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IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

25 marks

Referee errors: chi-squared, binomial and paired t-tests

Statistics and probability

(a)

2 marks

Pick the first team at random from the alphabetical list, for example with a random number from 1 to 7.
From there, take every $k$ th team down the list ( $30 \div 4 = 7.5$ , so take $k = 7$ ), wrapping round to the top if you run off the end, until you have four teams.

\frac{30}{4} = 7.5 ⟹ k = 7 e.g. start at team 3 : 3, 10, 17, 24

Mentor tip

Two ideas are needed: a random start, then a fixed gap. Describing only one of them leaves the method incomplete.

(b)(i)

2 marks

If the calls are spread evenly, the 156 calls share equally between the six 20-second intervals.

E = \frac{156}{6} = 26

(b)(ii)

1 mark

Six categories with one constraint (the totals must match), so subtract 1.

ν = 6 - 1 = 5

(b)(iii)

2 marks

Put the observed counts in one list and 26 six times in another, then run the goodness of fit test on the GDC with 5 degrees of freedom.

χ_{c a l c}^{2} = \sum \frac{( O - E ) ^{2}}{E} = \frac{1 5 ^{2} + 5 ^{2} + 3 ^{2} + 4 ^{2} + 6 ^{2} + 7 ^{2}}{26} = \frac{360}{26} = 13.8 (3 s.f.)

Mentor tip

The squared differences add to 360, which gives a quick hand check on the GDC value.

(b)(iv)

2 marks

Compare the test statistic with the critical value 11.1.
It is bigger, so it lies in the rejection region: reject $H_{0}$ . There is evidence that the calls are not spread evenly over the intervals.

13.8 > 11.1 ⟹ reject H_{0}

Mentor tip

You are given the critical value, so compare $χ_{c a l c}^{2}$ with 11.1 and say so. Quoting only the $p$ -value (0.0166) skips the comparison the part is built around.

(b)(v)

1 mark

Look at the pattern in the counts: they rise steadily as $t$ gets smaller (closer to the final buzzer), and the last 20 seconds has far more than any other interval, 41 against an expected 26.

19 < 20 < 22 < 23 < 31 < 41

Mentor tip

The chi-squared test only says the spread is uneven. It says nothing about direction, which is why the trend in the counts is the extra evidence.

(c)(i)

1 mark

A binomial model needs independent trials, so he is assuming each incorrect call is independent of every other call (with the same probability of favouring the home team each time).

X \sim B (n, p) needs independent trials

Mentor tip

Keep it in context: say the calls are independent, not just the trials.

(c)(ii)

3 marks

The number of trials is the total number of calls.
Let $X$ be the number benefitting the home team. Under $H_{0}$ , $X \sim B (1332, 0.5)$ .
The alternative is one-tailed (greater than), so the $p$ -value is the chance of 695 or more. Use the binomial cdf on the GDC.

n = 695 + 637 = 1332 p -value = P (X \geq 695) = 1 - P (X \leq 694) = 0.0592 (GDC)

Mentor tip

The classic slip is $1 - P (X \leq 695)$ , which leaves out 695 itself. For a "greater than or equal to" tail, subtract the cdf at one less.

(c)(iii)

2 marks

Compare the $p$ -value with the 5% significance level.
It is larger, so do not reject $H_{0}$ : there is not enough evidence that incorrect calls favour the home team.

0.0592 > 0.05 ⟹ do not reject H_{0}

Mentor tip

Write "do not reject" rather than claiming $H_{0}$ is true. A $p$ -value just above 5% means weak evidence, not proof that $H_{0}$ is true.

(d)(i)

2 marks

$H_{0}$ : which team (home or away) benefits from an incorrect call is independent of which team is playing at home.
$H_{1}$ : which team benefits from an incorrect call is not independent of (depends on) which team is playing at home.

H_{0} : benefitting team and home stadium are independent H_{1} : they are not independent

Mentor tip

"In context" means naming both variables. "The variables are independent" on its own is too vague.

