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AQA MATHEMATICAL STUDIES

Paper 2A

PREREQUISITES

Calculate mean of a data set

Find z-values for any normal distribution (Chapter 5)

Use Φ tables to find probabilities

CHAPTER 6

Confidence Intervals

AQA Mathematical Studies · Paper 2A

Point Estimates

Distribution of Sample Means X̄ ~ N(μ, σ²/n)

Standard Error σ/√n

90% / 95% / 99% CIs

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AQA MATHEMATICAL STUDIES · PAPER 2A

Chapter 6: Confidence Intervals — Overview

SECTION 6.1

Point Estimates

The sample mean x̄ is used as a point estimate for the population mean μ. A larger sample size n produces a more accurate estimate of the true population mean.

x̄ = Σx ÷ n

SECTION 6.2

Distribution of Sample Means

If X ~ N(μ, σ²) then the sample mean follows X̄ ~ N(μ, σ²/n). The standard error measures the spread of sample means around the population mean.

SE = σ / √n

SECTION 6.3

Confidence Intervals

Standard CIs use fixed z-values: 90% → z = 1.64 · 95% → z = 1.96 · 99%→ z = 2.58. Non-standard intervals use the formula Φ(z) = 0.5 + p/200 to find the required z-value.

CI: x̄ ± z · (σ/√n)

PREREQUISITES · CHAPTER 5

Skills Required

Students must be able to: calculate the mean of a data set; find z-values for any normal distribution; and use Φ tables to find probabilities from the standard normal N(0,1).

Φ(z) = P(Z ≤ z)

CHAPTER 6 · CONFIDENCE INTERVALS

EACH SECTION IS FOLLOWED BY A PAST PAPER QUESTION

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6.1 Point Estimates — Key Skill

SKILL GUIDE

DEFINITION

A point estimate is a single value used to estimate a population parameter. The sample mean x̄ is a point estimate for the population mean μ.

x̄ = Σx ÷ n

Sum of all sample values ÷ sample size

Key principle: Larger sample size n → more accurate point estimate for μ. A sample of n=25 gives a better estimate than n=5.

STEP-BY-STEP METHOD

Add all values in the sample: compute Σx

Divide by n

(sample size):

x̄ = Σx ÷ n

State x̄ as the point estimate for μ

Combining samples: total = n₁x̄₁ + n₂x̄₂, then divide by (n₁+n₂)

Normal distribution:

use

z = (x − μ) ÷ σ

then look up Φ(z) in tables

Key insight: Even though the machine fills jars to a mean of 260g (10g above the 250g label), more than 1 in 10 jars is still underweight due to natural variation. This is why quality control uses confidence intervals — not just the mean.

Example 1 — Yeast Extract Jars (Fully Worked)

Context:

Jars of yeast extract are filled by a machine. The weight X follows a normal distribution:

X ~ N(260, 8²)

— mean 260g, standard deviation 8g. The label on the jar says

250g

. Find

the probability that a randomly chosen jar is underweight (i.e. weighs less than 250g).

Standardise: z = (250 − 260) ÷ 8 = −10 ÷ 8 = −1.25

Use symmetry: P(X < 250) = P(Z < −1.25) = 1 − Φ(1.25)

Look up Φ(1.25) in tables: Φ(1.25) = 0.89435

P(X < 250) = 1 − 0.89435 = 0.10565 ≈ 10.6%

≈ 10.6% of jars are underweight

— despite mean being 10g above the label

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Exercise 6A — Questions

AQA CHAPTER 6.1 · POINT ESTIMATES

CONTEXT

A factory fills jars of wholegrain mustard. The weight X follows a normal distribution:

X ~ N(μ, σ²)

. Two inspectors each weigh a sample of jars to estimate the true mean weight μ. All data

needed to answer every question is provided below — no textbook required.

QUESTION 1 — INSPECTOR A

Find the point estimate from a sample of 5 jars

Inspector A weighs 5 jars of wholegrain mustard and records the following weights (in grams):

181 176 172 178 171

(a) Calculate the sample mean x̄ for Inspector A's sample.

(b) State what x̄ estimates and explain what it represents.

Hint: x̄ = Σx ÷ n. Add all 5 values, then divide by 5.

QUESTION 2 — INSPECTOR B

Compare accuracy of two point estimates

Inspector B weighs a larger sample of 25 jars and calculates a sample mean of x̄ = 174.5 g .

(a) Write down Inspector B's point estimate for μ.

(b) Which inspector's estimate is more accurate — A (n=5) or B (n=25)? Give a reason.

Hint: Think about how sample size affects the accuracy of a point estimate.

QUESTION 3 — COMBINED SAMPLE

Find the combined point estimate from both inspectors

Inspector A: n = 5 jars , x̄ = your answer from Q1. | Inspector B: n = 25 jars , x̄ = 174.5 g. | Total sample size = 30 jars .

(a) Find the total weight of all 30 jars combined.

