How to use this menu
Six original tasks: two for criterion A (a short check and an end-of-unit test), one investigation for B, one communication task for C, and two real-life tasks for D that also assess C. Set at least one task for each criterion across the unit; nobody needs all six. The first falls after lesson 3, the investigation after lesson 4 and the rest after lesson 5, so assessment is not left until the end.
Descriptors: Year 11 Extended is assessed with the MYP Year 5 descriptors, paraphrased below each task. Check them against your school's current guide.
The menu at a glance
| Task | Criteria | When | Time | Format |
|---|---|---|---|---|
| A1: Quick check: samples, covariance and r | A | After lesson 3 | 30 minutes | Individual, written, in class. 24 marks. |
| B1: Changing the units: what happens to the line? | B | Mid-unit, after lesson 4 | 50 minutes | In class. Students may collect results in pairs, but each writes up their own findings. |
| C1: A note for the stall owner | C | After lesson 5 | 35 minutes | Individual, written, in class or as homework. One page of words, a table and a sketch. |
| A2: End-of-unit test | A | End of unit, after lesson 5 | 50 minutes | Individual, written, in class, one lesson. 31 marks in three sections. |
| D1: Fertilizer and the rice harvest | D and C | End of unit, after lesson 5 | 50 minutes | Individual, in class. A written solution with explanations and a short note to the co-operative. |
| D2: Counting life on a reef | D and C | End of unit, after lesson 5 | 50 minutes | Individual, in class. A written solution with explanations and a short reply to the survey leader. |
D1 and D2 are alternatives. Both assess criteria D and C in a real-life setting, one about crops and one about a reef: set one, not both.
The unit's inquiry
- Key concept. Relationships: how a variable is linked to another, and how far a sample or a line can be trusted to describe the link.
- Related concepts. Models, Generalization, Representation
- Global context. Globalization and sustainability: using samples and measured links to make decisions about food, ecosystems and resources.
- Statement of inquiry. A sample, a correlation coefficient and a regression line are models of a population and of a relationship; judging how far each can be trusted is part of using them.
MYP Year 5 descriptors, paraphrased
Each task's clarification, on its student sheet below, says what these levels look like in that task.
| Level | Criterion A: Knowing and understanding |
|---|---|
| 1–2 | Selects suitable mathematics for simple problems in familiar situations and applies it with some success. |
| 3–4 | Selects suitable mathematics for more complex problems in familiar situations and applies it successfully. |
| 5–6 | Selects suitable mathematics for challenging problems in familiar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| 7–8 | Selects suitable mathematics for challenging problems in both familiar and unfamiliar situations, applies it successfully and generally solves these problems correctly in a range of contexts. |
| Level | Criterion B: Investigating patterns |
|---|---|
| 1–2 | With support, applies problem-solving techniques to find simple patterns and states predictions that fit them. |
| 3–4 | Applies problem-solving techniques to find simple patterns and suggests general rules that fit the findings. |
| 5–6 | Applies problem-solving techniques to find complex patterns, describes them as general rules that fit the findings and verifies that the rules hold. |
| 7–8 | Selects and applies problem-solving techniques to find complex patterns, describes them as general rules that fit correct findings, and proves, or verifies and justifies, those rules. |
| Level | Criterion C: Communicating |
|---|---|
| 1–2 | Uses limited mathematical language and limited forms of representation; the reasoning is hard to follow. |
| 3–4 | Uses some appropriate mathematical language and forms of representation; the reasoning can be followed and the work is adequately organized. |
| 5–6 | Usually uses appropriate language, moves between forms of representation with some success, and the reasoning is complete and coherent in a mostly logical structure. |
| 7–8 | Consistently uses appropriate language, moves effectively between forms of representation, and the reasoning is complete, coherent and concise in a logical structure. |
| Level | Criterion D: Applying mathematics in real-life contexts |
|---|---|
| 1–2 | Identifies some relevant elements of the real-life situation and selects, with some success, strategies to model it. |
| 3–4 | Identifies the relevant elements, selects adequate strategies to model the situation, applies them to reach a solution and discusses whether it makes sense. |
| 5–6 | Identifies the relevant elements, selects adequate modelling strategies and applies them to reach a valid solution, explains the degree of accuracy and explains whether the solution makes sense. |
| 7–8 | Identifies the relevant elements, selects appropriate modelling strategies and applies them successfully to reach a correct solution, justifies the degree of accuracy and justifies whether the solution makes sense. |
Task A1: Quick check: samples, covariance and r
Why set it. A short check of lessons 1 to 3 while there is still time to reteach: every possible sample, covariance from deviations, and r. The numbers are chosen so that r is exact by hand. Questions 5 and 6 are unfamiliar, so the top band is open to everyone.
When: After lesson 3. Time: 30 minutes. Format: Individual, written, in class. 24 marks.
The student sheet, with its task-specific clarification
When: After lesson 3. Time: 30 minutes. Calculator: no calculator. Criterion: A, knowing and understanding.
Show every step of your working: a correct method earns credit even if the final answer slips. Questions 5 and 6 are meant to make you think, so try something rather than leaving them blank. No calculator is needed.