(d)(ii)

2 marks

Enter the table as a $2 \times 4$ matrix and run the chi-squared test for independence on the GDC. The degrees of freedom are $(2 - 1) (4 - 1) = 3$ .
Compare the $p$ -value with 0.05.

χ_{c a l c}^{2} = 5.86, ν = 3 p -value = 0.119 (GDC) 0.119 > 0.05 ⟹ not significant at 5%

Mentor tip

Check the smallest expected count (about 14.4) is above 5 before trusting the test. It is, so no columns need combining.

(e)

4 marks

Write the hypotheses about the mean difference (home minus away). The test is one-tailed.
Work out the seven differences, home minus away.
Run a one-sample $t$ -test on the differences against 0 (greater than) on the GDC.
Compare the $p$ -value with 0.05 and conclude.

H_{0} : μ_{D} = 0, H_{1} : μ_{D} > 0 d : 1.9, 1.0, 0, - 0.1, - 1.1, 2.9, - 0.1 p -value = 0.131 (GDC) 0.131 > 0.05 ⟹ do not reject H_{0}

Mentor tip

A paired test works on the differences, so do not run a two-sample test on the two rows. Watch the signs: $- 2.3 - (- 1.2) = - 1.1$ , not $- 3.5$ .

(f)

1 mark

Draw on more than one of the home-advantage tests. None of the binomial, independence or paired tests is significant at 5%, so he cannot claim referees favour the home team late in close games.
A fair add-on: the binomial $p$ -value (0.059) is close to 5%, so there may be a small effect worth testing with more data, for example just the final 20 seconds.

0.0592, 0.119, 0.131 all > 0.05

Mentor tip

Repeating one test's conclusion is not enough here. Pull at least two of the results together into one overall judgement.

Question 2

30 marks

Modelling cooling water: quadratic and exponential regression

Functions
Calculus
Statistics and probability

(a)(i)

1 mark

The points fall and then rise again, with a single minimum near $t = 16$ . That parabola-like shape suggests a quadratic.

one turning point (a minimum) \Rightarrow quadratic

(a)(ii)

3 marks

Enter $t$ and $T$ as two lists (all nine points, including $t = 0$ ).
Run quadratic regression on the GDC and round each coefficient to 3 s.f.

T = p t^{2} + q t + r p = 0.0261025 \dots, q = - 1.01236 \dots, r = 16.3757 \dots (GDC) T = 0.0261 t^{2} - 1.01 t + 16.4

Mentor tip

Keep the full stored coefficients if you reuse the model later, and write the final equation with $T$ and $t$ , not $y$ and $x$ .

(b)

2 marks

Once the ice has melted, the water warms back towards room temperature and then levels off at a constant value.
A quadratic with a positive $t^{2}$ term keeps increasing without limit, so it would predict the water getting hotter and hotter.

0.0261 t^{2} \to \infty as t \to \infty

Mentor tip

Make both halves of the argument: what really happens (a plateau near room temperature) and what the model does (grows for ever).

(c)

1 mark

Substitute $t = 0$ and $T = 19$ . The term with $t^{\frac{1}{2}}$ is zero, leaving $c$ .

19 = c - a (0)^{\frac{1}{2}} e^{0} 19 = c - 0 ⟹ c = 19

Mentor tip

On a "show that", start from the data and end at $c = 19$ . Plugging $c = 19$ in to check proves nothing.

(d)(i)

3 marks

The constant 19 differentiates to zero.
Treat $a t^{\frac{1}{2}} e^{- b t}$ as a product and use the product rule. Remember the chain rule gives $\frac{d}{d t} e^{- b t} = - b e^{- b t}$ .

T = 19 - a t^{\frac{1}{2}} e^{- b t} \frac{d T}{d t} = - (\frac{1}{2} a t^{- \frac{1}{2}} e^{- b t} + a t^{\frac{1}{2}} (- b) e^{- b t}) \frac{d T}{d t} = - \frac{1}{2} a t^{- \frac{1}{2}} e^{- b t} + ab t^{\frac{1}{2}} e^{- b t}

Mentor tip

Keep the minus sign in front of the whole product. Losing it flips every term and still leads to the right $b$ , but the derivative itself would be wrong.