(b) Hence find the combined point estimate for μ using all 30 jars.

Hint: Total weight = n₁ × x̄₁ + n₂ × x̄₂. Then divide by (n₁ + n₂) = 30.

QUESTION 4 — NORMAL DISTRIBUTION PROBABILITY

Find the probability a jar is underweight

A different brand of mustard has weight

X ~ N(180, 3²)

— mean

180g

, standard deviation

. The label says

175g

. Find the probability that a randomly chosen jar weighs less than 175g.

(a) Standardise: find z = (175 − 180) ÷ 3.

(b) Use the symmetry rule: P(X < 175) = 1 − Φ(|z|). Look up Φ(|z|) in the table below.

Φ(1.667) ≈ 0.9522 | Φ(1.65) = 0.9505 | Φ(1.70) = 0.9554

Hint: z = (175 − 180) ÷ 3 = −1.667. Use P(Z < −1.667) = 1 − Φ(1.667).

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Exercise 6A — Answers

WORKED SOLUTIONS

Q1 — Inspector A (n = 5 jars: 181, 176, 172, 178, 171 g)

STEP 1 — SAMPLE MEAN

Σx = 181 + 176 + 172 + 178 + 171 = 878 g

x̄ = 878 ÷ 5 = 175.6 g

x̄ = 175.6 g

— point estimate for μ

STEP 2 — WHAT X̄ ESTIMATES

x̄ = 175.6 g is a point estimate for the true population mean weight μ of all jars produced by the machine.

Q2 — Inspector B (n = 25 jars, x̄ = 174.5 g)

STEP 1 — POINT ESTIMATE

x̄ = 174.5 g

— Inspector B's point estimate for μ

STEP 2 — WHICH IS MORE ACCURATE?

Inspector B (n = 25) gives a more accurate estimate than Inspector A (n = 5). A larger sample size reduces the variability of the sample mean, making it a more reliable estimate of the true population mean μ.

Q3 — Combined Sample (n = 30 jars total)

STEP 1 — TOTAL WEIGHT OF ALL 30 JARS

Inspector A total = 5 × 175.6 = 878 g

Inspector B total = 25 × 174.5 = 4,362.5 g

Combined total = 878 + 4,362.5 = 5,240.5 g

STEP 2 — COMBINED POINT ESTIMATE

x̄ = 5,240.5 ÷ 30 = 174.68 g

x̄ = 174.68 g

— combined point estimate for μ (n=30)

Q4 — X ~ N(180, 3²) · Find P(X < 175)

STEP 1 — STANDARDISE

z = (175 − 180) ÷ 3 = −5 ÷ 3 = −1.667

STEP 2 — FIND PROBABILITY USING SYMMETRY

P(X < 175) = P(Z < −1.667)

= 1 − Φ(1.667)

= 1 − 0.9522 = 0.0478

P(X < 175) ≈ 4.8%

— about 1 in 20 jars is underweight

Interpretation: Even though the mean is 5g above the label (180g vs 175g), approximately 4.8% of jars are still underweight due to natural variation (σ = 3g).

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Past Paper — Point Estimates (AQA 2011)

AQA MATHEMATICAL STUDIES · PAPER 2A · 2011 Q1

Source: AQA Mathematical Studies 1350 Paper 2A 2011 Q1

EXAM TIP

P(X = exact value) = 0 for ALL continuous distributions — this must be stated explicitly to secure the 1 mark in part (a)(i). Never write a non-zero probability for a single point on a continuous distribution.

QUESTION

The weight of a tin of baked beans is normally distributed with mean μ grams and standard deviation 2.5 grams:

X ~ N(421, 2.5²)

(a)(i)

Write down P(X = 421).

1 mark

(a)(ii)

Find P(X < 425).

2 marks

(a)(iii)

Find P(418 < X < 424).

2 marks

(b)

Find the value of x such that P(X < x) = 0.98.

2 marks

Use the standard normal table Φ(z) provided in the exam.

MARK SCHEME

(a)(i)

P(X = 421) = 0

Continuous distribution — probability of any single exact value is always zero.

(a)(ii)

P(X < 425) = Φ(1.6) = 0.9452

Standardise: z = (425 − 421) / 2.5 = 4 / 2.5 = 1.6 Look up Φ(1.6) from tables → 0.9452

(a)(iii)

P(418 < X < 424) = 0.7698

z₁ = (418−421)/2.5 = −1.2 | z₂ = (424−421)/2.5 = 1.2 By symmetry: 2Φ(1.2) − 1 = 2(0.8849) − 1 = 0.7698

(b)

x = 426.1 g

Φ(z) = 0.98 → z = 2.054 (from tables) x = 421 + 2.054 × 2.5 = 421 + 5.135 = 426.1 g

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6.2 Distribution of Sample Means — Key Skill

If X ~ N(μ, σ²) and samples of size n are taken, then X̄ ~ N(μ, σ²/n) | Standard Error SE = σ/√n | Larger n → smaller SE → means cluster tighter around μ

Core Result & Example 2 — Chocolates

KEY FORMULAE

X̄ ~ N(μ, σ²/n) SE = σ/√n z = (x̄ − μ) / (σ/√n)

Example 2

W ~ N(10, 2²) | Find P(9.8 < W < 10.2)

W = weight of a single chocolate (grams). μ = 10g, σ = 2g.

