Mathematics

Answer Key

Bivariate Statistics

Pack A — Answers

# Question Answer
1 A scatter plot shows that as $x$ increases, $y$ tends to $increase$. Describe the correlation. Positive correlation
2 A scatter plot shows random points with no clear trend. Describe the correlation. No correlation (or zero)
3 Points lie tightly along a straight line going up. Describe the strength and direction of correlation. Strong positive correlation
4 A study plots height (cm) against weight (kg). Which is the (a) independent variable and (b) dependent variable, by convention if predicting weight? (a) height (x-axis) (b) weight (y-axis)
5 A graph shows hours of sunshine vs ice cream sales. The points trend upward. Describe the relationship. Positive correlation: more sunshine, more sales
6 A line of best fit has equation $y = 3x + 2$. Predict $y$ when $x = 5$. 17
7 A line of best fit has equation $y = 4x + 7$. State the gradient and $y$-intercept. Gradient 4; y-intercept 7
8 A study finds a positive correlation between ice cream sales and shark attacks. Does eating ice cream cause shark attacks? No — both depend on a third variable (hot weather)
9 In a scatter of test scores vs revision hours, one point sits well above the line of best fit. What does this represent? A student who scored higher than predicted for their revision hours
10 A regression line of $y$ on $x$ is $y = 0.8 x + 5$. Predict $y$ for $x = 20$. $y = 21$
11 A line of best fit passes through $(0, 3)$ and $(5, 18)$. Find its equation. $y = 3x + 3$
12 A regression line for height ($y$, cm) on age ($x$, years) is $y = 6x + 70$. Predict the height of a 9-year-old. 124 cm
13 A regression line of test mark ($y$) on hours of revision ($x$) is $y = 5x + 40$. Interpret the gradient in context. Each extra hour of revision adds 5 marks (on average)
14 A regression line of test mark ($y$) on hours of revision ($x$) is $y = 5x + 40$. Interpret the $y$-intercept in context. Predicted mark with zero hours of revision = 40
15 For each correlation coefficient $r$, describe the correlation. $r = 0.85$. Strong positive correlation
16 A line of best fit is $y = 4x + 6$. Find $x$ when $y = 30$. $x = 6$
17 A regression line of $y$ on $x$ passes through the mean point $(\bar{x}, \bar{y})$. If $\bar{x} = 5$ and the line is $y = 3x + 4$, find $\bar{y}$. $\bar{y} = 19$
18 A regression equation was fitted using $x$-values from 5 to 20. Using the line, $y$ is predicted for $x = 10$ (case A) and $x = 30$ (case B). Which is interpolation, which is extrapolation? $x = 10$: interpolation; $x = 30$: extrapolation
19 Plot 1 has $r = 0.9$ and Plot 2 has $r = 0.2$. Which shows a stronger linear relationship? Plot 1 ($r$ closer to 1)
20 A scatter plot of time spent gaming vs test scores has line of best fit $y = -2x + 70$. Interpret in context. Each hour of gaming predicts a 2-mark decrease; with zero gaming the predicted score is 70
21 Five data points: (1, 3), (2, 5), (3, 7), (4, 9), (5, 11). Find the regression line by inspection. $y = 2x + 1$