(d)(ii)

4 marks

At a minimum the derivative is zero, so set $\frac{d T}{d t} = 0$ at $t = 16$ .
Use $1 6^{- \frac{1}{2}} = \frac{1}{4}$ and $1 6^{\frac{1}{2}} = 4$ .
Divide through by $a e^{- 16 b}$ , which is never zero because $a > 0$ .
Solve the simple equation that remains for $b$ .

- \frac{1}{2} a (\frac{1}{4}) e^{- 16 b} + ab (4) e^{- 16 b} = 0 a e^{- 16 b} (- \frac{1}{8} + 4 b) = 0 4 b = \frac{1}{8} ⟹ b = \frac{1}{32}

Mentor tip

Say why you may divide by $a e^{- 16 b}$ : an exponential is always positive and $a$ is positive. Do not substitute $b = \frac{1}{32}$ to check, as that runs the argument backwards.

(d)(iii)

2 marks

Substitute $t = 16$ , $T = 7.2$ and $b = \frac{1}{32}$ into the model with $c = 19$ , then solve for $a$ .

7.2 = 19 - a (16)^{\frac{1}{2}} e^{- \frac{16}{32}} 4 a e^{- 0.5} = 11.8 a = \frac{11.8}{4 e ^{- 0.5}} = 4.86

(d)(iv)

1 mark

Put $c = 19$ , $a = 4.86$ and $b = \frac{1}{32}$ into the model.

T = 19 - 4.86 t^{\frac{1}{2}} e^{- \frac{t}{32}}

(e)

1 mark

Use the definition of $x$ with $t = 16$ and $T = 7.2$ .

q = \frac{19 - 7.2}{1 6 ^{\frac{1}{2}}} = \frac{11.8}{4} = 2.95

(f)(i)

3 marks

Make a new list of $ln x$ values from the table (with $q = 2.95$ ). The $t = 0$ point is left out because $x$ is undefined there.
Run linear regression with $t$ as the independent variable and $ln x$ as the dependent one.

ln x : 1.482, 1.330, 1.182, 1.082, 0.900, 0.779, 0.678, 0.519 ln x = - 0.0338898 \dots t + 1.60391 \dots (GDC) ln x = - 0.0339 t + 1.60

Mentor tip

" $ln x$ on $t$ " means $ln x$ is the $y$ -variable. Swapping the lists gives a different line altogether.

(f)(ii)

4 marks

Take logs of $x = a e^{- b t}$ to get $ln x = ln a - b t$ , a straight line in $t$ .
Match it with the regression line: the intercept is $ln a$ and the gradient is $- b$ .
Exponentiate the intercept for $a$ , using the unrounded value.

ln x = ln a - b t ln a = 1.60391 \dots ⟹ a = e^{1.60391 \dots} = 4.97 - b = - 0.0338898 \dots ⟹ b = 0.0339

Mentor tip

Use the stored intercept, not 1.60: $e^{1.60}$ gives 4.95, a slightly different $a$ . And $b$ is positive, since the minus sign is already built into the model.

(f)(iii)

1 mark

Put $c = 19$ and the new $a$ and $b$ into the model.

T = 19 - 4.97 t^{\frac{1}{2}} e^{- 0.0339 t}

(g)(i)

2 marks

The quadratic is the only one whose value at $t = 0$ is about 16.4. The other two models both give exactly 19 there.
It is also the only symmetric, parabola-shaped curve that keeps climbing faster and faster after its minimum. So the quadratic is graph A.

T (0) = 16.4 (quadratic) vs T (0) = 19 (other two) Graph A

Mentor tip

The vertical intercept is the fastest check here. Always compare the models at an easy value like $t = 0$ .

