How many VCAA Specialist Mathematics questions cover Data analysis, probability and statistics?

AusGrader has 212 VCAA Specialist Mathematics questions on Data analysis, probability and statistics, all with instant AI grading and detailed marking feedback.

VCE Specialist Mathematics — Data analysis, probability and statistics Questions & Answers

Q2

2023

VCAA

Paper 1

4 marks

Q2

The time taken for Sam to vacuum the house, $V$ hours, is normally distributed with a mean of 1 hour and a standard deviation of 0.3 hours. The time taken for Sam to mop the floor, $M$ hours, is independent of the time taken to vacuum and is normally distributed with a mean of 2 hours and a standard deviation of 0.4 hours. The time taken for Sam to clean the house, $C$ hours, is the sum of the times taken to vacuum and to mop, $V + M = C$ .

Q2a

1 mark

Find the expected value of $C$ .

Reveal Answer

$E(C)=E(V+M)=1+2=3$

Marking Criteria

Descriptor	Marks
Correctly calculates the expected value of $C$ as 3	1

Q2b

1 mark

Show that the variance of $C$ is 0.25

Reveal Answer

$\operatorname{Var}(C)=\operatorname{Var}(V)+\operatorname{Var}(M)$
$=\frac{9}{100}+\frac{16}{100}$
$=\frac{25}{100}$
$=0.25$

Marking Criteria

Descriptor	Marks
Demonstrates that the variance of $C$ is the sum of the variances of $V$ and $M$ , showing the calculation $0.3^2 + 0.4^2 = 0.25$	1

Q2c

2 marks

Sam wants to clean the house before a friend visits at 1.00 pm.

If Sam starts cleaning at 9.00 am, what is the probability that the cleaning will not be finished by 1.00 pm? Use $\Pr(-2 \le Z \le 2) = 0.9545$ and give your answer correct to three decimal places.

Reveal Answer

$\Pr(V+M>4)=\Pr(C>4)$
$=\Pr\left(Z>\frac{4-3}{\frac{1}{2}}\right)$
$=\Pr(Z>2)$
$=\frac{1}{2}(1-0.9545)$
$=\frac{1}{2}(0.0455)$
$\approx 0.023$

Marking Criteria

Descriptor	Marks
Correctly formulates the probability expression $\Pr(C>4)$ and standardises to $\Pr(Z>2)$	1
Correctly evaluates the probability using the given standard normal values to obtain $0.023$	1

Q4

2025

QCAA

Paper 1

1 mark

Q4

1 mark

$X$ is a random variable with mean $\mu$ and standard deviation $\sigma$ .

From random samples of $X$ values, each of size $n$ , the sample mean is calculated. This sampling and calculation is repeated a large number of times.

The mean of the distribution of the sample means would be approximately

A

$\frac{\overline{x}}{n}$

B

$\frac{\mu}{\sqrt{n}}$

C

$\overline{x}$

D

$\mu$

Reveal Answer

A

$\frac{\overline{x}}{n}$

This formula does not represent the mean of the sampling distribution. The mean of the sample means is equal to the population mean and is not divided by the sample size $n$ .

B

$\frac{\mu}{\sqrt{n}}$

This incorrectly combines the population mean $\mu$ with the denominator for the standard error. The standard deviation of the sample means is $\frac{\sigma}{\sqrt{n}}$ , but the mean remains $\mu$ .

C

$\overline{x}$

$\overline{x}$ represents the mean of a single specific sample. The question asks for the mean of the distribution of all possible sample means, which is a population parameter.

D

$\mu$

Correct Answer

According to the properties of sampling distributions, the expected value (or mean) of the distribution of sample means is always exactly equal to the population mean $\mu$ .

Q19

2022

VCAA

Paper 2

1 mark

Q19

1 mark

A random sample of 100 oranges is selected from an orange orchard and their diameters are recorded. It is found, for the sample, the mean is 6 cm and the standard deviation is 0.8 cm.
Which one of the following statements is necessarily correct?

A

The means of random samples of 100 oranges would be normally distributed with a standard deviation of $\frac{0.8}{100}$ cm.

B

To obtain a 90% confidence interval for $\mu$ , a standardised $z$ value of 1.96 is used.

C

The mean diameter of all oranges grown at the orchard is 6 cm.