22 A regression line $y = 0.5x + 20$ was fitted to $x$-values from 10 to 50. (a) Predict $y$ at $x = 30$. (b) At $x = 80$ the prediction would be? (a) 35 (b) 60 (extrapolation — unreliable)
23 A car's mileage (km) vs its resale value (£). Suggested data has $r \approx -0.92$. Interpret. Strong negative: more mileage → lower value
24 A regression line has a strong outlier. Will removing it (a) increase or decrease $|r|$? (b) Why? (a) Increase $|r|$ — removes scatter (b) Outliers reduce correlation strength
25 Five data points: $\bar{x} = 5, \bar{y} = 12$; regression line has gradient $2$. State the regression equation. $y = 2x + 2$
26 Categorise the correlation based on $r$: (a) $r = 0.3$, (b) $r = 0.7$, (c) $r = -0.95$, (d) $r = 0.05$. (a) Weak positive (b) Strong positive (c) Very strong negative (d) None
27 Five data points have $\bar{x} = 4$ and $\bar{y} = 11$. Four are $(2, 7), (3, 9), (5, 13), (6, 15)$. Find the fifth. $(4, 11)$
28 Studies find $r = 0.9$ between countries' chocolate consumption and number of Nobel Prize winners. Does eating chocolate cause Nobel Prizes? No — confounding variable (national wealth) drives both
29 LOBF $y = 2x + 1$ predicts $y$ at $x = 5$. Actual measured value is 9. What is the residual? Residual = $-2$ (observed minus predicted)
30 A regression line predicts temperature in °C from altitude (m): $T = -0.006a + 15$. Predict $T$ at altitude 2000 m. $T = 3°$C
31 A regression has $r = 0.8$. What proportion of variation in $y$ is explained by $x$? 64% ($r^2 = 0.64$)
32 Two studies of "$y$ vs $x$" have regression lines: Study A: $y = 3x + 5$, $r = 0.6$. Study B: $y = 3x + 5$, $r = 0.9$. Which is a better fit? Why? B is better fit (higher $|r|$)
33 A regression line $y = 2x + 5$ has $r^2 = 0.7$. At $x = 10$, you predict $y = 25$. What does $r^2 = 0.7$ tell you? 70% of variation explained — moderate-strong predictive value
34 Three points and their LOBF predictions: $(1, 4) \to$ predicted 3.5; $(2, 6) \to$ predicted 5.5; $(3, 8) \to$ predicted 7.5. Find the sum of residuals. 1.5
35 Five students rank their preference for two subjects. The rank correlation is 0.8. Interpret. Strong positive rank agreement
36 A scatter plot looks like a U-shape. A student fits a straight LOBF and finds $r = 0$. What does this tell us about the relationship? There IS a relationship — but not linear
37 A regression line of "ice cream sales" on "temperature (°C)" gives $y = 2.5x + 5$ for $x$ in $[15, 35]$. At $x = -10$°C, the prediction is $-20$. Comment. Extrapolation — and negative sales is nonsensical
38 A dataset has $r = -0.7$ and $r^2 = 0.49$. Write a one-sentence summary for a non-technical audience. There is a fairly strong negative relationship; the model explains about 49% of the variation
39 For a sample of cars, fuel economy ($y$, mpg) vs engine size ($x$, L) gives $y = -8x + 50$. (a) Predict mpg for a 2 L engine. (b) Comment on $x = 5$ L. (a) 34 mpg (b) 10 mpg — likely extrapolation
40 A study finds $r = 0.95$ between heights of parent and child. Does this mean a tall parent guarantees a tall child? No — strong tendency, not certainty

