VCAA General Mathematics Data analysis, probability and statistics

15 sample questions with marking guides and sample answers

Q5
2025
VCAA
Paper 1
1 mark
Q5
1 mark

The heights of females aged between 16 and 18 years within a population are normally distributed.

Analysis of the heights of this group of females showed that:

  • 2.5% of the heights were greater than 178.9 cm
  • 16% of the heights were less than 157.6 cm.

Using the 68–95–99.7% rule, the mean and standard deviation of the heights of these females are respectively

A

150.5 and 21.3

B

154.9 and 7.1

C

164.7 and 7.1

D

171.8 and 14.2

Reveal Answer
A

150.5 and 21.3

This option incorrectly uses the difference between the two given values (3σ=21.33\sigma = 21.3) as the standard deviation instead of dividing by 3 to find σ\sigma.

B

154.9 and 7.1

While the standard deviation of 7.1 is correct, the mean is miscalculated. The mean should be found by adding the standard deviation to 157.6.

C

164.7 and 7.1

Correct Answer

Using the 68-95-99.7% rule, the bottom 16% corresponds to μ−σ=157.6\mu - \sigma = 157.6 and the top 2.5% corresponds to μ+2σ=178.9\mu + 2\sigma = 178.9. Solving this system of equations yields a standard deviation of σ=7.1\sigma = 7.1 and a mean of μ=164.7\mu = 164.7.

D

171.8 and 14.2

This option incorrectly calculates the standard deviation as 2σ=14.22\sigma = 14.2 and subsequently miscalculates the mean.

Q22
2020
QCAA
Paper 1
4 marks
Q22

A store asked its junior and senior staff whether or not they would like to change the store uniform.
The results are in the frequency table.

 Change uniformDo not change uniform
Junior staff9228
Senior staff2367
Q22a
2 marks

Convert the two-way table into a percentaged two-way frequency table using column totals.

Reveal Answer

Total # change uniform = 115
Total # do not change = 95

 Change uniformDo not change uniform
Junior staff80%29.5%
Senior staff20%70.5%
 100%100%
Marking Criteria
DescriptorMarks

Correctly determines column totals

1

Correctly represents the data in a percentaged two-way table

1
Q22b
2 marks

Explain whether there is an association between staff groups and a desire to change the store uniform.

Reveal Answer

There does appear to be an association between the staff groups and wanting to change the uniform.
The data suggests that junior staff want to change the uniform (80% as opposed to 20% of senior staff) and senior staff do not want to change (70.5% compared with 29.5% of junior staff).

Marking Criteria
DescriptorMarks

Suggests the presence of an association

1

Provides reasons to support conclusion

1
Q12
2025
VCAA
Paper 1
1 mark
Q12
1 mark

A least squares line, y=a+bxy = a + bx, is fitted to data for an explanatory variable xx and a response variable yy.

• The standard deviation of the explanatory variable data is twice that of the response variable data.
• 81% of the variation in the response variable can be explained by the variation in the explanatory variable.
• The correlation between xx and yy is negative.
• The residual associated with the data point (2, 5) is 0.5

What is the residual associated with the data point (3, 4)?

A

-0.05

B

-1.05

C

0.30

D

1.30

Reveal Answer
A

-0.05

Correct Answer

The slope is b=r(sy/sx)=−0.9(1/2)=−0.45b = r(s_y/s_x) = -0.9(1/2) = -0.45. Using the residual formula e=y−(a+bx)e = y - (a + bx) for the point (2, 5), the intercept aa is 5.45.4. For x=3x=3, the predicted value is 4.054.05, making the residual 4−4.05=−0.054 - 4.05 = -0.05.

B

-1.05

This incorrect value likely results from an arithmetic error when calculating the y-intercept or applying the residual formula e=y−y^e = y - \hat{y}.

C

0.30

This answer is incorrect and may stem from missing the negative sign on the correlation coefficient, which would incorrectly change the slope and intercept of the least squares line.

D

1.30

This incorrect residual could result from reversing the ratio of standard deviations when calculating the slope (sx/sys_x/s_y instead of sy/sxs_y/s_x), leading to an incorrect line equation.

Q1
2021
QCAA
Paper 1
1 mark
Q1
1 mark

The second smoothed value for the 3-point moving average is

Day1234567
Value5101832527090
A

32

B

25

C

20

D

18

Reveal Answer
A

32

This is the raw data value for Day 4, not the calculated moving average.

B

25

This value is incorrect; the average of the second window of data points (10, 18, and 32) is 20, not 25.

