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Worked solutions

IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2

Session: May 2024
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 2024 TZ2 paper, which your school or the IB can provide.

Practice mode

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Questions

Questions 1–2 · 55 marks

Question 1

24 marks

Darts averages: quartiles, a paired t-test, error types and a Wilcoxon test

Statistics and probability

(a)

3 marks

The Year 2 values in order are $65.0, 68.0, 68.0, 70.1, 70.6, 72.0, 74.9, 92.9$ .

a = \frac{70.1 + 70.6}{2} = 70.35 b = \frac{72.0 + 74.9}{2} = 73.45 c = 73.45 - 68 = 5.45

(b)

1 mark

Year 2 has a higher median (70.35 compared with 68.65) and higher quartiles, so the typical player has improved. The spread is about the same, slightly smaller (IQR 5.45 compared with 5.65).

(c)

1 mark

The differences in the means (Year 2 minus Year 1) must be normally distributed (at least approximately).

(d)(i)

2 marks

Let $μ_{d}$ be the mean of the Year 2 minus Year 1 differences.

H_{0} : μ_{d} = 0, H_{1} : μ_{d} > 0

(d)(ii)

2 marks

Enter the eight differences in a one-tailed paired $t$ -test.

p \approx 0.134

(d)(iii)

2 marks

Since $0.134 > 0.10$ there is not enough evidence to reject $H_{0}$ . The data do not show that the players' averages have increased.

(e)(i)

2 marks

A Type I error is rejecting $H_{0}$ when it is true. It can be made less likely by lowering the significance level (for example from 10% to 5%).

(e)(ii)

2 marks

A Type II error is not rejecting $H_{0}$ when it is false. It can be made less likely by increasing the sample size (or by raising the significance level).

(f)(i)

3 marks

Rank the absolute differences $0.2, 0.8, 1.2, 1.4, 1.8, 2.0, 3.5, 28$ from 1 to 8.
Check: 0.2 is rank 1, 0.8 is rank 2, 1.2 is rank 3, 1.4 is rank 4, 1.8 is rank 5, 2.0 is rank 6, 3.5 is rank 7, 28 is rank 8.

A = 6, B = 5, C = 7, D = 4

(f)(ii)

1 mark

The negative differences are $- 0.2$ , $- 0.8$ and $- 1.8$ , with ranks 1, 2 and 5.

\sum N = 1 + 2 + 5 = 8

(g)(i)

1 mark

$H_{1}$ : the population median of the three dart average increased from Year 1 to Year 2.

(g)(ii)

1 mark

The positive ranks add to $6 + 3 + 7 + 4 + 8 = 28$ and the negative ranks add to 8. The test statistic is the smaller.

T = 8

(g)(iii)

2 marks

The critical region is $T \leq 5$ and $8 > 5$ , so $T$ is not in the critical region. Do not reject $H_{0}$ . There is not enough evidence that the median of the players' averages increased.

(h)

1 mark

She could repeat the tests with a different sample of players, or repeat the tests the next year, and check whether she gets a similar result.

Question 2

31 marks

A predator-prey model: equilibrium points, linearising and a phase portrait

Calculus

(a)(i)

2 marks

Substitute $x = 5$ and $y = 5$ into $\frac{d x}{d t} = 2 x - \frac{x y}{2}$ .

\frac{d x}{d t} = 10 - \frac{25}{2} = - 2.5 (hundreds of geese per year)

(a)(ii)

1 mark

Substitute into $\frac{d y}{d t} = - 3 y + x y$ .

\frac{d y}{d t} = - 15 + 25 = 10 (hundreds of foxes per year)

(b)(i)

1 mark

At $(0, 0)$ both $\frac{d x}{d t}$ and $\frac{d y}{d t}$ are 0, so the populations do not change.

(b)(ii)

3 marks

Set both rates to zero with $x, y \neq = 0$ .

2 x - \frac{x y}{2} = 0 \Rightarrow y = 4, - 3 y + x y = 0 \Rightarrow x = 3 p = 3, q = 4

(c)(i)

4 marks

Separate the variables in $\frac{d x}{d t} = 2 x$ .

\int \frac{1}{x} d x = \int 2 d t \Rightarrow ln x = 2 t + c x = A e^{2 t}

(c)(ii)

1 mark

The same method for $\frac{d y}{d t} = - 3 y$ .

y = B e^{- 3 t}

(d)(i)

3 marks

Cube $x$ and square $y$ .
This is a constant, so $x^{3} y^{2} = k$ with $k = A^{3} B^{2} > 0$ .

x^{3} y^{2} = A^{3} e^{6 t} B^{2} e^{- 6 t} = A^{3} B^{2}

(d)(ii)

3 marks

The curve is $y = \frac{k}{x ^{3}}$ , which is decreasing and convex, close to the $y$ -axis for small $x$ and the $x$ -axis for large $x$ , without touching either.
Sketch: one such curve in the first quadrant, with an arrow along the curve in the direction of increasing $x$ and decreasing $y$ (since $x$ grows and $y$ shrinks with time).

