Time series and trends: VCE General Mathematics Unit 3 Data analysis

Syllabus dot point

“Qualitative features of time series plots (trend, seasonality, irregular fluctuations, possible outliers and structural change); numerical smoothing using moving means, with centring when appropriate, and graphical smoothing using moving medians over an odd number of points; and modelling trend by fitting a least squares line with time as the explanatory variable to make forecasts, considering the limitations of a linear model and of extending into the future”

13 min readUpdated 23 Sep 2026

Quick answer

A time series plots a variable against time. Describe it with the Study Design's terms: trend (long-term direction), seasonality (a repeating calendar pattern), irregular fluctuations (random short-term variation, always present), outliers, and structural change (a sudden shift). Smooth it to see the trend: an odd moving mean or moving median is centred on the middle value; an even moving mean falls between two times, so average two neighbouring means (centring). A moving mean over one full seasonal cycle (4 quarters, 12 months) removes seasonality. Fit a least squares line with time as the explanatory variable to forecast, remembering that forecasts are extrapolations.

Jump to a section

What this dot point is asking
The answer
Smoothing, seasonal indices and trend lines together
Exam-style questions
Practice questions

What this dot point is asking

In "Investigating and modelling time series data" the Study Design lists:

"qualitative features of time series plots; recognition of features such as trend (long-term direction), seasonality (systematic, calendar related movements) and irregular fluctuations (unsystematic, short-term fluctuations); possible outliers and their sources, including one-off real-world events, and signs of structural change such as a discontinuity in the time series";
"numerical smoothing of time series data using moving means with consideration of the number of terms required (using centring when appropriate) to help identify trends in time series plot with large fluctuations";
"graphical smoothing of time series plots using moving medians (involving an odd number of points only) to help identify long-term trends in time series with large fluctuations";
"modelling trend by fitting a least squares line to a time series with time as the explanatory variable ... and the use of the model to make forecasts ... including consideration of the possible limitations of fitting a linear model and the limitations of extending into the future".

(Seasonal indices have their own page.) Smoothing is a reliable multiple-choice pair each year: in 2025 Examination 1 a five-median ( $76\%$ ) and a centred four-mean ( $81\%$ ), and in 2023 a seven-median ( $68\%$ ) and a centred eight-mean ( $66\%$ ). Describing features in words is where marks are lost: $26\%$ in 2023 Examination 2 Question 4a and $32\%$ in 2025 Question 6a.

The answer

Features of a time series

Feature	What it looks like	Example
Trend	a long-term upward or downward drift	the margin falling over 18 weeks (2025 Exam 1 Q14)
Seasonality	a pattern that repeats at the same time each calendar period	ice cream peaks every June, 12 months apart (2023 Exam 2 Q4a)
Irregular fluctuations	unpredictable short-term ups and downs	present in every real series
Outliers	isolated values far from the pattern, often from one-off events	a spike from a one-day promotion
Structural change	a sudden change of level or pattern (a discontinuity)	sales jump when a competitor closes

When asked to "identify features", use these exact words and point to evidence on the plot (for example "peaks occur 12 months apart"). Do not write a definition in place of a feature, and do not claim a trend when the level stays roughly constant (2025 Examination 2 Question 6a had seasonality and irregular fluctuations but no trend).

Moving-mean smoothing

A moving mean replaces each value by the mean of the values around it, which averages out irregular fluctuations.

Odd number of terms (3, 5, 7): centred on the middle value.

\text{three-mean at } t = \frac{y_{t-1} + y_t + y_{t+1}}{3}, \qquad \text{five-mean at } t = \frac{y_{t-2} + y_{t-1} + y_t + y_{t+1} + y_{t+2}}{5}.

Even number of terms (2, 4, 8, 12): centring. The mean of four consecutive times, $t = 1$ to $4$ , belongs at $t = 2.5$ , between two time periods. To bring it onto a time period, average two neighbouring means:

\text{centred four-mean at } t = 3 = \frac{\text{mean}(t = 1 \text{ to } 4) + \text{mean}(t = 2 \text{ to } 5)}{2}.

The same idea works for eight-means (2023 Examination 1 Question 15) and 12-means. Smoothing cannot produce values at the ends of the series: a five-mean has no value for the first two and last two times.

How many terms? More terms give a smoother result but lose more values at the ends and react more slowly to change. For seasonal data, use one full cycle: 4 for quarterly data, 12 for monthly data. Each mean then contains every season once, so the seasonal pattern cancels and the trend shows.