D

The standard deviation of the diameters of all oranges grown is $\frac{0.8}{\sqrt{100}}$ cm.

E

The smallest value in the 99% confidence interval for $\mu$ would not be less than 5.79 cm.

Reveal Answer

A

The means of random samples of 100 oranges would be normally distributed with a standard deviation of $\frac{0.8}{100}$ cm.

The standard deviation of the sample means (standard error) is calculated as $\frac{s}{\sqrt{n}}$ , which would be $\frac{0.8}{\sqrt{100}}$ , not $\frac{0.8}{100}$ .

B

To obtain a 90% confidence interval for $\mu$ , a standardised $z$ value of 1.96 is used.

A 90% confidence interval uses a $z$ -score of approximately 1.645. A $z$ -score of 1.96 is used for a 95% confidence interval.

C

The mean diameter of all oranges grown at the orchard is 6 cm.

The value of 6 cm is the sample mean, which is only an estimate. The true population mean of all oranges is not necessarily exactly equal to the sample mean.

D

The standard deviation of the diameters of all oranges grown is $\frac{0.8}{\sqrt{100}}$ cm.

The expression $\frac{0.8}{\sqrt{100}}$ represents the standard error of the sample mean, not the standard deviation of the entire population of oranges.

E

The smallest value in the 99% confidence interval for $\mu$ would not be less than 5.79 cm.

Correct Answer

The lower bound of the 99% confidence interval is calculated as $\bar{x} - z \times \frac{s}{\sqrt{n}} = 6 - 2.576 \times \frac{0.8}{\sqrt{100}} \approx 5.794$ cm. Since 5.794 is greater than 5.79, this statement is necessarily correct.

Q6

2021

VCAA

Paper 2

10 marks

Q6

A real estate agent’s income is comprised of a fixed monthly payment of ＄5000 plus a commission of ＄4000 for each house sold. The agent may sell up to three houses in a month with the following probability distribution, where $H$ is the number of houses sold.

$h$	0	1	2	3
$\Pr(H=h)$	0.2	0.6	0.19	0.01

Q6c

To prepare a certain house for sale, the agent recommends that the garden be landscaped and replanted. The time taken for a contractor to landscape the garden is normally distributed with a mean of 18 hours and a standard deviation of four hours. The time taken for replanting is independent of the time taken for landscaping and is normally distributed with a mean of 24 hours and a standard deviation of six hours.

Q6a

2 marks

Find the agent’s expected total income each month. Give your answer in dollars, correct to the nearest dollar.

Reveal Answer

$5000 + 4000 \times 0.6 + 8000 \times 0.19 + 12000 \times 0.01 = ＄9040$

Marking Criteria

Descriptor	Marks
Sets up a correct expression to calculate the expected total income	1
Calculates the correct expected total income of ＄9040	1

Q6b

2 marks

What extra amount of commission should the agent request for each house sold so that their expected monthly income increases by ＄2000? Give your answer in dollars and cents, correct to the nearest cent.

Reveal Answer

$2000 = a \times 0.6 + 2a \times 0.19 + 3a \times 0.01$

$a = ＄1980.20$

Marking Criteria

Descriptor	Marks
Sets up a correct equation for the expected increase in income	1
Calculates the correct extra amount of commission of ＄1980.20	1

Q6c

2 marks

What is the probability, correct to four decimal places, that the garden can be landscaped and replanted within a week? Assume that the contractor will work eight hours per day for five days of the week.

Reveal Answer

$T \sim N(18 + 24, 4^2 + 6^2)$

$\Pr(T < 40) = 0.3908$

Marking Criteria

Descriptor	Marks
Determines the correct parameters for the normal distribution of the total time	1
Calculates the correct probability of $0.3908$	1

Q6d

1 mark

What period of time should the agent allow the contractor to complete the tasks of landscaping and replanting, in order to be 90% confident that these tasks will be completed within this period of time? Give your answer in hours, correct to two decimal places.

Reveal Answer

$\Pr(T < a) = 0.90$

$a = 51.24 \, (\text{h})$

Marking Criteria

Descriptor	Marks
Calculates the correct period of time of $51.24$ hours	1

Q6e

3 marks

Before the annual income review, the agent believes that the company’s remuneration level is below the industry average of ＄115 000 per annum, with an industry standard deviation of ＄12 000. To test this hypothesis, the agent takes a random sample of 36 of the company’s employees and finds their mean income to be ＄110 800 per annum.
Assuming an industry standard deviation of ＄12 000, carry out a one-tailed statistical test at the 5% level of significance to ascertain whether the sample mean of ＄110 800 per annum supports the agent’s belief.