Part 1. Samples
- Five ferry crossings were late by 2, 4, 6, 10 and 18 minutes. (a) Work out the mean lateness of the five crossings. (b) A sample of two crossings is chosen. List every possible sample of two and give the mean of each. (c) Give the lowest and the highest sample mean, and say how many of the samples have a mean within 2 minutes of the mean of all five crossings. (d) Without listing them, would samples of size 3 have a wider or a narrower spread of sample means than samples of size 2? Give a reason. (7 marks)
- A ferry operator has sold 600 adult and 400 child tickets and wants a stratified sample of 50 tickets. How many of each kind should it take? Give one reason for stratifying rather than choosing 50 tickets at random. (2 marks)
Part 2. Two variables
- A bakery records, on four days, x, the number of staff on shift, and y, the minutes a customer waits: x = 2, 4, 6, 8 and y = 9, 5, 7, 3. (a) Find the mean of x and the mean of y. (b) Write down dx and dy for each day, and the product dx × dy. (c) Work out the covariance. (d) What does the sign of the covariance tell the bakery? (e) The sum of dx2 is 20 and the sum of dy2 is 20. Work out r. (8 marks)
- The waiting times in question 3 are re-measured in seconds (every y multiplied by 60). What are the covariance and r now? (2 marks)
Part 3. Think it through
- Four points have x = −2, −1, 1, 2 and y = 4, 1, 1, 4. Show that their covariance is 0. Does that mean there is no relationship between x and y? Explain. (3 marks)
- Priya says: “Data set P has covariance 120 and data set Q has covariance 0.8, so the link in P is stronger.” Explain why she may be wrong, and what she should compare instead. (2 marks)
How your work is judged
| Level | Criterion A in task A1: what it looks like |
|---|---|
| 1–2 | You answer some of the questions in Parts 1 and 2 correctly, for example the mean of the five crossings or the deviations in the bakery table. |
| 3–4 | You get most of questions 1 to 3 right: the list of samples, the stratified split, the deviations and the covariance. Finding r in question 3 or the effect of changing units in question 4 is missing or has errors. |
| 5–6 | Questions 1 to 4 are right, with working that shows your method, and you make a sound attempt at question 5 or 6 that uses the figures. |
| 7–8 | Questions 1 to 4 are right with working, and 5 and 6 are reasoned from the figures: you show the covariance is 0 and say what that does and does not show, and you explain why covariances cannot be compared across different units. |
Markscheme
| Task A1 question | Answer and marking |
|---|---|
| 1(a) | 8 minutes (40 ÷ 5). 1 mark. |
| 1(b) | Ten samples with means 3, 4, 6, 10, 5, 7, 11, 8, 12, 14 (for 2 and 4, 2 and 6, 2 and 10, 2 and 18, 4 and 6, 4 and 10, 4 and 18, 6 and 10, 6 and 18, 10 and 18). 1 mark for a complete list of ten samples, 2 marks for all ten means right (1 mark for at most two slips). |
| 1(c) | Lowest 3, highest 14; four samples have a mean from 6 to 10 (means 6, 7, 8 and 10). 1 mark for the pair, 1 mark for 4. |
| 1(d) | Narrower. Each sample has more values in it, so a very low or very high crossing is averaged with more typical ones. Check: for size 3 the lowest mean is 4 and the highest is 11.33, a range of 7.33 against 11. 1 mark for narrower with a reason. |
| 2 | 30 adult and 20 child tickets (600 ÷ 1000 × 50 = 30). 1 mark for the numbers. 1 mark for a reason: the two kinds are different groups, so a random sample could by chance contain too few of one. |
| 3(a) | Mean x = 5, mean y = 6. 1 mark. |
| 3(b) | dx = −3, −1, 1, 3; dy = 3, −1, 1, −3; products −9, 1, 1, −9. 1 mark for the deviations, 1 mark for the products. Students who write 9 for the first product have lost a sign. |
| 3(c) | −4 (−16 ÷ 4). 1 mark for the sum −16, 1 mark for dividing by 4. A covariance of −16 has forgotten to divide. |
| 3(d) | Negative: on days with more staff the wait tends to be shorter. 1 mark; it must say more staff goes with shorter waits, not only “negative correlation”. |
| 3(e) | r = −16 ÷ √(20 × 20) = −16 ÷ 20 = −0.8. 1 mark for the quotient set up, 1 mark for the value with its sign. |
| 4 | Covariance −240; r is still −0.8. 1 mark each. The covariance carries the unit of y; r has no unit. |
| 5 | Mean x = 0 and mean y = 2.5; the products are −3, 1.5, −1.5 and 3, which add to 0. There is a clear relationship (y = x2, a U shape); covariance 0 only says there is no straight-line trend. 1 mark for the means, 1 for the products or sum, 1 for the interpretation. |
| 6 | Covariance has the units of x times y, so its size changes when units change; P and Q may be measured in different units. She should compare r, which has no units. 1 mark for each of the two ideas. |
| Marks to a level | Total 24. 1 to 6 marks: 1 to 2. 7 to 12: 3 to 4. 13 to 19: 5 to 6. 20 to 24, with question 5 or 6 fully correct: 7 to 8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A1 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Question 1: 1(a) right (1); 1(b) all ten samples but two means wrong (1 + 1 = 2); 1(c) lowest and highest right, count missing (1); 1(d) no reason (0); question 2 right with a reason (2). Question 3: means right (1), deviations and products right (2), sum of products found but not divided (1), sign explained (1), r not attempted (0). Questions 4 to 6 not attempted. 11 marks, and the profile matches 3 to 4: most of the first questions, nothing unfamiliar. |
| Level 7 to 8 answer to question 5 | “Mean of x is (−2 − 1 + 1 + 2) ÷ 4 = 0 and mean of y is 10 ÷ 4 = 2.5. dx × dy = (−2)(1.5) = −3, (−1)(−1.5) = 1.5, (1)(−1.5) = −1.5 and (2)(1.5) = 3. These add to 0, so the covariance is 0. But y = x2 exactly, so the two variables are closely related; they are not related by a straight line. Covariance and r only measure straight-line association.” |
Differentiation
A way in for anyone stuck
- Give the table for question 3 with the headings dx, dy and dx × dy and the means filled in, so the work is the products.
- Let students list the ten pairs for 1(b) in a systematic order (2 with each of the others, then 4 with each of the rest) before working out the means.
- Read questions 5 and 6 aloud on request and suggest sketching the four points. Reading is not what is being assessed.
Stretch
- Find the mean of all ten sample means in question 1(b). What do you notice, and is it a coincidence? (It is 8, the population mean: every value appears in the same number of samples.)
- Make up four points with covariance 0 where a straight line is not just a poor fit but the wrong shape, and sketch what a residual plot would show.
Students learning in English as an additional language
| Word in task A1 | Plain meaning |
|---|---|
| sample | a small group chosen from a bigger group to find out about the bigger group |
| stratified sample | a sample that takes the right number from each separate group |
| deviation | how far a value is above or below the mean |
| covariance | one number for whether two variables rise together (positive) or one rises as the other falls (negative) |
| unit | what a number is measured in, for example minutes or seconds |
Sentence starters for task A1:
- The sample means are lowest when … and highest when …
- The covariance is negative, so …
- I cannot compare the covariances because …
Task B1: Changing the units: what happens to the line?
Why set it. A real pattern hunt built on lesson 4's regression line. Lesson 3 showed that r does not depend on units; this task asks what does happen to the gradient and the intercept. The sheet never names the pattern: students organize the lines they find, state rules, test them on a change of their own and justify them.
When: Mid-unit, after lesson 4. Time: 50 minutes. Format: In class. Students may collect results in pairs, but each writes up their own findings.
The student sheet, with its task-specific clarification
In MYP Year 5, support you give (a prompt, a table outline) is allowed at 1 to 2 and 3 to 4: note it on the student's work.