Moving-median smoothing

A moving median replaces each value by the median of the values around it, always using an odd number of points (3, 5, 7) so the median is an actual value at the middle time. It is usually done graphically: for each point, look at it and its neighbours on the plot, pick the middle height, and plot that at the middle time. Medians resist outliers better than means.

For a three-median at $t$ , take $y_{t-1}$ , $y_t$ , $y_{t+1}$ and choose the middle-sized one; for a five-median, the middle of five. In 2025 Examination 1 Question 13, weeks 6 to 10 were 50, 38, 32, 35, 41, so the five-median for week 8 is 38.

Trend lines and forecasting

To model a trend, fit a least squares line with time as the explanatory variable (deseasonalise first if the series is seasonal). The slope is the average change per time period, and forecasts come from substituting future times. Forecasts are extrapolations, so treat them with caution: the trend may not stay linear, and structural change or one-off events can break the pattern. The further ahead, the less reliable.

Formula

Three-mean at $t$ : $\frac{y_{t-1} + y_t + y_{t+1}}{3}$ . Centred four-mean at $t$ : average the four-means of $t - 2$ to $t + 1$ and $t - 1$ to $t + 2$ . Moving medians use an odd number of points. A moving mean over one full seasonal cycle removes seasonality. Trend line: $y = a + b \times t$ , forecasts are extrapolations.

Worked examples: smoothing and forecasting

Three-mean, five-mean and moving medians

A series for $t = 1$ to $10$ is 12, 18, 11, 16, 25, 14, 20, 27, 19, 24.

Three-mean at $t = 5$: $\frac{16 + 25 + 14}{3} = 18.33$ .
Five-mean at $t = 5$: $\frac{11 + 16 + 25 + 14 + 20}{5} = \frac{86}{5} = 17.2$ .
Three-median at $t = 5$: 16, 25, 14 in order is 14, 16, 25, so 16.
Five-median at $t = 5$: 11, 16, 25, 14, 20 in order is 11, 14, 16, 20, 25, so 16.

The full three-median smoothed series ( $t = 2$ to $9$ ) is 12, 16, 16, 16, 20, 20, 20, 24: the irregular spikes at 25 and 27 disappear and a gentle upward trend remains.

A centred four-mean (2025 Examination 1 Question 15)

Step 1: the four-mean of days 6 to 9 is 134.25, centred at 7.5.
Step 2: the four-mean of days 7 to 10 is 145.75, centred at 8.5.
Step 3: average them: $\frac{134.25 + 145.75}{2} = 140$ , now centred at day 8. Answer: B.

A centred eight-mean (2023 Examination 1 Question 15)

The eight-means of days 2 to 9 (323) and days 3 to 10 (327) are centred at 5.5 and 6.5. Their average, 325, is the centred value for day 6. Answer: C.

Removing seasonality with a centred four-mean

Quarterly park visitors (thousands) over four years: 24, 31, 19, 14, 27, 35, 21, 17, 29, 38, 24, 18, 32, 40, 27, 21.

Step 1: the first two four-means: $t = 1$ to $4$ : $\frac{88}{4} = 22$ (centred at 2.5). $t = 2$ to $5$ : $\frac{91}{4} = 22.75$ (centred at 3.5).
Step 2: centre: At $t = 3$ : $\frac{22 + 22.75}{2} = 22.375$ .
Step 3: continue: The centred values for $t = 3$ to $14$ are 22.375, 23.25, 24.0, 24.625, 25.25, 25.875, 26.625, 27.125, 27.625, 28.25, 28.875, 29.625. The seasonal zigzag has gone and a steady upward trend of about 0.65 thousand visitors per quarter is visible (see the figure).

A trend line and its limits

A town's registered electric vehicles (hundreds) at the end of each year from 2017 ( $t = 1$ ) to 2024 ( $t = 8$ ) were 12, 15, 17, 22, 24, 29, 33, 35.

Step 1: fit the line: Linear regression with $t$ as the explanatory variable: vehicles $= 7.89 + 3.44t$ , $r^2 = 0.991$ .
Step 2: interpret: On average, the number of registered electric vehicles increases by 344 per year.
Step 3: forecast: 2028 is $t = 12$ : $7.89 + 3.44 \times 12 = 49.2$ , about 4920 vehicles.
Step 4: limitations: This is an extrapolation four years beyond the data. Growth like this often speeds up or levels off rather than staying linear, so the forecast for 2036 ( $t = 20$ , about 7670) would be much less reliable than the one for 2025.