C

20

Correct Answer

The second smoothed value is calculated by averaging the second set of three data points (Days 2, 3, and 4): 10+18+323=603=20\frac{10 + 18 + 32}{3} = \frac{60}{3} = 20.

D

18

This is the raw data value for Day 3, not the calculated moving average.

Q2
2024
VCAA
Paper 1
1 mark
Q2
1 mark

Freddie organised a function at work. He surveyed the staff about their preferences.

He asked them about their payment preference (cash or electronic payment) and their budget preference (less than $50 or more than $50).

The variables in this survey, payment preference and budget preference, are

A

both categorical variables.

B

both numerical variables.

C

categorical and numerical variables, respectively.

D

numerical and categorical variables, respectively.

Reveal Answer
A

both categorical variables.

Correct Answer

Both variables group responses into distinct categories ('cash' vs. 'electronic' and 'less than $50' vs. 'more than $50') rather than measuring specific numerical quantities.

B

both numerical variables.

Neither variable asks for a specific numerical measurement, such as an exact dollar amount. Since the responses are groups or labels, they are not numerical variables.

C

categorical and numerical variables, respectively.

While payment preference is categorical, budget preference is also categorical because it groups responses into ranges ('less than $50' or 'more than $50') rather than asking for an exact numerical value.

D

numerical and categorical variables, respectively.

Payment preference ('cash' or 'electronic') is clearly a categorical variable, not numerical. Budget preference is also categorical, making this option entirely incorrect.

Q1
2023
QCAA
Paper 1
1 mark
Q1
1 mark

A linear association with a correlation coefficient of 0.23 is best described as

A

weak positive.

B

weak negative.

C

strong positive.

D

strong negative.

Reveal Answer
A

weak positive.

Correct Answer

The correlation coefficient is positive (r>0r > 0) and closer to 00 than to 11, which indicates a weak positive linear association.

B

weak negative.

A negative association requires a correlation coefficient less than zero (r<0r < 0), but the given value is 0.230.23.

C

strong positive.

A strong positive association typically corresponds to an rr value closer to 11 (e.g., r>0.7r > 0.7), whereas 0.230.23 represents a weak relationship.

D

strong negative.

This option describes a correlation coefficient close to −1-1, but the given value is positive and indicates a weak relationship.

Q3
2024
QCAA
Paper 1
1 mark
Q3
1 mark

The coefficient of determination, R2R^2, is equal to 0.36 for the linear association between xx (explanatory variable) and yy (response variable).

Which statement is correct?

A

36% of the variation in xx can be explained by the variation in yy.

B

36% of the total variation can be explained by the linear association.

C

36% of the predicted outcomes can be explained by the variation in xx.

D

36% of the variation in xx can be predicted by the linear association.

Reveal Answer
A

36% of the variation in xx can be explained by the variation in yy.

This reverses the variables; R2R^2 measures the proportion of variation in the response variable (yy) explained by the explanatory variable (xx), not the variation in xx explained by yy.

B

36% of the total variation can be explained by the linear association.

Correct Answer

The coefficient of determination, R2R^2, is defined as the proportion of the total variation in the response variable (yy) that is explained by the linear relationship with the explanatory variable (xx).

C

36% of the predicted outcomes can be explained by the variation in xx.

R2R^2 measures the proportion of the variation in the observed response values (yy), not the predicted outcomes, that is explained by the model.

D

36% of the variation in xx can be predicted by the linear association.

This refers to the variation in the explanatory variable (xx), whereas R2R^2 specifically measures the explained variation in the response variable (yy).

Q14
2025
VCAA
Paper 1
1 mark
Q14
1 mark

Umbrella sales for 2022–2023 and seasonal indices are shown in the table below.

 SummerAutumnWinterSpring
Number of umbrellas sold814394691427611478
Seasonal index0.75 1.281.10

The seasonal index for summer is 0.75

This tells us that, on average, the number of umbrellas sold during summer is

A

25% less than the seasonal average.

B

75% less than the seasonal average.

C

25% more than the seasonal average.

D

75% more than the seasonal average.

Reveal Answer
A

25% less than the seasonal average.

Correct Answer

A seasonal index of 1.0 represents the seasonal average. An index of 0.75 means sales are 75% of the average, which is 25% less than the seasonal average.

B

75% less than the seasonal average.

This would correspond to a seasonal index of 0.25 (calculated as 1.0−0.751.0 - 0.75), not 0.75.

C

25% more than the seasonal average.

This would correspond to a seasonal index of 1.25 (calculated as 1.0+0.251.0 + 0.25), meaning sales are above average.

D

75% more than the seasonal average.

This would correspond to a seasonal index of 1.75 (calculated as 1.0+0.751.0 + 0.75), meaning sales are significantly above average.

