Normal Distribution — Edexcel A-level Mathematics (Applied)

SPEC 6.2 • STATISTICAL DISTRIBUTIONS

The Normal Distribution

Understanding continuous probability distributions, standardisation, and calculating probabilities using the normal model

Continuous Random Variables

The normal distribution models continuous random variables - quantities that can take any value in a range, not just whole numbers.

Separate bars, countable outcomes

(e.g., number of heads)

Smooth curve, any value possible

(e.g., height, weight, time)

Critical Concept

For continuous distributions: P(X = exact value) = 0

We can only find probabilities for ranges of values, e.g., P(X < 5) or P(3 < X < 7)

The Emergence of the Normal Curve

When we collect continuous data and draw histograms with increasingly narrow class widths, a familiar shape emerges:

Real-World Examples

Many natural phenomena follow the normal distribution:

Heights of adult humans

Weights of manufactured products

Blood pressure readings

IQ scores (designed to be normal)

Measurement errors in scientific experiments

Properties of Normal Distribution

Notation

X \sim N (μ, σ^{2})

X follows a Normal distribution with mean μ and variance σ²

Key Properties

Symmetric about μ

Bell-shaped curve

Mean = Median = Mode

Total area under curve = 1

68-95-99.7 Rule

68% within μ ± σ

95% within μ ± 2σ

99.7% within μ ± 3σ

Finding Probabilities from Properties

Use these key properties to find or simplify probabilities for any normal distribution:

P(X > a) = 1 − P(X ≤ a)

P(X < μ) = P(X > μ) = 0.5

Key Probability Relationships

For any value a:

P(X > a) = 1 − P(X < a)

P(X < a) + P(X > a) = 1

Using symmetry:

P(X < μ − a) = P(X > μ + a)

P(X > μ − a) = P(X < μ + a)

Finding P(a < X < b)

P(a < X < b) = P(X < b) − P(X < a)

Find the area up to b, then subtract the area up to a.

01Worked Example - Using Properties

X ~ N(50, 25). Given that P(X < 55) = 0.8413, find:

(a) P(X > 55) (b) P(X > 45) (c) P(X < 45) (d) P(45 < X < 55)

a)

Complement rule:

•

P(X > 55) = 1 − P(X < 55)

= 1 − 0.8413 = 0.1587

b)

Symmetry: 55 is 5 above μ, so 45 is 5 below μ

•

P(X > 45) = P(X < 55) = 0.8413

c)

Symmetry:

•

P(X < 45) = P(X > 55) = 0.1587

d)

Difference of areas:

•

P(45 < X < 55) = P(X < 55) − P(X < 45)

= 0.8413 − 0.1587 = 0.6826

The Standard Normal Distribution

The standard normal distribution has mean 0 and variance 1:

Standard Normal

Z \sim N (0, 1)

Any normal distribution can be standardised using:

Standardisation Formula

Z = \frac{X - μ}{σ}

This converts X ~ N(μ, σ²) to Z ~ N(0, 1)

Common Exam Error

In X ~ N(μ, σ²), the second parameter is the variance (σ²), not the standard deviation (σ)!

If X ~ N(50, 16), then μ = 50 and σ = √16 = 4

Why Standardise?

Tables and calculators give probabilities for Z ~ N(0, 1). Converting any normal to standard normal lets us use these tools.

Using the Normal Tables

The table gives Φ(z) = P(Z < z) for the standard normal distribution Z ~ N(0, 1).

Φ(z) = shaded area = P(Z < z)

Reading the Table

To find Φ(1.25):

Find row 1.2

Find column 0.05

Read: Φ(1.25) = 0.8944

Symmetry

Φ (- z) = 1 - Φ (z)

Φ(-1.25) = 1 - 0.8944 = 0.1056

NORMAL DISTRIBUTION TABLE (extract)

z0.000.05

1.00.84130.8531

1.20.88490.8944

1.40.91920.9265

1.60.94520.9505

The Standardisation Process

To find probabilities for any normal distribution X ~ N(μ, σ²), convert to standard normal:

Key Relationship

P (X < a) = P (Z < \frac{a - μ}{σ}) = Φ (\frac{a - μ}{σ})

Example: If X ~ N(50, 25), find P(X < 55)

Step 1: Identify parameters: μ = 50, σ² = 25 → σ = 5

Step 2: Standardise: z = (55 - 50)/5 = 1

Step 3: Use table: P(X < 55) = P(Z < 1) = Φ(1) = 0.8413

For P(X > a)

P(X > a) = 1 - P(X < a)

= 1 - Φ((a - μ)/σ)

For P(a < X < b)

P(a < X < b) = P(X < b) - P(X < a)

= Φ((b-μ)/σ) - Φ((a-μ)/σ)

Calculator Methods

Your calculator has built-in functions for normal distribution probabilities. Use the Normal CD (cumulative distribution) function.

