AQA MATHEMATICAL STUDIES

Level 3 Certificate  ·  Paper 2A

CHAPTER 5

The Normal Distribution

AQA Mathematical Studies  —  Paper 2A

5.1 Features of a Normal Distribution

5.2 Standard Normal N(0,1)

5.3 Calculating Probabilities (z-values)

Consolidation Exercise 5

5.1 Features of a Normal Distribution

AQA CH. 5

THE BELL CURVE — N(Μ, Σ²)

≈ 68%

within 1 standard deviation of the mean (μ ± σ)

≈ 95%

within 2 standard deviations of the mean (μ ± 2σ)

≈ 99.7%

within 3 standard deviations of the mean (μ ± 3σ)

KEY FEATURES & NOTATION

Bell-shaped and symmetrical about the mean — the curve rises to a single peak at μ and falls symmetrically on both sides

Mean = Median = Mode — all three averages coincide at the centre of the distribution

Think in standard deviations — express any value as a number of σ above or below μ. This is the key skill for all calculations.

Tails extend to infinity — the curve never quite touches the x-axis, but 99.7% of data lies within 3σ of the mean

X ~ N(μ, σ²)

EXAMPLE — N(10, 2²): MEAN = 10 CM, S.D. = 2 CM

NOTATION

μ = mean  |  σ² = variance  |  σ = standard deviation

⚠ The second parameter is the variance (σ²) , NOT the standard deviation σ

Example 1 & Activity 1: Normal Distribution in Context

Applying N(μ, σ²) to real-world data — motorway speeds and household waste

Example 1 — Motorway Speeds

400 million cars recorded on a motorway

N(70, 10²) — mean 70 mph, s.d. 10 mph

(A) FROM THE HISTOGRAM

1

Cars over 80 mph:

6×10 + 1×10 = 70 million

2

Percentage:

70 ÷ 400 × 100 =

17.5%

(B) FROM THE NORMAL DISTRIBUTION MODEL

3

80 mph = mean + 1 s.d. → more than 1 s.d. above mean

4

About ²⁄₃ within 1 s.d., so ¹⁄₃ outside. Split equally:

¹⁄₆ above ≈

16.7%

17.5% (data) ≈ 16.7% (model) — the normal distribution fits well

Activity 1 — Household Waste

100,000 households; council provides 90-litre bins

N(80, 5²) — mean 80 litres/week, s.d. 5 litres

(A) AT LEAST 10 LITRES SPARE CAPACITY

1

Spare ≥ 10L means waste ≤ 80L:

90 − 80 = 10 litres spare

2

80 litres = the mean → exactly 50% of households are at or below the mean

50,000 households have at least 10 litres spare capacity

(B) TOO MUCH WASTE (OVERFLOW)

3

90 litres =

80 + 2×5 = mean + 2 s.d.

4

About 2.5% of data lies more than 2 s.d. above the mean

2,500 households would produce too much waste for their bin

Skills for Exercise 5A — The 68-95-99.7 Rule

BEFORE EXERCISE 5A

THE EMPIRICAL RULE — NO TABLES NEEDED

68%

P(μ − σ < X < μ + σ) ≈ 0.68

About 2 in 3 values lie within 1 standard deviation of the mean. Each tail beyond ±1σcontains about 16% .

95%

P(μ − 2σ < X < μ + 2σ) ≈ 0.95

About 19 in 20 values lie within 2 standard deviations. Each tail beyond ±2σ contains about 2.5% .

99.7%

P(μ − 3σ < X < μ + 3σ) ≈ 0.997

Almost all values lie within 3 standard deviations. Each tail beyond ±3σ contains about 0.15% .

⚠ NOTATION — N(Μ, Σ²)

The second parameter is the variance (σ²) , NOT the standard deviation.

e.g. N(5, 36) → σ² = 36 → σ = √36 = 6  |  N(70, 100) → σ = √100 = 10

HOW TO USE THE RULE

METHOD — FINDING A PROBABILITY USING THE RULE

1

Identify μ and σ from the notation N(μ, σ²). Remember σ = √(variance).

2

Find how many standard deviations the value is from the mean: (x − μ) / σ

3

Apply the rule: 1 s.d. → 68%, 2 s.d. → 95%, 3 s.d. → 99.7%

4

Use symmetry: the remaining % is split equally between the two tails.

Beyond ±1σ: 32% total → 16% in each tail

Beyond ±2σ: 5% total → 2.5% in each tail

Beyond ±3σ: 0.3% total → 0.15% in each tail

WORKED EXAMPLE — MOTORWAY SPEEDS N(70, 10²): Μ = 70 MPH, Σ = 10 MPH

Q1

P(60 < speed < 80)?   60 = μ−σ, 80 = μ+σ → within 1 s.d.

≈ 68%

Q2

P(speed > 80)?   80 = μ+σ → upper tail beyond 1 s.d.

≈ 16%

Q3

P(speed > 90)?   90 = μ+2σ → upper tail beyond 2 s.d.

≈ 2.5%

Q4

P(speed < 50)?   50 = μ−2σ → lower tail beyond 2 s.d.

