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IB Maths AI HL · Paper 3

Mathematics: Applications & Interpretation, Higher Level

Timezone

TZ1TZ2 TZ3

Session: May 2025
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 2025 TZ1 paper, which your school or the IB can provide.

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Questions

Questions 1–2 · 55 marks

Question 1

29 marks

Customers moving between three stores: Markov chains, then arithmetic and geometric models

Number and algebra

(a)

3 marks

With columns for Aroma, Bodega and Clover (the stores customers start in) and rows for where they go:

T = 0.91 0.05 0.04 0.04 0.95 0.01 0.06 0.02 0.92

(b)

4 marks

After 5 years use $T^{5}$ , and read the entry for Clover customers who end up at Aroma (row 1, column 3).

(T^{5})_{13} \approx 0.219 21.9%

(c)

2 marks

Raise $T$ to a high power, or find the eigenvector for eigenvalue 1, scaled so the entries add to 1.

0.342 0.432 0.225

(d)

1 mark

In the long term, the proportion of customers who are at Aroma is the first entry.

34.2%

(e)

3 marks

Raise the entry for Clover to Aroma (and lower the entry for Clover staying by the same amount) until the steady state for Aroma is at least 0.40.
With 12%: steady state $(0.398, 0.456, 0.146)$ , which is below 0.40. With 13%, the matrix is $0.91 0.05 0.04 0.04 0.95 0.01 0.13 0.02 0.85$ and the steady state is
The first entry is at least 0.40. The minimum integer is 13%.

(0.403, 0.459, 0.138)

(f)

2 marks

Bodega's long-term share goes up from 0.432 to 0.459 when more Clover customers move to Aroma, so Bodega's manager would benefit.

(g)

1 mark

It takes many years for the steady state to be reached. During that time a new store could open or one store could change its advertising, so the percentages would change.

(h)

2 marks

Model 1 is an arithmetic sequence with $u_{1} = 600$ and $d = 30$ .

u_{10} = 600 + 9 (30) = 870

(i)

2 marks

In week 5 the number is $600 + 4 (30) = 720$ . In week 6 the number of new customers is 12% more than the 30 new customers of the week before.

720 + 30 (1.12) = 720 + 33.6 = 753.6

(j)

3 marks

The new customers each week form a geometric sequence $30 (1.12)^{k}$ after week 5. By week 10 there have been five more weeks.

720 + 30 (1.12 + 1.1 2^{2} + 1.1 2^{3} + 1.1 2^{4} + 1.1 2^{5}) \approx 933

(k)

6 marks

Model 1 in week $n$ is $600 + 30 (n - 1)$ . Model 2 in week $n$ is $720 + 30 \times 1.12 \times \frac{1.1 2 ^{n - 5} - 1}{0.12}$ .
Make a table of values. In week 17, the models give 1080 and 1531 (a difference of 451). In week 18 they give 1110 and 1662 (a difference of 552).

difference = Model 2 - Model 1 \geq 500 week 18

Question 2

26 marks

Farmer Giles's eggs: binomial, normal, t-tests, a confidence interval and an F-test

Statistics and probability

(a)

4 marks

Each egg has probability $P (X \geq 55) = 0.2087$ , so the number of heavy eggs is $Y \sim B (25, 0.2087)$ .

P (Y \geq 5) \approx 0.621

(b)

3 marks

The mean of 25 eggs is $\overset{ˉ}{X} \sim N (52, \frac{3. 7 ^{2}}{25})$ , with standard deviation $0.74$ .

P (51 < \overset{ˉ}{X} < 53) \approx 0.823

(c)

6 marks

$H_{0}$ : $μ = 52.0$ . $H_{1}$ : $μ > 52.0$ , where $μ$ is the population mean weight after the new food. Use a one-tailed one-sample $t$ -test, because the population standard deviation is not known, on the data (mean 52.66, standard deviation 1.55).
Since $0.105 > 0.05$ there is insufficient evidence to reject $H_{0}$ . The data do not show that the mean weight has risen above 52.0 g.

t \approx 1.35, p \approx 0.105

(d)

2 marks

Use a 99% $t$ -interval for the mean.

