AQA

LEVEL 3 MATHEMATICAL STUDIES

1h 30min

60 marks

Calculator allowed

CH 5

The Normal Distribution

CH 6

Confidence Intervals

CH 7

Correlation & Regression

AQA Mathematical Studies

Paper 2A— Revision

Chapters 5, 6 & 7 ·60 marks ·1h 30min ·Calculator allowed

OVERVIEW → SKILL → PAST PAPER → ANSWER

Paper 2A — What to Expect

EXAM INFO

1 hour 30 minutes

60 marks

Calculator allowed

Chapters 5, 6 & 7

The Normal Distribution

Bell-shaped, symmetrical distributions. Standardise using z = (x − μ)/σ, find probabilities with Φ(z), and solve reverse problems to find x given a probability.

N(μ, σ²) notation

z = (x−μ)/σ

Φ(z) tables

Reverse problems

Confidence Intervals

Sample means follow X̄ ~ N(μ, σ²/n). Use standard error σ/√n to construct 90%, 95%, and 99% confidence intervals from sample data.

Point estimates

Standard error σ/√n

90% / 95% / 99% CIs

Interpretation

Correlation & Regression

Scatter graphs, lines of best fit through (x̄, ȳ), regression lines y = a + bx from a calculator, and Pearson's PMCC r to measure correlation strength.

Lines of best fit

y = a + bx

PMCC r

Interpolation

Revision Structure

Each skill is taught step-by-step, immediately followed by a real AQA past paper question on that exact skill, then a fully worked answer slide.

① Skill explanation

② Past paper Q

③ Worked answer

Show all working

CHAPTER 5

CHAPTER 6

CHAPTER 7

THIS DECK

Exam strategy: Show all working, use your calculator for z-values and regression, and always interpret results in context.

Chapter 5 — The Normal Distribution: Key Skills

AQA Mathematical Studies

KEY Φ VALUES:

Φ(1.282) = 0.90

90th percentile

Φ(1.645) = 0.95

95th percentile

Φ(1.960) = 0.975

97.5th percentile

Φ(2.326) = 0.99

99th percentile

Φ(2.576) = 0.995

99.5th percentile

THE NORMAL DISTRIBUTION X ~ N(Μ, Σ²)

68% within ±1σ

95% within ±2σ

99.7% within ±3σ

COMMON MISTAKES TO AVOID

N(μ, σ²) — the second parameter is VARIANCE. Always take √ to find σ before standardising.

P(X = exact value) = 0 for all continuous distributions — state this explicitly for 1-mark questions.

Negative z: use symmetry Φ(−z) = 1 − Φ(z) — never look up a negative z directly in tables.

KEY FORMULAE & RULES

Notation:

X ~ N(μ, σ²)

σ² is VARIANCE — take √ to get σ

Standardise:

z = (x − μ) / σ

converts to N(0,1)

P(X < a):

Φ((a−μ)/σ)

P(X > a):

1 − Φ((a−μ)/σ)

P(a < X < b):

Φ(z₂) − Φ(z₁)

Symmetry:

Φ(−z) = 1 − Φ(z)

for negative z values

Reverse — find a:

Step 1: find z where Φ(z) = p  →  Step 2:

a = μ + z·σ

Past Paper — Normal Distribution (AQA 2011)

X ~ N(421, 2.5²)  |  7 marks total

Exam Tip: P(X = exact value) = 0 for ALL continuous distributions — always state this explicitly to secure the 1 mark. For part (a)(iii), use symmetry: since z₁ = −z₂, the probability equals 2Φ(z₂) − 1. For reverse problems, always convert to P(X < x) form first, then find z from tables.

QUESTION

The weight of a tin of baked beans is normally distributed with mean 421 g and standard deviation 2.5 g .

X ~ N(421, 2.5²)  →  μ = 421, σ = 2.5

(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 x such that P(X < x) = 0.98.

2 marks

MARK SCHEME

(a)(i)

Continuous distribution — probability of any exact value is zero.

P(X = 421) = 0

1 mark

(a)(ii)

Standardise: z = (425 − 421) / 2.5 = 4 / 2.5 = 1.6

Look up: Φ(1.6) = 0.9452

P(X < 425) = Φ(1.6) = 0.9452

2 marks

(a)(iii)

z₁ = (418 − 421) / 2.5 = −1.2    z₂ = (424 − 421) / 2.5 = 1.2

By symmetry: P = 2Φ(1.2) − 1 = 2(0.8849) − 1

P(418 < X < 424) = 0.7698

2 marks

(b)

Φ(z) = 0.98  →  z = 2.054 (from tables)

x = μ + z·σ = 421 + 2.054 × 2.5 = 421 + 5.135

x = 426.1 g

2 marks

Standardising & Finding Probabilities

Converting any normal distribution to Z ~ N(0,1) using the standardising formula

Step-by-Step Method

1

Identify

μ

and

σ

from the notation X ~ N(μ, σ²) — remember

σ = √(variance)