(g)(ii)

2 marks

The model from (d) was forced through the point $(16, 7.2)$ , so that is the curve that passes through that data point.
The model from (f) comes from least squares regression, so it balances the fit over all the points rather than hitting one exactly. That leaves the curve that lies closest to the data overall: graph B.

(d): through (16, 7.2) \Rightarrow C (f): least squares fit \Rightarrow Graph B

Mentor tip

"It fits best" alone is thin. Tie your reason to how each model was built: one uses a single point, the other minimises the squared residuals.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Statistics and probability carries the most marks on this paper.

Statistics and probability30 marks · 55%
Q1, Q2
Functions18 marks · 33%
Q2
Calculus7 marks · 13%
Q2

Mentor insights

Traps to watch for

01
Binomial tail: P(X ≥ 695), not 1 − P(X ≤ 695) · Q1
For an upper-tail $p$ -value you need $1 - P (X \leq 694)$ . Subtracting the cdf at 695 drops the observed value and gives the wrong $p$ -value.
02
Paired test on the differences · Q1
Work out home minus away for each opponent and run a one-sample $t$ -test on those seven values. A two-sample test on the two rows ignores the pairing.
03
Treating H0 as proven · Q1
A $p$ -value above 5% means there is not enough evidence to reject $H_{0}$ . It never proves $H_{0}$ is true.
04
Losing the minus sign in dT/dt · Q2
The model is $19$ minus a product, so the whole product-rule result must be negated. The chain rule on $e^{- b t}$ adds a factor of $- b$ .
05
Rounding ln a before exponentiating · Q2
Using the rounded intercept 1.60 gives $a = 4.95$ instead of 4.97. Keep the stored value until the final answer.

Solutions and tips are GeorgeTheTutor's own; the questions are not reproduced here. Not affiliated with or endorsed by the International Baccalaureate Organization.

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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1 TZ2TZ3

Session: May 2026
Marks: 55
Duration: 1 hour 15 min
Calculator: Allowed
Questions: 2

Q1 Q2

Keep your copy of the paper open alongside. We don't reproduce the questions: each part below is labelled to match the May 2026 TZ3 paper, which your school or the IB can provide.

Practice mode

Hide our working until you've had a go.

Questions

Questions 1–2 · 55 marks

Question 1

25 marks

Referee errors: chi-squared, binomial and paired t-tests

Statistics and probability

(a)

2 marks

Pick the first team at random from the alphabetical list, for example with a random number from 1 to 7.
From there, take every $k$ th team down the list ( $30 \div 4 = 7.5$ , so take $k = 7$ ), wrapping round to the top if you run off the end, until you have four teams.

\frac{30}{4} = 7.5 ⟹ k = 7 e.g. start at team 3 : 3, 10, 17, 24

Mentor tip

Two ideas are needed: a random start, then a fixed gap. Describing only one of them leaves the method incomplete.

(b)(i)

2 marks

If the calls are spread evenly, the 156 calls share equally between the six 20-second intervals.

E = \frac{156}{6} = 26

(b)(ii)

1 mark

Six categories with one constraint (the totals must match), so subtract 1.

ν = 6 - 1 = 5

(b)(iii)

2 marks

Put the observed counts in one list and 26 six times in another, then run the goodness of fit test on the GDC with 5 degrees of freedom.

χ_{c a l c}^{2} = \sum \frac{( O - E ) ^{2}}{E} = \frac{1 5 ^{2} + 5 ^{2} + 3 ^{2} + 4 ^{2} + 6 ^{2} + 7 ^{2}}{26} = \frac{360}{26} = 13.8 (3 s.f.)

Mentor tip

The squared differences add to 360, which gives a quick hand check on the GDC value.

(b)(iv)

2 marks

Compare the test statistic with the critical value 11.1.
It is bigger, so it lies in the rejection region: reject $H_{0}$ . There is evidence that the calls are not spread evenly over the intervals.