≈ 2.5%

Q5

In 400 million cars, how many travel > 80 mph?

≈ 64 million

SYMMETRY — TAIL PROBABILITIES

Exercise 5A — Questions & Values

AQA CH 5.1 · 68-95-99.7 RULE

Q4

Stratified Sampling

4 marks

A factory has 240 workers : Day shift 144 , Night shift 96 . Heights follow N(175, 6²) cm.

(a) A stratified sample of 30 workers is taken. How many from each shift?

(b) What proportion of all workers have height between 163 cm and 187 cm ?

(c) How many workers in the sample would you expect to be in this range?

Values: Day: 144/240 × 30 = 18 . Night: 96/240 × 30 = 12 . 163 = 175 − 2×6, 187 = 175 + 2×6 → 95% → 0.95 × 30 = 28.5 ≈ 29

KEY SKILLS FOR 5A

68% RULE

68% of data lies within μ ± 1σ

95% RULE

95% of data lies within μ ± 2σ

99.7% RULE

99.7% of data lies within μ ± 3σ

NOTATION

N(μ, σ²) — second parameter is variance , not SD. Take √ to find σ.

STRATIFIED

Sample size = (group ÷ total) × n

Q1

X ~ N(50, 4²)

3 marks

(a) Find P(42 < X < 58).

(b) Find P(X > 58).

(c) In 400 observations, how many would you expect to lie between 46 and 54?

Q2

Y ~ N(100, 225)

3 marks

(a) Find P(70 < Y < 130).  

(b) Find P(Y < 55).  

(c) Find P(85 < Y < 115).  

Q3

Weights of apples ~ N(180, 20²) g

4 marks

(a) What percentage of apples weigh between 140 g and 220 g ?

(b) In a box of 500 apples, how many weigh more than 220 g ?

(c) An apple is rejected if it weighs less than 120 g . What percentage are rejected?

 42 = 50 − 2×4, 58 = 50 + 2×4 → 95%​

​

 Above 2σ → (100 − 95)/2 = 2.5%

 46 = 50 − 1×4, 54 = 50 + 1×4 → 68% → 0.68 × 400 = 272

 [σ = √225 = 15]

70 = 100 − 2×15, 130 = 100 + 2×15 → 95%

55 = 100 − 3×15 → below 3σ → (100 − 99.7)/2 = 0.15%

85 = 100 − 1×15, 115 = 100 + 1×15 → 68%

 140 = 180 − 2×20, 220 = 180 + 2×20 → 95%

 2.5% × 500 = 12.5 ≈ 13 apples

 120 = 180 − 3×20 → 0.15%

5.2 The Standard Normal Distribution N(0,1)

AQA Mathematical Studies · Chapter 5

Q

Find the proportion of N(0,1) between z = −0.2 and z = 1.5 , i.e. P(−0.2 < z < 1.5)

1

We need Φ(1.5) − Φ(−0.2) — subtract the lower cumulative probability from the upper

2

From tables: Φ(1.5) = 0.93319

3

Apply symmetry rule: Φ(−0.2) = 1 − Φ(0.2) = 1 − 0.57926 = 0.42074

4

Subtract: 0.93319 − 0.42074 = 0.51245

P(−0.2 < z < 1.5) = 0.51245 — so 51.245% of the distribution lies between z = −0.2 and z = 1.5

ACTIVITY 3 — KEY POINTS

P(z = 1) = 0 — exact values have zero probability in continuous distributions.  |  P(−0.2 ≤ z ≤ 1.5) = 0.51245 — strict and non-strict inequalities give the same result.

Skills for Exercise 5B — Using Φ(z) Tables

BEFORE EXERCISE 5B

THE STANDARD NORMAL N(0,1) AND Φ(Z)

DEFINITION OF Φ(Z)

Φ(z) = P(Z < z)

Φ(z) gives the area to the left of z under the N(0,1) curve. Tables only show positive z values —use the symmetry rule for negative z.

Φ(−z) = 1 − Φ(z)

PROBABILITY BETWEEN TWO VALUES

P(a < Z < b) = Φ(b) − Φ(a)

e.g. P(−1 < Z < 1.3) = Φ(1.3) − Φ(−1) = 0.9032 − 0.1587 = 0.7445

REVERSE PROBLEMS — FINDING Z GIVEN PROBABILITY

If P(Z < a) = p, find a by looking up p in the Φ table.

If p < 0.5: use symmetry — find z where Φ(z) = 1−p, then a = −z.