(a)

Single chocolate: z₁ = (9.8−10)/2 = −0.1 , z₂ = 0.1

P = Φ(0.1) − Φ(−0.1) = 0.5398 − 0.4602 = 0.07966

(b)

Sample of n = 25: SE = 2/√25 = 0.4 → W̄ ~ N(10, 0.4²)

z₁ = (9.8−10)/0.4 = −0.5 , z₂ = 0.5

P = Φ(0.5) − Φ(−0.5) = 0.6915 − 0.3085 = 0.38292

COMPARISON

0.07966

Single (n = 1)

0.38292

Sample mean (n = 25)

Nearly 5× higher probability for the sample mean — larger n reduces spread

Example 3 — Combined Weight of Students

Male student weight: X ~ N(70, 5²) | μ = 70 kg, σ = 5 kg

Question: Find P(combined weight of 4 students < 260 kg)

Step-by-Step Solution

Convert total to mean: P(total < 260) = P(X̄ < 260 ÷ 4) = P(X̄ < 65)

Find Standard Error (n = 4):

SE = σ/√n = 5/√4 = 5/2 = 2.5

Distribution of sample mean:

X̄ ~ N(70, 2.5²)

Standardise:

z = (65 − 70) / 2.5 = −5/2.5 = −2

Find probability:

P(X̄ < 65) = P(Z < −2) = 1 − Φ(2) = 1 − 0.97725 = 0.02275

FINAL ANSWER

P(total < 260 kg) = 0.02275

Only a 2.3% chance — very unlikely for 4 students to total under 260 kg

AQA Mathematical Studies — Chapter 6.2 | Examples 2 & 3

2 & 3

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Past Paper — Distribution of Sample Means

AQA Mathematical Studies · Paper 2A · 2013

PAST PAPER QUESTION

The diameter of a disc produced by a machine has a normal distribution with mean μ cm and variance 25 cm² . A random sample of 100 discs is taken and the sample mean is found to be 40.5 cm .

(a)

Calculate a

95% confidence interval

for μ.

3 marks

(b)

Comment on whether the machine produces discs with

mean diameter 40 cm

1 mark

MARK SCHEME — PART (A)

3 MARKS

Variance = 25, so

σ = √25 = 5

← CRITICAL STEP

Standard Error:

SE = σ/√n = 5/√100 = 0.5

95% CI uses

z = 1.96

x̄ ± 1.96 × SE = 40.5 ± 1.96 × 0.5 = 40.5 ±

0.98

CI = [39.52, 41.48] cm

MARK SCHEME — PART (B)

1 MARK

Check whether

40 cm

lies inside the confidence interval

[39.52, 41.48]

Since 39.52 < 40 < 41.48 , the value 40 cm lies inside the CI.

No evidence to doubt the claim of μ = 40 cm

State explicitly: 40 cm is inside the CI, so there is no evidence to suggest the machine is NOT producing discs with mean diameter 40 cm.

EXAM TIP — MOST COMMON SOURCE OF LOST MARKS

When a question states X ~ N(μ, 25) , the 25 is the VARIANCE , not the standard deviation. You MUST compute σ = √25 = 5 before finding the standard error SE = σ/√n. Writing SE = 25/√100 = 2.5 instead of 0.5 will cost you all 3 marks in part (a).

Source: AQA Mathematical Studies 1350 Paper 2A 2013 Q (Distribution of Sample Means)

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6.3 Confidence Intervals — Key Skill

SKILL REFERENCE

Source: AQA Chapter 6.3 — Key Formulae and Examples 4 & 5

Confidence Level	z-value	Formula
95% CI	1.96	x̄ ± 1.96 × σ/√n
99% CI	2.58	x̄ ± 2.58 × σ/√n
90% CI	1.64	x̄ ± 1.64 × σ/√n
Non-standard p%	Φ(z) = 0.5 + p/200	x̄ ± z × σ/√n

CONFIDENCE INTERVAL FORMULAE — CI = X̄ ± Z × Σ/√N

Interpretation: A 95% CI means that if 100 samples were taken and 100 CIs calculated, approximately 95 of them would contain the true population mean μ. The CI does not mean there is a 95% chance μ lies in this interval.

6-STEP METHOD

Find x̄ — if given total, divide by n

Extract σ from N(μ, σ²) — 2nd param is VARIANCE, take √

SE = σ/√n — standard error of the mean

Find z — use table for required confidence level

CI = x̄ ± z × SE — compute lower and upper bounds

Comment — is claimed value inside or outside the CI?