Pack B — Answers

# Question Answer
1 A scatter plot shows that as $x$ increases, $y$ tends to $decrease$. Describe the correlation. Negative correlation
2 A scatter plot shows random points with no clear trend. Describe the correlation. No correlation
3 Points lie tightly along a straight line going up. Describe the strength and direction of correlation. Strong negative correlation
4 A study plots height (cm) against weight (kg). Which is the (a) independent variable and (b) dependent variable, by convention if predicting weight? (a) hours studied (b) test score
5 A graph shows hours of sunshine vs ice cream sales. The points trend upward. Describe the relationship. Positive correlation: more revision, higher score
6 A line of best fit has equation $y = 2x + 10$. Predict $y$ when $x = 8$. 26
7 A line of best fit has equation $y = -2x + 15$. State the gradient and $y$-intercept. Gradient $-2$; y-intercept 15
8 A study finds a positive correlation between ice cream sales and shark attacks. Does eating ice cream cause shark attacks? No — both depend on size of fire
9 In a scatter of test scores vs revision hours, one point sits well above the line of best fit. What does this represent? A student who scored lower than predicted
10 A regression line of $y$ on $x$ is $y = 0.8 x + 5$. Predict $y$ for $x = 20$. $y = 21$
11 A line of best fit passes through $(0, 4)$ and $(6, 22)$. Find its equation. $y = 3x + 4$
12 A regression line for height ($y$, cm) on age ($x$, years) is $y = 6x + 70$. Predict the height of a 9-year-old. 120 cm
13 A regression line of test mark ($y$) on hours of revision ($x$) is $y = 5x + 40$. Interpret the gradient in context. Each extra hour of revision adds 8 marks (on average)
14 A regression line of test mark ($y$) on hours of revision ($x$) is $y = 5x + 40$. Interpret the $y$-intercept in context. Predicted mark with zero revision = 50
15 For each correlation coefficient $r$, describe the correlation. $r = -0.42$. Moderate negative correlation
16 A line of best fit is $y = 3x + 5$. Find $x$ when $y = 32$. $x = 9$
17 A regression line of $y$ on $x$ passes through the mean point $(\bar{x}, \bar{y})$. If $\bar{x} = 8$ and the line is $y = 2x + 6$, find $\bar{y}$. $\bar{y} = 22$
18 A regression equation was fitted using $x$-values from 5 to 20. Using the line, $y$ is predicted for $x = 10$ (case A) and $x = 30$ (case B). Which is interpolation, which is extrapolation? $x = 4$: interpolation; $x = 20$: extrapolation
19 Plot 1 has $r = 0.9$ and Plot 2 has $r = 0.2$. Which shows a stronger linear relationship? Plot 1 ($|r| = 0.85$ vs 0.1)
20 A scatter plot of time spent gaming vs test scores has line of best fit $y = -2x + 70$. Interpret in context. Each hour of gaming predicts a 1.5-mark decrease; intercept 80
21 Five data points: (1, 3), (2, 5), (3, 7), (4, 9), (5, 11). Find the regression line by inspection. $y = 3x + 4$
22 A regression line $y = 0.5x + 20$ was fitted to $x$-values from 10 to 50. (a) Predict $y$ at $x = 30$. (b) At $x = 80$ the prediction would be? (a) 40 (b) 70 (extrapolation)
23 A car's mileage (km) vs its resale value (£). Suggested data has $r \approx -0.92$. Interpret. Strong positive: older athletes have slower sprints
24 A regression line has a strong outlier. Will removing it (a) increase or decrease $|r|$? (b) Why? (a) Possibly decrease $|r|$ (b) An aligned outlier may anchor a strong correlation
25 Five data points: $\bar{x} = 5, \bar{y} = 12$; regression line has gradient $3$. State the regression equation. $y = 3x - 3$
26 Categorise the correlation based on $r$: (a) $r = 0.3$, (b) $r = 0.7$, (c) $r = -0.95$, (d) $r = 0.05$. (a) Weak negative (b) Strong positive (c) None (d) Moderate positive
27 Five data points have $\bar{x} = 4$ and $\bar{y} = 11$. Four are $(2, 7), (3, 9), (5, 13), (6, 15)$. Find the fifth. $(5, 14)$
28 Studies find $r = 0.9$ between countries' chocolate consumption and number of Nobel Prize winners. Does eating chocolate cause Nobel Prizes? No — population size drives both
29 LOBF $y = 2x + 1$ predicts $y$ at $x = 5$. Actual measured value is 9. What is the residual? Residual = 0
30 A regression line predicts temperature in °C from altitude (m): $T = -0.006a + 15$. Predict $T$ at altitude 2000 m. $T = -6°$C
31 A regression has $r = 0.8$. What proportion of variation in $y$ is explained by $x$? 36%
32 Two studies of "$y$ vs $x$" have regression lines: Study A: $y = 3x + 5$, $r = 0.6$. Study B: $y = 3x + 5$, $r = 0.9$. Which is a better fit? Why? B is better fit
33 A regression line $y = 2x + 5$ has $r^2 = 0.7$. At $x = 10$, you predict $y = 25$. What does $r^2 = 0.7$ tell you? 40% of variation explained — weak predictive value
34 Three points and their LOBF predictions: $(1, 4) \to$ predicted 3.5; $(2, 6) \to$ predicted 5.5; $(3, 8) \to$ predicted 7.5. Find the sum of residuals. 1.5
35 Five students rank their preference for two subjects. The rank correlation is 0.8. Interpret. Moderate negative rank agreement
36 A scatter plot looks like a U-shape. A student fits a straight LOBF and finds $r = 0$. What does this tell us about the relationship? Strong non-linear relationship hidden by $r$
37 A regression line of "ice cream sales" on "temperature (°C)" gives $y = 2.5x + 5$ for $x$ in $[15, 35]$. At $x = -10$°C, the prediction is $-20$. Comment. Same — extrapolation gives nonsense
38 A dataset has $r = -0.7$ and $r^2 = 0.49$. Write a one-sentence summary for a non-technical audience. There is a moderate positive relationship; the model explains about 25% of the variation
39 For a sample of cars, fuel economy ($y$, mpg) vs engine size ($x$, L) gives $y = -8x + 50$. (a) Predict mpg for a 2 L engine. (b) Comment on $x = 5$ L. (a) 45 mpg (b) 5 mpg — extrapolation
40 A study finds $r = 0.95$ between heights of parent and child. Does this mean a tall parent guarantees a tall child? No — moderate tendency only

