Unit 3 Review — General Maths

This review covers all four topics of Unit 3: interpreting and using regression lines; time series smoothing, seasonal indices, and forecasting; arithmetic and geometric sequences and their financial applications; and earth geometry including distances on spheres, time zones, and the International Date Line. Allow approximately 90–120 minutes for this review.

Bivariate Data Analysis

Q1 — Interpreting the correlation coefficient
Fluency
A study finds the Pearson correlation coefficient between daily hours of screen time and sleep quality score is r = −0.85.

(a) Describe the strength and direction of the linear relationship.

(b) Can we conclude that screen time causes lower sleep quality? Explain.

(a) r = −0.85 indicates a strong negative linear relationship. As screen time increases, sleep quality scores tend to decrease.

(b) No — correlation does not imply causation. There may be confounding variables (e.g. stress, age, exercise) that affect both screen time and sleep quality. The data shows an association, not a cause-and-effect relationship.
Q2 — Using the regression line for prediction
Fluency
The least squares regression line for predicting plant height (y cm) from weekly rainfall (x mm) is:

ŷ = 8.5 + 0.24x

The data was collected for rainfall between 10 mm and 80 mm per week.

(a) Predict the plant height when weekly rainfall is 50 mm.

(b) Interpret the slope of the regression line in context.

(c) Is a prediction at x = 150 mm reliable? Explain.

(a) ŷ = 8.5 + 0.24(50) = 8.5 + 12.0 = 20.5 cm

(b) For each additional millimetre of weekly rainfall, plant height is predicted to increase by an average of 0.24 cm.

(c) Not reliable. x = 150 mm lies well outside the data range (10–80 mm). This is extrapolation — the linear model may not hold beyond the observed range, and predictions could be significantly inaccurate.
Q3 — Residuals
Fluency
Using the regression line from Q2 (ŷ = 8.5 + 0.24x), an actual data point has rainfall x = 50 mm and height y = 23 cm.

(a) Find the residual for this data point.

(b) Is the actual value above or below the regression line?

(a) Predicted: ŷ = 8.5 + 0.24(50) = 20.5 cm
Residual = actual − predicted = 23 − 20.5 = +2.5 cm

(b) The residual is positive, so the actual value (23 cm) is above the regression line.
Q4 — Finding the regression equation from summary statistics
Understanding
Summary statistics from a study relating study time (x hours) to exam score (y):
- x̄ = 5.0 hours, ȳ = 68
- Standard deviation of x: s_x = 2.0; Standard deviation of y: s_y = 10.0
- Pearson correlation: r = 0.80
(a) Find the slope b and intercept a of the least squares regression line.

(b) Write the regression equation.

(c) Use the equation to predict the score for a student who studied 8 hours.

(a) Slope: b = r × (s_y/s_x) = 0.80 × (10.0/2.0) = 0.80 × 5 = 4.0

Intercept: a = ŷ − b × x̄ = 68 − 4.0 × 5.0 = 68 − 20 = 48

(b) ŷ = 48 + 4.0x

(c) ŷ = 48 + 4.0(8) = 48 + 32 = 80 marks
Q5 — Bivariate data analysis in context
Problem Solving
A researcher studies crop yield (y tonnes/ha) versus daily sunlight hours (x). Summary: r = 0.92, regression line ŷ = 1.2 + 0.35x, data range: x = 4 to 10 hours.

(a) Predict yield for 7 hours of sunlight.

(b) An actual farm has 7 hours of sunlight and produces 5.0 t/ha. Find the residual and comment on whether this farm over- or under-performs relative to the model.

(c) A second researcher reports r = 0.45 for a similar study in a different region. Compare the usefulness of the two regression models for prediction.

(a) ŷ = 1.2 + 0.35(7) = 1.2 + 2.45 = 3.65 t/ha

(b) Residual = 5.0 − 3.65 = +1.35 t/ha. The farm outperforms the model — it produces 1.35 t/ha more than predicted. This positive residual means the actual data point is above the regression line.

(c) r = 0.92 indicates a strong linear relationship — the regression line is a reliable predictor within the data range. r = 0.45 indicates only a weak-to-moderate relationship; the regression line explains little of the variation in yield, and predictions would be much less reliable. The first model is significantly more useful for prediction.

Time Series Analysis

Q6 — Centred moving average
Fluency
Monthly electricity usage (kWh): Jan=320, Feb=280, Mar=250, Apr=220, May=200, Jun=180.