EXAMPLE 4

Breakfast Cereal — 95% CI

X ~ N(μ, 1²) | n = 200 | x̄ = 340 g | Find a 95% confidence interval for μ.

σ:

σ = √1 = 1 (variance = 1²)

SE:

σ/√n = 1/√200 = 0.0707

z = 1.96 (95% CI)

CI:

340 ± 1.96 × 0.0707 = 340 ± 0.1386

95% CI

[339.86, 340.14] g

EXAMPLE 5

Biscuits — 99% CI from Raw Data

X ~ N(μ, 5²) | n = 10 | Data: 442, 450, 447, 446, 453, 449, 444, 454, 443, 457 | Claimed mean: 450 g

x̄:

Σx = 4485 → x̄ = 4485 ÷ 10 = 448.5 g

σ:

σ = √25 = 5 (variance = 5² = 25)

SE:

5/√10 = 1.581

z = 2.58 (99% CI)

CI:

448.5 ± 2.58 × 1.581 = 448.5 ± 4.08

99% CI

[444.4, 452.6] g

450 g lies inside [444.4, 452.6] — no reason to doubt the claimed mean.

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Past Paper — 98% CI (AQA 2013 Sand Bags)

AQA 2013 · PAPER 2A · [3+1 MARKS]

QUESTION

Sand bags are filled by a machine. The weight of sand in a bag is normally distributed:

X ~ N(μ, 0.4²)

A random sample of

25 bags

is taken and the

total weight is 497.5 kg

(a)

Construct a

98%

confidence interval

for μ.

(b)

Comment on whether the machine fills bags with a mean weight of

20 kg

MARK SCHEME — PART (A): STEP-BY-STEP

Part (b):

The claimed mean of

20 kg

lies

within

the 98% CI

[19.71, 20.09] kg

—

there is

no evidence

to suggest the machine is not filling bags with a mean weight of 20 kg.

EXAM TIP — NON-STANDARD CI

For a

98% CI

, always show:

Φ(z) = 0.5 + 98/200 = 0.99 → z = 2.33

This Φ(z) line earns the method mark — never skip it. You must look up z from tables; it is not one of the standard

values (1.645 or 1.96).

Source: AQA Mathematical Studies 1350 Paper 2A 2013

SAMPLE MEAN

x̄ = 497.5 ÷ 25

= 19.9 kg

STANDARD ERROR

SE = σ/√n = 0.4/√25

= 0.4/5 = 0.08

FIND Z (NON-STANDARD!)

Φ(z) = 0.5 + 98/200

Φ(z) = 0.99 → z = 2.33

CONFIDENCE INTERVAL

19.9 ± 2.33 × 0.08

19.9 ± 0.1864

CI = [19.71, 20.09] kg

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Past Paper — 98% CI (AQA 2011 Guillemots)

AQA 2011 · Paper 2A · 5 marks

QUESTION

The wingspan of guillemots is normally distributed with mean

μ mm

and standard deviation

σ = 69 mm

A researcher measures the wingspans of

9 guillemots

and finds the

total

wingspan is 8514 mm

Construct a

98% confidence interval

for μ.

5 marks

MARK SCHEME — FULL WORKING

SAMPLE MEAN

x̄ = 8514 ÷ 9

= 946 mm

STANDARD ERROR — SIMPLIFY √N FIRST

SE = σ/√n = 69/√9 = 69/3

= 23 mm

FIND Z FOR 98% CI

Φ(z) = 0.5 + 98/200 = 0.99 → z

z = 2.33

CONSTRUCT CI

946 ± 2.33 × 23 = 946 ± 53.59

✓

98% CONFIDENCE INTERVAL FOR Μ

[ 892.4 , 999.6 ] mm

Width = 107.2 mm | Centre = 946 mm

KEY VALUES

946

x̄ (mm)

sample mean

SE (mm)

69 ÷ 3

2.33

z-value

Φ(z) = 0.99

53.59

margin of error

2.33 × 23

n = 9 (this question)

SE = 23 mm → wide CI

n = 100 (larger sample)

SE = 6.9 mm → narrow CI

Source: AQA Mathematical Studies 1350 Paper 2A 2011

EXAM TIP — SMALL N GIVES A WIDE CI

Here n = 9 , so SE = 69/3 = 23 mm — relatively large, producing a wide interval of 107.2 mm . A larger sample (e.g. n = 100 ) would give SE = 6.9 mm and a much narrower, more precise interval. Always simplify √n when possible — √9 = 3 exactly saves calculation time.

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Past Paper — Non-Standard 96% CI (AQA 2016 Rice)

AQA Mathematical Studies 1350 · Paper 2A · 2016

PAST PAPER QUESTION

The weight of rice in a packet is normally distributed with mean μ grams and standard deviation 1.94 g . A random sample of 50 packets is taken and the sample mean is 251.1 g .

