IB · Mathematics: Applications & Interpretation SL & HL · Statistics & probability

Confidence intervals: IB Mathematics: Applications & Interpretation questions with answers

Free Confidence intervals questions for IB Mathematics: Applications & Interpretation (SL & HL), HL only, part of Statistics & probability. Each question has the answer and the marking points an examiner would look for.

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Confidence intervals practice questions

Question 1Structured · 5 marksA* challenge

The lengths of leaves are normally distributed with standard deviation 4 mm. A random sample of 36 leaves has mean 59.5 mm.

(a)

Find a 95% confidence interval for the mean length.

[3]

(b)

State how many leaves would be needed to halve the width of the interval.

[1]

(c)

A botanist claims the mean is 61.81 mm. Comment on this claim.

[1]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	58.19 < μ < 60.81 z-interval (M1) 58.19 A1 60.81 A1	3
(b)	144 144 A1	1
(c)	outside the interval the value lies outside the interval, so the claim is not supported R1	1

Question 2Structured · 5 marks

The masses, in grams, of a random sample of 12 bags of flour are shown.
359, 341, 347, 365, 354, 348, 338, 340, 342, 363, 335, 360

(a)

Find the sample mean and the unbiased estimate of the population variance.

[2]

(b)

Find a 95% confidence interval for the population mean, assuming the masses are normally distributed.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	349; 110 x̄ = 349 A1 s_n−1² = 110 A1	2
(b)	342.7 < μ < 356.0 t-interval (σ unknown) (M1) 11 degrees of freedom (A1) 342.7 < μ < 356.0 A1	3

Question 3Structured · 4 marks

A random sample of 40 bags of rice has mean mass 54.2 kg and standard deviation 6.8 kg.

Find a 95% confidence interval for the mean mass of all bags, and state whether the manufacturer’s claim of 57 kg is supported.

[4]

Show answers and mark scheme

Part	Answer and marking points	Marks
	(52.1, 56.3) GDC interval (M1) (52.1, 56.3) A2 57 is outside the interval R1	4

Question 4Structured · 13 marks

The volume of drink in bottles has mean 250 ml and standard deviation 20 ml. The distribution of the volumes is not known.

(a)

A sample of 36 bottles is taken. State the approximate distribution of the sample mean X̄, giving a reason.

[3]

(b)

Find the probability that the sample mean is less than 248 ml.

[2]

(c)

A different sample of 36 bottles has mean 245 ml. Find a 95% confidence interval for the population mean, assuming σ = 20.

[3]

(d)

Hence comment on the claim that the mean volume is 250 ml.

[2]

(e)

Find the least sample size needed for the width of a 95% confidence interval to be at most 4 ml.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	X̄ ~ N(250, 11.1) central limit theorem, n large R1 mean 250 A1 variance 20²36 = 11.1 A1	3
(b)	0.274 normal CDF with standard error (M1) 0.274 A1	2
(c)	238.5 < μ < 251.5 z-interval (M1) 245 ± 1.96 × 20√36 (A1) 238.5 < μ < 251.5 A1	3
(d)	supported 250 is in the interval R1 the claim is supported A1	2
(e)	385 2 × 1.96 × 20√n ≤ 4 (M1) solving for n (M1) 385 A1	3

Question 5Structured · 5 marksA* challenge

The lengths of leaves are normally distributed with standard deviation 8 mm. A random sample of 36 leaves has mean 54.6 mm.

(a)

Find a 95% confidence interval for the mean length.

[3]

(b)

State how many leaves would be needed to halve the width of the interval.

[1]

(c)

A botanist claims the mean is 58.21 mm. Comment on this claim.

[1]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	51.99 < μ < 57.21 z-interval (M1) 51.99 A1 57.21 A1	3
(b)	144 144 A1	1
(c)	outside the interval the value lies outside the interval, so the claim is not supported R1	1

Question 6Structured · 5 marks

The masses, in grams, of a random sample of 12 bags of flour are shown.
509, 498, 506, 502, 508, 507, 490, 489, 510, 491, 502, 515

(a)

Find the sample mean and the unbiased estimate of the population variance.

[2]

(b)

Find a 95% confidence interval for the population mean, assuming the masses are normally distributed.

[3]

Show answers and mark scheme

Part	Answer and marking points	Marks
(a)	502; 73.5 x̄ = 502 A1 s_n−1² = 73.5 A1	2
(b)	496.8 < μ < 507.7 t-interval (σ unknown) (M1) 11 degrees of freedom (A1) 496.8 < μ < 507.7 A1	3

More Statistics & probability questions

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in