For symmetric intervals: P(−b < Z < b) = p → Φ(b) = 0.5 + p/2

KEY Φ VALUES & WORKED EXAMPLES

KEY Φ(Z) VALUES — MEMORISE THESE

Φ(0.53)

= 0.7019

Φ(1.00)

= 0.8413

Φ(1.20)

= 0.8849

Φ(1.28)

= 0.9000

Φ(1.30)

= 0.9032

Φ(1.50)

= 0.9332

Φ(1.60)

= 0.9452

Φ(1.645)

= 0.9500

Φ(1.70)

= 0.9554

Φ(1.960)

= 0.9750

Φ(2.00)

= 0.9772

Φ(2.326)

= 0.9900

Φ(2.50)

= 0.9938

Φ(3.00)

= 0.9987

WORKED EXAMPLES — N(0,1)

(a)

P(Z < 1.3) = Φ(1.3)

= 0.9032

(b)

P(Z > 1.5) =

= 0.0668

(c)

P(Z < −1.2) =

= 0.1151

(d)

P(−1 < Z < 1.3) =

= 0.7445

(e)

Find a: P(Z < a) =

a = 1.30

(f)

Symmetric: P(−b < Z < b) =

b = 1.645

SYMMETRY RULE — FOR NEGATIVE Z VALUES

e.g. Φ(−1.3) = 1 − Φ(1.3) = 1 − 0.9032 = 0.0968

Also: Φ(0) = 0.5 (by symmetry — half the distribution is below the mean)

1 − Φ(1.5) = 1 − 0.9332

1 − Φ(1.2) = 1 − 0.8849

Φ(1.3) − Φ(−1) = 0.9032 − 0.1587

0.9032 → look up 0.9032 in table

0.9 → Φ(b) = 0.95 → b = 1.645

Exercise 5B — Using Φ(z) Tables

AQA CH 5.2 · STANDARD NORMAL N(0,1)

Φ(Z) REFERENCE VALUES

Symmetry rule:

Φ(−z) = 1 − Φ(z)

e.g. Φ(−1.5) = 1 − 0.9332 = 0.0668

P(Z > z):

= 1 − Φ(z)

e.g. P(Z > 1.96) = 1 − 0.9750 = 0.0250

Q1

Z ~ N(0, 1)

4 marks

(a)

P(Z < 1.5)  

= Φ(1.5)

= 0.9332

(b)

P(Z > 1.5)  

= 1 − Φ(1.5)

= 0.0668

(c)

P(Z < −1.5)  

= 1 − Φ(1.5)

= 0.0668

(d)

P(−1.5 < Z < 1.5)  

= 2Φ(1.5) − 1

= 0.8664

Q2

Z ~ N(0, 1)

4 marks

(a)

P(Z < 2.0)  

= Φ(2.0)

= 0.9772

(b)

P(Z > 2.0)  

= 1 − 0.9772

= 0.0228

(c)

P(−1 < Z < 2)  

= Φ(2) − Φ(−1) = 0.9772 − 0.1587

=

0.8185

(d)

Find a such that P(Z < a) = 0.9000  

Φ(a) = 0.9000

a = 1.28

Q3

Z ~ N(0, 1) — Reverse Problems

4 marks

(a)

Find b such that P(Z < b) = 0.9750  

b = 1.96

(b)

Find c such that P(Z > c) = 0.0668  

P(Z < c) = 0.9332

c =

1.50

(c)

Find d such that P(Z < d) = 0.25  

Φ(−d) = 0.75 → d = −0.674

d = −0.674

(d)

Find e such that P(−e < Z < e) = 0.95  

Φ(e) = 0.975

e = 1.96

Q4

Z ~ N(0, 1) — Mixed

4 marks

(a)

P(0 < Z < 1.28)  

= Φ(1.28) − 0.5 = 0.9000 − 0.5

=

0.4000

(b)

P(Z < −2.33)  

= 1 − Φ(2.33) = 1 − 0.9901

= 0.0099

(c)

P(1.0 < Z < 2.0)  

= Φ(2.0) − Φ(1.0) = 0.9772 − 0.8413

=

0.1359

(d)

Find f such that P(Z > f) = 0.0500  

P(Z < f) = 0.95 → Φ(f) =

0.95

f = 1.645

5.3 Calculating Probabilities — z-values & Standardising

Converting any normal distribution N(μ, σ²) to the standard normal N(0,1)

Core Concept & Example 4

STANDARDISING FORMULA

z = (x − μ) / σ

Converts any value x to a standard normal z-score

⚠ NOTATION

N(μ, σ²) — the second number is the variance (σ²) , NOT the standard deviation σ.

e.g. N(159, 11² ) means μ = 159, σ = 11

Example 4 — Heights of 13-year-old boys: N(159, 11²)

(a)

P(h ≤ 165): z = (165 − 159) / 11 = 6/11 ≈ 0.545

Φ(0.545) ≈ 0.71 → P(h ≤ 165) = 0.71

(b)

Expected number in 1000: 1000 × 0.71 = 710 boys

P = 0.71 → 710 boys

Examples 5 & 6 — Worked Solutions

Example 5 — Weights of honey jars: N(454, 4²)

(a)

450g → z = (450 − 454) / 4 = −1

455.5g → z = (455.5 − 454) / 4 = 0.375

(b)

P(450 < w < 455.5) = Φ(0.375) − Φ(−1)

= 0.64617 − (1 − 0.84134) = 0.64617 − 0.15866

P ≈ 0.487

Example 6 — Pizza delivery time: N(30, 64) → σ = √64 = 8 min

Goal

95% chance of arriving within 60 min (by 7:00 pm)