Calculate the 3-point centred moving averages for March, April, and May.

March: (Jan + Feb + Mar) / 3 = (320 + 280 + 250) / 3 = 850/3 = 283.3 kWh

April: (Feb + Mar + Apr) / 3 = (280 + 250 + 220) / 3 = 750/3 = 250.0 kWh

May: (Mar + Apr + May) / 3 = (250 + 220 + 200) / 3 = 670/3 = 223.3 kWh

The moving averages show a downward trend, removing month-to-month fluctuations.
Q7 — Seasonal indices
Fluency
Quarterly sales ($000s) over two years are shown below. The overall mean of quarterly averages is $14 500.

Quarter Yr 1 Yr 2 Average

Q1 10 12 11

Q2 14 16 15

Q3 18 20 19

Q4 12 14 13

Calculate the seasonal index for each quarter. (Overall mean = 14.5)

Seasonal Index = Quarterly Average ÷ Overall Mean

SI(Q1) = 11 ÷ 14.5 = 0.759

SI(Q2) = 15 ÷ 14.5 = 1.034

SI(Q3) = 19 ÷ 14.5 = 1.310

SI(Q4) = 13 ÷ 14.5 = 0.897

Verify: 0.759 + 1.034 + 1.310 + 0.897 = 4.000 ✓

Q3 has the highest seasonal index (sales 31% above average); Q1 has the lowest (sales 24% below average).
Q8 — Deseasonalising and forecasting
Understanding
Use the seasonal indices from Q7 to:

(a) Deseasonalise Year 2’s actual Q3 sales of $20 000.

(b) The trend equation for quarterly sales is T = 11.5 + 0.5n, where n = 1 is Q1 Year 1. Forecast actual (seasonalised) sales for Q3 Year 3 (n = 11).

(a) Deseasonalise:
Deseasonalised = Actual ÷ Seasonal Index = 20 ÷ 1.310 = $15.27 thousand
This removes the seasonal effect, revealing the underlying trend value.

(b) Forecast Q3 Year 3 (n = 11):
Trend: T = 11.5 + 0.5(11) = 11.5 + 5.5 = $17 thousand
Seasonalised forecast = Trend × Seasonal Index = 17 × 1.310 = $22.27 thousand
Q9 — Interpreting a time series
Problem Solving
A cafe records quarterly revenue ($000s) over three years:

Year Q1 Q2 Q3 Q4

1 18 24 30 20

2 22 28 35 25

3 26 32 40 29

(a) Identify the trend and seasonal pattern from the data.

(b) Q4 has seasonal index 0.82. Deseasonalise Year 3 Q4 sales of $29 000.

(c) Year 3 annual average = $31 750. Year 2 annual average = $27 500. Using these, estimate Year 4 Q3 revenue using the seasonal index 1.25 for Q3.

(a) The data shows an upward trend: revenue increases each year across all quarters. There is also a clear seasonal pattern: Q3 is consistently the highest revenue quarter and Q1 the lowest, suggesting a summer peak (Q3 = Jan–Mar in Australian quarters) and a winter trough.

(b) Deseasonalised Y3 Q4:
= Actual ÷ SI = 29 ÷ 0.82 = $35.37 thousand

(c) Forecast Y4 Q3:
Annual trend growth = 31.75 − 27.5 = $4.25 thousand/year.
Estimated Y4 annual average = 31.75 + 4.25 = $36 thousand.
Y4 Q3 forecast = Trend × SI(Q3) = 36 × 1.25 = $45 thousand

Quarter	Yr 1	Yr 2	Average
Q1	10	12	11
Q2	14	16	15
Q3	18	20	19
Q4	12	14	13

Year	Q1	Q2	Q3	Q4
1	18	24	30	20
2	22	28	35	25
3	26	32	40	29

Growth and Decay in Sequences

Q10 — Arithmetic sequence
Fluency
An arithmetic sequence has first term a = 5 and common difference d = 4.

(a) Write the first five terms.

(b) Find t₁₂ (the 12th term).

(c) Find S₂₀ (the sum of the first 20 terms).