2

Standardise:

z = (x − μ) / σ

— this converts any value to a standard

normal score

3

Look up

Φ(z)

in tables or use calculator to get P(Z < z)

4

Apply symmetry if z is negative:

Φ(−z) = 1 − Φ(z)

5

For intervals P(a < X < b):

Φ(z₂) − Φ(z₁)

KEY Φ VALUES FOR EXAM

Worked Example — Heights N(170, 8²)

Heights of adults: H ~ N(170, 8²)  →  μ = 170 cm , σ = 8 cm

FIND P(H > 178)

Step 2:

z = (178 − 170) / 8 = 1.00

Step 3:

Φ(1.00) = 0.8413

P(H>178):

1 − Φ(1.00) = 1 − 0.8413 = 0.1587

ANSWER A:

P(H > 178) = 0.1587 ≈ 15.9%

FIND P(160 < H < 180)

z₁:

(160 − 170) / 8 = −1.25

z₂:

(180 − 170) / 8 = +1.25

Step 5:

Φ(1.25) − Φ(−1.25) = 2Φ(1.25) − 1

Result:

2 × 0.8944 − 1 = 0.7887

ANSWER B:

P(160 < H < 180) = 0.7887 ≈ 78.9%

Chapter 5 · AQA Mathematical Studies

Past Paper — Standardising (AQA)

AQA Mathematical Studies · Paper 2A

PAST PAPER QUESTION

The heights of adult males are normally distributed with mean 175 cm and standard deviation 7 cm .

H ~ N(175, 7²)

(a)

Find P(H > 182).

2 marks

(b)

Find P(168 < H < 189).

3 marks

(c)

A doorway is designed so that 95% of adult males can pass through without ducking. Find the minimum height of the doorway.

3 marks

USEFUL Φ VALUES

Φ(1.0) = 0.84134

Φ(1.645) = 0.95000

Φ(1.5) = 0.93319

Φ(2.0) = 0.97725

Φ(1.96) = 0.97500

Φ(2.326) = 0.99000

Symmetry rule: Φ(−z) = 1 − Φ(z)

MARK SCHEME

PART (A) — P(H > 182)

2 marks

Standardise: z = (182 − 175) / 7 = 1

Probability: P = 1 − Φ(1) = 1 − 0.84134

P(H > 182) ≈ 0.159

PART (B) — P(168 < H < 189)

3 marks

Lower z: z₁ = (168 − 175) / 7 = −1

Upper z: z₂ = (189 − 175) / 7 = 2

Apply: Φ(2) − Φ(−1) = Φ(2) − (1 − Φ(1))

= 0.97725 − (1 − 0.84134) = 0.97725 − 0.15866

P(168 < H < 189) ≈ 0.819

PART (C) — MINIMUM DOORWAY HEIGHT

3 marks

Need: P(H < h) = 0.95

Find z: Φ(z) = 0.95 → z = 1.645

Reverse: h = μ + z·σ = 175 + 1.645 × 7

h = 175 + 11.515

Minimum height h = 186.5 cm

EXAM TIP

For part (c), always interpret the answer in context : "The doorway must be at least 186.5 cm so that 95% of adult males can pass through without ducking." Answers without context lose the final mark.

CHAPTER 5 · KEY SKILL

Reverse Normal Problems

Finding x given a probability — step-by-step method with worked examples

Step-by-Step Method

1

Write in the form

P(X < a) = p

. Rearrange if given P(X > a) or a

symmetric interval.

2

If

p > 0.5

: look up z directly —

Φ(z) = p

If

p < 0.5

: use symmetry —

z = −Φ⁻¹(1 − p)

 (z is negative)

3

Compute the value:

a = μ + z · σ

4

Symmetric interval

P(−b < Z < b) = p

: use

Φ(b) = 0.5 + p/2

KEY Z-VALUES FOR REVERSE PROBLEMS

Worked Examples — Flour Bags: W ~ N(500, 8²)

Context: μ = 500 g, σ = 8 g  (σ² = 64, so σ = √64 = 8)

EXAMPLE 1 — P > 0.5 (Z IS POSITIVE)

Find the weight below which 90% of bags fall.

i.e. find a such that P(W < a) = 0.90

▶

p = 0.90 > 0.5 → look up directly: Φ(z) = 0.90

▶

From table: z = 1.282

▶

a = 500 + 1.282 × 8 = 500 + 10.256 = 510.3 g

EXAMPLE 2 — P < 0.5 (Z IS NEGATIVE)

Find the weight below which only 5% of bags fall.

i.e. find a such that P(W < a) = 0.05

▶

p = 0.05 < 0.5 → use symmetry: z = −Φ⁻¹(1 − 0.05)

▶

Φ⁻¹(0.95) = 1.645, so z = −1.645

▶

a = 500 + (−1.645) × 8 = 500 − 13.16 = 486.8 g

Exam Tip: Always convert to P(X < a) = p form first. If p < 0.5, the answer a will be below the mean — z must be negative. Check: is your answer on the correct side ofμ?