Problems — Worked Solutions

1

**Hours studied and exam mark.** A teacher records the number of hours that 10 students revised, and their exam marks (out of 100): | Hours | 2 | 3 | 5 | 6 | 7 | 8 | 9 | 10 | 12 | 15 | |-------|---|---|---|---|---|---|---|----|----|----| | Mark | 35 | 40 | 50 | 55 | 60 | 65 | 70 | 70 | 80 | 90 | (a) Sketch a scatter plot. Describe the correlation. (b) Find the means $\bar{x}$ and $\bar{y}$. (c) Estimate the gradient of the line of best fit and write its equation, passing through the mean. (d) Predict the score for a student who revises 4 hours. Is this reliable?

Answer

(a) Strong positive (b) $\bar{x} = 7.7$, $\bar{y} = 61.5$ (c) Slope ≈ 4.3; $y \approx 4.3x + 28$ (d) ≈ 45.4 (interpolation, reasonable)

(a) Strong positive correlation: marks increase steadily with hours. (b) $\bar{x} = (2+3+5+6+7+8+9+10+12+15)/10 = 77/10 = 7.7$. $\bar{y} = (35+40+50+55+60+65+70+70+80+90)/10 = 615/10 = 61.5$. (c) Roughly, from (2, 35) to (15, 90): gradient $\approx (90 - 35)/(15 - 2) = 55/13 \approx 4.23$. Using the mean point: $y - 61.5 = 4.23(x - 7.7) \Rightarrow y \approx 4.23x + 28.93$. Round to $y \approx 4.3x + 29$. (d) At $x = 4$: $y \approx 4.3(4) + 29 = 46.2$. Since 4 is within the data range (2–15), this is **interpolation** and reasonably reliable.
2

**Spurious correlation.** Between 2000 and 2019, the number of films Nicolas Cage starred in correlates strongly with the number of swimming pool drownings in the USA ($r \approx 0.67$). (a) Does this mean Nicolas Cage films cause drownings? (b) Suggest two explanations for the correlation. (c) What does this example teach us about interpreting correlation?

Answer

No — coincidence / lurking variables. The lesson: correlation does not imply causation.

(a) **No.** This is a famous example of a **spurious correlation** — two unrelated time series can correlate by chance over a small range of years. (b) Two explanations: - **Coincidence.** With many possible variables, some pairs will appear correlated purely by chance over short time windows. - **Confounding by time.** Both quantities may simply increase or decrease over time (population growth, more films and pools available). The correlation reflects time, not a causal link. (c) Correlation does **not** imply causation. Strong $r$ alone is not evidence of a causal mechanism; you need: - A plausible mechanism. - Replication across data. - Controlling for confounding variables. This is why scientists do **controlled experiments**, not just correlations.
3

**Predicting temperature.** A dataset of altitude (m) and temperature (°C) at noon yields the regression line $$T = -0.0065 a + 15.$$ (a) Interpret the gradient and intercept. (b) Predict $T$ at altitude 1500 m. (c) Predict $T$ at altitude 9000 m (Everest). Is this reliable?

Answer

(a) Each 1 m up loses 0.0065°C; sea level baseline 15°C (b) 5.25°C (c) ≈ −43.5°C — probably extrapolation

(a) Gradient $-0.0065$: temperature drops by 0.0065 °C per metre of altitude (i.e., about 6.5 °C per km). Intercept 15 °C: predicted temperature at sea level ($a = 0$). (b) $T = -0.0065(1500) + 15 = -9.75 + 15 = 5.25$°C. (c) $T = -0.0065(9000) + 15 = -58.5 + 15 = -43.5$°C. This is **extrapolation** way beyond the typical data range (most measurements would be from $a$ ≈ 0 to 4000 m). The atmosphere at 9000 m has different physics (lapse rate varies with altitude); the linear model is probably less reliable. The actual temperature near the summit of Everest is around $-30°$ to $-40°$C — the prediction is in the right region.
4

**Identifying the outlier.** Eight data points $(x, y)$: $$(1, 3), (2, 5), (3, 7), (4, 8), (5, 10), (6, 12), (7, 25), (8, 16)$$ (a) Identify the outlier. (b) Compute the LOBF gradient with and without the outlier (informally). (c) Comment on the effect of the outlier.