When: Mid-unit, after lesson 4. Time: 50 minutes. Calculator: gdc optional. Criterion: B, investigating patterns.
A gardener measures the height of a bean seedling at the end of each of five weeks. In week 1, 2, 3, 4 and 5 the heights were 3, 4, 6, 7 and 10 cm. Write x for the week number and y for the height. You will be finding regression lines for this set of data and for changed versions of it, and looking for what the changes do to the line.
Part 1. The starting line
- Find the regression line of y on x, and r. Check that the line passes through the mean point.
Part 2. Four changed versions
- Find the regression line of y on x, and r, for each of these changed versions of the data, and record every line and r in a table you design yourself. (i) Every height written in millimetres instead of centimetres. (ii) Time in half-weeks instead of weeks, so week 1 becomes 2, week 2 becomes 4, and so on. (iii) The pot stands on a table and every height is measured from the table top, which is 5 cm below the soil surface, so every height is 5 cm more. (iv) Time counted in weeks from the first measurement, so the first measurement is at 0 and the last at 4.
Part 3. Describe what you find
- Look across your five lines. Describe in words, as a rule another student could use, what the changes do to the gradient, to the intercept and to r.
Part 4. Test your rules
- Choose a change of your own to the data that is not one of the four above. Use your rules to predict the new line before you work it out, then work it out and compare.
Part 5. Predict, then check
- A report needs heights in millimetres, time in half-weeks, and heights measured from the table top, which is 20 mm below the soil surface. Predict the regression line for this version of the data without working out any new sums. Then check your prediction using the new data values.
Part 6. Explain
- Explain why your rules must hold for any data set and not only this one.
How your work is judged
| Level | Criterion B in task B1: what it looks like |
|---|---|
| 1–2 | With help, you find the line for the original data and for at least one changed version, and you say something you notice about the numbers. |
| 3–4 | You find the line for all five versions and suggest a rule for what happens to the gradient or to the intercept for at least one kind of change, and it fits your results. |
| 5–6 | You find all five lines accurately, describe rules for both the gradient and the intercept for all four changes, and verify them on your own change in Part 4 and on the prediction in Part 5. |
| 7–8 | As the level before, and your rules are written in general form for any change of the type new value = multiplier times old value plus a shift, verified, and you justify from y = a + bx (or from the sums) why they must always hold. |
Markscheme
| Task B1 question | Answer and marking |
|---|---|
| The pattern (do not share it before the task) | Write the original line as y = a + bx. If y′ = p y + q and x′ = m x + n, the new gradient is p b ÷ m and the new intercept is p a + q − p b n ÷ m. So multiplying heights by p multiplies both a and b by p; adding q to heights adds q to a; multiplying x by m divides b by m and leaves a; subtracting 1 from x raises the intercept by the old gradient. r never changes (lesson 3), apart from its sign if p and m have opposite signs. |
| 1 | y = 0.9 + 1.7x, r = 0.98. Mean point (3, 6), sum of products 17, sum of squares of dx 10, sum of squares of dy 30, r = 17 ÷ √300. The line passes through (3, 6): 0.9 + 1.7 × 3 = 6. |
| 2(i) | y = 9 + 17x, r = 0.98. |
| 2(ii) | y = 0.9 + 0.85x, r = 0.98. |
| 2(iii) | y = 5.9 + 1.7x, r = 0.98. |
| 2(iv) | y = 2.6 + 1.7x, r = 0.98. The intercept is now the predicted height at the first measurement (the end of week 1), which is the old line at x = 1. |
| 3 | Any correct, precise description: heights times 10 makes gradient and intercept ten times as big; doubling x halves the gradient and leaves the intercept; adding 5 to every height adds 5 to the intercept; taking 1 off every x raises the intercept by 1.7, the old gradient. r is 0.98 every time. A rule for one kind of change only is evidence for 3 to 4. |
| 4 | Check against the general result. For example, heights halved: y = 0.45 + 0.85x. Negative multipliers give a negative gradient and the opposite sign of r: every y times −1 gives y = −0.9 − 1.7x with r = −0.98. Credit a prediction made first and then checked. |
| 5 | y′ = 29 + 8.5x′. The new values are x′ = 2, 4, 6, 8, 10 and y′ = 50, 60, 80, 90, 120. Mean point (6, 80), sum of products 340, sum of squares of dx 40, gradient 8.5, intercept 80 − 8.5 × 6 = 29. From the rules: gradient 1.7 × 10 ÷ 2 = 8.5, intercept 0.9 × 10 + 20 = 29. |
| 6 | At 7 to 8: write y = a + bx for the original, then y′ = p y + q and x′ = m x + n, and substitute: y′ = p(a + bx) + q = (pa + q) + (pb)x with x = (x′ − n) ÷ m, which gives the general gradient and intercept. An equivalent argument from the sums (deviations are multiplied by m and by p, so the sum of products is multiplied by pm and the sum of squares of dx by m2) also earns the top band. |
Exemplar answers
| Task B1 response | What it looks like |
|---|---|
| Level 3 to 4 | “Millimetres: the line is y = 9 + 17x, so the numbers are 10 times as big. Half-weeks: y = 0.9 + 0.85x, the gradient is smaller. Adding 5: y = 5.9 + 1.7x. When I make the numbers bigger, the line numbers go bigger.” All four lines right and a rule suggested for the first, but the rules for time and for shifting x are not stated, no change of the student's own is tested, and Part 5 is not attempted. |
| Level 7 to 8 | “Heights times 10 make both a and b ten times as big. Doubling x halves b and leaves a. Adding 5 to heights adds 5 to a. Subtracting 1 from x adds b to a, because the old line at x = 1 is the new intercept. For y′ = p y + q and x′ = m x + n: b′ = pb ÷ m and a′ = pa + q − pbn ÷ m. Test: heights ÷ 2 gives 0.45 + 0.85x, which is what the sums give. Part 5: p = 10, q = 20, m = 2, n = 0, so b′ = 17 ÷ 2 = 8.5 and a′ = 9 + 20 = 29, and the new data gave the same. Why: y′ = p(a + bx) + q and x = x′ ÷ m, so the x′ coefficient is pb ÷ m.” |
Differentiation
A way in for anyone stuck
- Give a table outline for Parts 1 and 2 with columns for x, y, the two means, the sum of products and the sum of squares of dx, and let the student write the headings of the last columns.
- Suggest working out the sums once for the original data and then asking what each change does to the deviations before recalculating.