Common traps

Forgetting to centre an even moving mean: The two uncentred means were the distractors in both 2023 and 2025.
Using the wrong window: A five-point window for $t$ runs from $t - 2$ to $t + 2$ ; a centred four-mean at $t$ uses $t - 2$ to $t + 2$ with two four-means.
Giving the actual value instead of the smoothed one: In 2025 Examination 1 Question 13, 32 was the week 8 value, not its five-median.
Taking the median of a time series in time order: For an ordinary median (2023 Examination 1 Question 14), order all the values first.
Defining instead of identifying: "Seasonality is a repeating pattern" earns nothing; "peaks occur every 12 months, at months 6, 18 and 30" does.
Claiming a trend that is not there: A series that repeats at a constant level has seasonality but no trend.
Moving medians with an even number of points: They are defined for odd numbers only.
Trusting long-range forecasts: Every forecast from a trend line is an extrapolation.

Exam technique

For smoothing multiple choice, write the window as a list of the values used, then compute; the distractors are usually the neighbouring window or the uncentred mean. For graphical median smoothing, cover everything outside the window with your fingers and pick the middle-height point. For feature questions, use the Study Design words and give evidence from the plot (a time, a value or a spacing).

Smoothing, seasonal indices and trend lines together

The three time-series tools answer different questions. Smoothing (means or medians) removes irregular fluctuations, and a mean over a full seasonal cycle also removes seasonality, so the trend can be seen. Seasonal indices measure the seasonal pattern and let you deseasonalise individual values. A least squares trend line on the (deseasonalised) data models the trend and gives forecasts, which you then reseasonalise. In an extended-response question you may need all three in sequence.

Note

A time series is a graph of something measured over time. Look for four things: an overall rise or fall (trend), a pattern that repeats at the same time each year (seasonality), random wobbles (irregular fluctuations), and sudden jumps (structural change). To see the trend more clearly, smooth the graph: replace each point with the average, or the middle value, of it and its neighbours. If you average an even number of points, average two of those averages so the result lines up with a real time. To forecast, fit a straight line to the data against time, but trust it less the further ahead you go.

Exam-style questions

Questions in the style of VCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2023 VCAA Exam 1 Q15Past paper1 mark

The number of visitors to a public library each day for 10 consecutive days was recorded. These results are shown in the table below. Day number 1 to 10; number of visitors: 337, 317, 313, 335, 322, 335, 322, 338, 302, 349. The eight-mean smoothed number of visitors with centring for day number 6 is A. 323 B. 324 C. 325 D. 326 E. 327

Show worked answer →

An eight-mean is centred halfway between two days, so average the two eight-means either side of day 6.

Days 2 to 9 (centred at day 5.5): $\frac{317 + 313 + 335 + 322 + 335 + 322 + 338 + 302}{8} = \frac{2584}{8} = 323$ .
Days 3 to 10 (centred at day 6.5): $\frac{313 + 335 + 322 + 335 + 322 + 338 + 302 + 349}{8} = \frac{2616}{8} = 327$ .

Centred value for day 6: $\frac{323 + 327}{2} = 325$ , option C ( $66\%$ correct). The most common wrong answer was E ( $15\%$ ), 327, the uncentred eight-mean of days 3 to 10.

Source: VCAA 2023 General Mathematics Examination 1, Question 15, and the 2023 examination report.

2025 VCAA Exam 1 Q15Past paper1 mark

The number of drinks sold daily by a juice bar, drinks, over a 10-day period is shown in the table below. Day 1 to 10; drinks: 146, 98, 120, 176, 145, 78, 187, 106, 166, 124. The four-mean smoothed number of drinks, with centring, sold on day 8 is closest to A. 134.25 B. 140.0 C. 142.75 D. 145.75

Show worked answer →

Days 6 to 9 (centred at 7.5): $\frac{78 + 187 + 106 + 166}{4} = 134.25$ .
Days 7 to 10 (centred at 8.5): $\frac{187 + 106 + 166 + 124}{4} = 145.75$ .

Centred value for day 8: $\frac{134.25 + 145.75}{2} = 140$ , option B ( $81\%$ correct). Options A and D are the two uncentred four-means.