Answer

(a) $(7, 25)$ is the outlier (b) With: ≈ 2.5; Without: ≈ 2.0 (c) Outlier inflates gradient and reduces $r$

(a) Plot the points: (1,3), (2,5), (3,7) ... all roughly follow $y = 2x + 1$. Point $(7, 25)$ is way above this trend (expected $y \approx 15$). **(7, 25) is the outlier.** (b) With outlier (8 points): gradient via approximation through start and end: $(16 - 3)/(8 - 1) = 13/7 ≈ 1.86$. With outlier averaging, the gradient is pulled up. Without (7, 25): remaining 7 points fit $y \approx 2x + 1$ closely → gradient ≈ 2. (c) The outlier: - **Distorts** the LOBF (raises gradient slightly). - **Reduces** $r$ — adds scatter. - Could be a data-entry error or a legitimate anomaly. Always check the source before deleting.
5

**Causation vs correlation.** For each pair, suggest whether the correlation likely reflects causation, confounding, or coincidence. (a) Hours of sleep per night and student exam performance. (b) Sales of sunscreen and number of drownings, weekly. (c) Daily temperature in London and your favourite football team's wins. (d) Number of fire trucks at a fire and damage caused.

Answer

(a) Plausibly causal (sleep affects cognition) (b) Confounding (hot weather) (c) Coincidence (d) Confounding (size of fire)

(a) **Plausibly causal**: well-rested students perform better on cognitive tasks. Evidence from controlled studies supports a direct effect, though revision habits matter too. (b) **Confounding by weather**: hot/sunny weeks mean more sunscreen *and* more swimming (hence drownings). Sunscreen doesn't cause drownings. (c) **Coincidence**: no plausible mechanism linking London weather and a football team's results. Likely spurious. (d) **Confounding by fire size**: bigger fires call for more trucks AND cause more damage. The trucks don't cause damage; fire size does.
6

**Effect of changing units.** A regression of weight (kg) on height (m) gives $W = 50 H + 5$. (a) Predict the weight of a 1.7 m person. (b) If height is now measured in cm instead of m, what is the new regression equation? (c) Verify your equation gives the same prediction at height 170 cm.

Answer

(a) 90 kg (b) $W = 0.5 H + 5$ (c) ✓ — same 90 kg

(a) $W = 50(1.7) + 5 = 85 + 5 = 90$ kg. (b) Convert: $H_{\text{m}} = H_{\text{cm}}/100$. So $W = 50(H_{\text{cm}}/100) + 5 = 0.5 H_{\text{cm}} + 5$. (c) At $H_{\text{cm}} = 170$: $W = 0.5(170) + 5 = 90$ kg ✓. **Lesson:** the gradient depends on units, but the intercept is unchanged (when the conversion is purely multiplicative on $x$).
7

**Time-series correlation.** Over 30 years, global $CO_2$ levels and average global temperature both increased. Linear regression gives $r = 0.95$. (a) Does this prove $CO_2$ causes warming? (b) What other evidence would strengthen a causal claim? (c) Could the correlation be coincidence?

Answer

(a) No, alone (b) Physical mechanism, experiments, models (c) Unlikely given mechanism evidence

(a) No — correlation alone never proves causation, no matter how strong. (b) Strengthening evidence: - **Physical mechanism**: $CO_2$ absorbs infrared radiation, a known physical effect (greenhouse effect). - **Predictive models**: climate models incorporating the greenhouse effect reproduce observed temperature changes. - **Experimental confirmation**: lab experiments demonstrate the greenhouse effect of $CO_2$. - **Multiple datasets**: temperature increase verified across many independent sources. (c) Given the strong physical mechanism *and* the consistency of the data across many measurements, coincidence is implausible. This is a case where the correlation **plus** a well-understood mechanism gives high confidence in causation.
8

**Inverse trend.** A scatter plot has $r = -0.85$. (a) Describe the relationship. (b) The LOBF passes through $(\bar{x}, \bar{y}) = (10, 20)$ with gradient $-2$. State its equation. (c) Predict $y$ at $x = 6$.

Answer

(a) Strong negative (b) $y = -2x + 40$ (c) $y = 28$

(a) Strong negative correlation: as $x$ increases, $y$ decreases, with relatively little scatter. (b) Through $(10, 20)$ with gradient $-2$: $y - 20 = -2(x - 10) \Rightarrow y = -2x + 40$. (c) $y(6) = -2(6) + 40 = 28$.
9

**Reading a scatter.** Estimate the correlation coefficient $r$ for each described scatter: (a) Points lie on a perfect straight line going up. (b) Points form a cloud with no trend. (c) Most points cluster around a line going down, but with notable scatter. (d) Points lie on a perfect curve (parabola), symmetric.