σ = 1.94 g

n = 50

x̄ = 251.1 g

(a)

Construct a

96% confidence interval

for μ, giving limits to 1 d.p.

3 marks

(b)

Comment on the manufacturer's claim that the mean weight

> 250 g

1 mark

96% CI VS CLAIMED VALUE (250 G)

PART (B) — COMMENT ON MANUFACTURER'S CLAIM

The entire CI [250.5, 251.7] g lies above 250 g — the manufacturer's claim that the mean weight is greater than 250 g is supported by the data . There is no evidence to doubt the claim.

MARK SCHEME — STEP-BY-STEP

STANDARD ERROR

SE = σ/√n = 1.94/√50 = 1.94/7.071 = 0.2744

FIND Z FOR 96% CI

NON-STANDARD

Φ(z) = 0.5 + 96/200 = 0.5 + 0.48 = 0.98

→ z = 2.054

CALCULATE MARGIN OF ERROR

E = z × SE = 2.054 × 0.2744 = 0.564

CONSTRUCT CI (TO 1 D.P.)

CI = 251.1 ± 0.564

= [250.5, 251.7] g

FINAL ANSWER (1 D.P.)

96% CI = [250.5, 251.7] g

CLAIM: Μ > 250 G

Supported

EXAM TIP — NON-STANDARD CI METHOD

Always show the Φ(z) line:

Φ(z) = 0.5 + 96/200 = 0.98 → z = 2.054

— this earns the

method mark. If the

entire CI lies above

(or below) a claimed value, the claim is supported (or

contradicted). Never skip the Φ(z) step for non-standard CIs.

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Exercise 6B — Questions

Key Formula: X̄ ~ N(μ, σ²/n) | SE = σ/√n | z = (x̄ − μ) / (σ/√n)

QUESTION 1

Normal distribution — P(X̄ = 62) and P(X̄ > 62)

X ~ N(60, 4²) | n = 15

(a) What is P(X̄ = 62)?

(b) What is P(X̄ > 62)?

Recall: X̄ ~ N(60, 4²/15). For part (a), think about continuous distributions.

QUESTION 2

Variance given directly — find probabilities for X̄

X ~ N(100, 80) | n = 5 | σ = √80 ≈ 8.944

(a) Find P(X̄ > 107)

(b) Find P(101 < X̄ < 109)

Note: variance = 80 is given directly, so σ = √80. SE = σ/√n = √80/√5 = √16 = 4.

QUESTION 3

Daily miles — show weekly total probability < 20%

X ~ N(90, 90²) | n = 7 days | Weekly total > 840 miles

(a) Show that P(motorist drives > 840 miles in a week) < 20%

(b) State any assumptions you make

Hint: Total > 840 miles in 7 days means X̄ > 840/7 = 120 miles/day.

QUESTION 4

Insect body length — compare P for individual vs. sample means

X ~ N(5.1, 0.2²) | Length in mm

(a) Find P(length < 5.08 mm) for a randomly chosen insect

(b) Find P(X̄ < 5.08 mm) for the mean of a sample of n = 10

For (a): use X ~ N(5.1, 0.2²) directly.

For (b): X̄ ~ N(5.1, 0.2²/10) → SE = 0.2/√10 ≈ 0.0632.

For (c): X̄ ~ N(5.1, 0.2²/100) → SE = 0.2/√100 = 0.02.

Notice how the probability changes as n increases — larger n → smaller SE → probability moves further into the tail.

QUESTION 5

Plant heights — find sample size n from a given probability

H ~ N(20, 3²) | P(19 < H̄ < 21) = 0.68268

(a) Explain why P(H̄ < 21) = 0.84134

(b) Hence show that 21 = 20 + 3/√n (i.e. z = 1 at x̄ = 21)

Hint for (a): The interval is symmetric about μ = 20, so P(H̄ < 21) = 0.5 + 0.68268/2.

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Exercise 6B — Answers

AQA CH.6 · DISTRIBUTION OF SAMPLE MEANS

Q1 — N(60, 4²), n = 15

SE = 4/√15 = 1.033 | X̄ ~ N(60, 1.033²)

STEP 1 — P(X̄ = 62)

Key:

Continuous distribution

P(X̄ = 62) = 0 exactly

STEP 2 — P(X̄ > 62)

z =

(62 − 60) / 1.033 = 1.936

P =

1 − Φ(1.936)

P(X̄ > 62) ≈ 0.0265

Q2 — N(100, 80), n = 5

σ = √80 = 8.944 | SE = 8.944/√5 = 4.0

STEP 1 — P(X̄ > 107)

z =

(107 − 100) / 4 = 1.75

P =

1 − Φ(1.75) = 1 − 0.95994

P(X̄ > 107) = 0.04006

STEP 2 — P(101 < X̄ < 109)

z₁ =

0.25 z₂ = 2.25

P =

Φ(2.25) − Φ(0.25)

0.98778 − 0.59871

P = 0.38907

Q3 — N(90, 90²), n = 7 days

Total > 840 miles ⟹ X̄ > 120 (÷ 7)