- Prompt with a question, never the rule: “What is the new mean point, and what does the line have to pass through?”
Stretch
- Every y is multiplied by −1 and every x is left alone. Predict the new line and r, then check. What is different about r?
- A data set has regression line y = 4 − 2x. The x values are now measured in units half the size (every x doubled) and every y is raised by 6. Write the new line without any new sums. (y = 10 − x.)
Students learning in English as an additional language
| Word in task B1 | Plain meaning |
|---|---|
| regression line | the straight line that fits the data best, written y = a + bx |
| gradient | how much y changes when x increases by 1 |
| intercept | the value of y where the line meets the y-axis, when x is 0 |
| rule | a sentence that tells you what happens every time |
| predict | say what will happen before you check |
| verify | check that something is true by testing it |
Sentence starters for task B1:
- When every height is multiplied by 10, the line …
- My rule is: …
- I tested my rule on … and it worked / did not work because …
Task C1: A note for the stall owner
Why set it. Criterion C on its own: notation, a table and a sketch used together, and a line of reasoning a reader with no statistics can follow, built around five real errors from lessons 2 to 5.
When: After lesson 5. Time: 35 minutes. Format: Individual, written, in class or as homework. One page of words, a table and a sketch.
The student sheet, with its task-specific clarification
When: After lesson 5. Time: 35 minutes. Calculator: no calculator. Criterion: C, communicating.
Zoe runs a market stall. For five weeks she handed out leaflets, and counted visitors to the stall. Her friend Tomas worked out the link and the regression line, and gave her this working. It has mistakes in it.
Data. x is the number of leaflets handed out, in tens: 1, 2, 3, 4, 5. y is the number of visitors, in tens: 2, 4, 3, 6, 5.
Tomas’s working
Mean x = 3, mean y = 4. dx = −2, −1, 0, 1, 2 and dy = −2, 0, −1, 2, 1.
dx × dy = 4, 0, 0, 2, 2, which add to 8, so the covariance is 8.
The sum of dx2 is 10 and the sum of dy2 is 10, so r = 8 ÷ (10 + 10) = 0.4.
Gradient b = 8 ÷ 10 = 0.8. Intercept a = 3 − 0.8 × 4 = −0.2. Line: y = −0.2 + 0.8x.
For 12 tens of leaflets the line gives y = 9.4, so about 94 visitors, and r shows that leaflets cause the visitors.
Write a one-page note for Zoe. She has not studied statistics, and should be able to follow it on her own.
Part 1. Put it right
- Find every mistake in Tomas’s working. For each one, say what he did, why it is wrong, and write the correct working.
Part 2. Show the data
- Make a table that shows the deviations and products, and a small sketch of the five points with the corrected line. Mark the mean point on your sketch.
Part 3. Tell Zoe what it means
- In plain words, say what the corrected r and the corrected line tell Zoe, and one thing neither of them tells her. Use the words deviation, correlation and gradient in a way she could follow.
- Give a prediction for 3 tens of leaflets and say whether Zoe can trust a prediction for 12 tens, and why.
How your work is judged
| Level | Criterion C in task C1: what it looks like |
|---|---|
| 1–2 | Some of your note can be followed, but it is hard to tell which of Tomas’s steps you are talking about, the table or sketch is missing or confusing, and key words are missing. |
| 3–4 | You explain several of Tomas’s mistakes with some correct words and notation, and your steps are in a sensible order, so Zoe could follow most of it, with gaps. |
| 5–6 | You deal with every step in order, with reasons; your table and sketch match your words; you usually use the right notation (for example dx, dy, r, b and a) and the key words; and your plain-words paragraph says what the results mean and do not mean. |
| 7–8 | Each correction sits beside the step it corrects, and your table, sketch and words agree with each other; the notation is correct throughout, the key words are used as Zoe would need them, and the plain-words part says in a few precise sentences what can and cannot be concluded. |
Markscheme
| Task C1 question | Answer and marking |
|---|---|
| Covariance | He added the products and did not divide by n = 5: the covariance is 8 ÷ 5 = 1.6. The sum 8 is correct. |
| r | He added the two sums of squares. The formula multiplies them and takes the square root: r = 8 ÷ √(10 × 10) = 8 ÷ 10 = 0.8, not 0.4. |
| Intercept | He swapped the means. The intercept is mean y minus gradient times mean x: a = 4 − 0.8 × 3 = 1.6, so the line is y = 1.6 + 0.8x. Check: it passes through the mean point (3, 4). |
| Prediction | At 12 tens the corrected line gives 1.6 + 0.8 × 12 = 11.2, not 9.4. More importantly 12 is far outside the data (1 to 5), so this is extrapolation and cannot be trusted. |
| Cause | r = 0.8 measures how closely the dots follow a rising straight line. It does not show that leaflets cause visitors: something else, such as the weather or the day, could affect both. |
| Own prediction (Part 3) | At x = 3 the line gives 1.6 + 0.8 × 3 = 4.0 tens, the mean of y, because the line passes through the mean point. Accept any prediction at an x from 1 to 5 that matches the line. |
| How to level | Level the note as a whole against the clarification, not by counting corrections. The table, sketch and wording decide between 5 to 6 and 7 to 8 more often than the corrections do. Mathematical slips count only where they break the line of reasoning Zoe has to follow; accuracy itself is criterion A. |
Exemplar answers
| Task C1 response | What it looks like |
|---|---|
| Level 3 to 4 extract | “Tomas did not divide the 8 by 5 so the covariance is 1.6. For r he put 10 + 10 and it is the square root of the two times, so 0.8. The intercept is wrong, it is 4 − 0.8 × 3 = 1.6. 12 tens is a long way from the others so I would not trust it.” The mistakes are found and the words are mostly right, but there is no table or sketch, the steps are not labelled with Tomas’s lines, and nothing is said about cause. |
| Level 7 to 8 extract | “Tomas’s line 3. Covariance is the mean of the five products, so 8 ÷ 5 = 1.6 (he stopped at the total). Line 4. r = Σdxdy ÷ √(Σdx2 × Σdy2) = 8 ÷ √100 = 0.8; he added the two sums instead of multiplying them. What this means for Zoe: the visitors tend to rise along a fairly straight line as leaflets increase (r = 0.8, quite strong). What it does not mean: that leaflets cause the visitors, or that the line holds at 12 tens, which lies far outside the 1 to 5 she has tried.” Each correction sits beside the line it corrects, with a table and sketch, and the last sentence sets the limits. |
Differentiation
A way in for anyone stuck
- Give the table outline with the columns x, y, dx, dy, dx × dy, dx2 and dy2, and the totals row left empty.