Answer

(a) $r = 1$ (b) $r ≈ 0$ (c) $r$ around $-0.7$ (d) $r ≈ 0$

(a) $r = +1$ (perfect positive linear). (b) $r \approx 0$ (no linear trend). (c) Negative correlation, moderately strong — $r$ around $-0.7$ to $-0.8$. (d) **$r \approx 0$** because $r$ measures *linear* association. A symmetric parabola has $r = 0$ even though there's a clear (non-linear) relationship. **This is a key warning: a low $r$ doesn't rule out a relationship — only a linear one.**
10

**Predict and assess.** A regression of car price (£) on age (years) is $P = -500a + 8000$ for cars aged 0–10 years. (a) Predict price at age 0 and at age 10. (b) Predict price at age 30. Is this reliable? (c) Below what age would the model predict zero price? Is this realistic?

Answer

(a) £8000 and £3000 (b) £-7000 — nonsensical and extrapolation (c) 16 years — partly realistic (very old cars near 0 value)

(a) Age 0: $P = 8000$ (new car). Age 10: $P = -5000 + 8000 = 3000$. (b) Age 30: $P = -15000 + 8000 = -7000$ — a **negative** price, which is nonsensical. The model breaks down because it's a linear extrapolation outside the fitting range (0–10). (c) Set $P = 0$: $0 = -500a + 8000 \Rightarrow a = 16$ years. At 16 years old the model predicts zero value. Some cars do retain very low value at this age, so the prediction is **plausible** for typical cars at the boundary, but the model can't extend further: a 30-year-old classic car may even *appreciate*, not depreciate further.
11

**Comparing methods.** Two students fit lines of best fit to the same scatter plot of 8 points. - Anya draws the line by eye. - Bea uses a calculator to compute the least-squares regression line. Compare and contrast the two approaches: accuracy, repeatability, suitability.

Answer

See working — Bea's method is more accurate and repeatable.

**Anya (by eye).** - ✗ **Less accurate**: humans pick lines based on visual judgement; bias can creep in. - ✗ **Not repeatable**: two students may draw slightly different lines. - ✓ **Quick** for rough estimates. - ✓ **Robust to outliers** if drawn carefully. **Bea (calculator regression).** - ✓ **More accurate**: minimises sum of squared residuals, mathematically optimal for linear fit. - ✓ **Repeatable**: same data → same line every time. - ✓ **Can be extended** to compute $r$, $r^2$, residuals, etc. - ✗ **Sensitive to outliers**: a single bad point can pull the line strongly. - ✗ Requires understanding when linear regression is appropriate. **For a research report**, use the least-squares regression. For quick sketches, by-eye is fine.
12

**Choosing a model.** A scatter plot of plant height (cm) vs days since planting (days) has the following data: | Days | 5 | 10 | 15 | 20 | 30 | 50 | 80 | |--------|----|----|----|----|----|----|----| | Height | 2 | 6 | 12 | 20 | 32 | 50 | 65 | (a) Plot the points (sketch). (b) Is a linear model appropriate? Justify by considering the shape of the data. (c) Describe how the rate of growth changes with time.

Answer

(a) See sketch (b) No — growth slows over time (c) Initially rapid, then slows (sub-linear)

(a) Sketch the points. Initially they rise steeply, but the rate of increase slows down (the curve flattens). (b) **Not entirely appropriate.** The data show a clear non-linear pattern: the gradient decreases as days increase. A linear fit would over-predict heights at early days and at late days (it would be straightest in the middle). Computing increments per day-range: - Days 5–20 (15 days): height +18, so +1.2 cm/day. - Days 20–50 (30 days): height +30, so +1 cm/day. - Days 50–80 (30 days): height +15, so +0.5 cm/day. The rate is **clearly decreasing**. (c) The plant grows rapidly when young (about 1.2 cm/day) and slows as it ages (down to 0.5 cm/day by days 50–80). This is sub-linear (logarithmic-style) growth — characteristic of biological growth approaching a maximum height. **Better model**: a logistic or square-root function. A linear fit would still capture the general trend but with poor predictive accuracy near the extremes.