Past Paper — Reverse Normal (AQA)

W ~ N(500, 8²)  |  Total: 7 marks

AQA MATHEMATICAL STUDIES — PAPER 2A QUESTION

The weights of bags of flour are normally distributed with mean 500 g and standard deviation 8 g , so W ~ N(500, 8²) .

(a)

Find the weight below which 90% of bags fall.

2 marks

(b)

Find the weight above which only 2% of bags fall.

3 marks

(c)

Bags weighing less than 485 g are rejected. What percentage are

rejected?

2 marks

Exam Tip: For part (b), "above which only 2% fall" means P(W > w) = 0.02 . Always convert to P(X < a) form first before looking up z — so P(W < w) = 1 − 0.02 = 0.98, then find z such that Φ(z) = 0.98.

(A)

90TH PERCENTILE

2 MARKS

SET UP:

Find w such that P(W < w) = 0.90

FIND Z:

Φ(z) = 0.90 → z = 1.282

APPLY FORMULA:

w = μ + z·σ

w = 500 + 1.282 × 8

w = 500 + 10.256

w = 510.3 g

(B)

TOP 2% THRESHOLD

3 MARKS

CONVERT:

"Above which only 2% fall" means

P(W > w) = 0.02

∴ P(W < w) = 0.98

FIND Z:

Φ(z) = 0.98 → z = 2.054

APPLY FORMULA:

w = 500 + 2.054 × 8

w = 500 + 16.432

w = 516.4 g

(C)

QUALITY CONTROL

2 MARKS

SET UP:

Find P(W < 485)

STANDARDISE:

z = (485 − 500) / 8

z = −15 / 8 = −1.875

SYMMETRY RULE:

Φ(−1.875) = 1 − Φ(1.875)

= 1 − 0.9696 = 0.0304

3.04% of bags are rejected

Distribution of Sample Means

Chapter 6.2 — Key Skill & Worked Example

Core Theory & Formulae

MAIN RESULT

If X ~ N(μ, σ²) and sample size n:

X̄ ~ N(μ, σ²/n)

STANDARD ERROR (SE)

SE = σ / √n

STANDARDISE

z = (x̄ − μ) / (σ / √n)

Larger n → smaller SE → sample means cluster more tightly around μ, giving a more precise estimate.

P(X̄ = exact value) = 0 for any continuous distribution — always use an interval.

KEY WARNING: N(μ, σ²) — the second parameter is VARIANCE . Always take√ to find σ before calculating SE.

Worked Example — Chocolates

W ~ N(10, 2²) — weight of a chocolate (grams). Find P(9.8 < W < 10.2) for:

(a) Single chocolate

σ = 2  |  z₁ = (9.8 − 10) / 2 = −0.1 ,  z₂ = 0.1

P = Φ(0.1) − Φ(−0.1) = 0.07966

(b) Sample mean of 25 chocolates

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) = 0.38292

P rises from 0.07966 → 0.38292 — nearly 5× higher for the sample mean. Larger n concentrates the distribution, making the interval much more likely to contain the mean.

Past Paper — Distribution of Sample Means

AQA Mathematical Studies 1350 · Paper 2A · 2013

Exam Tip — Most Common Mistake: When the question states X ~ N(μ, 25), the 25 is the VARIANCE , not the standard deviation. Always compute σ = √25 = 5 before finding SE = σ/√n. Forgetting this step is the most frequent source of lost marks on this topic.

QUESTION

The diameter of a disc produced by a machine has a normal distribution with mean μcm and variance 25 cm² .

A random sample of 100 discs is taken and the sample mean diameter 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)

1

Variance = 25 , so σ = √25 = 5

⚠ Second parameter is VARIANCE — must take square root first

M1

2

Standard Error: SE = σ/√n = 5/√100 = 0.5

M1

3

95% CI uses z = 1.96 :

x̄ ± 1.96 × SE = 40.5 ± 1.96 × 0.5 = 40.5 ± 0.98

A1

✓

95% CI = [39.52, 41.48] cm

Answer

Part (b) — Comment [1 mark]

40 cm lies inside the CI [39.52, 41.48] — there is no evidence to suggest the machine is NOT producing discs with mean diameter 40 cm.

Chapter 6 — Confidence Intervals: Key Skill

AQA Level 3 Mathematical Studies

CI FORMULAE REFERENCE

Non-Standard p% CI

Find z such that Φ(z) = 0.5 + p/200

Then use: CI = x̄ ± z × σ/√n  |  Example: 98% CI → Φ(z) = 0.5 + 98/200 = 0.99 → z = 2.33

WHAT DOES A 95% CI MEAN?

If 100 samples were taken and 100 confidence intervals calculated, approximately 95 of them would contain the true population mean μ. The CI does not mean there is a 95% chance μ is in this specific interval.

