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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 TZ1 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

26 marks

Cracked components: a chi-squared test, a binomial test and a two-sample t-test

Statistics and probability

(a)(i)

1 mark

The total number of components with no cracks is 141, and 88 of them were made by the current technique.

a = 141 - 88 = 53

(a)(ii)

1 mark

The 120 components made with the new technique are split between none, minor and major cracks.

b = 120 - 53 - 54 = 13

(b)

2 marks

150 components had minor cracks and 54 of them were made with the new technique.

P = \frac{54}{54 + 96} = \frac{54}{150} = 0.36

(c)(i)

1 mark

$H_{0}$ : the level of cracking (none, minor, major) is independent of the production technique.
$H_{1}$ : the level of cracking is not independent of the production technique.

(c)(ii)

2 marks

Enter the table into the calculator for a $χ^{2}$ test (the smallest expected count is 23, so the test is valid). The degrees of freedom are $(3 - 1) (2 - 1) = 2$ .

χ^{2} \approx 8.15, p \approx 0.0170

(c)(iii)

2 marks

Since $0.0170 < 0.05$ we reject $H_{0}$ . There is sufficient evidence that the amount of cracking depends on the production technique.

(d)

1 mark

For the current technique, $96 + 56 = 152$ of the 240 components developed cracks.

\frac{152}{240} = \frac{19}{30}

(e)

1 mark

The distribution $B (120, \frac{19}{30})$ assumes that the components are independent of each other (whether one cracks does not affect whether another does) and that each has the same probability of cracking.

(f)

5 marks

The number of components with cracks among the 120 new ones is $X = 54 + 13 = 67$ . Under $H_{0}$ , $X \sim B (120, \frac{19}{30})$ , and the alternative is that fewer crack.
$H_{0}$ : $p = \frac{19}{30}$ , $H_{1}$ : $p < \frac{19}{30}$ .
Since $0.0549 > 0.05$ we do not reject $H_{0}$ . There is insufficient evidence at the 5% level that the new technique reduces the proportion of components that develop cracks.

P (X \leq 67) \approx 0.0549

(g)(i)

1 mark

The test in (f) is one-tailed and answers the question that matters here directly: whether the new technique reduces the number of components that crack. The test in (c) only says there is some difference.

(g)(ii)

1 mark

The test in (f) treats the proportion for the current technique (19/30) as an exact known value, but it is only an estimate from a sample, so $H_{0}$ could be wrong. It also lumps minor and major cracks together, losing information that the test in (c) uses.

(h)

7 marks

Let $μ_{1}$ and $μ_{2}$ be the population mean times before cracking for the new and the current technique.
Use a one-tailed two-sample $t$ -test with the sample sizes 100 and 200, means 250.1 and 249.1 and standard deviations 3.8 and 3.8 (pooled, since the standard deviations are equal).
Since $0.0162 < 0.05$ , reject $H_{0}$ . There is sufficient evidence that the new technique increases the mean time before cracks appear.

H_{0} : μ_{1} = μ_{2}, H_{1} : μ_{1} > μ_{2} t \approx 2.15, p \approx 0.0162

(i)

1 mark

Although statistically significant, the mean time has only gone up by about 1 hour out of about 250, a tiny effect. And a company would care about the smallest times before cracking (when parts need replacing) as well as the mean, so the statement overstates what the test shows.

Question 2

29 marks

A T-shirt cannon: projectile motion and the boundary of all reachable points

Calculus

(a)(i)

2 marks

Each row is 1.0 m further and 0.5 m higher. Row 5 is four rows behind row 1.

A 5 = (30 + 4, 2.1 + 4 \times 0.5) = (34, 4.1)

(a)(ii)

3 marks

Row $n$ is $n - 1$ rows behind row 1.

A_{n} = (30 + (n - 1), 2.1 + 0.5 (n - 1)) = (29 + n, 1.6 + 0.5 n)

(b)(i)

3 marks

Integrate the acceleration $(0, - 9.8)$ and use the initial velocity as the constant.

(\overset{x}{˙} \overset{y}{˙}) = (29.4 cos θ 29.4 sin θ - 9.8 t)

(b)(ii)

3 marks

A vertical launch has $θ = 9 0^{\circ}$ , so $\overset{y}{˙} = 29.4 - 9.8 t$ .
The maximum height occurs when the vertical velocity is zero.