Probability TypeLower BoundUpper Bound

P(X < a)-10⁹⁹ or -1×10⁹⁹a

P(X > b)b10⁹⁹ or 1×10⁹⁹

P(a < X < b)ab

CASIO: Menu → Statistics → DIST → NORM → Ncd

Enter: Lower, Upper, σ, μ

Tables vs Calculator

Tables: You MUST standardise first (convert to z), then look up Φ(z)

Calculator: Enter μ and σ directly - it standardises automatically!

02Worked Example - Calculator Use

The time T minutes to complete an exam is normally distributed with T ~ N(90, 225).

Find: (a) P(T < 80) (b) P(T > 100) (c) P(85 < T < 95)

1

First identify parameters: μ = 90, σ² = 225, so σ = 15

a)

P(T < 80): Lower = -10⁹⁹, Upper = 80, σ = 15, μ = 90

•

Calculator gives: 0.2525 (4 s.f.)

b)

P(T > 100): Lower = 100, Upper = 10⁹⁹, σ = 15, μ = 90

•

Calculator gives: 0.2525 (4 s.f.)

c)

P(85 < T < 95): Lower = 85, Upper = 95, σ = 15, μ = 90

•

Calculator gives: 0.2611 (4 s.f.)

03Worked Example - "Or" Probability

The mass M of bags of sugar is normally distributed with M ~ N(1000, 100), measured in grams.

Find the probability that a randomly selected bag has mass less than 985g or more than 1015g.

1

Identify parameters: μ = 1000, σ² = 100, so σ = 10

2

Standardise both boundaries:

•

For M = 985:

z = \frac{985 - 1000}{10} = \frac{- 15}{10} = - 1.5

•

For M = 1015:

z = \frac{1015 - 1000}{10} = \frac{15}{10} = 1.5

3

Find probability using tables:

•

P(985 < M < 1015) = P(−1.5 < Z < 1.5)

•

= Φ(1.5) − Φ(−1.5)

•

= Φ(1.5) − [1 − Φ(1.5)] (using symmetry: Φ(−z) = 1 − Φ(z))

•

= 2Φ(1.5) − 1 = 2(0.9332) − 1 = 0.8664

4

Use complement: P(M < 985 or M > 1015) = 1 − 0.8664 = 0.1336

04Worked Example - Normal + Binomial

IQ scores are normally distributed with mean 100 and standard deviation 15. A person is classified as a "genius" if their IQ exceeds 145.

(a) Find the probability that a randomly selected person is a genius.

(b) In a group of 500 people, find the probability that exactly 2 are geniuses.

a)

X ~ N(100, 225) where σ = 15

•

P(X > 145): Calculator with Lower = 145, Upper = 10⁹⁹

•

P(X > 145) = 0.001349... ≈ 0.00135

b)

Let p = 0.001349 (probability of being a genius)

•

Let Y = number of geniuses in 500 people

•

Y ~ B(500, 0.001349)

•

P(Y = 2) = ⁵⁰⁰C₂ × 0.001349² × 0.998651⁴⁹⁸

•

P(Y = 2) = 0.1247 (4 s.f.)

Inverse Normal Distribution

Sometimes we need to find the value that corresponds to a given probability. This is the inverse of the usual problem.

Forward Problem

Inverse Problem

Given p, find x such that P(X < x) = p

Calculator Method

CASIO: Menu → Statistics → DIST → NORM → InvN

Enter: Area (probability to the LEFT), σ, μ

Critical Point

The inverse normal function gives the value x such that P(X < x) = p

If asked for "top 10%", you need P(X < x) = 0.90, not 0.10!

Percentage Points Table

The percentage points table gives z-values for common probabilities. It shows values of z such that P(Z > z) = p.

PERCENTAGE POINTS TABLE

pzpz

0.50000.00000.05001.6449

0.40000.25330.02501.9600

0.30000.52440.01002.3263

0.20000.84160.00502.5758

0.15001.03640.00103.0902

0.10001.28160.00053.2905

How to Read

P(Z > 1.96) = 0.025

So P(Z < 1.96) = 0.975

P(Z > 1.6449) = 0.05

So P(Z < 1.6449) = 0.95

Using Symmetry

For negative z-values:

P(Z < -1.96) = 0.025

P(Z > -1.96) = 0.975

When Finding Unknown Parameters

Use (x - μ)/σ = z where z comes from this table. For simultaneous equations (both μ and σ unknown), you'll need two probability conditions.