Reveal Answer

$H_0: \mu = 115\,000, \quad H_1: \mu < 115\,000$

$\overline{X} \sim N\left(115\,000, \frac{12000^2}{36}\right)$

$p = \Pr(\overline{X} < 110\,800 \mid \mu = 115\,000) = 0.0179$

$p < 0.05$

So, there is strong evidence to support the agent's belief.

Marking Criteria

Descriptor	Marks
States the correct null and alternative hypotheses: $H_0: \mu = 115\,000$ and $H_1: \mu < 115\,000$	1
Calculates the correct $p$ -value of $0.0179$	1
Compares the $p$ -value to the significance level ( $0.0179 < 0.05$ ) and states a correct conclusion supporting the agent's belief	1

Q20

2021

VCAA

Paper 2

1 mark

Q20

1 mark

An office has two coffee machines that operate independently of each other. The time taken for each machine to produce a cup of coffee is normally distributed with a mean of 30 seconds and a standard deviation of 5 seconds. On a particular morning, a cup of coffee is produced from each machine.

The probability that the time taken by each coffee machine to produce one cup of coffee will differ by less than 3 seconds is closest to

A

0.164

B

0.236

C

0.329

D

0.451

E

0.671

Reveal Answer

A

0.164

This is the probability that the first machine takes up to 3 seconds longer than the second machine, $P(0 < X_1 - X_2 < 3)$ , rather than the absolute difference.

B

0.236

This incorrect value results from adding the standard deviations ( $5 + 5 = 10$ ) instead of the variances when finding the distribution of the difference.

C

0.329

Correct Answer

The difference in times $D = X_1 - X_2$ is normally distributed with mean $0$ and variance $5^2 + 5^2 = 50$ . The probability $P(-3 < D < 3)$ is approximately $0.329$ .

D

0.451

This assumes the standard deviation of the difference is $5$ , failing to account for the combined variance of both independent machines.

E

0.671

This is the probability that the times differ by more than 3 seconds, which is the complement of the correct answer ( $1 - 0.329 = 0.671$ ).

Q10

2022

QCAA

Paper 2

1 mark

Q10

1 mark

In a town, the mean number of residents per household is 3.79 people with a standard deviation of 1.47 people.
Using a random sample of 45 households from the town, determine the probability that the mean number of residents per household will be more than 4.

A

0.17

B

0.33

C

0.83

D

0.96

Reveal Answer

A

0.17

Correct Answer

First, calculate the z-score: $z = \frac{4 - 3.79}{1.47/\sqrt{45}} \approx 0.96$ . The probability $P(Z > 0.96)$ is $1 - 0.8315 \approx 0.17$ .

B

0.33

This value does not result from the standard normal distribution calculation using the Central Limit Theorem parameters provided.

C

0.83

This represents the probability that the mean is less than 4 ( $P(Z < 0.96) \approx 0.83$ ). You must subtract this from 1 to find the probability of being more than 4.

D

0.96

This value is the calculated z-score ( $z \approx 0.96$ ), not the probability associated with that z-score.

Q1

2021

QCAA

Paper 2

1 mark

Q1

1 mark

The time taken to complete orders at a pizza store is normally distributed with a mean time ( $\mu$ ) of 10 minutes.
The owner of the pizza store records the time taken to complete orders for a random sample of 20 pizzas each day over a 30-day period. From this data, an approximate 90% confidence interval for $\mu$ is calculated at the end of each day.
How many of these confidence intervals would be expected to contain $\mu$ ?

A

3

B

18

C

27

D

30

Reveal Answer

A

3

This represents $10\%$ of the 30 days ( $0.10 \times 30 = 3$ ). This is the expected number of intervals that would \textit{fail} to contain the mean, not the number that would contain it.

B

18

This represents only $60\%$ of the 30 days ( $0.60 \times 30 = 18$ ). Given a $90\%$ confidence level, the expected number of successful intervals should be higher.

C

27

Correct Answer

By definition, a $90\%$ confidence interval is expected to contain the true population parameter $90\%$ of the time in repeated sampling. Therefore, the expected number is $0.90 \times 30 = 27$ .