Step 1

Φ(z) = 0.95 → z = 1.64

Step 2

Max time = μ + z·σ = 30 + 1.64 × 8 = 30 + 13.12 ≈ 43 min

Step 3

7:00 pm − 43 min = Phone at 6:17 pm

Call by 6:17 pm for 95% on-time

Skills for Exercise 5C — Standardising & Reverse Problems

BEFORE EXERCISE 5C

THE Z-SCORE FORMULA — STANDARDISING ANY NORMAL DISTRIBUTION

z = (x − μ) / σ

METHOD — FINDING P(X < A) FOR X ~ N(Μ, Σ²)

1

Find σ from the notation: σ = √(variance). e.g. N(50, 16) → σ = 4

2

Calculate z = (a − μ) / σ

3

If z > 0: P(X < a) = Φ(z) directly from tables

4

If z < 0: P(X < a) = 1 − Φ(|z|) using symmetry rule

5

For P(X > a): answer = 1 − P(X < a)

6

For P(a < X < b): standardise both → Φ(z₂) − Φ(z₁)

⚠ CRITICAL — N(Μ, Σ²) NOTATION

The second parameter is the variance (σ²) , NOT the standard deviation.

N(30, 64) → σ = √64 = 8  |  N(100, 225) → σ = √225 = 15

REVERSE PROBLEMS & WORKED EXAMPLES

REVERSE PROBLEMS — FINDING X GIVEN A PROBABILITY

x = μ + z · σ

Step 1: Find z from tables: Φ(z) = given probability.

Step 2: If p < 0.5, use symmetry: z is negative → z = −Φ⁻¹(1−p).

Step 3: Calculate x = μ + z·σ.

Symmetric interval: P(−b < Z < b) = p → Φ(b) = 0.5 + p/2

COMMON REVERSE Φ VALUES

P = 0.75

z = 0.674

P = 0.80

z = 0.842

P = 0.90

z = 1.282

P = 0.95

z = 1.645

P = 0.975

z = 1.960

P = 0.99

z = 2.326

WORKED EXAMPLES

Heights of boys N(159, 11²) — find P(h ≤ 165)

z = (165 − 159) / 11 = 6/11 ≈ 0.545

P(h ≤ 165) = Φ(0.545) ≈ 0.71 (71%)

Honey jars N(454, 4²) — find P(450 < w < 455.5)

z₁ = (450−454)/4 = −1  |  z₂ = (455.5−454)/4 = 0.375

P = Φ(0.375) − Φ(−1) = 0.646 − 0.159 ≈ 0.487

Reverse — Pizza delivery N(30, 64): find time for 95% on-time

σ = √64 = 8  |  Φ(z) = 0.95 → z = 1.645

Max time = 30 + 1.645 × 8 = 43.2 min → phone at 6:17 pm

STANDARDISING FORMULA

Converts any value x from N(μ, σ²) into a standard normal z-score.

Then look up Φ(z) in the standard normal table.

AQA CHAPTER 5.3 — REVERSE PROBLEMS

Example 7 & Activity 4

Example 7 — Resting Heart Rate  

PART (A) — FIND P(HR > 100)

Standardise: z = (100 − 70) / 12 = 30/12 = 2.5

P(z > 2.5) = 1 − Φ(2.5) = 1 − 0.99379

P(HR > 100) = 0.00621 ≈ 0.6% — very rare

PART (B) — 90TH PERCENTILE (REVERSE PROBLEM)

Need Φ(z) = 0.9 → from tables: z = 1.28

Unstandardise: x = μ + z·σ = 70 + 1.28 × 12

x = 70 + 15.36

90th percentile = 85.4 bpm

Activity 4 — Mensa IQ Check & Exercise 5C

ACTIVITY 4 — IQ ~ N(100, 15²)  |  MENSA: IQ > 130 = TOP 2%?

Standardise: z = (130 − 100) / 15 = 30/15 = 2

P(IQ > 130) = 1 − Φ(2) = 1 − 0.97725 = 0.02275 ≈2.3%

Key insight: IQ > 130 gives ≈ 2.3%, not exactly 2%. Mensa's claim is approximately correct — close but not exact.

EXERCISE 5C — Q1: IDENTIFY STANDARD DEVIATION

N(5, 7²)

σ = 7

N(5, 36)

σ = √36 = 6

N(17, 40)

σ = √40 ≈ 6.32

EXERCISE 5C — Q2: CALCULATE Z-VALUES  Z = (X − Μ) / Σ

(A) X=9, Μ=6, Σ=5

z = 3/5 = 0.6

(B) X=7, Μ=8, Σ=16

z = −1/16 = −0.0625

(C) X=−1 IN N(2,9)

z = −3/3 = −1

(D) X=6 IN N(4,9)

z = 2/3 = 0.667

(E) X=4 IN N(3,7)

z = 1/√7 = 0.378

(F) X=3 IN N(5,5)

z = −2/√5 = −0.894

N(70, 12²)

Exercise 5C — Q1 to Q4 (with Values)

AQA CH 5.3 · STANDARDISING Z = (X − Μ)/Σ

Standardise:

z = (x − μ) / σ

P(X < a):

Φ(z)

P(X > a):

1 − Φ(z)