Source: VCAA 2025 General Mathematics Examination 1, Question 15, and the 2025 examination report.

2025 VCAA Exam 1 Q13Past paper1 mark

The following time series graph shows the margin, in points, for a team winning all of its games in the first 18 weeks of a season. The team plays one game per week. [The margins for weeks 6 to 10 are 50, 38, 32, 35 and 41 points; overall the margins fall from about 58 in week 1 to about 15 in week 18, with irregular ups and downs.] The time series graph is smoothed using five-median smoothing. The smoothed value for the margin in week 8, in points, is A. 32 B. 35 C. 38 D. 41

Show worked answer →

A five-median for week 8 uses weeks 6 to 10: 50, 38, 32, 35, 41. In order: 32, 35, 38, 41, 50. The median is 38, option C ( $76\%$ correct).

Option A (32) is the actual week 8 value, not the smoothed one.

Source: VCAA 2025 General Mathematics Examination 1, Question 13, and the 2025 examination report.

2025 VCAA Exam 1 Q14Past paper1 mark

[For the same graph of the margin over the first 18 weeks.] Which one of the following options best describes the qualitative features of the time series graph above? A. decreasing trend only B. irregular fluctuations only C. structural change only D. decreasing trend with irregular fluctuations

Show worked answer →

The margins fall steadily over the season (a decreasing trend) with week-to-week rises and falls that follow no calendar pattern (irregular fluctuations). There is no sudden jump or change of level (structural change) and no repeating pattern (seasonality).

The answer is D ( $85\%$ correct). Every real time series has irregular fluctuations, so "trend only" (A, $13\%$ ) is incomplete.

Source: VCAA 2025 General Mathematics Examination 1, Question 14, and the 2025 examination report.

2023 VCAA Exam 1 Q13 and Q14Past paper2 marks

The graph shows one winning time for the women's 800 m track event for each calendar year from 2000 to 2022. [Times in seconds read from the graph, 2000 to 2022: 116.4, 118.3, 115.2, 117.9, 116.4, 116.8, 117.2, 116.0, 116.0, 117.6, 118.0, 118.3, 117.6, 118.0, 117.6, 116.8, 117.6, 115.6, 114.4, 118.2, 117.6, 115.2, 116.4.] Question 13: The time series is smoothed using seven-median smoothing. The smoothed value for the winning time in 2006, in seconds, is closest to A. 116.0 B. 116.4 C. 116.8 D. 117.2 E. 117.6. Question 14: The median winning time, in seconds, for all the calendar years from 2000 to 2022 is closest to A. 116.8 B. 117.2 C. 117.6 D. 118.0 E. 118.3

Show worked answer →

Question 13. A seven-median for 2006 uses 2003 to 2009: 117.9, 116.4, 116.8, 117.2, 116.0, 116.0, 117.6. In order: 116.0, 116.0, 116.4, 116.8, 117.2, 117.6, 117.9. The smoothed value is 116.8 s, option C ( $68\%$ correct).

Question 14. This is an ordinary median of 23 values, so order them first: the median is the 12th value in order, 117.2 s (the 2006 time), option B. Only $37\%$ were correct; $30\%$ chose E (118.3), the 12th value in time order (2011), which is not the median.

Source: VCAA 2023 General Mathematics Examination 1, Questions 13 and 14, and the 2023 examination report.

2023 VCAA Exam 2 Q4aPast paper1 mark

The time series plot below shows the average monthly ice cream consumption recorded over three years, from January 2010 to December 2012. The data for the graph was recorded in the Northern Hemisphere. In this graph, month number 1 is January 2010, month number 2 is February 2010 and so on. [The plot rises to peaks at months 6, 18 and 30 and falls to lows late in each year.] Identify a feature of this plot that is consistent with this time series having a seasonal component.

Show worked answer →

The maximum values occur 12 months apart (at months 6, 18 and 30, each June), so the pattern repeats every year.

Only $26\%$ were correct: many students wrote a definition of seasonality instead of pointing to a feature of this plot. Name something you can see, with its timing.

Source: VCAA 2023 General Mathematics Examination 2, Question 4, and the 2023 examination report.

2025 VCAA Exam 2 Q6aPast paper2 marks

The time series plot below shows the number of homes sold in a town each month over a four-year period. Month 1 is January 2016 and month 48 is December 2019. [The monthly sales repeat a similar pattern of highs and lows each year, with no overall rise or fall across the four years.] Excluding any possible outliers, identify two qualitative features of the time series plot.