D

30

This assumes that every single interval will contain the mean ( $100\%$ ). While possible, the expected value is determined by the specific confidence level of $90\%$ , not $100\%$ .

Q6

2025

VCAA

Paper 2

10 marks

Q6

A manufacturer produces a bicycle in three stages. The times taken to complete each stage, $T_1$ , $T_2$ and $T_3$ hours, may be assumed to be independent and normally distributed.
The mean and standard deviation of each stage are shown in the following table, where $a$ is a real constant.

Stage	Mean (hours)	Standard deviation (hours)
1	5	1
2	8	$a$
3	3	0.2

The standard deviation of the total processing time to produce a bicycle is 1.5 hours.

Q6c

The heights of people in Australia may be assumed to be normally distributed, with a standard deviation of 10 cm. The bicycles are designed for people of varying heights. The manufacturer takes a random sample of 150 people, the mean height for which is 169 cm.

Q6f

Bicycle tyre life is defined as the number of kilometres for which a bicycle tyre is safe to use. The life of bicycle tyres may be assumed to be normally distributed.
The manufacturer claims that the mean life of bicycle tyres is 3200 km and the standard deviation is 450 km. Over the years a large number of customers reported that their bicycle tyres had a tyre life of more than 3200 km. To assess these reports, the manufacturer performs a one-sided statistical test at the 5% level of significance.
A random sample of 100 tyres is taken and the mean bicycle tyre life is found to be 3300 km. Assume that the standard deviation is still 450 km.

Q6a

1 mark

Find the value of $a$ .

Reveal Answer

$a = 1.1$

Marking Criteria

Descriptor	Marks
States the correct value of $a = 1.1$	1

Q6b

1 mark

What is the probability that the total time taken to produce a bicycle exceeds 20 hours?
Give your answer correct to four decimal places.

Reveal Answer

$T \sim N(16, 1.5^2)$

$\Pr(T > 20) = 0.0038$

Marking Criteria

Descriptor	Marks
Calculates the correct probability of $0.0038$	1

Q6c

1 mark

Find an approximate 99% confidence interval for the mean height of people in Australia.
Give your answer in centimetres rounded to one decimal place.

Reveal Answer

CI is $(166.9, 171.1)$

Marking Criteria

Descriptor	Marks
States the correct confidence interval $(166.9, 171.1)$	1

Q6d

1 mark

If the manufacturer took 200 such samples, how many of the 99% confidence intervals would be expected to contain the value of the population mean?

Reveal Answer

$200 \times 0.99 = 198$

Marking Criteria

Descriptor	Marks
Calculates the correct expected number of $198$	1

Q6e

1 mark

The manufacturer wants to get a better estimate of the population mean by decreasing the width of the confidence interval found in part c by at least 60%.
What minimum sample size should be taken to achieve this?

Reveal Answer

Width of CI is $2 \times 2.5758 \times \frac{10}{\sqrt{150}}$

$2 \times 2.5758 \times \frac{10}{\sqrt{150}} \times 0.4 = 2 \times 2.5758 \times \frac{10}{\sqrt{n}}$

$n = 938$

Marking Criteria

Descriptor	Marks
Calculates the correct minimum sample size of $n = 938$	1

Q6f

1 mark

Write down the null and alternative hypotheses for this test.

Reveal Answer

$H_0: \mu = 3200, \ H_1: \mu > 3200$

Marking Criteria

Descriptor	Marks
States the correct null and alternative hypotheses: $H_0: \mu = 3200, \ H_1: \mu > 3200$	1

Q6g

1 mark

Find the $p$ value for the test. Give your answer correct to four decimal places.

Reveal Answer

$p = 0.0131$

Marking Criteria

Descriptor	Marks
Calculates the correct $p$ value of $0.0131$	1

Q6h

1 mark

At the 5% level of significance, should the manufacturer change the claim that the mean bicycle tyre life is 3200 km? Give a reason for your conclusion.

Reveal Answer

Yes, as $p < 0.05$

Marking Criteria

Descriptor	Marks
Concludes 'Yes' and provides the correct reason that $p < 0.05$	1

Q6i

2 marks

Suppose that the true mean bicycle tyre life is 3350 km.
Find the probability, correct to three decimal places, that the manufacturer will still conclude that the mean bicycle tyre life is 3200 km.
Assume a 5% level of significance, a sample size of 100 and that $\sigma = 450 \text{ km}$ .