- Let students explain aloud to a partner first, then write down what they said.
- Offer the frame: “Tomas wrote …. This is wrong because …. The correct working is ….”
Stretch
- Write a second version of the plain-words paragraph for a reader who is a statistician. What did you take out, and what did you add?
- Add a sixth week to Tomas’s data of 6 tens of leaflets and 7 tens of visitors, and say what your note would need to change.
Students learning in English as an additional language
| Word in task C1 | Plain meaning |
|---|---|
| correlation | how closely two variables follow a straight line together |
| mistake | a step that is not correct |
| cause | make something happen |
Sentence starters for task C1:
- Tomas wrote …, but …
- The correct working is …, because …
- This does not show that … because …
Task A2: End-of-unit test
Why set it. The whole unit at three levels of demand. Section 3 is unfamiliar on purpose: it is where the evidence for 7 to 8 comes from.
When: End of unit, after lesson 5. Time: 50 minutes. Format: Individual, written, in class, one lesson. 31 marks in three sections.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 5. Time: 50 minutes. Calculator: gdc optional. Criterion: A, knowing and understanding.
Three sections, from familiar questions to ones you have not met in this form. Show your working everywhere: it earns credit even when an answer slips. A GDC is optional; every number can be found by hand.
Section 1. Samples
- A population has four values: 1, 3, 5 and 7. List every possible sample of size 2 and the mean of each. (2 marks)
- Give the range of the sample means for samples of size 2. (1 mark)
- List every sample of size 3 and its mean. Give the range of these sample means. (2 marks)
- What happens to the range as the sample size increases? What is the range for samples of size 4, and why? (1 mark)
- A survey of 600 residents has 240 in the north, 160 in the centre and 200 in the south. A stratified sample of 30 is taken. How many come from each area? (2 marks)
Section 2. A linear model
- A kiosk records the outside temperature, x in degrees Celsius, and the number of hot drinks sold, y, on five days: x = 3, 5, 7, 9, 11 and y = 14, 10, 4, 6, 6. Write down the mean point. (1 mark)
- Work out dx, dy and dx × dy for each day, and the covariance. (3 marks)
- Work out r to two decimal places and describe the link in words. (2 marks)
- Find the regression line of y on x. (3 marks)
- Say what the gradient and the intercept mean in this context, and whether the intercept makes sense. (2 marks)
- Use the line to predict the number of drinks sold at 8 degrees and at 20 degrees. Comment on each. (2 marks)
Section 3. Judging the model
- Work out the five residuals for the line in Section 2. Check that they add to 0. (2 marks)
- Use the residuals and r to say whether a straight line is a suitable model for this kiosk. (2 marks)
- A sixth day is added at 7 degrees with 8 drinks sold. Without redoing the whole calculation, say what happens to the covariance, to r and to the regression line. Explain using the sums from Section 2. (4 marks)
- Samira says: “A stratified sample of 30 is always better than a simple random sample of 30.” Describe a situation in which stratifying would make almost no difference, and explain why. (2 marks)
How your work is judged
| Level | Criterion A in task A2: what it looks like |
|---|---|
| 1–2 | You answer some of the questions in Sections 1 and 2 correctly, for example the sample means, the mean point or the deviations. |
| 3–4 | You answer most of Sections 1 and 2 correctly, with a method visible, but finding r, the regression line or the residuals has errors or is missing. |
| 5–6 | You answer Sections 1 and 2 correctly with working, find and use the residuals, and make a sound attempt at one question in Section 3 that goes beyond the lessons. |
| 7–8 | You answer Sections 1 and 2 correctly with working, and in Section 3 you reason correctly from the sums and the residuals in the unfamiliar questions, including the new point. |
Markscheme
| Task A2 question | Answer and marking |
|---|---|
| 1 and 2 | Six samples with means 2, 3, 4, 4, 5, 6 (1 and 3, 1 and 5, 1 and 7, 3 and 5, 3 and 7, 5 and 7). 1 mark for six samples, 1 mark for the means. Range: 6 − 2 = 4 (1 mark). |
| 3 | Four samples with means 3, 3.67, 4.33, 5 (sums 9, 11, 13 and 15, divided by 3). Range 2. 1 mark for the samples and means, 1 for the range. |
| 4 | The range gets narrower as the sample gets bigger; size 4 is the whole population, so there is one sample, with mean 4, and the range is 0. 1 mark for both ideas. |
| 5 | 12 north, 8 centre, 10 south (240 ÷ 600 × 30, 160 ÷ 600 × 30, 200 ÷ 600 × 30). 1 mark for the method, 1 for the three numbers. |
| 6 | (7, 8). 1 mark. |
| 7 | dx = −4, −2, 0, 2, 4; dy = 6, 2, −4, −2, −2; products −24, −4, 0, −4, −8, sum −40, covariance −8. 1 mark for dx and dy, 1 mark for the products and sum, 1 mark for −8. Covariance of −40 has forgotten to divide by 5. |
| 8 | Sum of dx2 = 40 and of dy2 = 64, so r = −40 ÷ √2560 = −0.79: a fairly strong negative link; warmer days go with fewer hot drinks. 1 mark for the value, 1 for the description. |
| 9 | Gradient b = −40 ÷ 40 = −1, intercept a = 8 − (−1) × 7 = 15, line y = 15 − x. 1 mark each. |
| 10 | Each extra degree goes with one fewer hot drink sold; at 0 degrees the line gives 15 drinks, which is outside the range of temperatures observed (3 to 11), so the intercept is an extrapolation and not a measured value. 1 mark each. |
| 11 | At 8 degrees: 7 drinks, inside the data, so interpolation and reasonably reliable. At 20 degrees: −5, impossible, because 20 is far outside the data. 1 mark each. |
| 12 | Predicted 12, 10, 8, 6, 4 against actual 14, 10, 4, 6, 6: residuals 2, 0, −4, 0, 2, which add to 0. 1 mark for the residuals, 1 for the check. |
| 13 | The residuals make a U shape (positive, zero, negative, zero, positive) and not a random scatter, which suggests a curve; r = −0.79 is fairly strong, but it rests on only five points. A straight line is not strongly supported. 1 mark for using the residuals, 1 for using r. |
| 14 | The new point is the mean point (7, 8), so its dx and dy are 0 and the three sums stay −40, 40 and 64. So r is unchanged (−0.79) and the line is unchanged (y = 15 − x), since the gradient is −40 ÷ 40 and it still passes through (7, 8). The covariance changes to −40 ÷ 6 = −6.67, because the total is now shared among six days. 1 mark each for r, the line, the covariance, and 1 for the reasoning. |
| 15 | A situation where the strata do not differ in the thing measured, for example the preference between two flavours when all areas have the same tastes: then each random sample already has about the right mix whatever the strata, so there is little gain. 1 mark for a situation, 1 for the reason. |
| Marks to a level | Total 31. 1 to 7 marks: 1 to 2. 8 to 16: 3 to 4. 17 to 25: 5 to 6. 26 to 31, with question 14 fully correct: 7 to 8. Use the marks as a guide, then check the student's profile against the clarification before settling the level. |
Exemplar answers
| Task A2 response | What it looks like |
|---|---|
| Level 3 to 4 profile | Section 1: samples of 2 and of 3 right but the range for size 3 left out (4 marks), trend comment (1), strata right (2); Section 2: mean point (1), deviations and products right with the covariance left as −40 (2), no r, gradient found but intercept wrong (1); nothing in Section 3. 11 marks, and the profile matches 3 to 4: methods known, sustained errors, nothing unfamiliar. |
| Level 7 to 8 answer to the sixth-day question | “The point (7, 8) is the mean point, so dx = 0 and dy = 0 and it adds nothing to any of the three sums: −40, 40, 64. So r = −40 ÷ √(40 × 64) is the same. The gradient is −40 ÷ 40 = −1 as before, and the mean is still (7, 8), so the line is the same. Only the covariance changes, because it is the sum divided by the number of points: −40 ÷ 6 = −6.67 instead of −8.” |
Differentiation
A way in for anyone stuck
- Give the table outline for Section 2 with x, y, dx, dy, dx × dy, dx2 and dy2 and the totals row blank.