Show worked answer →

Seasonality (the same pattern of highs and lows repeats each year) and irregular fluctuations. There is no increasing or decreasing trend.

$32\%$ scored both marks. The report stresses using the formal terms named in the Study Design (trend, seasonality, irregular fluctuations, structural change, outliers) and notes that irregular fluctuations are present in every time series.

Source: VCAA 2025 General Mathematics Examination 2, Question 6, and the 2025 examination report.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A time series of a shop's weekly sales shows sales rising over three years, with a peak every December and random week-to-week variation. In March of the second year sales suddenly drop to a new, lower level and then keep rising from there. Name the qualitative features present.

Show worked solution →

Increasing trend, seasonality (the December peaks), irregular fluctuations, and structural change (the sudden drop to a new level in March of year 2). (2 marks: 1 for three features, 1 for all four.)

foundation2 marks

A time series for

t = 1

6

is 40, 46, 43, 51, 48, 55. Find the three-mean smoothed values for

t = 3

and

t = 4

, to two decimal places.

Show worked solution →

$t = 3$ : $\frac{46 + 43 + 51}{3} = 46.67$ . (1 mark)

$t = 4$ : $\frac{43 + 51 + 48}{3} = 47.33$ . (1 mark)

core1 mark

A time series for

t = 1

7

is 15, 22, 9, 30, 18, 12, 25. Find the five-median smoothed value for

t = 4

Show worked solution →

Use $t = 2$ to $6$ : 22, 9, 30, 18, 12. In order: 9, 12, 18, 22, 30. The smoothed value is 18. (1 mark)

core2 marks

A time series for

t = 1

7

is 60, 72, 58, 81, 66, 79, 70. Find the four-mean smoothed value with centring for

t = 4

Show worked solution →

Four-mean of $t = 2$ to $5$ (centred at 3.5): $\frac{72 + 58 + 81 + 66}{4} = 69.25$ . Four-mean of $t = 3$ to $6$ (centred at 4.5): $\frac{58 + 81 + 66 + 79}{4} = 71$ . (1 mark)

Centred value: $\frac{69.25 + 71}{2} = 70.125$ . (1 mark)

core2 marks

(a) Why does a four-mean need centring, but a five-mean does not? (b) Why is moving-median smoothing done with an odd number of points only?

Show worked solution →

(a) The middle of four consecutive times is halfway between two of them (for example $t = 2.5$ ), so the mean does not line up with a time period. Averaging two neighbouring four-means moves it onto a whole time period. A five-mean is already centred on the middle (third) value. (1 mark)

(b) With an odd number of points the median is an actual data point at the middle time period, so it can be read off the graph directly. (1 mark)

exam3 marks

The number of members of a sports club at the end of each year is modelled by members

= 1480 + 36.5 \times t

, where

t = 1

is 2012 and the data runs to 2024 (

t = 13

). (a) Interpret the slope. (b) Forecast the membership at the end of 2026. (c) A committee member uses the line to forecast membership for 2041. Give two reasons to treat this with caution.

Show worked solution →

(a) On average, membership increases by 36.5 members per year. (1 mark)

(b) 2026 is $t = 15$ : $1480 + 36.5 \times 15 = 2027.5$ , about 2028 members. (1 mark)

(c) 2041 ( $t = 30$ ) is far outside the data (extrapolation), and the trend may not stay linear for so long: the club may reach a capacity limit, or a structural change could alter the pattern. (1 mark)

exam2 marks

A time series of monthly electricity use shows strong seasonality. (a) How many terms should a moving mean use to remove the seasonal pattern, and does it need centring? (b) What would the smoothed series then show?

Show worked solution →

(a) 12 terms (one full year of months), with centring because 12 is even. (1 mark)

(b) Each 12-mean contains every month once, so the seasonal highs and lows cancel and the smoothed series shows the underlying trend. (1 mark)

Practise this

Sources & how we know this

VCE General Mathematics Examination 1 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 2 (2023) and examination report — VCAA (2023)
VCE General Mathematics Examination 1 (2025) and examination report — VCAA (2025)
VCE General Mathematics Examination 2 (2025) and examination report — VCAA (2025)
VCE Mathematics Study Design: General Mathematics (from 2023) — VCAA
VCE General Mathematics: examination specifications, past examinations and reports — VCAA