SE =

90 / √7 = 34.02

z =

(120 − 90) / 34.02 = 0.882

P =

1 − Φ(0.882) ≈ 0.189

P = 18.9% < 20% ✓ Shown

ASSUMPTION

Daily distances are independent and identically distributed

Q4 — N(5.1, 0.2²) insect length (mm)

Find P(length < 5.08) for single insect and sample means

STEP 1 — SINGLE INSECT

z =

(5.08 − 5.1) / 0.2 = −0.1

P = Φ(−0.1) = 0.46017

STEP 2 — N = 10 SE = 0.2/√10 = 0.06325

z =

(5.08 − 5.1) / 0.06325 = −0.316

P = Φ(−0.316) = 0.37600

STEP 3 — N = 100 SE = 0.2/√100 = 0.02

z =

(5.08 − 5.1) / 0.02 = −1.0

P = Φ(−1.0) = 0.15866

Q5 — N(20, 3²) plant heights (cm) | P(19 < H̄ < 21) = 0.68268

STEP 1 — EXPLAIN P(H̄ < 21) = 0.84134

By symmetry:

P(H̄ < 21) = 0.5 + 0.68268/2

= 0.5 + 0.34134 = 0.84134

Distribution is symmetric about μ = 20

STEP 2 — SHOW 21 = 20 + 3/√N (I.E. Z = 1 AT X̄ = 21)

Φ(z) =

0.84134 ⟹ z = 1

So:

(21 − 20) / (3/√n) = 1

3/√n = 1 ⟹ √n = 3

Shown: SE = 3/√n = 1 when √n = 3

STEP 3 — FIND N

From (b):

√n = 3

n = 3² = 9

n = 9

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Consolidation Exercise 6 — Questions (Q1–Q8)

AQA PAST PAPERS 2011–2016

AQA 2011

7 marks

X ~ N(421, 2.5²)

(a)(i) Find P(X = 421). (ii) Find P(X < 425). (iii) Find P(418 < X < 424).

(b) Find the value of x such that P(X < x) = 0.98.

AQA 2013

3 marks

Y ~ N(μ, 25) cm²

A random sample of n = 100 is taken. The sample mean is x̄ = 40.5 cm² .

(a) Calculate a 95% confidence interval for μ.

Note: variance = 25, so σ = √25 = 5. Use SE = σ/√n.

AQA 2011

5 marks

X ~ N(μ, 69²) g

The weights of guillemots follow N(μ, 69²) g. A sample of 9 guillemots has a total weight of 8514 g .

(a) Construct a 98% confidence interval for μ.

Hint: z = 2.33 for a 98% CI. Find x̄ = total ÷ n first.

AQA 2013

4 marks

X ~ N(μ, 0.4²) kg

Bags of flour follow N(μ, 0.4²) kg. A sample of n = 25 bags has a total weight of 497.5 kg .

(a) Construct a 98% CI for μ. (b) Comment on the claim that the mean weight is 20 kg .

AQA

4 marks

X ~ N(μ, 8.3²) g

Pear weights follow N(μ, 8.3²) g. A sample of 6 pears has weights:

160.6, 155.4, 181.3, 176.2, 162.3, 172.8 g

(a) Find a 95% CI for the mean weight. (b) Comment on the suspicion that the mean is less than 175 g .

AQA 2016

4 marks

X ~ N(μ, 1.94²) g

Rice bags follow N(μ, 1.94²) g. A sample of n = 50 bags gives x̄ = 251.1 g .

(a) Construct a 96% CI (give limits to 1 d.p.). (b) Comment on the claim that the mean weight is greater than 250 g .

AQA

4 marks

X ~ N(μ, 10²) ml

Orange juice cartons follow N(μ, 10²) ml. A sample of n = 12 cartons:

763, 769, 746, 765, 756, 755, 756, 750, 758, 758, 765, 755 ml

Calculate a 95% CI . Should the manufacturer be concerned about over-filling ? (Target = 750 ml)

AQA 2013

4 marks

X ~ N(μ, 22) cm²

Grass snake lengths follow N(μ, 22) cm². A sample of n = 10 snakes:

102, 87, 109, 93, 98, 112, 86, 105, 97, 89 cm

Calculate a 95% CI for μ.

Note: variance = 22 (given directly), so σ = √22 ≈ 4.690.