Reverse:

x = μ + z·σ

N(μ, σ²):

σ = √(variance)

Q1

Flour bags ~ N(500, 8²) g

6 marks

(a)

P(weight < 490 g)

= 0.10565

z = (490−500)/8 = −1.25 → P = 1 − Φ(1.25) = 1 − 0.89435 = 0.10565

(b)

P(weight > 510 g)

= 0.10565

z = (510−500)/8 = 1.25 → P = 1 − Φ(1.25) = 0.10565

(c)

P(490 < weight < 510)

= 0.78870

= 1 − 2 × 0.10565 = 0.78870

(d)

In 1000 bags: expected number between 490 g and 510 g

= 789 bags

0.78870 × 1000 = 788.7 ≈ 789 bags

Q2

Bolts ~ N(3.5, 0.04²) cm

6 marks

(a)

P(length < 3.42 cm)

= 0.02275

z = (3.42−3.5)/0.04 = −2 → P = 1 − Φ(2) = 1 − 0.97725 = 0.02275

(b)

P(length > 3.58 cm)

= 0.02275

z = (3.58−3.5)/0.04 = 2 → P = 1 − Φ(2) = 0.02275

(c)

P(3.42 < length < 3.58)

= 0.95450

= 1 − 2 × 0.02275 = 0.95450

(d)

In 5000 bolts: number rejected (too short or too long)

= 114 bolts

2 × 0.02275 × 5000 = 0.04550 × 5000 = 227.5 → rejected = 5000 × 0.04550 = 227.5 ≈ 228... wait: each tail = 0.02275 × 5000 = 113.75 ≈ 114 per tail →total = 228

Q3

Heights ~ N(170, 7²) cm

5 marks

(a)

P(height > 180 cm)

= 0.07656

z = (180−170)/7 = 1.429 → P = 1 − Φ(1.43) ≈ 1 − 0.9236 = 0.0764

(b)

P(height < 160 cm)

= 0.07656

z = (160−170)/7 = −1.429 → P = 1 − Φ(1.43) ≈ 0.0764 (symmetric)

(c)

Find height h such that P(H > h) = 0.10 (tallest 10%)

h ≈ 179.0 cm

Φ(z) = 0.90 → z = 1.282 → h = 170 + 1.282 × 7 = 170 + 8.97 = 178.97 ≈ 179.0 cm

Q4

Exam marks ~ N(55, 15²)

6 marks

(a)

P(mark > 70)

= 0.15866

z = (70−55)/15 = 1 → P = 1 − Φ(1) = 1 − 0.84134 = 0.15866

(b)

P(mark < 40)

= 0.15866

z = (40−55)/15 = −1 → P = 1 − Φ(1) = 0.15866 (symmetric)

(c)

Pass mark if 80% of students pass (find mark m)

m ≈ 42.4

P(X > m) = 0.80 → P(X < m) = 0.20 → Φ(z) = 0.20 → z = −0.842 → m = 55 + (−0.842)×15 = 55 − 12.63 = 42.37 ≈ 42

(d)

Grade A if top 10%: find mark a

a ≈ 74.2

Φ(z) = 0.90 → z = 1.282 → a = 55 + 1.282 × 15 = 55 + 19.23 = 74.23 ≈ 74

Exercise 5C — Q5 to Q8 (with Values)

AQA CH 5.3 · REVERSE PROBLEMS: X = Μ + Z·Σ

Reverse method: find z from Φ(z) = p, then

x = μ + z·σ

Top 5%:

Φ(z) = 0.95 → z = 1.645

Top 10%:

Φ(z) = 0.90 → z = 1.282

Bottom 5%:

z = −1.645

Middle 90%:

z =±1.645

Q5

Bulbs ~ N(1200, 100²) hours

REVERSE

6 marks

(a)

P(lifetime > 1350 hours)

= 0.06681

z = (1350−1200)/100 = 1.5 → P = 1 − Φ(1.5) = 1 − 0.93319 = 0.06681

(b)

P(lifetime < 1050 hours)

= 0.06681

z = (1050−1200)/100 = −1.5 → P = 1 − Φ(1.5) = 0.06681 (symmetric)

(c)

P(1050 < lifetime < 1350)

= 0.86638

= 1 − 2 × 0.06681 = 0.86638

(d)

Guarantee period so that only 5% fail: find t

t = 1035.5 hours

Φ(z) = 0.05 → z = −1.645 → t = 1200 + (−1.645)×100 = 1200 − 164.5 = 1035.5 hours

Q6

Exam marks ~ N(55, 15²)

REVERSE

6 marks

(a)

P(mark > 70)

= 0.15866

z = (70−55)/15 = 1 → P = 1 − Φ(1) = 0.15866

(b)

P(mark < 40)

= 0.15866

z = (40−55)/15 = −1 → P = 0.15866 (symmetric)

(c)

Pass mark if 80% pass: find m

m ≈ 42

Φ(z) = 0.20 → z = −0.842 → m = 55 + (−0.842)×15 = 55 − 12.63 = 42.37 ≈ 42

(d)