- Provide the formulas for covariance, r, gradient and intercept on a reference card, as in lessons 2 to 4. Recall is not the target; choosing and using them is.
- Let students list the samples for Section 1 in pairs as a systematic grid, and read the Section 3 questions aloud on request.
Stretch
- Show that for any data set a new point at the mean point leaves r and the regression line unchanged, using the formulas for r, b and a.
- A seventh day is added at 9 degrees with 6 drinks sold. Does r go up or down? Use the sums to decide before you recalculate.
Students learning in English as an additional language
| Word in task A2 | Plain meaning |
|---|---|
| population | the whole group you want to know about |
| sample mean | the mean of the values in one sample |
| residual | the actual value of y minus the value the line predicts |
| extrapolation | using a line to predict outside the range of the data |
| interpolation | using a line to predict inside the range of the data |
Sentence starters for task A2:
- The range of the sample means is … because …
- The residuals show a pattern, which means …
- The new point does not change … because …
Task D1: Fertilizer and the rice harvest
Why set it. A regression line, its limits and a stratified sample inside one real decision. An alternative to D2: set one of them.
When: End of unit, after lesson 5. Time: 50 minutes. Format: Individual, in class. A written solution with explanations and a short note to the co-operative.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 5. Time: 50 minutes. Calculator: gdc optional. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A farmers’ co-operative in a rice-growing region tried fertilizer on five equal trial plots. The amount of fertilizer, in kilograms per hectare, and the yield, in tonnes per hectare, were:
| Fertilizer, x (kg per hectare) | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|
| Yield, y (tonnes per hectare) | 3.0 | 3.6 | 4.0 | 4.2 | 4.2 |
Part 1. The model
- Find the regression line of yield on fertilizer, and r. What does the gradient say, in words with its units?
Part 2. Is a straight line the right model?
- Work out the residuals and say what they show about how the yield responds to more fertilizer.
Part 3. Using the line
- Predict the yield for 70 kg per hectare. How accurate should you say it is?
- An adviser says: “The line gives 5.3 tonnes for 160 kg, so tell farmers to use 160 kg.” Reply.
Part 4. Testing the soil
- The co-operative has 1000 farms in three districts, with 450, 300 and 250 farms. It will soil-test 40 farms. How many from each district should be chosen if the sample is stratified? Say why it is better than asking for volunteers.
Part 5. A note
- Write a short note (about 100 words) to the co-operative: how much fertilizer to try next season, and what the trial results can and cannot tell them.
How your work is judged
| Level | Criterion D in task D1: what it looks like |
|---|---|
| 1–2 | You identify some of the quantities involved and find a regression line with help, but your answers are not related to the farms, so it is unclear whether they make sense. |
| 3–4 | You choose a regression line, residuals and a stratified allocation as the methods, apply them with some errors, and comment on whether the results suit the situation when prompted. |
| 5–6 | You choose suitable methods, apply them correctly, and justify the accuracy of your prediction and whether 160 kg is sensible, using the residuals and the range of the data. |
| 7–8 | As the level before, and you also use the shape of the residuals to criticize the linear model, say what an alternative should show, and justify your recommendation in the context of the harvest. |
| Level | Criterion C in task D1: what it looks like |
|---|---|
| 1–2 | Your working can sometimes be followed, but the layout is hard to read and the notation or words are often missing. |
| 3–4 | Your working is mostly in order, with some correct notation and words, so a reader could follow most of your solution. |
| 5–6 | Your solution is organized by part, uses correct notation and key words, and your note is clear and complete enough for a reader outside statistics. |
| 7–8 | Your solution and note are concise and logical, move between table, graph and words, and every claim in the note is backed by a figure from your working. |
Markscheme
| Task D1 question | Answer and marking |
|---|---|
| 1 | Mean point (60, 3.8); sum of products 60; sum of squares of dx 4000; gradient 60 ÷ 4000 = 0.015 tonnes per kg; intercept 3.8 − 0.015 × 60 = 2.9; line y = 2.9 + 0.015x. Sum of squares of dy 1.04, so r = 60 ÷ √4160 = 0.93. Meaning: every extra kilogram of fertilizer goes with 0.015 t more, which is 0.3 t for each extra 20 kg, over the range tested. |
| 2 | Residuals −0.2, 0.1, 0.2, 0.1, −0.2 (predictions 3.2, 3.5, 3.8, 4.1, 4.4), adding to 0. They are negative, positive, positive, positive, negative: an arch. The yield rises quickly and then flattens, so a straight line fits well only in the middle and over-predicts at the ends. r = 0.93 is strong but not enough: the residual pattern shows a curve. |
| 3 | 2.9 + 0.015 × 70 = 3.95 t per hectare, say about 4 tonnes: inside the range tested, but the data are to one decimal place and the residuals are about 0.2, so a second decimal place claims too much. |