13.8 > 11.1 ⟹ reject H_{0}

Mentor tip

You are given the critical value, so compare $χ_{c a l c}^{2}$ with 11.1 and say so. Quoting only the $p$ -value (0.0166) skips the comparison the part is built around.

(b)(v)

1 mark

Look at the pattern in the counts: they rise steadily as $t$ gets smaller (closer to the final buzzer), and the last 20 seconds has far more than any other interval, 41 against an expected 26.

19 < 20 < 22 < 23 < 31 < 41

Mentor tip

The chi-squared test only says the spread is uneven. It says nothing about direction, which is why the trend in the counts is the extra evidence.

(c)(i)

1 mark

A binomial model needs independent trials, so he is assuming each incorrect call is independent of every other call (with the same probability of favouring the home team each time).

X \sim B (n, p) needs independent trials

Mentor tip

Keep it in context: say the calls are independent, not just the trials.

(c)(ii)

3 marks

The number of trials is the total number of calls.
Let $X$ be the number benefitting the home team. Under $H_{0}$ , $X \sim B (1332, 0.5)$ .
The alternative is one-tailed (greater than), so the $p$ -value is the chance of 695 or more. Use the binomial cdf on the GDC.

n = 695 + 637 = 1332 p -value = P (X \geq 695) = 1 - P (X \leq 694) = 0.0592 (GDC)

Mentor tip

The classic slip is $1 - P (X \leq 695)$ , which leaves out 695 itself. For a "greater than or equal to" tail, subtract the cdf at one less.

(c)(iii)

2 marks

Compare the $p$ -value with the 5% significance level.
It is larger, so do not reject $H_{0}$ : there is not enough evidence that incorrect calls favour the home team.

0.0592 > 0.05 ⟹ do not reject H_{0}

Mentor tip

Write "do not reject" rather than claiming $H_{0}$ is true. A $p$ -value just above 5% means weak evidence, not proof that $H_{0}$ is true.

(d)(i)

2 marks

$H_{0}$ : which team (home or away) benefits from an incorrect call is independent of which team is playing at home.
$H_{1}$ : which team benefits from an incorrect call is not independent of (depends on) which team is playing at home.

H_{0} : benefitting team and home stadium are independent H_{1} : they are not independent

Mentor tip

"In context" means naming both variables. "The variables are independent" on its own is too vague.

(d)(ii)

2 marks

Enter the table as a $2 \times 4$ matrix and run the chi-squared test for independence on the GDC. The degrees of freedom are $(2 - 1) (4 - 1) = 3$ .
Compare the $p$ -value with 0.05.

χ_{c a l c}^{2} = 5.86, ν = 3 p -value = 0.119 (GDC) 0.119 > 0.05 ⟹ not significant at 5%

Mentor tip

Check the smallest expected count (about 14.4) is above 5 before trusting the test. It is, so no columns need combining.

(e)

4 marks

Write the hypotheses about the mean difference (home minus away). The test is one-tailed.
Work out the seven differences, home minus away.
Run a one-sample $t$ -test on the differences against 0 (greater than) on the GDC.
Compare the $p$ -value with 0.05 and conclude.

H_{0} : μ_{D} = 0, H_{1} : μ_{D} > 0 d : 1.9, 1.0, 0, - 0.1, - 1.1, 2.9, - 0.1 p -value = 0.131 (GDC) 0.131 > 0.05 ⟹ do not reject H_{0}

Mentor tip

A paired test works on the differences, so do not run a two-sample test on the two rows. Watch the signs: $- 2.3 - (- 1.2) = - 1.1$ , not $- 3.5$ .

(f)

1 mark

Draw on more than one of the home-advantage tests. None of the binomial, independence or paired tests is significant at 5%, so he cannot claim referees favour the home team late in close games.
A fair add-on: the binomial $p$ -value (0.059) is close to 5%, so there may be a small effect worth testing with more data, for example just the final 20 seconds.

0.0592, 0.119, 0.131 all > 0.05

Mentor tip

Repeating one test's conclusion is not enough here. Pull at least two of the results together into one overall judgement.