6-STEP METHOD

1

Find x̄ — if given total, divide by n

2

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

3

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

4

Find z — use table for required confidence level

5

CI = x̄ ± z × SE — compute both limits

6

Comment — is claimed value inside or outside CI?

Key Warning: N(μ, σ²) — the second parameter is the VARIANCE , not the standard deviation. Always take √ to find σ before calculating SE. This is the most common source of lost marks.

WORKED EXAMPLE

Biscuits — N(μ, 5²), n = 10

Data (grams): 442, 450, 447, 446, 453, 449, 444, 454, 443, 457  |  Claimed mean: 450 g  | Construct a 99% CI.

Step 1

x̄ = 4485 ÷ 10

= 448.5 g

Step 2

N(μ, 5²) → σ = √25

= 5

Step 3

SE = 5 ÷ √10

= 1.581

Step 4

99% CI → z

= 2.58

Step 5

448.5 ± 2.58 × 1.581 = 448.5 ± 4.08

= [444.4, 452.6]

Step 6 — Comment: The claimed mean of 450 g lies inside the 99% CI [444.4, 452.6] g —there is no reason to doubt the claim.

Past Paper — Confidence Intervals

AQA 2013 · PAPER 2A

EXAM TIP

For a 98% CI , always show: Φ(z) = 0.5 + 98/200 = 0.99 → z = 2.33 . This is a non-standard CI — you must look up z from tables. Writing the Φ(z) line earns the method mark — never skip it . Also: σ = 0.4 is given directly here (not variance), so no square root needed.

QUESTION

AQA Mathematical Studies 1350 — Paper 2A, 2013

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

with mean

μ kg

and standard deviation

0.4 kg

, so

X ~ N(μ, 0.4²)

A random sample of 25 bags is taken and the total weight of sand is 497.5 kg .

(a)

Construct a 98% confidence interval for μ.

3 marks

(b)

Comment on whether the machine is filling bags with a mean weight of 20 kg .

1 mark

GIVEN INFORMATION

σ = 0.4

std deviation (kg)

n = 25

sample size

Σx = 497.5

total weight (kg)

MARK SCHEME

PART (A) — CONSTRUCT 98% CI

3 MARKS

Find sample mean: x̄ = 497.5 ÷ 25 = 19.9 kg

Standard error: SE = σ/√n = 0.4/√25 = 0.4/5 = 0.08

98% CI → Φ(z) = 0.5 + 98/200 = 0.99 → z = 2.33

CI = 19.9 ± 2.33 × 0.08 = 19.9 ± 0.1864

98% CI = [19.71, 20.09] kg

PART (B) — COMMENT

1 MARK

The claimed mean of 20 kg lies inside 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.

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

AQA PAPER 2A · 2016

Exam Tip — Non-Standard CI Method:

For a 96% CI, always show the full line:

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

.

This

Φ(z)

calculation earns the

method mark

— never

skip it.

Standard values: 95% → z = 1.96  |  99% → z = 2.576  |  90% → z = 1.645  |  Non-standard → use

Φ(z) = 0.5 + p/200

.

QUESTION

The weight of rice in a packet is normally distributed:

X ~ N(μ, 1.94²)

mean μ grams, standard deviation σ = 1.94 g

Sample size

n = 50

Sample mean

x̄ = 251.1 g

σ (given directly)

1.94 g

CI level

96%

(a)

Construct a 96% confidence interval for μ, giving limits to 1 decimal place .

3 marks

(b)

Comment on the manufacturer's claim that the mean weight > 250 g .

1 mark

MARK SCHEME

PART (A) — 96% CONFIDENCE INTERVAL

1

Standard Error: SE = σ/√n = 1.94/√50 = 0.2744

2

Find z (non-standard): Φ(z) = 0.5 + 96/200 = 0.98 → z = 2.054

3

CI calculation: 251.1 ± 2.054 × 0.2744 = 251.1 ± 0.564

4

Final answer (1 d.p.):

96% CI =

[250.5, 251.7] g

PART (B) — COMMENT ON CLAIM

The entire confidence interval [250.5, 251.7] g lies above 250 g .

∴ The manufacturer's claim that the mean weight is > 250 g is supported by the data.

AQA Mathematical Studies 1350 Paper 2A 2016

Lines of Best Fit

CHAPTER 7.1 — KEY SKILL

Exam Tip: Always calculate the mean point first and ensure your line passes through (x̄, ȳ). The gradient tells you the rate of change — always interpret it in context with units. P(X̄ = exact value) = 0 for continuous distributions.

4-Step Method

1

Plot all data points on a scatter graph with sensible scales and labelled axes.

2

Calculate the mean of x-values (x̄) and mean of y-values (ȳ) .

3

Plot the mean point (x̄, ȳ) clearly — your line MUST pass through it.

4

Draw a straight line through (x̄, ȳ) with roughly equal points above and below .