| 4 | 160 kg is well outside the 20 to 100 tested, so it is extrapolation. The residuals already show the yield levelling off, so the line (5.3 t at 160 kg) is likely to overestimate. Trial intermediate amounts such as 120 kg before recommending anything. |
| 5 | 18, 12 and 10 farms (450 ÷ 1000 × 40, 300 ÷ 1000 × 40, 250 ÷ 1000 × 40). Each district appears in proportion; volunteers might be the farmers who are most interested and not typical. |
| 6 (criterion C) | A note that says what to trial (for example 80 to 120 kg, since yields appear to flatten after 80), what the data support (the yield rose with fertilizer from 20 to 100 kg), and what they do not (anything beyond 100 kg, or that fertilizer alone causes the rise, as the plots were not the same in every other way unless that was controlled). |
| How to level | D: the justification of the accuracy and of whether 160 kg makes sense separates 5 to 6 from 7 to 8. C: the note, with every claim supported by a figure from the working, separates the top bands. Calculation slips are assessed in criterion D only when they change the conclusion. |
Exemplar answers
| Task D1 response | What it looks like |
|---|---|
| Level 3 to 4 | “The line is y = 2.9 + 0.015x and r = 0.93, so it is a strong link. At 70 kg the yield is 3.95. At 160 kg it is 5.3 so use 160.” Method and calculations correct for Parts 1 and 3(a); the residuals are not used, the adviser is agreed with, and the note is a single line. |
| Level 7 to 8 | “The residuals are −0.2, 0.1, 0.2, 0.1, −0.2: a shape, not scatter. The line is a fair model between 40 and 80 kg but yields are levelling off, so it will over-predict at the ends. At 70 kg about 4 t is reasonable. At 160 kg the line says 5.3 t but this is far beyond the 100 kg tested and the curve is flattening, so I would not recommend 160 kg: I would trial 80 to 120 kg.” The residual shape is used to criticize the model, and the recommendation follows from it. |
Differentiation
A way in for anyone stuck
- Give the table outline for dx, dy, products and squares, with the means filled in.
- Suggest sketching the five points and the line first, so the arch in the residuals can be seen before it is calculated.
- Give a frame for the note: “I recommend … because …. The data show … but cannot show ….”
Stretch
- The co-operative adds a sixth plot with 120 kg and a yield of 4.1 t. Does the regression line get steeper or flatter, and does the residual pattern change?
- Sketch the curve you think the yield follows from 0 to 200 kg and say what evidence you would need to trust it.
Students learning in English as an additional language
| Word in task D1 | Plain meaning |
|---|---|
| fertilizer | something put on the soil to help plants grow |
| yield | the amount of crop that grows on a piece of land |
| hectare | a unit of land area, 10 000 square metres |
| trial plot | a small piece of land used to test something |
| co-operative | a group of farmers who work together |
Sentence starters for task D1:
- The residuals show …, so the line …
- I do not trust the prediction for 160 kg because …
- I recommend … because …
Task D2: Counting life on a reef
Why set it. Every possible sample, a stratified plan and a regression line inside one ecological survey. An alternative to D1: set one of them.
When: End of unit, after lesson 5. Time: 50 minutes. Format: Individual, in class. A written solution with explanations and a short reply to the survey leader.
The student sheet, with its task-specific clarification
When: End of unit, after lesson 5. Time: 50 minutes. Calculator: gdc optional. Criteria: D, applying mathematics in real-life contexts; C, communicating.
A marine survey team studies a section of coral reef. It has eight equal quadrats (square marked areas): four in the shallow zone and four in the deep zone. A diver can survey only two quadrats on one trip. The number of sea urchins in each quadrat is:
| Zone | Quadrat 1 | Quadrat 2 | Quadrat 3 | Quadrat 4 |
|---|---|---|---|---|
| Shallow | 8 | 10 | 12 | 14 |
| Deep | 2 | 4 | 4 | 6 |
Separately, the team counted the number of coral species at five depths: at 2, 4, 6, 8 and 10 metres they found 14, 11, 9, 5 and 6 species.
Part 1. Planning the dive
- Work out the mean number of sea urchins per quadrat for the whole reef section.
- A trip surveys two quadrats chosen at random. How many different pairs are possible? Find the lowest and the highest mean number of sea urchins that a trip could report, and how many pairs have both quadrats in the same zone.
- A second plan takes one quadrat from each zone. Find the lowest and highest mean it could report. Which plan would you use, and why?
Part 2. Species and depth
- Find the regression line of species on depth and r. Say what the gradient means.
- Work out the residuals. Is a straight line a reasonable model?
Part 3. Using the line
- Predict the number of species at 7 metres. A volunteer wants to use the line to predict the number of species at 15 metres. Explain what goes wrong.
Part 4. A reply
- The survey leader asks whether the team should use the one-quadrat-from-each-zone plan and the line in its report. Write a short reply (about 100 words), with figures from your working.