Question 2

30 marks

Modelling cooling water: quadratic and exponential regression

Functions
Calculus
Statistics and probability

(a)(i)

1 mark

The points fall and then rise again, with a single minimum near $t = 16$ . That parabola-like shape suggests a quadratic.

one turning point (a minimum) \Rightarrow quadratic

(a)(ii)

3 marks

Enter $t$ and $T$ as two lists (all nine points, including $t = 0$ ).
Run quadratic regression on the GDC and round each coefficient to 3 s.f.

T = p t^{2} + q t + r p = 0.0261025 \dots, q = - 1.01236 \dots, r = 16.3757 \dots (GDC) T = 0.0261 t^{2} - 1.01 t + 16.4

Mentor tip

Keep the full stored coefficients if you reuse the model later, and write the final equation with $T$ and $t$ , not $y$ and $x$ .

(b)

2 marks

Once the ice has melted, the water warms back towards room temperature and then levels off at a constant value.
A quadratic with a positive $t^{2}$ term keeps increasing without limit, so it would predict the water getting hotter and hotter.

0.0261 t^{2} \to \infty as t \to \infty

Mentor tip

Make both halves of the argument: what really happens (a plateau near room temperature) and what the model does (grows for ever).

(c)

1 mark

Substitute $t = 0$ and $T = 19$ . The term with $t^{\frac{1}{2}}$ is zero, leaving $c$ .

19 = c - a (0)^{\frac{1}{2}} e^{0} 19 = c - 0 ⟹ c = 19

Mentor tip

On a "show that", start from the data and end at $c = 19$ . Plugging $c = 19$ in to check proves nothing.

(d)(i)

3 marks

The constant 19 differentiates to zero.
Treat $a t^{\frac{1}{2}} e^{- b t}$ as a product and use the product rule. Remember the chain rule gives $\frac{d}{d t} e^{- b t} = - b e^{- b t}$ .

T = 19 - a t^{\frac{1}{2}} e^{- b t} \frac{d T}{d t} = - (\frac{1}{2} a t^{- \frac{1}{2}} e^{- b t} + a t^{\frac{1}{2}} (- b) e^{- b t}) \frac{d T}{d t} = - \frac{1}{2} a t^{- \frac{1}{2}} e^{- b t} + ab t^{\frac{1}{2}} e^{- b t}

Mentor tip

Keep the minus sign in front of the whole product. Losing it flips every term and still leads to the right $b$ , but the derivative itself would be wrong.

(d)(ii)

4 marks

At a minimum the derivative is zero, so set $\frac{d T}{d t} = 0$ at $t = 16$ .
Use $1 6^{- \frac{1}{2}} = \frac{1}{4}$ and $1 6^{\frac{1}{2}} = 4$ .
Divide through by $a e^{- 16 b}$ , which is never zero because $a > 0$ .
Solve the simple equation that remains for $b$ .

- \frac{1}{2} a (\frac{1}{4}) e^{- 16 b} + ab (4) e^{- 16 b} = 0 a e^{- 16 b} (- \frac{1}{8} + 4 b) = 0 4 b = \frac{1}{8} ⟹ b = \frac{1}{32}

Mentor tip

Say why you may divide by $a e^{- 16 b}$ : an exponential is always positive and $a$ is positive. Do not substitute $b = \frac{1}{32}$ to check, as that runs the argument backwards.

(d)(iii)

2 marks

Substitute $t = 16$ , $T = 7.2$ and $b = \frac{1}{32}$ into the model with $c = 19$ , then solve for $a$ .

7.2 = 19 - a (16)^{\frac{1}{2}} e^{- \frac{16}{32}} 4 a e^{- 0.5} = 11.8 a = \frac{11.8}{4 e ^{- 0.5}} = 4.86

(d)(iv)

1 mark

Put $c = 19$ , $a = 4.86$ and $b = \frac{1}{32}$ into the model.