Grade A if top 10%: find mark a

a ≈ 74

Φ(z) = 0.90 → z = 1.282 → a = 55 + 1.282×15 = 55 + 19.23 = 74.23 ≈ 74

Q7

Reaction times ~ N(0.8, 0.1²) seconds

REVERSE

5 marks

(a)

P(time > 1.0 s)

= 0.02275

z = (1.0−0.8)/0.1 = 2 → P = 1 − Φ(2) = 1 − 0.97725 = 0.02275

(b)

P(time < 0.6 s)

= 0.02275

z = (0.6−0.8)/0.1 = −2 → P = 0.02275 (symmetric)

(c)

Fastest 5% of people: find reaction time t

t = 0.6355 s

Fastest 5% → P(T < t) = 0.05 → Φ(z) = 0.05 → z = −1.645 → t = 0.8 + (−1.645)×0.1 = 0.8 − 0.1645 = 0.6355 s

Q8

Apples ~ N(180, 20²) g

REVERSE

5 marks

(a)

P(weight > 200 g)

= 0.15866

z = (200−180)/20 = 1 → P = 1 − Φ(1) = 0.15866

(b)

P(weight < 150 g)

= 0.06681

z = (150−180)/20 = −1.5 → P = 1 − Φ(1.5) = 0.06681

(c)

Middle 90% of apples: find weight range [w₁, w₂]

[147.1, 212.9] g

z = ±1.645 → w₁ = 180 − 1.645×20 = 180 − 32.9 = 147.1 g

w₂ = 180 + 1.645×20 = 180 + 32.9 = 212.9 g

Consolidation Exercise 5 — Q1 to Q4

All z-values and probabilities provided for reference  |  Use Φ(z) from standard normal tables  |  Recall: z = (x − μ) / σ

Q1 — BABY WEIGHTS

N(3.5, 0.5²) kg

(A) P(WEIGHT < 2.5)

z = (2.5 − 3.5) / 0.5 = −2

P = 1 − Φ(2) = 1 − 0.97725

P = 0.02275 ≈ 2.3%

(B) P(WEIGHT > 4.5)

z = (4.5 − 3.5) / 0.5 = 2

P = 1 − Φ(2) = 1 − 0.97725

P = 0.02275 ≈ 2.3%

(C) P(2.5 < W < 4.5)

P = 1 − 2 × 0.02275

P = 0.95450 ≈ 95.5%

(D) IN 1000 BIRTHS

Under 2.5 kg: 23 babies

Over 4.5 kg: 23 babies

Between: 955 babies

Q2 — BODY TEMPERATURE

N(98.3, 0.8²) °F

(A) P(TEMP > 100)

z = (100 − 98.3) / 0.8 = 2.125

P = 1 − Φ(2.125) ≈ 1 − 0.9832

P ≈ 0.0168 (1.7%)

(B) P(TEMP < 97)

z = (97 − 98.3) / 0.8 = −1.625

P = 1 − Φ(1.625) ≈ 1 − 0.9479

P ≈ 0.0521 (5.2%)

(C) NORMAL RANGE — MIDDLE 95%: Z = ±1.96

Lower: 98.3 − 1.96 × 0.8 = 96.73 °F  |  Upper: 98.3 + 1.96 × 0.8 = 99.87 °F

Normal range: (96.73, 99.87) °F

Q3 — BOYS' HEIGHTS (AGE 14)

N(149.3, 12.7²) cm

(A) P(HEIGHT > 160)

z = (160 − 149.3) / 12.7 = 0.843

P = 1 − Φ(0.843) ≈ 1 − 0.8003

P ≈ 0.1997 (20%)

(B) P(HEIGHT < 130)

z = (130 − 149.3) / 12.7 = −1.52

P = 1 − Φ(1.52) ≈ 1 − 0.9357

P ≈ 0.0643 (6.4%)

(C) TALLEST 10%: Φ(Z) = 0.90 → Z = 1.282

Height = 149.3 + 1.282 × 12.7 = 149.3 + 16.3 = 165.6 cm

Tallest 10% are above 165.6 cm

Q4 — CROSSWORD TIME

N(65, 20²) minutes

(A) P(TIME < 45)

z = (45 − 65) / 20 = −1

P = 1 − Φ(1) = 1 − 0.84134

P = 0.15866 (15.9%)

(B) P(TIME > 90)

z = (90 − 65) / 20 = 1.25

P = 1 − Φ(1.25) = 1 − 0.89435

P = 0.10565 (10.6%)

(C) FASTEST 25%: Φ(Z) = 0.25 → Z = −0.674

Time = 65 − 0.674 × 20 = 65 − 13.48 = 51.52 minutes

Fastest 25% complete in under 51.5 minutes

Consolidation Exercise 5 — Q5 to Q8

ALL VALUES PROVIDED

Q5

N(48, 20²)

Electricity meter readings (kWh/day)

(a)

P(reading > 70): z = (70−48)/20 = 1.1 → P = 1−Φ(1.1) = 1−0.86433

= 0.13567

(b)

P(reading < 20): z = (20−48)/20 = −1.4 → P = 1−Φ(1.4) = 1−0.91924

= 0.08076

(c)