How your work is judged
| Level | Criterion D in task D2: what it looks like |
|---|---|
| 1–2 | You pick out some of the quantities in the reef survey and carry out a calculation or two with help, but you do not check your answers against the situation. |
| 3–4 | You choose a method for each part (the mean, listing pairs, the regression line), apply them mostly correctly, and say something about whether the results are sensible when prompted. |
| 5–6 | You choose suitable methods, apply them correctly, and justify the plan you prefer and the limits of the line using your figures, including why 15 metres is a problem. |
| 7–8 | As the level before, and you compare the two sampling plans using the full list of results, judge the line using the residuals, and your recommendation to the leader follows from both and says how far each can be trusted. |
| Level | Criterion C in task D2: what it looks like |
|---|---|
| 1–2 | Parts of your reply can be followed, but the order is hard to follow and key figures or words are missing. |
| 3–4 | Your solution is mostly in order, with some correct notation and words, and your reply gives a view with some figures behind it. |
| 5–6 | Your solution is set out by part with correct notation and key words, and your reply is clear, complete and uses figures from the working. |
| 7–8 | Your solution and reply are concise and logically ordered, move between table, graph and words, and every claim in the reply is tied to a specific figure. |
Markscheme
| Task D2 question | Answer and marking |
|---|---|
| 1 | (8 + 10 + 12 + 14 + 2 + 4 + 4 + 6) ÷ 8 = 60 ÷ 8 = 7.5 per quadrat. |
| 2 | 28 pairs (8 × 7 ÷ 2). Lowest mean (2 + 4) ÷ 2 = 3, highest (12 + 14) ÷ 2 = 13. Both quadrats in the same zone: 6 shallow pairs and 6 deep pairs, so 12. Those 12 have means from 9 to 13 or from 3 to 5, so they are far from 7.5. |
| 3 | 16 pairs, lowest (8 + 2) ÷ 2 = 5, highest (14 + 6) ÷ 2 = 10. Every stratified pair has a mean within 2.5 of 7.5, while a random pair can be as far as 5.5 away (a mean of 13). The mean of all 16 results is exactly 7.5, so it centres on the truth, as in lesson 1. The stratified plan is better, because the zones differ a lot. |
| 4 and 5 | Mean point (6, 9); sum of products −44; sum of squares of dx 40; gradient −1.1; intercept 9 + 1.1 × 6 = 15.6; line y = 15.6 − 1.1x. Sum of squares of dy 54, r = −44 ÷ √2160 = −0.95. Residuals 0.6, −0.2, 0, −1.8, 1.4, adding to 0, with no clear pattern: a straight line is reasonable. Meaning: about one fewer species for each extra metre. |
| 6 | At 7 m: 15.6 − 1.1 × 7 = 7.9, about 8 species. At 15 m: 15.6 − 16.5 = −0.9, a negative number of species: 15 m is outside the 2 to 10 m measured, and the line cannot keep going down at that rate. |
| 7 (criterion C) | A reply that recommends the one-from-each-zone plan (range 5 to 10 against 3 to 13), uses the line only between about 2 and 10 m, and says that only five depths were measured so the gradient is approximate. Credit any accurate use of figures. |
| How to level | D: whether the choice of plan is justified from the full list of results, and whether the limits of the line are tied to the range of the data, decides between 5 to 6 and 7 to 8. C: every claim in the reply tied to a figure. Calculation slips are assessed in criterion D only when they change the conclusion. |
Exemplar answers
| Task D2 response | What it looks like |
|---|---|
| Level 3 to 4 | “The mean is 7.5. There are 28 pairs. The lowest is 3 and the highest is 13. For the second plan it is 5 and 10, which is closer, so use the second plan. The line is y = 15.6 − 1.1x. At 7 m it is 7.9. At 15 m it is −0.9 which is wrong.” The calculations are right, but the residuals are not worked out, the number of same-zone pairs is missing, and the reply gives a preference without saying what it rests on. |
| Level 7 to 8 | “Of the 28 random pairs, 12 have both quadrats in one zone and their means are 3 to 5 or 9 to 13, far from the true mean of 7.5, and a random trip can report as low as 3 or as high as 13. With one quadrat from each zone, all 16 results are between 5 and 10 and average 7.5. So I recommend the stratified plan. The line y = 15.6 − 1.1x has r = −0.95 and residuals with no pattern, so it is a fair model from 2 to 10 m. At 15 m it predicts −0.9 species, which is impossible, so the report should state the line for 2 to 10 m only.” |
Differentiation
A way in for anyone stuck
- Give a grid of the eight quadrats with the pairs to cross off, so the 28 pairs can be listed systematically.
- Give the table outline for dx, dy, products and squares for Part 2, with the means filled in.
- Give a frame for the reply: “I recommend … because the results of the plan range from … to …. The line holds between … and ….”
Stretch
- Find the mean of all 28 random results. How does it compare with the mean of all 16 stratified results, and why?
- A fifth depth of 12 m is measured and 3 species are found. Would the line need to change? Work out its residual from the original line before deciding.
Students learning in English as an additional language
| Word in task D2 | Plain meaning |
|---|---|
| reef | a long line of rock or coral near the surface of the sea |
| quadrat | a square area marked out to count the living things in it |
| zone | one part of a larger area |
| species | one kind of living thing |
| depth | how far down from the surface |
Sentence starters for task D2:
- The pairs from the same zone give means of …, which are …
- I prefer the plan that … because …
- The line should only be used between … and … because …
Practicalities
Every task is original and written for this unit; none is copied or adapted from a published book.
The student sheets, with their task-specific clarifications, are public on the unit page so students know what each level looks like before they start. Markschemes and exemplars are only on this page.
To print a sheet, open it on the unit page, then print the page. A1 and C1 need no calculator. The others say “GDC optional”: every number can be found by hand, and a GDC may be used to check or to save time.
The sheets are public, so a student may have seen A1 and A2. For an unseen A1 use this parallel version (same marks, same markscheme notes). Population of lateness 1, 3, 5, 7 and 14 minutes (mean 6). Q1 all ten samples of size 2 (means 2, 3, 4, 7.5, 4, 5, 8.5, 6, 9.5, 10.5), lowest 2 and highest 10.5, five samples within 2 minutes of 6, and for size 3 the lowest is 3 and the highest 8.67, a narrower spread than 8.5. Q2 800 adult and 200 child tickets, sample 50: 40 and 10. Q3 x = 1, 3, 5, 7 and y = 3, 7, 5, 9 (means 4 and 6, products 9, −1, −1, 9, covariance 4, r = 16 ÷ 20 = 0.8). Q4 y in seconds, multiplied by 60: covariance 240, r still 0.8. Q5 x = −3, −1, 1, 3 and y = 9, 1, 1, 9 (covariance 0, a U shape). Q6 as written, with covariances 45 and 0.3.
For an unseen A2 use this parallel version. Q1 population 2, 4, 6, 8 (mean 5): size 2 means 3, 4, 5, 5, 6, 7 (range 4), size 3 means 4, 4.67, 5.33, 6 (range 2), size 4 one sample. Q2 360, 240 and 200 of 800, sample 40: 18, 12, 10. Q3 distance from the coast x = 2, 4, 6, 8, 10 km and bird species y = 15, 13, 6, 9, 7: mean point (6, 10), products −20, −6, 0, −2, −12, total −40, covariance −8, r = −40 ÷ √2400 = −0.82, line y = 16 − x, 11 at 5 km, −4 at 20 km. Q4 residuals 1, 1, −4, 1, 1. Q5 one dip with no smooth curve: ask for judgement from r and residuals. Q6 the point (6, 10): sums unchanged, covariance −40 ÷ 6 = −6.67. Q7 as written.
The Year 5 descriptor bands on this page are band summaries written for this site; check them against the current IB MYP Mathematics guide before relying on them for reporting.