Reveal Answer

Right tail 5% cut-off for $\bar{L}$ is $3274$

$\Pr(\bar{L} < 3274 \,|\, \mu = 3350) = 0.046$

Marking Criteria

Descriptor	Marks
Calculates the correct right tail 5% cut-off for $\bar{L}$ as $3274$	1
Calculates the correct probability of $0.046$	1

Q20

2024

VCAA

Paper 2

1 mark

Q20

1 mark

The masses of avocados in a crop may be assumed to be normally distributed, with a mean of $200$ grams and a standard deviation of $7.5$ grams.

After an avocado of mass $M$ grams is peeled and the stone is removed, the mass of edible flesh $F$ grams is given by $F = 0.70M$ . Four avocados are randomly selected from the crop.

What is the probability, correct to four decimal places, that a total of more than $570$ grams of edible flesh is obtained?

A

$0.0868$

B

$0.1705$

C

$0.2128$

D

$0.3170$

Reveal Answer

A

$0.0868$

This answer is incorrect and likely results from an error in calculating the standard deviation of the combined mass of the four avocados.

B

$0.1705$

Correct Answer

The total edible mass $T$ of 4 avocados has a mean of $4 \times (0.70 \times 200) = 560$ g and a variance of $4 \times (0.70 \times 7.5)^2 = 110.25$ . Calculating $P(T > 570)$ gives $P(Z > \frac{570 - 560}{\sqrt{110.25}}) = P(Z > 0.9524) \approx 0.1705$ .

C

$0.2128$

This is incorrect. It stems from misapplying the properties of variance when combining independent normally distributed variables.

D

$0.3170$

This is the probability that 4 times the mass of a single avocado is greater than $570$ grams. It incorrectly uses $Var(4F) = 16Var(F)$ instead of the correct sum of independent variances $Var(F_1+F_2+F_3+F_4) = 4Var(F)$ .

Q19

2021

QCAA

Paper 2

7 marks

Q19

7 marks

Consider the following information.

	mean	variance
Continuous random variable $X$	$E(X) = \mu = \int_{-\infty}^{\infty} x p(x)dx$	$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x)dx$

The waiting time (minutes) until workers at a certain call centre receive their $n$ th phone call, where $n \in Z^+$ , is a random variable $T$ with probability density function

$f(t) = \begin{cases} \frac{k^n t^{n-1}}{(n-1)!} e^{-\frac{t}{3}}, & t \ge 0 \\ 0 & , \text{otherwise} \end{cases}$

where $k$ is a positive constant.

The waiting time until workers receive their 5th call is collected from a random sample of 80 workers.
Determine the probability that the mean waiting time from this sample is more than 16 minutes.

Reveal Answer

Using the property of a PDF
$\int_{-\infty}^{\infty} p(x) dx = 1$
Using $n=5$ in the given PDF
$\int_0^\infty \frac{k^5 t^4}{4!} e^{-\frac{t}{3}} dt = 1$

Solving the equation: $k = \frac{1}{3}$
Mean of distribution for waiting time until 5th call, $\mu$
$E(X) = \int_{-\infty}^{\infty} x p(x) dx$
$\mu = \int_0^\infty t \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = \int_0^\infty \frac{\left(\frac{1}{3}\right)^5 t^5}{4!} e^{-\frac{t}{3}} dt$
$= 15 \text{ minutes}$

Variance of distribution for 5th call
$Var(X) = \int_{-\infty}^{\infty} (x-\mu)^2 p(x) dx$
$= \int_0^\infty (t-15)^2 \frac{\left(\frac{1}{3}\right)^5 t^4}{4!} e^{-\frac{t}{3}} dt = 45 \text{ minutes}^2$
$\therefore \sigma = \sqrt{45} \text{ minutes}$

Consider the distribution of the sample mean of the waiting time until the 5th phone call is received, $\bar{T}$ .
As the sample size is large, the distribution of $\bar{T}$ can be considered normal.
$\mu_{\bar{T}} = 15$ and $\sigma_{\bar{T}} = \frac{\sigma}{\sqrt{n}} = \frac{\sqrt{45}}{\sqrt{80}} = 0.75$