(e)

3 marks

Substitute $x = 3 + X$ and $y = 4 + Y$ into $\frac{d x}{d t} = 2 x - \frac{x y}{2}$ , so $\dot{X} = \frac{d x}{d t}$ .
For small $X$ and $Y$ the term $X Y$ is much smaller than the others and is ignored.

\dot{X} = 2 (3 + X) - \frac{1}{2} (3 + X) (4 + Y) = 6 + 2 X - \frac{1}{2} (12 + 3 Y + 4 X + X Y) \dot{X} = 6 + 2 X - 6 - \frac{3}{2} Y - 2 X - \frac{1}{2} X Y = - \frac{3}{2} Y - \frac{1}{2} X Y \dot{X} \approx - \frac{3}{2} Y

(f)

1 mark

Write the two linear approximations as a matrix equation.

(\dot{X} \dot{Y}) = (04 - \frac{3}{2} 0) (X Y), M = (04 - \frac{3}{2} 0)

(g)

4 marks

Find the eigenvalues from $det (M - λ I) = 0$ .
The eigenvalues are purely imaginary, so near $B$ the trajectories are closed loops (ellipses) around $B$ . The populations of geese and foxes go round the loops repeatedly, anticlockwise, without settling down or dying out.

λ^{2} + 6 = 0 \Rightarrow λ = \pm i 6

(h)

3 marks

Sketch: the equilibrium points $A (0, 0)$ and $B (3, 4)$ are marked. Near $A$ the trajectories come in along the $y$ -axis and leave along the $x$ -axis, as in part (d)(ii). Around $B$ there is a family of closed loops, all in the first quadrant, with arrows going anticlockwise.

(i)

2 marks

The keeper's assumption is not supported. At $(5, 5)$ the geese are decreasing ( $\overset{x}{˙} = - 2.5$ ) but the foxes increase quickly ( $\overset{y}{˙} = 10$ ). The trajectory through $(5, 5)$ is one of the closed loops around $B$ , so the goose population falls for a while and then recovers, and it does not die out.

Our analysis

Where the marks are

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

Calculus31 marks · 56%
Q2
Statistics and probability24 marks · 44%
Q1

Mentor insights

Traps to watch for

01
Quartiles · Q1
For 8 values, take the median of the lower four and the median of the upper four.
02
Paired t-test · Q1
It tests the mean of the differences, and needs the differences to be roughly normally distributed.
03
Equilibrium · Q2
Both rates of change are zero.
04
Linearisation · Q2
Near an equilibrium, substitute $x = p + X$ , $y = q + Y$ and ignore the product of the small terms $X Y$ .
05
Eigenvalues of the linear system · Q2
Purely imaginary eigenvalues give closed loops around the equilibrium point.

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

TZ1TZ2

Session: May 2024
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 2024 TZ2 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

24 marks

Darts averages: quartiles, a paired t-test, error types and a Wilcoxon test

Statistics and probability

(a)

3 marks

The Year 2 values in order are $65.0, 68.0, 68.0, 70.1, 70.6, 72.0, 74.9, 92.9$ .

a = \frac{70.1 + 70.6}{2} = 70.35 b = \frac{72.0 + 74.9}{2} = 73.45 c = 73.45 - 68 = 5.45

(b)

1 mark

Year 2 has a higher median (70.35 compared with 68.65) and higher quartiles, so the typical player has improved. The spread is about the same, slightly smaller (IQR 5.45 compared with 5.65).

(c)

1 mark

The differences in the means (Year 2 minus Year 1) must be normally distributed (at least approximately).

(d)(i)

2 marks

Let $μ_{d}$ be the mean of the Year 2 minus Year 1 differences.

H_{0} : μ_{d} = 0, H_{1} : μ_{d} > 0

(d)(ii)

2 marks

Enter the eight differences in a one-tailed paired $t$ -test.

p \approx 0.134

(d)(iii)

2 marks

Since $0.134 > 0.10$ there is not enough evidence to reject $H_{0}$ . The data do not show that the players' averages have increased.

(e)(i)

2 marks

A Type I error is rejecting $H_{0}$ when it is true. It can be made less likely by lowering the significance level (for example from 10% to 5%).

(e)(ii)

2 marks

A Type II error is not rejecting $H_{0}$ when it is false. It can be made less likely by increasing the sample size (or by raising the significance level).

(f)(i)

3 marks

Rank the absolute differences $0.2, 0.8, 1.2, 1.4, 1.8, 2.0, 3.5, 28$ from 1 to 8.
Check: 0.2 is rank 1, 0.8 is rank 2, 1.2 is rank 3, 1.4 is rank 4, 1.8 is rank 5, 2.0 is rank 6, 3.5 is rank 7, 28 is rank 8.

A = 6, B = 5, C = 7, D = 4

(f)(ii)

1 mark

The negative differences are $- 0.2$ , $- 0.8$ and $- 1.8$ , with ranks 1, 2 and 5.