29.4 - 9.8 t = 0 \Rightarrow t = 3 seconds

(c)

2 marks

Use the displacement formula with $θ = 9 0^{\circ}$ and $t = 3$ .

y = 29.4 (3) - 4.9 (3)^{2} = 88.2 - 44.1 = 44.1 m

(d)(i)

3 marks

The T-shirt lands when $y = 0$ .
Substitute into $x = 29.4 cos θ t$ .

29.4 sin θ t - 4.9 t^{2} = 0 \Rightarrow t = 6 sin θ (t \neq = 0) x = 29.4 cos θ \times 6 sin θ = 176.4 sin θ cos θ

(d)(ii)

2 marks

Since $sin θ cos θ = \frac{1}{2} sin 2 θ$ has maximum $\frac{1}{2}$ at $θ = 4 5^{\circ}$ .

x_{m a x} = 176.4 \times \frac{1}{2} = 88.2 m

(e)(i)

1 mark

The vertex $V$ is the highest point reached by any T-shirt, which is the vertical launch.

c = 44.1

(e)(ii)

2 marks

The vertex is on the $y$ -axis, so the axis of symmetry is $x = 0$ .

- \frac{b}{2 a} = 0 \Rightarrow b = 0

(e)(iii)

3 marks

The furthest point reached on the ground is $(88.2, 0)$ , which is on the boundary.

0 = a (88.2)^{2} + 44.1 \Rightarrow a = - \frac{44.1}{88. 2 ^{2}} = - \frac{5}{882} \approx - 0.00567

(f)

5 marks

Seat $A 40$ has $n = 40$ , so it is at
The boundary curve $y = - \frac{5}{882} x^{2} + 44.1$ at $x = 69$ is
The seat is at height 21.6, which is above 17.1, so it is outside the region that any T-shirt can reach. The spectator can never get a T-shirt.

(29 + 40, 1.6 + 0.5 \times 40) = (69, 21.6) y = 44.1 - \frac{5}{882} (69)^{2} \approx 17.1

Our analysis

Where the marks are

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

Calculus29 marks · 53%
Q2
Statistics and probability26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Chi-squared test for independence · Q1
Expected count $= \frac{row total \times column total}{grand total}$ , with $ν = (r - 1) (c - 1)$ .
02
Reading a test · Q1
Compare the p-value with the significance level. Statistical significance is not the same as practical importance.
03
Which technique is which · Q1
The question is whether the new technique reduces cracking, so a one-tailed test uses the lower tail.
04
Integrate acceleration · Q2
Velocity is the integral of acceleration, with the initial velocity as the constant.
05
Boundary curve · Q2
The parabola passes through the highest point $(0, 44.1)$ and the furthest landing point $(88.2, 0)$ , with its vertex on the $y$ -axis.

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 TZ1 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

26 marks

Cracked components: a chi-squared test, a binomial test and a two-sample t-test

Statistics and probability

(a)(i)

1 mark

The total number of components with no cracks is 141, and 88 of them were made by the current technique.

a = 141 - 88 = 53

(a)(ii)

1 mark

The 120 components made with the new technique are split between none, minor and major cracks.

b = 120 - 53 - 54 = 13

(b)

2 marks

150 components had minor cracks and 54 of them were made with the new technique.

P = \frac{54}{54 + 96} = \frac{54}{150} = 0.36

(c)(i)

1 mark

$H_{0}$ : the level of cracking (none, minor, major) is independent of the production technique.
$H_{1}$ : the level of cracking is not independent of the production technique.

(c)(ii)

2 marks

Enter the table into the calculator for a $χ^{2}$ test (the smallest expected count is 23, so the test is valid). The degrees of freedom are $(3 - 1) (2 - 1) = 2$ .

χ^{2} \approx 8.15, p \approx 0.0170

(c)(iii)

2 marks

Since $0.0170 < 0.05$ we reject $H_{0}$ . There is sufficient evidence that the amount of cracking depends on the production technique.

(d)

1 mark

For the current technique, $96 + 56 = 152$ of the 240 components developed cracks.

\frac{152}{240} = \frac{19}{30}

(e)

1 mark

(f)

5 marks

The number of components with cracks among the 120 new ones is $X = 54 + 13 = 67$ . Under $H_{0}$ , $X \sim B (120, \frac{19}{30})$ , and the alternative is that fewer crack.
$H_{0}$ : $p = \frac{19}{30}$ , $H_{1}$ : $p < \frac{19}{30}$ .
Since $0.0549 > 0.05$ we do not reject $H_{0}$ . There is insufficient evidence at the 5% level that the new technique reduces the proportion of components that develop cracks.