52.66 \pm 1.589 = (51.07, 54.25)

(e)(i)

1 mark

A paired $t$ -test.

(e)(ii)

1 mark

The differences in the weights are normally distributed (or the weights from different chickens are independent).

(f)

6 marks

Let $μ_{d}$ be the mean of the differences, after minus before.
Run a one-tailed paired $t$ -test with the eight pairs.
Since $0.0338 < 0.05$ , there is sufficient evidence that the mean weight of the chickens' eggs has increased after the new feed.

H_{0} : μ_{d} = 0, H_{1} : μ_{d} > 0 t \approx 2.16, p \approx 0.0338

(g)

1 mark

The unbiased estimate of the variance from Table 1.

s_{G}^{2} \approx 2.39

(h)

2 marks

The test statistic is $F = \frac{s _{R}^{2}}{s _{G}^{2}}$ and the t-test is valid when $F < 3.1789$ .

s_{R}^{2} < 3.1789 \times 2.3916 \approx 7.60 maximum s_{R}^{2} \approx 7.60

Our analysis

Where the marks are

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

Number and algebra29 marks · 53%
Q1
Statistics and probability26 marks · 47%
Q2

Mentor insights

Traps to watch for

01
Transition matrix · Q1
Each column holds the yearly probabilities of moving from one store to each store. Columns add to 1.
02
Steady state · Q1
A high power of $T$ , or the eigenvector for 1 scaled to add to 1.
03
Two models · Q1
An arithmetic sequence adds a fixed amount each week. After the loyalty scheme the new customers form a geometric sequence.
04
Mean of a sample · Q2
The mean of 25 eggs has standard deviation $\frac{3.7}{25}$ .
05
Paired data · Q2
Eggs before and after from the same chicken are linked, so use a paired $t$ -test on the differences.

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 TZ3

Session: May 2025
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 2025 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

29 marks

Customers moving between three stores: Markov chains, then arithmetic and geometric models

Number and algebra

(a)

3 marks

With columns for Aroma, Bodega and Clover (the stores customers start in) and rows for where they go:

T = 0.91 0.05 0.04 0.04 0.95 0.01 0.06 0.02 0.92

(b)

4 marks

After 5 years use $T^{5}$ , and read the entry for Clover customers who end up at Aroma (row 1, column 3).

(T^{5})_{13} \approx 0.219 21.9%

(c)

2 marks

Raise $T$ to a high power, or find the eigenvector for eigenvalue 1, scaled so the entries add to 1.

0.342 0.432 0.225

(d)

1 mark

In the long term, the proportion of customers who are at Aroma is the first entry.

34.2%

(e)

3 marks

Raise the entry for Clover to Aroma (and lower the entry for Clover staying by the same amount) until the steady state for Aroma is at least 0.40.
With 12%: steady state $(0.398, 0.456, 0.146)$ , which is below 0.40. With 13%, the matrix is $0.91 0.05 0.04 0.04 0.95 0.01 0.13 0.02 0.85$ and the steady state is
The first entry is at least 0.40. The minimum integer is 13%.

(0.403, 0.459, 0.138)

(f)

2 marks

Bodega's long-term share goes up from 0.432 to 0.459 when more Clover customers move to Aroma, so Bodega's manager would benefit.

(g)

1 mark

It takes many years for the steady state to be reached. During that time a new store could open or one store could change its advertising, so the percentages would change.

(h)

2 marks

Model 1 is an arithmetic sequence with $u_{1} = 600$ and $d = 30$ .

u_{10} = 600 + 9 (30) = 870

(i)

2 marks

In week 5 the number is $600 + 4 (30) = 720$ . In week 6 the number of new customers is 12% more than the 30 new customers of the week before.