\sum N = 1 + 2 + 5 = 8

(g)(i)

1 mark

$H_{1}$ : the population median of the three dart average increased from Year 1 to Year 2.

(g)(ii)

1 mark

The positive ranks add to $6 + 3 + 7 + 4 + 8 = 28$ and the negative ranks add to 8. The test statistic is the smaller.

T = 8

(g)(iii)

2 marks

The critical region is $T \leq 5$ and $8 > 5$ , so $T$ is not in the critical region. Do not reject $H_{0}$ . There is not enough evidence that the median of the players' averages increased.

(h)

1 mark

She could repeat the tests with a different sample of players, or repeat the tests the next year, and check whether she gets a similar result.

Question 2

31 marks

A predator-prey model: equilibrium points, linearising and a phase portrait

Calculus

(a)(i)

2 marks

Substitute $x = 5$ and $y = 5$ into $\frac{d x}{d t} = 2 x - \frac{x y}{2}$ .

\frac{d x}{d t} = 10 - \frac{25}{2} = - 2.5 (hundreds of geese per year)

(a)(ii)

1 mark

Substitute into $\frac{d y}{d t} = - 3 y + x y$ .

\frac{d y}{d t} = - 15 + 25 = 10 (hundreds of foxes per year)

(b)(i)

1 mark

At $(0, 0)$ both $\frac{d x}{d t}$ and $\frac{d y}{d t}$ are 0, so the populations do not change.

(b)(ii)

3 marks

Set both rates to zero with $x, y \neq = 0$ .

2 x - \frac{x y}{2} = 0 \Rightarrow y = 4, - 3 y + x y = 0 \Rightarrow x = 3 p = 3, q = 4

(c)(i)

4 marks

Separate the variables in $\frac{d x}{d t} = 2 x$ .

\int \frac{1}{x} d x = \int 2 d t \Rightarrow ln x = 2 t + c x = A e^{2 t}

(c)(ii)

1 mark

The same method for $\frac{d y}{d t} = - 3 y$ .

y = B e^{- 3 t}

(d)(i)

3 marks

Cube $x$ and square $y$ .
This is a constant, so $x^{3} y^{2} = k$ with $k = A^{3} B^{2} > 0$ .

x^{3} y^{2} = A^{3} e^{6 t} B^{2} e^{- 6 t} = A^{3} B^{2}

(d)(ii)

3 marks

The curve is $y = \frac{k}{x ^{3}}$ , which is decreasing and convex, close to the $y$ -axis for small $x$ and the $x$ -axis for large $x$ , without touching either.
Sketch: one such curve in the first quadrant, with an arrow along the curve in the direction of increasing $x$ and decreasing $y$ (since $x$ grows and $y$ shrinks with time).

(e)

3 marks

Substitute $x = 3 + X$ and $y = 4 + Y$ into $\frac{d x}{d t} = 2 x - \frac{x y}{2}$ , so $\dot{X} = \frac{d x}{d t}$ .
For small $X$ and $Y$ the term $X Y$ is much smaller than the others and is ignored.

\dot{X} = 2 (3 + X) - \frac{1}{2} (3 + X) (4 + Y) = 6 + 2 X - \frac{1}{2} (12 + 3 Y + 4 X + X Y) \dot{X} = 6 + 2 X - 6 - \frac{3}{2} Y - 2 X - \frac{1}{2} X Y = - \frac{3}{2} Y - \frac{1}{2} X Y \dot{X} \approx - \frac{3}{2} Y

(f)

1 mark

Write the two linear approximations as a matrix equation.

(\dot{X} \dot{Y}) = (04 - \frac{3}{2} 0) (X Y), M = (04 - \frac{3}{2} 0)

(g)

4 marks

Find the eigenvalues from $det (M - λ I) = 0$ .
The eigenvalues are purely imaginary, so near $B$ the trajectories are closed loops (ellipses) around $B$ . The populations of geese and foxes go round the loops repeatedly, anticlockwise, without settling down or dying out.

λ^{2} + 6 = 0 \Rightarrow λ = \pm i 6

(h)

3 marks

(i)

2 marks

Our analysis

Where the marks are

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

Calculus31 marks · 56%
Q2
Statistics and probability24 marks · 44%
Q1

Mentor insights

Traps to watch for

01
Quartiles · Q1
For 8 values, take the median of the lower four and the median of the upper four.
02
Paired t-test · Q1
It tests the mean of the differences, and needs the differences to be roughly normally distributed.
03
Equilibrium · Q2
Both rates of change are zero.
04
Linearisation · Q2
Near an equilibrium, substitute $x = p + X$ , $y = q + Y$ and ignore the product of the small terms $X Y$ .
05
Eigenvalues of the linear system · Q2
Purely imaginary eigenvalues give closed loops around the equilibrium point.

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

Want someone to go through your paper with you?

Our maths mentors took these exams recently. One-to-one, they'll find where your marks are going and fix it before May.

Find your maths mentor More past papers