Middle 80%: z = ±1.282 → Range = 48 ± 1.282×20

(22.4, 73.6) kWh

Middle 80% of daily readings fall between 22.4 and 73.6 kWh

Q6

N(184.5, 13.6²)

Wrist circumferences (mm)

(a)

P(wrist > 200): z = (200−184.5)/13.6 = 1.14 → P = 1−Φ(1.14)

≈ 0.127

(b)

P(wrist < 160): z = (160−184.5)/13.6 = −1.80 → P = 1−Φ(1.80)

= 0.0359

(c)

Largest 5%: Φ(z) = 0.95 → z = 1.645 → Wrist = 184.5 + 1.645×13.6

= 206.9 mm

Largest 5% of wrist circumferences exceed 206.9 mm

Q7

N(68, 13²)

Pulse rates (bpm)

(a)

P(pulse > 90): z = (90−68)/13 = 1.69 → P = 1−Φ(1.69)

≈ 0.0455

(b)

P(pulse < 50): z = (50−68)/13 = −1.38 → P = 1−Φ(1.38)

≈ 0.0838

(c)

Bradycardia (pulse < 60): z = (60−68)/13 = −0.615 → P = 1−Φ(0.615)

≈ 0.269

About 27% of people have bradycardia (resting pulse below 60 bpm)

Q8

N(31.6, 4.3²)

Ear plug lengths (mm)

(a)

P(length > 38): z = (38−31.6)/4.3 = 1.49 → P = 1−Φ(1.49)

≈ 0.0681

(b)

P(length < 25): z = (25−31.6)/4.3 = −1.535 → P = 1−Φ(1.535)

≈ 0.0624

(c)

Middle 90%: z = ±1.645 → Range = 31.6 ± 1.645×4.3

(24.5, 38.7) mm

Middle 90% of ear plug lengths fall between 24.5 and 38.7 mm

Consolidation Exercise 5 — Q9 to Q12

All working values shown  ·  z = (x − μ) / σ

Q9

N(55, 1.5²) mm

Lengths of nails

(a)

P(length < 52):

z = (52−55)/1.5 = −2

→ P =

0.02275

(b)

P(length > 57):

z = (57−55)/1.5 = 1.333

→ P = 1−Φ(1.333) =

0.0912

(c)

P(52 < length < 57): 1 − 0.02275 − 0.0912 = 0.886

(d)

In 10,000 nails:

228

912

8,860

Q10

N(270, 15²) mm

Adult male foot lengths

(a)

P(foot > 300):

z = (300−270)/15 = 2

→ P =

0.02275

(b)

P(foot < 240):

z = (240−270)/15 = −2

→ P =

0.02275

(c)

Shoe size 9 fits feet 265–275 mm

z₁ = (265−270)/15 = −0.333

z₂ = (275−270)/15 = 0.333

P = 2Φ(0.333)−1 = 2(0.6304)−1 = 0.2608

About 26% of men take shoe size 9

Q11

N(185, 10²) cm

Heights of sunflowers

(a)

P(height > 200):

z = 1.5

→ P =

0.0668

(b)

P(height < 170):

z = −1.5

→ P =

0.0668

(c)

Tallest 20%:

z = 0.842

→ height = 185 + 0.842×10 =

193.4 cm

(d)

Middle 50%:

z = ±0.674

→ 185 ± 6.74

Middle 50% range: (178.3, 191.7) cm

Q12

N(140, 2.5²) mm

Diameters of tennis balls

(a)

P(diameter < 136):

z = (136−140)/2.5 = −1.6

P = 1−Φ(1.6) = 1−0.9452 = 0.0548

(b)

P(diameter > 144):

z = (144−140)/2.5 = 1.6

→ P =

0.0548

(c)

Acceptable range — middle 95%:

z = ±1.96

140 ± 1.96 × 2.5 = 140 ± 4.9

Acceptable range: (135.1, 144.9) mm

Too short

Too long

Acceptable ✓

Exercise 5B — Reverse Problems

Q3 to Q6 — Worked Solutions

Q3

Find a such that P( z < a ) = 0.9032

Step 1: Write Φ( a ) = 0.9032

Step 2: 0.9032 > 0.5 → look up directly in tables

Step 3: Φ(1.30) = 0.9032 ✓

ANSWER:

a = 1.30

Q4

Find b such that P( z > b ) = 0.0668

Step 1: Convert: P( z < b ) = 1 − 0.0668 = 0.9332

Step 2: 0.9332 > 0.5 → look up directly

Step 3: Φ(1.50) = 0.9332 ✓

ANSWER:

b = 1.50

Q5

Find c such that P(− c < z < c ) = 0.80

Step 1: Symmetric interval → remaining = 0.20, split equally = 0.10 each tail

Step 2: Φ( c ) = 0.5 + 0.80/2 = 0.90

Step 3: Φ(1.282) = 0.90 ✓

ANSWER:

c = 1.282

Q6

Find d such that P( z < d ) = 0.25

Step 1: 0.25 < 0.5 → use symmetry

Step 2: Φ(− d ) = 1 − 0.25 = 0.75 → Φ(0.674) = 0.75

Step 3: So d = −0.674 (negative, below mean)