KEY RULES

Gradient = Δy ÷ Δx — always state units and interpret in context

Interpolation (within data range) = reliable

Extrapolation (outside data range) = unreliable

Worked Example — Pocket Money vs Age

DATA (AGE 5–16, WEEKLY POCKET MONEY)

STEP 2 — CALCULATE MEAN POINT

x̄ = (5+6+…+16) ÷ 12 = 10.5  |  ȳ = (3.28+4.00+…+10.27) ÷ 12 = £6.33

Mean point: (10.5, £6.33) — mark clearly and draw line through it.

STEP 4 — CALCULATE GRADIENT

Gradient = (10.27 − 3.28) ÷ (16 − 5) = 6.99 ÷ 11 ≈ £0.64 per year

Interpretation: Children receive approximately 64p more pocket money per week for each year they get older.

CORRELATION

Strong positive correlation — as age increases, pocket money increases. State "strong" or "weak" when describing correlation.

Past Paper — Lines of Best Fit (AQA)

AQA PAPER 2A · EXERCISE 7A

QUESTION

The table shows average weekly pocket money (£) for children aged 5–16 in 2015.

(a) Draw a scatter graph and a line of best fit. [2 marks]

(b) Describe the correlation. [1 mark]

(c) How much extra pocket money per year older? [2 marks]

DATA: AVERAGE WEEKLY POCKET MONEY 2015

(A) MEAN POINT

2 MARKS

Plot all 12 points. Calculate mean point: (10.5, £6.33) . Draw line through this point with roughly equal points above and below.

(B) CORRELATION

1 MARK

Strong positive correlation— as age increases, pocket money increases.

(C) GRADIENT

2 MARKS

Gradient = (10.27 − 3.28) ÷(16 − 5) = 6.99 ÷ 11 ≈£0.64/year . Children receive ~64p more per week for each year older.

⚠ EXAM TIP

Always calculate the mean point first and ensure your line passes through it. State 'strong' or 'weak' when describing correlation. Interpret gradient in context with units .

Scatter Graph with Line of Best Fit

Data points

Line of best fit

Mean point (10.5, £6.33)

Chapter 7 — Regression Lines: Key Skill

AQA Mathematical Studies · Section 7.2 · Finding and interpreting y = a + bx

4-Step Method

y = a + bx

1

Enter data into calculator — input all x-values and y-values into the statistics/regression mode.

2

Use regression function — select linear regression to obtain values of a and b from the calculator.

3

Write the equation — state y = a + bx with values rounded to 3 significant figures.

4

Plot using mean point — the line MUST pass through (x̄, ȳ); use one other calculated point to draw it.

B = GRADIENT

Increase in y per unit increase in x. Always state units and context.

A = Y-INTERCEPT

Value of y when x = 0. May not be meaningful — check if x = 0 is realistic.

Worked Example — Oral & Written Marks

Context: Students sit an oral exam (x) and a written exam (y). Data for 10 students is entered into a calculator. The regression line of y on x is found to be:

y = 10.8 + 0.574x

KEY VALUES FROM CALCULATOR

Mean point: (x̄, ȳ) = (26.1, 25.8) — line passes through this point ✓

Gradient b = 0.574 — for each extra oral mark, written mark increases by 0.574

Intercept a = 10.8 — predicted written mark when oral = 0 (not meaningful here)

Ed's prediction (oral mark = 34):

y = 10.8 + 0.574 × 34 = 10.8 + 19.516 = 30.316 ≈ 30 marks

Interpolation vs Extrapolation: Only substitute x-values within the data range (interpolation = reliable). Predicting outside the range (extrapolation) is unreliable— state this in exam answers.

Past Paper — Regression Lines (AQA 2013 Sample)

CHAPTER 7 · PAPER 2A

QUESTION

The table shows oral and written marks for 12 students. Ed scored 34 on oral but was absent for written.

(a)

Find the equation of the regression line of written on oral

(ignore Ed).

2

(b) Draw a scatter graph and the regression line.

(c)

Predict Ed's written mark.

1

STUDENT DATA (ORAL / WRITTEN MARKS)

MARK SCHEME ANSWERS

(a) Regression line: y = 10.8 + 0.574x  [2 marks] —use calculator for a and b

Mean point: (26.1, 25.8) — line must pass through this point

(c) Ed's prediction: x = 34 → y = 10.8 + 0.574 × 34 = 30.3 ≈ 30 marks [1 mark]

Exam Tip: Always exclude incomplete data (Ed) when finding the regression line. Interpolation (x=34 is within range 16–37) is reliable. State the prediction clearly: "Ed's predicted written mark is approximately 30."