Q6
2025
VCAA
Paper 1
1 mark
Q6
1 mark

The distribution of the weights of eggs produced at a chicken farm is assumed to be normally distributed.

Analysis of 2000 eggs showed that:
• 16% of the eggs had a weight more than 96 grams
• 97.5% of the eggs had a weight more than 51 grams.

Using the 68–95–99.7% rule, the mean and standard deviation of the weights of these eggs are respectively

A

45 and 6

B

62.25 and 11.25

C

73.5 and 11.25

D

81 and 15

Reveal Answer
A

45 and 6

Incorrect. If μ=45\mu = 45 and σ=6\sigma = 6, then μ+σ=51\mu + \sigma = 51, which would mean only 16% of eggs weigh more than 51 grams, contradicting the given 97.5%.

B

62.25 and 11.25

Incorrect. These values do not satisfy the system of equations derived from the 68-95-99.7% rule: μ+σ=96\mu + \sigma = 96 and μ−2σ=51\mu - 2\sigma = 51.

C

73.5 and 11.25

Incorrect. If μ=73.5\mu = 73.5 and σ=11.25\sigma = 11.25, then μ+2σ=96\mu + 2\sigma = 96, which would mean only 2.5% of eggs weigh more than 96 grams, not the stated 16%.

D

81 and 15

Correct Answer

Correct. The 68-95-99.7% rule indicates that 96 is one standard deviation above the mean (μ+σ=96\mu + \sigma = 96) and 51 is two standard deviations below the mean (μ−2σ=51\mu - 2\sigma = 51). Solving this system of equations yields μ=81\mu = 81 and σ=15\sigma = 15.

Q11
2021
QCAA
Paper 1
1 mark
Q11
1 mark

Which option is an example of bivariate data?

A

The rating given to a brand of meat pies as poor, fair or good.

B

The number of people in a household and amount of water used.

C

The number of cars passing through a particular set of traffic lights.

D

The time a person spends using a mobile phone on a Friday evening.

Reveal Answer
A

The rating given to a brand of meat pies as poor, fair or good.

This is an example of univariate data because it involves only one variable (the rating) for each observation.

B

The number of people in a household and amount of water used.

Correct Answer

This is bivariate data because it involves two distinct variables (household size and water usage) collected for each household to analyze the relationship between them.

C

The number of cars passing through a particular set of traffic lights.

This is univariate data because it records only a single variable (the count of cars) at a specific location.

D

The time a person spends using a mobile phone on a Friday evening.

This is univariate data because it measures only one variable (time spent) for each person observed.

Q24
2025
QCAA
Paper 1
5 marks
Q24

The average weekly earnings for Australian workers from 2013 to 2023 are modelled by the least-squares line equation y=49x−97140y = 49x - 97 140, where xx is the year and yy is average weekly earnings ($).

The coefficient of determination, R2R^2, is 0.997.

Q24a
1 mark

State the percentage of the variation in yy that is explained by the linear relationship.

Reveal Answer

0.997×100%=99.7%0.997 \times 100\% = 99.7\%

Marking Criteria
DescriptorMarks

correctly states the percentage

1
Q24b
2 marks

Identify and interpret the slope of the least-squares line.

Reveal Answer

slope = 49

Australia's average weekly earnings increased each year by $49.

Marking Criteria
DescriptorMarks

correctly identifies the slope value as 49

1

correctly interprets the slope as an increase per year

1
Q24c
2 marks

Use the equation of the least-squares line to predict the average weekly earnings in 2035.

Reveal Answer

Substitute x=2035x = 2035 into y=49x−97140y = 49x - 97 140:

y=49(2035)−97140=2575y = 49(2035) - 97 140 = 2575

Predicted value for average weekly earnings in 2035 is $2575.

Marking Criteria
DescriptorMarks

correctly substitutes into the least-squares line equation

1

predicts value

1
Q16
2025
VCAA
Paper 1
1 mark
Q16
1 mark

The seasonal index for the number of meat pie sales in winter is 1.75

To correct for seasonality, the actual number of meat pie sales for winter should be reduced, to the nearest whole percentage, by

A

25%

B

43%

C

57%

D

75%

Reveal Answer
A

25%

Incorrect. A 25% reduction would mean multiplying the actual sales by 0.75, which does not match the required deseasonalizing factor of 1/1.75≈0.571/1.75 \approx 0.57.

B

43%

Correct Answer

Correct. To correct for seasonality, you divide the actual sales by the seasonal index (1/1.75≈0.57141/1.75 \approx 0.5714). This means the deseasonalized value is about 57% of the actual value, requiring a reduction of 100%−57%=43%100\% - 57\% = 43\%.