720 + 30 (1.12) = 720 + 33.6 = 753.6

(j)

3 marks

The new customers each week form a geometric sequence $30 (1.12)^{k}$ after week 5. By week 10 there have been five more weeks.

720 + 30 (1.12 + 1.1 2^{2} + 1.1 2^{3} + 1.1 2^{4} + 1.1 2^{5}) \approx 933

(k)

6 marks

Model 1 in week $n$ is $600 + 30 (n - 1)$ . Model 2 in week $n$ is $720 + 30 \times 1.12 \times \frac{1.1 2 ^{n - 5} - 1}{0.12}$ .
Make a table of values. In week 17, the models give 1080 and 1531 (a difference of 451). In week 18 they give 1110 and 1662 (a difference of 552).

difference = Model 2 - Model 1 \geq 500 week 18

Question 2

26 marks

Farmer Giles's eggs: binomial, normal, t-tests, a confidence interval and an F-test

Statistics and probability

(a)

4 marks

Each egg has probability $P (X \geq 55) = 0.2087$ , so the number of heavy eggs is $Y \sim B (25, 0.2087)$ .

P (Y \geq 5) \approx 0.621

(b)

3 marks

The mean of 25 eggs is $\overset{ˉ}{X} \sim N (52, \frac{3. 7 ^{2}}{25})$ , with standard deviation $0.74$ .

P (51 < \overset{ˉ}{X} < 53) \approx 0.823

(c)

6 marks

$H_{0}$ : $μ = 52.0$ . $H_{1}$ : $μ > 52.0$ , where $μ$ is the population mean weight after the new food. Use a one-tailed one-sample $t$ -test, because the population standard deviation is not known, on the data (mean 52.66, standard deviation 1.55).
Since $0.105 > 0.05$ there is insufficient evidence to reject $H_{0}$ . The data do not show that the mean weight has risen above 52.0 g.

t \approx 1.35, p \approx 0.105

(d)

2 marks

Use a 99% $t$ -interval for the mean.

52.66 \pm 1.589 = (51.07, 54.25)

(e)(i)

1 mark

A paired $t$ -test.

(e)(ii)

1 mark

The differences in the weights are normally distributed (or the weights from different chickens are independent).

(f)

6 marks

Let $μ_{d}$ be the mean of the differences, after minus before.
Run a one-tailed paired $t$ -test with the eight pairs.
Since $0.0338 < 0.05$ , there is sufficient evidence that the mean weight of the chickens' eggs has increased after the new feed.

H_{0} : μ_{d} = 0, H_{1} : μ_{d} > 0 t \approx 2.16, p \approx 0.0338

(g)

1 mark

The unbiased estimate of the variance from Table 1.

s_{G}^{2} \approx 2.39

(h)

2 marks

The test statistic is $F = \frac{s _{R}^{2}}{s _{G}^{2}}$ and the t-test is valid when $F < 3.1789$ .

s_{R}^{2} < 3.1789 \times 2.3916 \approx 7.60 maximum s_{R}^{2} \approx 7.60

Our analysis

Where the marks are

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

Number and algebra29 marks · 53%
Q1
Statistics and probability26 marks · 47%
Q2

Mentor insights

Traps to watch for

01
Transition matrix · Q1
Each column holds the yearly probabilities of moving from one store to each store. Columns add to 1.
02
Steady state · Q1
A high power of $T$ , or the eigenvector for 1 scaled to add to 1.
03
Two models · Q1
An arithmetic sequence adds a fixed amount each week. After the loyalty scheme the new customers form a geometric sequence.
04
Mean of a sample · Q2
The mean of 25 eggs has standard deviation $\frac{3.7}{25}$ .
05
Paired data · Q2
Eggs before and after from the same chicken are linked, so use a paired $t$ -test on the differences.

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