(a) 5, 9, 13, 17, 21

(b) t_n = a + (n−1)d = 5 + 11 × 4 = 5 + 44 = 49

(c) S_n = n/2 × (2a + (n−1)d) = 20/2 × (2 × 5 + 19 × 4) = 10 × (10 + 76) = 10 × 86 = 860
Q11 — Geometric sequence
Fluency
A geometric sequence begins: 3, 6, 12, 24, …

(a) State the first term a and common ratio r.

(b) Find t₈.

(c) Find S₆ (sum of the first 6 terms).

(a) a = 3, r = 2

(b) t_n = arⁿ⁻¹ = 3 × 2⁷ = 3 × 128 = 384

(c) S_n = a(rⁿ−1)/(r−1) = 3(2⁶−1)/(2−1) = 3 × 63 = 189
Q12 — Compound interest as a recurrence relation
Understanding
A savings account starts with $4000 and earns interest at 3% per annum compounded quarterly (0.75% per quarter).

(a) Write a recurrence relation for the balance after n quarters: V₀ = … and V_n+1 = …

(b) Find the account balance after 2 years (8 quarters).

(c) How much interest has been earned over the 2 years?

(a) V₀ = 4000; V_n+1 = 1.0075 × V_n

(b) V₈ = 4000 × (1.0075)⁸ = 4000 × 1.06136 = $4245.44

(c) Interest = 4245.44 − 4000 = $245.44
Q13 — Comparing arithmetic and geometric growth
Understanding
Two employees both start on $40 000 per year.
- Employee A receives a fixed raise of $2500 per year (arithmetic growth).
- Employee B receives a 5% raise each year (geometric growth).
(a) Write a recurrence relation for each employee’s annual salary.

(b) Find each employee’s salary in Year 10.

(c) In which year does Employee B first earn more than Employee A?

(a)
Employee A: S₀ = 40 000; S_n+1 = S_n + 2500
Employee B: S₀ = 40 000; S_n+1 = 1.05 × S_n

(b) Year 10 salaries (after 10 raises, i.e. entering year 11):
Employee A: 40 000 + 10 × 2500 = $65 000
Employee B: 40 000 × (1.05)¹⁰ = 40 000 × 1.62889 = $65 156

(c) B overtakes A in Year 10 (after the 10th raise, B earns $65 156 vs A’s $65 000 — B is fractionally ahead). For Year 9: A = $62 500; B = 40 000 × (1.05)⁹ = 40 000 × 1.5513 = $62 052. A is still ahead in Year 9. So B first earns more than A in Year 10.
Q14 — Geometric growth in context
Problem Solving
A rabbit population doubles every 3 months. The initial population is 20 rabbits.

(a) Write a rule for the population after n three-month periods.

(b) Find the population after 2 years.

(c) The island can support at most 1000 rabbits. After how many three-month periods does the population first exceed this limit?

(a) P_n = 20 × 2ⁿ

(b) 2 years = 8 three-month periods.
P₈ = 20 × 2⁸ = 20 × 256 = 5120 rabbits

(c) Need 20 × 2ⁿ > 1000 → 2ⁿ > 50.
Test: n=5: 2⁵=32 < 50 → P₅=640 ✕
n=6: 2⁶=64 > 50 → P₆=20×64=1280 > 1000 ✓
After 6 three-month periods (18 months), the population first exceeds 1000.

Earth Geometry and Time Zones

Q15 — Time zones
Fluency
Sydney operates on UTC+10 and Los Angeles operates on UTC−8.

(a) How many hours behind Sydney is Los Angeles?

(b) When it is 3:00 pm Thursday in Sydney, what time and day is it in Los Angeles?

(a) Difference = 10 − (−8) = 18 hours. Los Angeles is 18 hours behind Sydney.

(b) Sydney 3:00 pm Thursday → UTC = 3:00 pm − 10 h = 5:00 am Thursday UTC.
Los Angeles = 5:00 am − 8 h = 9:00 pm Wednesday.
Q16 — Distance along a meridian
Fluency
Two cities are both on the meridian of 153°E: City P at latitude 27°S and City Q at latitude 57°S. Take the Earth’s radius as 6400 km.

(a) Find the angular difference between the two cities.

(b) Calculate the great-circle distance between them.

(a) Angular difference = 57° − 27° = 30°

(b) d = (30/360) × 2π × 6400 = (1/12) × 40 212 = 3351 km
Q17 — Distance along a parallel of latitude
Understanding
Two cities are both at latitude 40°N and are separated by a longitude difference of 60°. Take R = 6400 km.