Scatter Graph: Oral vs Written Marks — with Regression Line y = 10.8 + 0.574x

Chapter 7 — Pearson's PMCC: Key Skill

USE CALCULATOR

FORMULA & PROPERTIES

AQA Ch 7.3

PMCC Formula

Pearson's r

r = s xy / (s x · s y )

Do NOT calculate by hand — use your calculator

Range:

−1 ≤ r ≤ +1 always

Sign:

Same as gradient of regression line

Strength:

|r| close to 1 = strong

Weak:

|r| close to 0 = weak / no correlation

Always interpret direction AND strength in context

State r to 3 significant figures

INTERPRETATION SCALE

r from −1 to +1

Strength Guide

Direction & Magnitude

r = +1

Perfect positive

+0.7 to +1

Strong positive

0 to +0.7

Moderate / weak +

r = 0

No linear correlation

−0.7 to 0

Moderate / weak −

−1 to −0.7

Strong negative

r = −1

Perfect negative

As one variable increases, the other increases (positive) or decreases (negative)

WORKED EXAMPLE

Airliners Data

5-Step Method

Calculator approach

1

Enter x and y data into calculator

2

Use regression function to find a and b

3

Use PMCC function to find r

4

State r to 3 significant figures

5

Interpret — state direction AND strength in context

Airliners — Length vs Wingspan

r = 0.960

Strong positive correlation

As length increases, wingspan also increases strongly.

Light Aircraft — Length vs Wingspan

r = 0.625

Moderate positive correlation

Less strongly correlated than airliners.

Past Paper — Pearson's PMCC (AQA)

CHAPTER 7 · PAPER 2A

QUESTION — HOUSE PRICES & RENTS, ENGLAND 2011

MARK SCHEME

(a)

Regression line:

R = 55.1 + 0.107P

(3 s.f.) — passes

through (228.6, 76.7)

2 marks

(b)

PMCC:

r = 0.981

(3 s.f.) — very strong positive

correlation

1 mark

(c)

Plot 9 data points; draw regression line through mean point

(228.6, 76.7)

2 marks

SCATTER DIAGRAM — R ON P WITH REGRESSION LINE

EXAM TIP — INTERPRETING PMCC

Always state the PMCC value AND interpret it in context: "r = 0.981 shows a very strong positive correlation between house prices and weekly rents — regions with higher house prices tend to have higher rents."

Weekly Rent (£R) vs House Price (£P thousands)

Data points

Regression line

Mean point

Exam Technique — How to Score Maximum Marks

PAPER 2A

Show All Working

Method marks are awarded even with a wrong final answer. Never write only the answer — show every step clearly.

N(μ, σ²): Always Take √ of Variance

The second parameter is

variance

, not σ. Always

compute

σ = √(σ²)

before any standardisation.

Most common source of lost marks.

P(X = exact value) = 0

For continuous distributions, state this explicitly. A 1-

mark question expects the statement

P(X = k) = 0

written out — do not skip it.

Non-Standard CI: Show Φ(z) = 0.5 + p/200

Always write out the full calculation

Φ(z) = 0.5 +

p/200

to earn the method mark — even if you know

the z-value directly.

Interpret in Context

For regression gradient, PMCC, and CI conclusions, always state what the numbers mean using the actual variable names from the question.

Use Your Calculator

Find regression coefficients

a

and

b

, and PMCC

r

, using the calculator. Never calculate these by

hand in the exam.

COMMON MISTAKES

Confusing variance and σ in N(μ,σ²)

Forgetting to comment on CI claim

Stating r without interpretation

Extrapolating beyond data range

Complete Formulae Reference — Paper 2A

AQA MATHEMATICAL STUDIES

⚠ CRITICAL EXAM WARNING:

N(μ, σ²)

— the second parameter is VARIANCE, NOT standard deviation. Always take √ before any calculation.  | 

P(X = exact) = 0

for continuous

distributions — state this explicitly.  | 

Always interpret r and regression gradient

in context

using actual variable names.

CHAPTER 5 — NORMAL DISTRIBUTION

NOTATION & STANDARDISING

Distribution notation

X ~ N(μ, σ²)

⚠ Second parameter is VARIANCE — take √ to get σ

Standardise to z-score

z = (x − μ) / σ

PROBABILITY RULES

Less than / greater than

P(X < a) = Φ(z) P(X > a) = 1 − Φ(z)

Symmetry property

Φ(−z) = 1 − Φ(z)

Reverse (find x from p)

a = μ + z · σ

KEY Φ VALUES

CHAPTER 6 — CONFIDENCE INTERVALS

DISTRIBUTION OF SAMPLE MEAN

Sample mean distribution

X̄ ~ N(μ, σ²/n)

σ²/n is the variance of X̄ — take √ to get SE

Standard error

SE = σ / √n

Standardise sample mean

z = (x̄ − μ) / (σ / √n)

CONFIDENCE INTERVAL FORMULAE

90% CI (z = 1.645)

x̄ ± 1.645 · σ/√n

95% CI (z = 1.960)

x̄ ± 1.96 · σ/√n

99% CI (z = 2.576)

x̄ ± 2.576 · σ/√n

NON-STANDARD CI

Find z for p% CI

Φ(z) = 0.5 + p/200

e.g. 96% CI: Φ(z) = 0.5 + 96/200 = 0.98 → z = 2.054

CHAPTER 7 — CORRELATION & REGRESSION

LINE OF BEST FIT

Must pass through mean point

(x̄, ȳ) — always on the line

Gradient from graph

gradient = Δy / Δx

REGRESSION LINE

Equation form (use calculator)

y = a + bx

b = gradient = increase in y per unit x increase

Interpolation vs Extrapolation

Within range → reliable Outside range →unreliable

PMCC

Pearson's PMCC (use calculator)

r = s_xy / (s_x · s_y)