P (X \leq 67) \approx 0.0549

(g)(i)

1 mark

(g)(ii)

1 mark

(h)

7 marks

Let $μ_{1}$ and $μ_{2}$ be the population mean times before cracking for the new and the current technique.
Use a one-tailed two-sample $t$ -test with the sample sizes 100 and 200, means 250.1 and 249.1 and standard deviations 3.8 and 3.8 (pooled, since the standard deviations are equal).
Since $0.0162 < 0.05$ , reject $H_{0}$ . There is sufficient evidence that the new technique increases the mean time before cracks appear.

H_{0} : μ_{1} = μ_{2}, H_{1} : μ_{1} > μ_{2} t \approx 2.15, p \approx 0.0162

(i)

1 mark

Question 2

29 marks

A T-shirt cannon: projectile motion and the boundary of all reachable points

Calculus

(a)(i)

2 marks

Each row is 1.0 m further and 0.5 m higher. Row 5 is four rows behind row 1.

A 5 = (30 + 4, 2.1 + 4 \times 0.5) = (34, 4.1)

(a)(ii)

3 marks

Row $n$ is $n - 1$ rows behind row 1.

A_{n} = (30 + (n - 1), 2.1 + 0.5 (n - 1)) = (29 + n, 1.6 + 0.5 n)

(b)(i)

3 marks

Integrate the acceleration $(0, - 9.8)$ and use the initial velocity as the constant.

(\overset{x}{˙} \overset{y}{˙}) = (29.4 cos θ 29.4 sin θ - 9.8 t)

(b)(ii)

3 marks

A vertical launch has $θ = 9 0^{\circ}$ , so $\overset{y}{˙} = 29.4 - 9.8 t$ .
The maximum height occurs when the vertical velocity is zero.

29.4 - 9.8 t = 0 \Rightarrow t = 3 seconds

(c)

2 marks

Use the displacement formula with $θ = 9 0^{\circ}$ and $t = 3$ .

y = 29.4 (3) - 4.9 (3)^{2} = 88.2 - 44.1 = 44.1 m

(d)(i)

3 marks

The T-shirt lands when $y = 0$ .
Substitute into $x = 29.4 cos θ t$ .

29.4 sin θ t - 4.9 t^{2} = 0 \Rightarrow t = 6 sin θ (t \neq = 0) x = 29.4 cos θ \times 6 sin θ = 176.4 sin θ cos θ

(d)(ii)

2 marks

Since $sin θ cos θ = \frac{1}{2} sin 2 θ$ has maximum $\frac{1}{2}$ at $θ = 4 5^{\circ}$ .

x_{m a x} = 176.4 \times \frac{1}{2} = 88.2 m

(e)(i)

1 mark

The vertex $V$ is the highest point reached by any T-shirt, which is the vertical launch.

c = 44.1

(e)(ii)

2 marks

The vertex is on the $y$ -axis, so the axis of symmetry is $x = 0$ .

- \frac{b}{2 a} = 0 \Rightarrow b = 0

(e)(iii)

3 marks

The furthest point reached on the ground is $(88.2, 0)$ , which is on the boundary.

0 = a (88.2)^{2} + 44.1 \Rightarrow a = - \frac{44.1}{88. 2 ^{2}} = - \frac{5}{882} \approx - 0.00567

(f)

5 marks

Seat $A 40$ has $n = 40$ , so it is at
The boundary curve $y = - \frac{5}{882} x^{2} + 44.1$ at $x = 69$ is
The seat is at height 21.6, which is above 17.1, so it is outside the region that any T-shirt can reach. The spectator can never get a T-shirt.

(29 + 40, 1.6 + 0.5 \times 40) = (69, 21.6) y = 44.1 - \frac{5}{882} (69)^{2} \approx 17.1

Our analysis

Where the marks are

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

Calculus29 marks · 53%
Q2
Statistics and probability26 marks · 47%
Q1

Mentor insights

Traps to watch for

01
Chi-squared test for independence · Q1
Expected count $= \frac{row total \times column total}{grand total}$ , with $ν = (r - 1) (c - 1)$ .
02
Reading a test · Q1
Compare the p-value with the significance level. Statistical significance is not the same as practical importance.
03
Which technique is which · Q1
The question is whether the new technique reduces cracking, so a one-tailed test uses the lower tail.
04
Integrate acceleration · Q2
Velocity is the integral of acceleration, with the initial velocity as the constant.
05
Boundary curve · Q2
The parabola passes through the highest point $(0, 44.1)$ and the furthest landing point $(88.2, 0)$ , with its vertex on the $y$ -axis.

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