number of students”
The final answer has to be a whole number of students.
What makes this simple familiar

Each part uses the distribution of the sample mean in a standard way: a probability, an inverse probability, then working backwards to the sample size.

If I had to guess where you might lose marks, it would be using $\sigma=4.7$ instead of $\frac{4.7}{\sqrt{120}}$ in parts a) and b), and the tail direction in part b). In part c), don’t stop at $35.3$. Round it to a whole number of students.

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 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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