T = 19 - 4.86 t^{\frac{1}{2}} e^{- \frac{t}{32}}

(e)

1 mark

Use the definition of $x$ with $t = 16$ and $T = 7.2$ .

q = \frac{19 - 7.2}{1 6 ^{\frac{1}{2}}} = \frac{11.8}{4} = 2.95

(f)(i)

3 marks

Make a new list of $ln x$ values from the table (with $q = 2.95$ ). The $t = 0$ point is left out because $x$ is undefined there.
Run linear regression with $t$ as the independent variable and $ln x$ as the dependent one.

ln x : 1.482, 1.330, 1.182, 1.082, 0.900, 0.779, 0.678, 0.519 ln x = - 0.0338898 \dots t + 1.60391 \dots (GDC) ln x = - 0.0339 t + 1.60

Mentor tip

" $ln x$ on $t$ " means $ln x$ is the $y$ -variable. Swapping the lists gives a different line altogether.

(f)(ii)

4 marks

Take logs of $x = a e^{- b t}$ to get $ln x = ln a - b t$ , a straight line in $t$ .
Match it with the regression line: the intercept is $ln a$ and the gradient is $- b$ .
Exponentiate the intercept for $a$ , using the unrounded value.

ln x = ln a - b t ln a = 1.60391 \dots ⟹ a = e^{1.60391 \dots} = 4.97 - b = - 0.0338898 \dots ⟹ b = 0.0339

Mentor tip

Use the stored intercept, not 1.60: $e^{1.60}$ gives 4.95, a slightly different $a$ . And $b$ is positive, since the minus sign is already built into the model.

(f)(iii)

1 mark

Put $c = 19$ and the new $a$ and $b$ into the model.

T = 19 - 4.97 t^{\frac{1}{2}} e^{- 0.0339 t}

(g)(i)

2 marks

The quadratic is the only one whose value at $t = 0$ is about 16.4. The other two models both give exactly 19 there.
It is also the only symmetric, parabola-shaped curve that keeps climbing faster and faster after its minimum. So the quadratic is graph A.

T (0) = 16.4 (quadratic) vs T (0) = 19 (other two) Graph A

Mentor tip

The vertical intercept is the fastest check here. Always compare the models at an easy value like $t = 0$ .

(g)(ii)

2 marks

The model from (d) was forced through the point $(16, 7.2)$ , so that is the curve that passes through that data point.
The model from (f) comes from least squares regression, so it balances the fit over all the points rather than hitting one exactly. That leaves the curve that lies closest to the data overall: graph B.

(d): through (16, 7.2) \Rightarrow C (f): least squares fit \Rightarrow Graph B

Mentor tip

"It fits best" alone is thin. Tie your reason to how each model was built: one uses a single point, the other minimises the squared residuals.

Our analysis

Where the marks are

Each part sorted by the topic it mostly tests. Statistics and probability carries the most marks on this paper.

Statistics and probability30 marks · 55%
Q1, Q2
Functions18 marks · 33%
Q2
Calculus7 marks · 13%
Q2

Mentor insights

Traps to watch for

01
Binomial tail: P(X ≥ 695), not 1 − P(X ≤ 695) · Q1
For an upper-tail $p$ -value you need $1 - P (X \leq 694)$ . Subtracting the cdf at 695 drops the observed value and gives the wrong $p$ -value.
02
Paired test on the differences · Q1
Work out home minus away for each opponent and run a one-sample $t$ -test on those seven values. A two-sample test on the two rows ignores the pairing.
03
Treating H0 as proven · Q1
A $p$ -value above 5% means there is not enough evidence to reject $H_{0}$ . It never proves $H_{0}$ is true.
04
Losing the minus sign in dT/dt · Q2
The model is $19$ minus a product, so the whole product-rule result must be negated. The chain rule on $e^{- b t}$ adds a factor of $- b$ .
05
Rounding ln a before exponentiating · Q2
Using the rounded intercept 1.60 gives $a = 4.95$ instead of 4.97. Keep the stored value until the final answer.

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