Range & interpretation

−1 ≤ r ≤ +1 Sign = direction; |r| = strength

Always state direction AND strength in context

Practice Questions — Chapters 5 & 6 (with Answers)

CH 5: NORMAL DISTRIBUTION

CH 6: CONFIDENCE INTERVALS

CHAPTER 5 — NORMAL DISTRIBUTION

Q1

X ~ N(50, 4²). Find P(X > 54).

ANSWER

z = (54 − 50) / 4 = 1

P(X > 54) = 1 − Φ(1) = 1 − 0.84134 = 0.159

Q2

X ~ N(100, 25). Find x such that P(X < x) = 0.95.

⚠ Variance = 25, so σ = √25 = 5 (not 25!)

ANSWER

Φ(z) = 0.95 → z = 1.645

x = μ + z·σ = 100 + 1.645 × 5 = 108.2

Q3

X ~ N(70, 9). Find P(64 < X < 76).

⚠ Variance = 9, so σ = √9 = 3

ANSWER

z₁ = (64 − 70)/3 = −2, z₂ = (76 − 70)/3 = 2

P = 2Φ(2) − 1 = 2(0.97725) − 1 = 0.9545

CHAPTER 6 — CONFIDENCE INTERVALS

⭐ KEY EXAM TIP

N(μ, σ²) — the second parameter is ALWAYS the variance . Take √ to find σ before any calculation. This is the most common source of lost marks.

Q4

X ~ N(μ, 16), n = 64, x̄ = 42.5. Construct a 95% CI for μ.

⚠ Variance = 16, so σ = √16 = 4

ANSWER

SE = σ/√n = 4/√64 = 4/8 = 0.5

95% CI = x̄ ± 1.96 × SE = 42.5 ± 1.96 × 0.5

= 42.5 ± 0.98 = [41.52, 43.48]

Q5

X ~ N(μ, 0.36), n = 25, total = 497.5. Construct a 99% CI for μ.

⚠ Variance = 0.36, so σ = √0.36 = 0.6

ANSWER

x̄ = 497.5 / 25 = 19.9

SE = 0.6/√25 = 0.6/5 = 0.12

99% CI = 19.9 ± 2.58 × 0.12 = 19.9 ± 0.31

= [19.59, 20.21]

QUICK REFERENCE

Key z-values & CI formulae

z-values:

Φ(1.645) = 0.95 → 90% CI

Φ(1.960) = 0.975 → 95% CI

Φ(2.326) = 0.99 → 98% CI

Φ(2.576) = 0.995 → 99% CI

Formulae:

z = (x − μ) / σ

SE = σ / √n

CI = x̄ ± z·(σ/√n)

x̄ = Σx / n

Practice Questions — Chapter 7

Correlation & Regression

Questions & Answers

Q1 — DESCRIBING CORRELATION

A scatter graph shows the relationship between hours of sunshine and ice cream sales. Describe the expected correlation and explain what it means.

✓ ANSWER

Strong positive correlation — as hours of sunshine increase, ice cream sales also increase. More sunshine leads to more ice cream sold.

Q4 — INTERPRETING PMCC

The PMCC for two variables is r = −0.87. Describe the correlation fully.

✓ ANSWER

Strong negative correlation — r = −0.87 is close to −1, so as one variable increases, the other decreases strongly. State both direction AND strength.

Q2 — MEAN POINT ERROR

The mean of x-values is 12 and the mean of y-values is 35. A student draws a line of best fit that does NOT pass through (12, 35). What mistake have they made?

✓ ANSWER

The line of best fit must always pass through the mean point (x̄, ȳ) = (12, 35). The student's line is incorrect.

Q5 — EXTRAPOLATION PROBLEM

A regression line is used to predict y when x = 150, but the data only covers x from 20 to 80. What is the problem with this prediction?

✓ ANSWER

Extrapolation — x = 150 is outside the data range (20–80). Predictions outside the data range are unreliable . Only interpolation (within range) is reliable.

Q3 — GRADIENT INTERPRETATION

The regression line of y on x is y = 8.5 + 2.3x. (a) What is the gradient? (b) Interpret the gradient if x = hours of study and y = exam mark.

✓ ANSWER

(a) Gradient = 2.3 .  (b) For each extra hour of study, the exam mark increases by 2.3 marks . Always interpret in context using variable names.