(a) Is the 40°N parallel a great circle? Explain.

(b) Calculate the distance between the two cities along the 40°N parallel.

(a) No — the 40°N parallel is a small circle. Only the equator (0° latitude) is a great circle because it is the only parallel that passes through the centre of the Earth. All other parallels have a smaller radius than the Earth’s radius.

(b) d = (60/360) × 2π × 6400 × cos(40°)
= (1/6) × 40 212 × 0.766
= 6702 × 0.766
= 5133 km
Q18 — International flight and time zones
Understanding
A flight departs Dubai (UTC+4) at 11:00 pm on a Friday. The flight takes 7 hours to reach London (UTC+0).

(a) What UTC time does the flight depart?

(b) What UTC time does the flight arrive?

(c) What is the local London time and day when the plane lands?

(a) Departure: Friday 11:00 pm Dubai (UTC+4) = Friday 11:00 pm − 4 h = Friday 7:00 pm UTC

(b) Flight time = 7 hours. Arrival UTC = Friday 7:00 pm + 7 h = Saturday 2:00 am UTC

(c) London is UTC+0, so London time = Saturday 2:00 am
Q19 — Combined earth geometry calculation
Understanding
A ship starts at (25°S, 148°E) and sails due south along the meridian to (55°S, 148°E). It then sails due east along the 55°S parallel for 2500 km. Take R = 6400 km.

(a) How far did the ship travel heading south?

(b) Find the longitude of the ship’s final position. (cos 55° ≈ 0.574)

(a) Angular change in latitude = 55° − 25° = 30°
d = (30/360) × 2π × 6400 = (1/12) × 40 212 = 3351 km

(b) Distance along 55°S parallel = (Δλ/360) × 2πR × cos(55°)
2500 = (Δλ/360) × 40 212 × 0.574
2500 = (Δλ/360) × 23 082
Δλ = 2500 × 360 ÷ 23 082 = 39.0°
Final longitude = 148°E + 39° = 187°E = 360° − 187° = 173°W (crosses the International Date Line)
Q20 — Time zone scheduling problem
Problem Solving
A student in Perth (UTC+8) wants to video call a friend in New York (UTC−5). The student is available from 6:00 pm to 10:00 pm Perth time each evening.

(a) Convert the student’s available window to New York time.

(b) The friend in New York is available from 8:00 am to 2:00 pm New York time. What is the common window when both can speak? Give the times in both time zones.

(c) The student’s family also wants to join from Tokyo (UTC+9). Convert the common window found in (b) into Tokyo time. Can the whole family join?

(a) Perth (UTC+8) to New York (UTC−5): difference = 13 hours. Perth is 13 hours ahead.
Student’s window in NY time: 6:00 pm − 13 h = 5:00 am; 10:00 pm − 13 h = 9:00 am.
Student’s window in New York = 5:00 am to 9:00 am

(b) Friend available 8:00 am–2:00 pm NY. Student corresponds to 5:00 am–9:00 am NY.
Common window (NY time): 8:00 am to 9:00 am
In Perth time: 8:00 am + 13 h = 9:00 pm to 10:00 pm Perth

(c) Tokyo (UTC+9) is 1 hour ahead of Perth.
Common window in Tokyo time: 9:00 pm Perth + 1 h = 10:00 pm to 11:00 pm Tokyo.
Yes, the Tokyo family can join — 10:00 pm–11:00 pm is a reasonable evening hour.

Unit 3 Review — General Maths

Bivariate Data Analysis

Q1 — Interpreting the correlation coefficient

Q2 — Using the regression line for prediction

Q3 — Residuals

Q4 — Finding the regression equation from summary statistics

Q5 — Bivariate data analysis in context

Time Series Analysis

Q6 — Centred moving average

Q7 — Seasonal indices

Q8 — Deseasonalising and forecasting

Q9 — Interpreting a time series

Growth and Decay in Sequences

Q10 — Arithmetic sequence

Q11 — Geometric sequence

Q12 — Compound interest as a recurrence relation

Q13 — Comparing arithmetic and geometric growth

Q14 — Geometric growth in context

Earth Geometry and Time Zones

Q15 — Time zones

Q16 — Distance along a meridian

Q17 — Distance along a parallel of latitude

Q18 — International flight and time zones

Q19 — Combined earth geometry calculation

Q20 — Time zone scheduling problem