C

57%

Incorrect. This is the percentage that the deseasonalized sales represent of the actual sales (1/1.75≈57%1/1.75 \approx 57\%), rather than the percentage by which the actual sales must be reduced.

D

75%

Incorrect. While a seasonal index of 1.75 means sales are 75% above the seasonal average, reversing this increase requires dividing by 1.75, not subtracting 75%.

Q2
2024
VCAA
Paper 2
4 marks
Q2

In the year 2021 there were 12 days where a new moon appeared in Sydney Harbour.
For these 12 days the occurrence of the highest high tides (HHTHHT) and lowest low tides (LLTLLT) occurred before, on or after the appearance of the new moon.
These records are displayed in the table below.

Table 3

 onbeforebeforeafterafterafter
LLTafterafterafterafterafterafter
 onbeforebeforebeforeafterafter
HHTafterononafterafterafter
Q2a
2 marks

Use the data from Table 3 to complete the following frequency table.

Table 4

 Frequency 
OccurrenceLLTLLTHHTHHT
before  
on  
after  
Total1212
Reveal Answer

OccurrenceLLTHHT
before23
on13
after96
TOTAL1212
Marking Criteria
DescriptorMarks

Correctly completes all six entries in the frequency table

2

Correctly completes at least three entries in the frequency table, or correctly completes one entire column

1

None of the above

0
Q2b
2 marks

Does the data in Table 4 support the contention that the occurrence of the LLTLLT is associated with the appearance of the new moon?
Explain your conclusion by comparing the values of appropriate percentages.

Reveal Answer

Yes, as the percentages differ from 'before' to 'on' to 'after'.

The LLTLLT occurred after the New Moon on 75% of the days compared to 8.3% on the New moon and 16.7% before the New moon.

Marking Criteria
DescriptorMarks

Correctly concludes that the data supports the contention ('Yes') and justifies this by calculating and comparing appropriate percentages (e.g., 75% after, 8.3% on, 16.7% before)

2

Correctly concludes 'Yes' with a partial explanation, OR calculates appropriate percentages without a clear conclusion

1

None of the above

0
Q4
2021
QCAA
Paper 1
1 mark
Q4
1 mark

A confounding variable is a variable that

A

can only take on a certain number of values.

B

remains constant throughout a statistical investigation.

C

is used to predict a difference in the response variable.

D

other than the explanatory variable, influences the response variable.

Reveal Answer
A

can only take on a certain number of values.

This describes a discrete variable, which is defined by having a countable number of possible values, rather than a confounding variable.

B

remains constant throughout a statistical investigation.

A value that remains constant is simply a constant or a controlled variable, whereas a confounding variable varies and affects the outcome.

C

is used to predict a difference in the response variable.

This describes the explanatory (or independent) variable, which is the specific factor the researcher is studying to see if it causes a change.

D

other than the explanatory variable, influences the response variable.

Correct Answer

A confounding variable is an outside influence that affects the response variable and is related to the explanatory variable, making it difficult to determine the true cause of the observed effect.

Q7
2021
QCAA
Paper 2
6 marks
Q7
6 marks

The table shows the total number of times a new song is played on a music service in the days following its first release.

Number of days since first release5101520
Total number of times played ('000s)8121827

The songwriter is paid 0.175 cents every time their song is played and will be paid after 60 days. They predict that by that time, they will be owed at least $1000.

Given that the number of times the song is played is increasing exponentially, evaluate the reasonableness of this prediction.

Reveal Answer

Let n=# of days5n = \frac{\# \text{ of days}}{5}
Let tn=t_n = the total number of plays

∴t1=8\therefore t_1 = 8

r=128r = \frac{12}{8}
=1.5= 1.5

∴tn=8×1.5(n−1)\therefore t_n = 8 \times 1.5^{(n-1)}
At 60 days
n=605n = \frac{60}{5}
=12= 12

Total number of plays (in 1000s)
∴t12=8×1.511\therefore t_{12} = 8 \times 1.5^{11}
=691.98= 691.98

Total predicted income
Income =0.175×691 980= 0.175 \times 691\,980
=121 096.5 cents= 121\,096.5 \text{ cents}
=$1210.97= $1210.97

At least $1000$1000 is a reasonable prediction if plays continue as a geometric progression.

Marking Criteria
DescriptorMarks

correctly defines the variables

1

correctly determines the parameter rr

1

correctly determines a geometric (exponential) model

1

determines total number of plays

1

determines income

1

evaluates reasonableness of solution

1

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