ANSWER:

d = −0.674

(use symmetry: Φ(0.674) = 0.75)

REVERSE PROBLEM METHOD

1

Write P(Z < z) = required probability

2

If p > 0.5 → look up Φ⁻¹(p) directly in tables

3

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

4

Symmetric interval: Φ(z) = 0.5 + P/2

KEY Φ VALUES FOR REVERSE PROBLEMS

Φ(0.674) = 0.75

(top 25%)

Φ(0.842) = 0.80

(top 20%)

Φ(1.282) = 0.90

(top 10%)

Φ(1.645) = 0.95

(top 5%)

Φ(1.960) = 0.975

(top 2.5%)

Φ(2.326) = 0.99

(top 1%)

SYMMETRY RULE

Φ(−z) = 1 − Φ(z)

If probability < 0.5, find the positive z for (1 − p), then negate it.

P(Z = k) = 0 for any exact value k

Standard Normal Table — Key Φ Values

Φ(z) = P(Z < z) for Z ~ N(0, 1)  |  AQA Chapter 5 Reference

Φ(Z) VALUES — USE THESE IN ALL EXERCISES

Key Rules & Reverse Values

The standard normal Z ~ N(0, 1) has mean 0 and variance 1. Standardise any normal using z = (x −μ) / σ, then look up Φ(z).

Φ(−z) = 1 − Φ(z)

COMMON REVERSE VALUES

SYMMETRY RULE

Use this when z is negative — look up the positive value and subtract from 1.

Chapter 5 Summary — Key Concepts & Formulae

REVIEW

Remember: N(μ, σ²) — the second parameter is always variance (σ²) , not standard deviation. Always square-root to get σ before standardising.

Features of a Normal Distribution

Bell-shaped and symmetrical about the mean. Notation:

N(μ, σ²)

— second parameter is

variance

.

•

68% of values lie within

±1σ

of the mean

•

95% of values lie within

±2σ

of the mean

•

99.7% of values lie within

±3σ

of the mean

Standard Normal N(0, 1)

The standard normal has mean 0 and variance 1. Values are read from the Φ table.

Φ(z) = P(Z < z) Φ(0) = 0.5 Φ(−z) = 1 − Φ(z) ← symmetry rule

Standardising — z-Score

Convert any

N(μ, σ²)

to the standard normal

N(0,1)

using:

z = (x − μ) / σ

•

Always check: is σ² or σ given?

•

For negative z, use the symmetry rule

Finding Probabilities

Standardise x, then look up Φ in tables:

P(X < a) = Φ((a−μ)/σ) P(X > a) = 1 − Φ((a−μ)/σ) P(a < X < b) = Φ(b*) − Φ(a*)

•

where a* and b* are the standardised z-values

Reverse Problems

Given a probability, find the original value x:

•

Step 1: Find z such that

Φ(z) = p

•

Step 2: Rearrange:

x = μ + z·σ

•

Symmetric interval:

Φ(b) = 0.5 + p/2

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

Key Reminders

Common values to remember from tables:

Φ(1.282) = 0.90 → top 10% Φ(1.645) = 0.95 →top 5% Φ(1.960) = 0.975 → top 2.5% Φ(2.326) = 0.99 → top 1%

•

P(Z = k) = 0

for any exact value k

Interactive Quiz — Chapter 5: The Normal Distribution

30S TIMER

1

2

3

4

5

Q 1 / 5

Next

TIME REMAINING

Reveal Answer

QUESTION 1

A distribution is N(50, 4²). What percentage of values lie between 42 and 58?

Hint: 42 and 58 are both 2 standard deviations from the mean.

29

ANSWER

95%

Using the 68–95–99.7 rule: 42 = 50 − 2×4 and 58 = 50 + 2×4, so both are exactly ±2 s.d. from the mean. Therefore 95% of values lie in this range.

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Ch. 5

Normal Dist.

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Chapter 5 Complete — Well Done!

AQA MATHEMATICAL STUDIES · THE NORMAL DISTRIBUTION

TOPICS MASTERED

Chapter 5: The Normal Distribution

68–95–99.7 rule and bell-curve features of N(μ, σ²)

Standard Normal N(0,1) and Φ(z) notation with symmetry Φ(−z) = 1 − Φ(z)

Standardising using z = (x − μ) / σ to calculate probabilities

Reverse problems: finding x given a probability using a = μ + z·σ

WHAT'S NEXT

Consolidate & Practise

Normal distribution features

Standard Normal N(0,1) & Φ(z)

z = (x − μ) / σ formula

Reverse problems & consolidation

RECOMMENDED NEXT STEPS

1

Practise AQA past exam questions — focus on standardising and reverse problems from Paper 2A.

2

Use the standard normal table confidently — look up Φ(z) values and apply the symmetry rule for negative z.

3

Review the key formulae card — ensure z = (x − μ)/σ and a = μ + z·σ are committed to memory.

You have completed all core topics in Chapter 5. You are now ready to tackle normal distribution questions with confidence in your AQA exam.