Using normal cdf on GDC: $P(\bar{T} > 16) \approx 0.09$

Marking Criteria

Descriptor	Marks
Correctly determines equation in terms of k	1
Solves equation to determine k	1
Determines population mean	1
Determine population variance	1
Justifies that the distribution of T can be considered normal	1
Determines mean and standard deviation of the sample mean	1
Determines required probability	1

Q2

2024

QCAA

Paper 2

1 mark

Q2

1 mark

Rounded to two decimal places, the z-value used in the calculation of an approximate 95% confidence interval for $\mu$ is

A

0.95

B

1.64

C

1.96

D

2.58

Reveal Answer

A

0.95

This value represents the confidence level itself (0.95), not the critical z-score derived from the standard normal distribution.

B

1.64

This z-value (approximately 1.645) is typically used for a 90% confidence interval, corresponding to a tail area of 0.05.

C

1.96

Correct Answer

For a 95% confidence interval, the significance level is $\alpha = 0.05$ . The critical value $z_{\alpha/2}$ leaves $0.025$ in the upper tail, which corresponds to $1.96$ .

D

2.58

This z-value is typically used for a 99% confidence interval, corresponding to a tail area of 0.005.

Q3

2021

VCAA

Paper 1

5 marks

Q3

A company produces a particular type of light globe called Shiny. The company claims that the lifetime of these globes is normally distributed with a mean of 200 weeks and it is known that the standard deviation of the lifetime of Shiny globes is 10 weeks. Customers have complained, saying Shiny globes were lasting less than the claimed 200 weeks. It was decided to investigate the complaints. A random sample of 36 Shiny globes was tested and it was found that the mean lifetime of the sample was 195 weeks.

Use $\Pr(-1.96 < Z < 1.96) = 0.95$ and $\Pr(-3 < Z < 3) = 0.9973$ to answer the following questions.

Q3a

1 mark

Write down the null and alternative hypotheses for the one-tailed test that was conducted to investigate the complaints.

Reveal Answer

$H_0 : \mu = 200$
$H_1 : \mu < 200$

Marking Criteria

Descriptor	Marks
States the correct null and alternative hypotheses ( $H_0 : \mu = 200$ and $H_1 : \mu < 200$ )	1

Q3b (i)

2 marks

Determine the $p$ value, correct to three decimal places, for the test.

Reveal Answer

$\text{Pr}\left(\bar{X} < 195 | \mu = 200\right)$
$= \text{Pr}\left(Z < \frac{195 - 200}{\frac{5}{3}}\right)$
$= \text{Pr}(Z < -3)$
$= 0.001$

Marking Criteria

Working

Descriptor	Marks
Calculates the correct standardised $Z$ value or sets up the correct probability expression (e.g., $\text{Pr}(Z < -3)$ )	1

Answer

Descriptor	Marks
Evaluates the correct $p$ value, correct to three decimal places ( $0.001$ )	1

Q3b (ii)

1 mark

What should the company be told if the test was carried out at the 1% level of significance?

Reveal Answer

Reject the null hypothesis.

Marking Criteria

Descriptor	Marks
States the correct conclusion to reject the null hypothesis	1

Q3c

1 mark

The company decided to produce a new type of light globe called Globeplus.

Find an approximate 95% confidence interval for the mean lifetime of the new globes if a random sample of 25 Globeplus globes is tested and the sample mean is found to be 250 weeks. Assume that the standard deviation of the population is 10 weeks. Give your answer correct to two decimal places.

Reveal Answer

$\left(250 - 1.96 \times \frac{10}{5}, 250 + 1.96 \times \frac{10}{5}\right) = (246.08, 253.92)$

Marking Criteria

Descriptor	Marks
Calculates the correct 95% confidence interval ( $(246.08, 253.92)$ )	1

Q19

2024

VCAA

Paper 2

1 mark

Q19

1 mark

When grown under the same conditions, two varieties of tomato plant have yields (the quantity of tomatoes produced) that are independent and approximately normally distributed.

Variety A has a mean yield of 6 kg per plant, with a standard deviation of 1.2 kg.

Variety B has a mean yield of 5 kg per plant, with a standard deviation of 1.5 kg.

