Mathematics • Year 10 • Unit 4 • Lesson 14
Bivariate Data Review — Skill Drill
Build fluency with the Lesson 14 review routine for ANY bivariate data set: (1) plot a scatter plot, (2) describe correlation (direction + strength + shape), (3) fit a line of best fit, (4) predict using interpolation. The "from start to finish" workflow from Lessons 12-13 combined.
1. I do — fully worked example
Read every step. Each one has a short reason on the right so you can see why, not just what.
Problem. Complete a full bivariate analysis of: (x = minutes of cardio per day, y = resting pulse in bpm) for 6 people:
(10, 78), (20, 74), (30, 70), (40, 66), (50, 62), (60, 58).
Step 1 — Sketch a scatter plot (mentally or on grid paper).
Six points sliding from top-left (10, 78) to bottom-right (60, 58).
Reason: Lesson 14 says the FIRST step in bivariate analysis is to plot.
Step 2 — Describe the correlation.
Strong negative linear correlation: as cardio rises, resting pulse falls; points sit almost on a line.
Reason: use direction + strength + shape (Lesson 12).
Step 3 — Fit a line of best fit.
Mean point: x̄ = 35, ȳ = 68. Slope m = (58 − 78)/(60 − 10) = −20/50 = −0.4.
68 = −0.4 × 35 + c → c = 68 + 14 = 82. Equation: y = −0.4x + 82.
Reason: a good line of best fit passes through (x̄, ȳ) — Lesson 13.
Step 4 — Predict using interpolation.
At x = 25 min: y = −0.4 × 25 + 82 = −10 + 82 = 72 bpm. Inside data range (10–60), so reliable.
Final answer: The data shows a strong negative linear correlation. Line of best fit: y = −0.4x + 82. Predicted resting pulse at 25 min cardio/day ≈ 72 bpm.
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Bivariate data: (x = study hours per week, y = test score out of 100) for 5 students:
(2, 50), (4, 60), (6, 70), (8, 75), (10, 85).
Step 1 — Describe the correlation. Direction: ____________; strength: ____________; shape: ____________.
Step 2 — Mean point. x̄ = (2+4+6+8+10)/5 = ________; ȳ = (50+60+70+75+85)/5 = ________.
Step 3 — Slope and equation. Using (2, 50) and (10, 85):
m = (85 − 50)/(10 − 2) = ________ / ________ = ________.
c using (x̄, ȳ) = (________, ________): c = ________.
Equation: y = ________ x + ________.
Step 4 — Predict at x = 7 hours (interpolation).
y = ____________________________ ≈ ________ %.
3. You do — independent practice
For each task, complete the step indicated. The first four are foundation (one step). The middle two are standard (two steps). The last two are extension (full bivariate workflow with traps).
Foundation — single step
3.1 A scatter plot shows 12 points clustered tightly around an upward-sloping straight line. Describe the correlation. 1 mark
3.2 Find the mean point of (1, 4), (2, 8), (3, 12), (4, 16). 1 mark
3.3 The line of best fit is y = −0.5x + 20. Predict y when x = 10. 1 mark
3.4 Match each scatter plot description to the most likely correlation: (a) points trending upward in a tight line, (b) points all over the plot with no trend, (c) points trending downward but scattered loosely. Use labels: strong positive linear / weak negative linear / no correlation. 1 mark
Standard — describe + fit OR fit + predict
3.5 For the data (1, 3), (2, 5), (3, 7), (4, 9), (5, 11): (a) describe the correlation in one sentence; (b) find the equation of the line of best fit by computing the slope and using the mean point. 2 marks
3.6 A line of best fit y = 1.2x + 5 fits data with x range 0–20. (a) Predict y at x = 12. (b) Predict y at x = 30. (c) Which prediction is interpolation? Which is more reliable? 2 marks
Extension — full bivariate workflow
3.7 Full analysis. Data: (x = minutes practice per day, y = darts score out of 50) for 5 players:
(10, 22), (20, 30), (30, 36), (40, 42), (50, 50).
(a) Describe the correlation. (b) Find the line of best fit. (c) Predict y at x = 25 (interpolation). (d) Comment on whether predicting at x = 200 is realistic — quote a Lesson 13 misconception. 3 marks
3.8 A research paper claims "study hours and exam score have r = +0.7, so studying more CAUSES higher scores." Using Lesson 12's correlation-vs-causation warning, write one sentence describing the relationship correctly and naming a possible confounding variable. 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (study vs score)
Step 1: Positive, strong, linear.
Step 2: x̄ = 30/5 = 6; ȳ = 340/5 = 68.
Step 3: m = 35/8 = 4.375. Using (6, 68): 68 = 4.375 × 6 + c → c = 68 − 26.25 = 41.75. Equation: y = 4.375x + 41.75.
Step 4: At x = 7, y = 4.375 × 7 + 41.75 = 30.625 + 41.75 = 72.375 ≈ 72.4 %.
3.1
Strong positive linear correlation.
3.2 — Mean point
x̄ = 10/4 = 2.5; ȳ = 40/4 = 10. Mean point (2.5, 10).
3.3 — Predict
y = −0.5 × 10 + 20 = 15.
3.4 — Match
(a) strong positive linear, (b) no correlation, (c) weak negative linear.
3.5 — Describe and fit
(a) Strong (perfect) positive linear correlation.
(b) m = (11 − 3)/(5 − 1) = 8/4 = 2. Mean point: x̄ = 15/5 = 3; ȳ = 35/5 = 7. 7 = 2 × 3 + c → c = 1. Equation: y = 2x + 1.
3.6 — Predict + classify
(a) y = 1.2 × 12 + 5 = 19.4 (interpolation).
(b) y = 1.2 × 30 + 5 = 41 (extrapolation, outside 0–20).
(c) x = 12 is interpolation and more reliable.
3.7 — Full analysis (darts)
(a) Strong positive linear correlation.
(b) Mean point: x̄ = 150/5 = 30; ȳ = 180/5 = 36. m = (50 − 22)/(50 − 10) = 28/40 = 0.7. 36 = 0.7 × 30 + c → c = 15. Equation: y = 0.7x + 15.
(c) At x = 25: y = 0.7 × 25 + 15 = 17.5 + 15 = 32.5 ≈ 33 (interpolation).
(d) At x = 200: y = 0.7 × 200 + 15 = 155 — impossible because the maximum score is 50. Lesson 13 misconception: "A trend will always continue in the same direction" is WRONG — trends hit ceilings.
3.8 — Correlation ≠ causation
r = +0.7 shows a moderately strong positive correlation between study hours and exam score, but this does NOT prove causation. A possible confounding variable is academic motivation: motivated students study more AND aim higher in exams, so motivation could drive both.