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Consolidation Exercise 6 — Answers (Q1–Q4)

AQA PAST PAPER SOLUTIONS

X ~ N(421, 2.5²) — Probabilities & Reverse (AQA 2011, 7 marks)

STEP 1 — P(X = 421)

P(X = 421) = 0

— continuous distribution, P(X = exact value) = 0

STEP 2 — P(X < 425)

z =

(425 − 421) / 2.5 = 4 / 2.5 = 1.6

P =

Φ(1.6) = 0.9452

P(X < 425) = 0.9452

STEP 3 — P(418 < X < 424)

z₁, z₂ =

±(3/2.5) = ±1.2

P =

2Φ(1.2) − 1 = 2(0.8849) − 1 = 0.7698

P(418 < X < 424) = 0.7698

STEP 4 — FIND X : P(X < X) = 0.98

z =

Φ⁻¹(0.98) = 2.054

x =

421 + 2.054 × 2.5 = 421 + 5.135

x = 426.1 g

Y ~ N(μ, 25) cm², n = 100, x̄ = 40.5 — 95% CI (AQA 2013, 3 marks)

Note: Second parameter is variance = 25, so σ = √25 = 5

σ =

√25 = 5

SE =

σ / √n = 5 / √100 = 5 / 10 = 0.5

z* =

1.96 (for 95% CI)

Margin =

1.96 × 0.5 = 0.98

CI =

[40.5 − 0.98, 40.5 + 0.98]

95% CI = [39.52, 41.48] cm²

Guillemots ~ N(μ, 69²) g, n = 9, Total = 8514 g — 98% CI (AQA 2011, 5 marks)

Hint: z = 2.33 for 98% CI

x̄ =

8514 / 9 = 946 g

SE =

69 / √9 = 69 / 3 = 23

z* =

2.33 (for 98% CI)

Margin =

2.33 × 23 = 53.59

CI =

[946 − 53.59, 946 + 53.59]

98% CI = [892.4, 999.6] g

X ~ N(μ, 0.4²) kg, n = 25, Total = 497.5 kg — 98% CI (AQA 2013, 4 marks)

STEP 1 — CONSTRUCT 98% CI

x̄ =

497.5 / 25 = 19.9 kg

SE =

0.4 / √25 = 0.4 / 5 = 0.08

z* =

2.33 (for 98% CI)

Margin =

2.33 × 0.08 = 0.1864

CI =

[19.9 − 0.1864, 19.9 + 0.1864]

98% CI = [19.71, 20.09] kg

STEP 2 — COMMENT ON CLAIM THAT MEAN = 20 KG

20 kg lies inside CI [19.71, 20.09]

— no reason to doubt the claim

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Consolidation Exercise 6 — Answers

Q5 – Q8

Q5 (AQA, 4 marks)

STEP 1 — SAMPLE MEAN

x̄ = (160.6 + 155.4 + 181.3 + 176.2 + 162.3 + 172.8) / 6

x̄ = 1008.6 / 6 = 168.1 g

STEP 2 — STANDARD ERROR

SE = 8.3 / √6 = 8.3 / 2.449 = 3.389

STEP 3 — 95% CI (Z = 1.96)

168.1 ± 1.96 × 3.389 = 168.1 ± 6.64

95% CONFIDENCE INTERVAL

CI = [161.5, 174.7] g

Comment: Entire CI lies below 175 g — supports the suspicion that the mean weight is less than 175 g.

Q6 (AQA 2016, 4 marks)

STEP 1 — STANDARD ERROR

SE = 1.94 / √50 = 1.94 / 7.071 = 0.2744

STEP 2 — Z-VALUE FOR 96% CI

Φ(z) = 0.5 + 0.96/2 = 0.98 → z = 2.054

STEP 3 — 96% CI

251.1 ± 2.054 × 0.2744 = 251.1 ± 0.564

96% CONFIDENCE INTERVAL (TO 1 D.P.)

CI = [250.5, 251.7] g

Comment: Entire CI lies above 250 g — manufacturer's claim that mean > 250 g is supported.

Q7 (AQA, 4 marks)

STEP 1 — SAMPLE MEAN

Σx = 763+769+746+765+756+755+756+750+758+758+765+755 = 9096

x̄ = 9096 / 12 = 758 ml

STEP 2 — STANDARD ERROR & 95% CI (Z = 1.96)

SE = 10 / √12 = 10 / 3.464 = 2.887

758 ± 1.96 × 2.887 = 758 ± 5.66

95% CONFIDENCE INTERVAL

CI = [752.3, 763.7] ml

Comment: Entire CI lies above 750 ml — manufacturer should be concerned about over-filling.

Q8 (AQA 2013, 4 marks)

⚠ Variance = 22 given directly — take σ = √22

STEP 1 — Σ AND SAMPLE MEAN

σ = √22 = 4.690

Σx = 102+87+109+93+98+112+86+105+97+89 = 978

x̄ = 978 / 10 = 97.8 cm

STEP 2 — STANDARD ERROR & 95% CI (Z = 1.96)

SE = 4.690 / √10 = 4.690 / 3.162 = 1.483

97.8 ± 1.96 × 1.483 = 97.8 ± 2.906

95% CONFIDENCE INTERVAL

CI = [94.9, 100.7] cm

Key note: Second parameter in N(μ, 22) is the variance — always take σ = √22 before computing SE.