The probability that a randomly selected tomato plant of variety B will have a higher yield than a randomly selected plant of variety A is closest to

A

0.108

B

0.133

C

0.301

D

0.393

E

0.699

Reveal Answer

A

0.108

Incorrect. This is an incorrect probability that likely stems from miscalculating the mean or standard deviation of the difference between the two distributions.

B

0.133

Incorrect. This value is obtained by incorrectly subtracting the variances ( $1.5^2 - 1.2^2 = 0.81$ ) instead of adding them to find the standard deviation of the difference.

C

0.301

Correct Answer

Correct. The difference in yields $X_B - X_A$ is normally distributed with mean $5 - 6 = -1$ and standard deviation $\sqrt{1.5^2 + 1.2^2} \approx 1.92$ . The probability $P(X_B - X_A > 0)$ is $P(Z > 1/1.92) \approx 0.301$ .

D

0.393

Incorrect. This results from forgetting to take the square root of the sum of the variances, incorrectly using the variance ( $3.69$ ) as the standard deviation to find the z-score.

E

0.699

Incorrect. This is the probability that variety A has a higher yield than variety B ( $P(X_A > X_B) \approx 0.699$ ), which is the complement of the correct answer.

Q15

2023

QCAA

Paper 2

7 marks

Q15

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.

Q15a

2 marks

Determine $P(\bar{X}_1 \leq 25)$ .

Reveal Answer

Given $\mu_{\bar{x}} = 25.2$
$\sigma_{\bar{x}_1} = \frac{\sigma}{\sqrt{n}} = \frac{4.7}{\sqrt{120}}$
$= 0.429 \text{ minutes}$

Using GDC
$P(\bar{X}_1 \le 25) = 0.32$

Marking Criteria

Descriptor	Marks
correctly calculates $\sigma_{\bar{x}}$ for the first sample	1
calculates required probability	1

Q15b

1 mark

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

Reveal Answer

$P(\bar{X}_1 > k) = 0.9$
Using GDC
$k = 24.65$ minutes

Marking Criteria

Descriptor	Marks
calculates $k$	1

Q15c

4 marks

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.

Given $P(\bar{X}_2 \leq 25) \approx 0.4$ , determine the number of students in the second sample.

Reveal Answer

$P(z \le z_1) \approx 0.4 \Rightarrow z_1 = -0.253$
$z = \frac{\bar{X}_2 - \mu}{\frac{\sigma}{\sqrt{n}}}$
$-0.253 = \frac{25 - 25.2}{\frac{4.7}{\sqrt{n}}}$

Using GDC
$n \approx 35.3$

The sample size is 35.

Marking Criteria

Descriptor	Marks
correctly calculates the z-value based on given probability	1
determines an equation in terms of the sample size (n)	1
determines an approximate value of n	1
evaluates the reasonableness of the solution by rounding n to an integer value	1

Q11

2024

QCAA

Paper 2

4 marks

Q11

A company claims that the mean battery life of their latest model of smartphone is 9.5 hours.
To test this claim, the battery lives of a random sample of 40 of the smartphones were measured.
A sample mean of 9.31 hours and a standard deviation of 0.52 hours were calculated from this data.

Q11a

1 mark

Determine an approximate 95% confidence interval for $\mu$ . Give your answer to at least two decimal places.

Reveal Answer

Given $n=40, \bar{x}=9.31$ and $s=0.52$

Using GDC

$CI(95\%) = (9.15, 9.47)$ hours

Marking Criteria

Descriptor	Marks
correctly calculates 95% confidence interval to at least two decimal places	1

Q11b

1 mark

Determine an approximate 99% confidence interval for $\mu$ . Give your answer to at least two decimal places.

Reveal Answer

Using GDC

$CI(99\%) = (9.10, 9.52)$ hours

Marking Criteria

Descriptor	Marks
correctly calculates 99% confidence interval to at least two decimal places	1

Q11c

2 marks

A manager comments that either confidence interval could be used to support the company’s claim.
Use your results from Questions 11a) and 11b) to evaluate the reasonableness of the manager’s comment. Justify your decision using mathematical reasoning.

Reveal Answer

The 95% confidence interval does not include the claimed mean battery life of 9.5 hours, although the 99% CI does.

So the comment is not reasonable.

Marking Criteria

Descriptor	Marks
justifies decision using mathematical reasoning	1
provides appropriate statement of reasonableness	1

VCAA Specialist Mathematics Data analysis, probability and statistics

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