Line of best fit: must pass through (x̄, ȳ)

PMCC: −1 ≤ r ≤ +1  |  sign = direction, |r| = strength

Gradient: increase in y per unit increase in x

Interpolation (within range) = reliable  | Extrapolation = unreliable

Interactive Revision Quiz — Paper 2A

30-Second Timer

1

2

3

4

5

Question 1 of 5

CHAPTER 5 — NORMAL DISTRIBUTION

X ~ N(60, 16). Find P(X > 64).

⚠ Variance Warning: N(60, 16 ) — the second parameter is the variance . Take √16 = 4 to find σ before standardising.

TIME REMAINING

24

Reveal Answer

Restart Quiz

Exam Tip: Show all working — method marks are available even if the final answer is wrong.

KEY REMINDERS

N(μ, σ²) — always take √ of variance to find σ

Use calculator for regression a, b and PMCC r

SE = σ/√n for sample means (Ch6)

Interpret r: direction AND strength in context

QUESTION 1

AQA MATHEMATICAL STUDIES — PAPER 2A

Good Luck in Paper 2A!

FINAL REVISION REMINDERS

N(μ,σ²): ALWAYS take √ of the variance to find σ before any calculation — most common error.

P(X = exact) = 0 for continuous distributions — state this explicitly for 1-mark questions.

Standardise: z = (x − μ)/σ for single values; z = (x̄ − μ)/(σ/√n) for sample means.

Non-standard CI: Always show Φ(z) = 0.5 + p/200 to earn the method mark.

Line of best fit must always pass through the mean point (x̄, ȳ) — never draw it elsewhere.

Regression y = a + bx: Use your calculator for a and b — never calculate by hand.

PMCC r: Use calculator; always interpret direction AND strength in context of the variables.

Interpolation (within data range) is reliable; extrapolation (outside range) is unreliable.

Show ALL working — even with a wrong answer, method marks are still available to earn.

Interpret in context: Use actual variable names — never just quote a number without explanation.

You've got this! 🎯

Chapters 5, 6 & 7 — Normal Distribution · Confidence Intervals · Correlation & Regression

Chapter 5 — Standardising: Worked Examples

Always standardise first: z = (x − μ) / σ

KEY FORMULA

z = (x − μ) / σ   |   P(X > a) = 1 − Φ(z)   |   P(a < X < b) = 2Φ(z) − 1  (symmetric)

Φ(1.00) = 0.8413  |  Φ(1.25) = 0.8944  |  Φ(1.333) = 0.9088  |  Φ(1.645) = 0.9500  | Φ(2.00) = 0.9773

EXAMPLE A — HEIGHTS

X ~ N(170, 8²)  |  μ = 170, σ = 8

Heights (cm) of adults. Find probabilities for specific height ranges.

PART 1: P(HEIGHT > 178)

STANDARDISE

z = (178 − 170) / 8 = 1.00

APPLY Φ

P(X > 178) = 1 − Φ(1.00)

= 1 − 0.84134

P(height > 178) = 0.1587 ≈ 15.9%

PART 2: P(160 < HEIGHT < 180)

TWO Z-SCORES

z₁ = (160−170)/8 = −1.25

z₂ = (180−170)/8 = +1.25

SYMMETRY RULE

P = 2 Φ(1.25) − 1

= 2(0.89435) − 1

P(160 < X < 180) = 0.7887

EXAMPLE B — HONEY JARS

X ~ N(454, 3²)  |  μ = 454, σ = 3

Weight (g) of honey jars. Find probability below 450g and 5th percentile.

PART 1: P(JAR < 450G)

STANDARDISE

z = (450 − 454) / 3 = −1.333

APPLY Φ (SYMMETRY)

P = Φ(−1.333) = 1 − Φ(1.333)

= 1 − 0.9088 = 0.0912

P(jar < 450g) = 0.0912 ≈ 9.1%

PART 2: FIND 5TH PERCENTILE (P(X < A) = 0.05)

P < 0.5 → Z IS NEGATIVE

Φ(z) = 0.05 → z = −1.645

REVERSE STANDARDISE

a = μ + z·σ = 454 + (−1.645)(3)

= 454 − 4.935

a = 449.1 g

EXAMPLE C — PIZZA DELIVERY

X ~ N(30, 5²)  |  μ = 30, σ = 5

Delivery time (min). Find P(>40 min) and P(25 < t < 35).

PART 1: P(DELIVERY > 40 MIN)

STANDARDISE

z = (40 − 30) / 5 = 2.00

APPLY Φ

P(X > 40) = 1 − Φ(2.00)

= 1 − 0.97725 = 0.02275

P(delivery > 40 min) = 0.0228 ≈ 2.3%

PART 2: P(25 < DELIVERY < 35)

TWO Z-SCORES

z₁ = (25−30)/5 = −1.00

z₂ = (35−30)/5 = +1.00

SYMMETRY RULE

P = 2 Φ(1.00) − 1

= 2(0.84134) − 1 = 0.6827

P(25 < X < 35) = 0.6827 ≈ 68.3%

ANSWER

ANSWER

ANSWER

ANSWER — 5TH PERCENTILE

ANSWER

ANSWER