Pears ~ N(μ, 8.3²) g | n = 6 | 95% CI

Rice ~ N(μ, 1.94²) g | n = 50, x̄ = 251.1 g | 96% CI

Orange juice ~ N(μ, 10²) ml | n = 12 | 95% CI | Target = 750 ml

Grass snakes ~ N(μ, 22) cm² | n = 10 | 95% CI

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Activity 5 — Tomatoes: 90% Confidence Intervals

N(μ, 4.8²) | z = 1.64

Distribution: Damaged tomatoes per crate ~ N(μ, 4.8²)

90% CI formula: x̄ ± 1.64 × σ/√n

SE formula: σ/√n = 4.8/√n

z-value: 1.64 (90% CI)

(a) Sample A

n = 9

Total:

201 damaged

x̄ =

201 ÷ 9 = 22.33

SE =

4.8 ÷ √9 = 4.8 ÷ 3 = 1.6

Margin =

1.64 × 1.6 = 2.624

90% CONFIDENCE INTERVAL

[19.71, 24.95]

Width = 5.248 ← widest

(b) Sample B

n = 16

Total:

362 damaged

x̄ =

362 ÷ 16 = 22.625

SE =

4.8 ÷ √16 = 4.8 ÷ 4 = 1.2

Margin =

1.64 × 1.2 = 1.968

90% CONFIDENCE INTERVAL

[20.66, 24.59]

Width = 3.936

n = 25

Total:

201 + 362 = 563

x̄ =

563 ÷ 25 = 22.52

SE =

4.8 ÷ √25 = 4.8 ÷ 5 = 0.96

Margin =

1.64 × 0.96 = 1.574

90% CONFIDENCE INTERVAL

[20.95, 24.09]

Width = 3.148 ← narrowest

KEY OBSERVATION

Larger sample size n reduces the Standard Error (SE = σ/√n).

SE = 1.6 (n=9)

SE = 1.2 (n=16)

SE = 0.96 (n=25)

Smaller SE → narrower CI → more precise estimate of μ.

Combining samples A+B gives n = 25 , producing the narrowest CI of width 3.148 .

CI Width Comparison — Larger Sample → Narrower Interval

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Key Formulae Summary & Exam Technique

CHAPTER 6 — CONFIDENCE INTERVALS

1. POINT ESTIMATE

x̄ = Σx / n

Sample mean estimates population mean μ. Larger n→ more accurate estimate.

2. SAMPLE MEAN DISTRIBUTION

X̄ ~ N(μ, σ²/n)

If X~N(μ,σ²) and samples of size n taken, then X̄ follows this distribution.

3. STANDARD ERROR

SE = σ / √n

Standard deviation of the sample mean. Decreases as sample size n increases.

4. STANDARDISING X̄

z = (x̄ − μ) / (σ/√n)

Converts sample mean to a standard normal z-score for probability calculations.

5. 95% CONFIDENCE INTERVAL

x̄ ± 1.96 · σ/√n

z = 1.96 for 95% CI. Most commonly used in AQA exam questions.

6. 99% CONFIDENCE INTERVAL

x̄ ± 2.58 · σ/√n

z = 2.58 for 99% CI. Wider interval — greater confidence, less precision.

7. 90% CONFIDENCE INTERVAL

x̄ ± 1.64 · σ/√n

z = 1.64 for 90% CI. Narrower interval — less confidence, more precision.

8. NON-STANDARD CI

Φ(z) = 0.5 + p/200

For a p% CI, find z from the Φ table using this formula, then apply x̄ ± z·SE.

6-STEP EXAM METHOD

Find x̄ — if given total, divide by n

Extract σ —2nd param is VARIANCE, take √

SE = σ/√n —calculate standard error

Choose z —1.96 / 2.58 / 1.64 or Φ table

CI = x̄ ± z·SE— compute both limits

Comment —is claimed value inside or outside CI?

COMMON MISTAKES TO AVOID

Do NOT use σ directly as the standard error — always divide by √n to get SE = σ/√n.

Do NOT say "95% probability that μ is in this interval" — μ is a fixed constant, not random.

Do NOT forget to take √ of the variance — N(μ, 25) means σ = √25 = 5, not σ = 25.

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Interactive Quiz — Chapter 6

TIMED QUIZ

AQA Level 3 Mathematical Studies — Chapter 6: Confidence Intervals

Revealed: 0 / 5

Question 1 of 5

Q1 — DISTRIBUTION OF SAMPLE MEAN

A population has distribution N(50, 4²). Random samples of size 16 are taken. What is the distribution of the sample mean X̄?

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ALL QUESTIONS

N(50, 4²), n=16. What is the distribution of X̄?

Flour bags N(μ, 3²)g, n=9, x̄=498g. Find 95% CI for μ.

N(μ, 10²), n=25, x̄=340. Find 99% CI for μ.

Plants N(μ, 5²)cm, n=100, x̄=42.3cm. 95% CI — does μ=43cm?

90% CI is [18.4, 21.6], n=9, σ=3. What was x̄?