05Worked Example - Inverse Normal

The lifetime L hours of a certain light bulb is normally distributed with L ~ N(2000, 160000).

(a) Find the lifetime exceeded by 90% of bulbs.

(b) Find the lifetime that only 5% of bulbs exceed.

1

Parameters: μ = 2000, σ² = 160000, so σ = 400

a)

"90% exceed" means P(L > x) = 0.9, so P(L < x) = 0.1

•

Use Inverse Normal with Area = 0.1, σ = 400, μ = 2000

•

x = 1487 hours (to nearest hour)

b)

"Only 5% exceed" means P(L > x) = 0.05, so P(L < x) = 0.95

•

Use Inverse Normal with Area = 0.95, σ = 400, μ = 2000

•

x = 2658 hours (to nearest hour)

Worked Example - Finding μ06

The weights of apples from an orchard are normally distributed with standard deviation 12g. Given that 20% of apples weigh more than 95g, find the mean weight.

1

Let X = weight. X ~ N(μ, 144) where σ = 12

2

Given: P(X > 95) = 0.20, so P(X < 95) = 0.80

3

Standardise: P(Z < (95 - μ)/12) = 0.80

4

From inverse normal: z = 0.8416 (for area 0.80)

5

So (95 - μ)/12 = 0.8416

6

95 - μ = 10.099

7

μ = 95 - 10.099 = 84.9g (3 s.f.)

Worked Example - Finding σ07

The reaction times of athletes are normally distributed with mean 0.18 seconds. Given that 16% of athletes have reaction times greater than 0.20 seconds, find the standard deviation.

1

Let X = reaction time. X ~ N(0.18, σ²)

2

Given: P(X > 0.20) = 0.16, so P(X < 0.20) = 0.84

3

Note: 16% ≈ half of 32%, and by the 68% rule, 16% lies beyond μ + σ

4

This suggests 0.20 ≈ μ + σ (quick check method)

5

Using exact method: z = 0.9945 for area 0.84

6

(0.20 - 0.18)/σ = 0.9945

7

σ = 0.02/0.9945 = 0.0201 seconds (3 s.f.)

Worked Example - Finding μ and σ08

The masses of parcels are normally distributed. Given that 10% of parcels weigh less than 2kg and 5% weigh more than 8kg, find the mean and standard deviation.

1

Let X ~ N(μ, σ²). We have two conditions:

2

P(X < 2) = 0.10 → z₁ = -1.2816 (from inverse normal)

•

So: (2 - μ)/σ = -1.2816 → 2 = μ - 1.2816σ ... (1)

3

P(X > 8) = 0.05 → P(X < 8) = 0.95 → z₂ = 1.6449

•

So: (8 - μ)/σ = 1.6449 → 8 = μ + 1.6449σ ... (2)

4

(2) - (1): 6 = 2.9265σ → σ = 2.05 kg

5

Substitute into (1): μ = 2 + 1.2816(2.05) = 4.63 kg

Practice

Question 1: The volume V ml of juice in cartons is normally distributed with V ~ N(330, 25).

(a) Find P(V < 325)

(b) Find P(V > 338)

(c) Find the probability that a carton contains between 327ml and 335ml.

1

Parameters: μ = 330, σ² = 25, so σ = 5

a)

P(V < 325): Calculator gives 0.1587

b)

P(V > 338): Calculator gives 0.0548

c)

P(327 < V < 335): Calculator gives 0.5704

Practice

Question 2: The length L cm of bolts produced by a machine is normally distributed with mean 5.0 cm and standard deviation 0.02 cm. Bolts are rejected if they are shorter than 4.96 cm or longer than 5.04 cm.

(a) Find the probability that a randomly selected bolt is rejected.

(b) In a batch of 200 bolts, find the expected number of rejected bolts.

1

L ~ N(5.0, 0.0004) where σ = 0.02

a)

P(rejected) = P(L < 4.96) + P(L > 5.04) = 1 - P(4.96 < L < 5.04)

•

P(4.96 < L < 5.04) = 0.9545 (this is μ ± 2σ)

•

P(rejected) = 1 - 0.9545 = 0.0455

b)

Expected rejections = 200 × 0.0455 = 9.1 bolts

Practice

Question 3: Scores on a test are normally distributed with mean μ and standard deviation 12. The top 15% of students score above 78.

(a) Find the mean score μ.

(b) Find the score that separates the bottom 30% of students.

a)

P(X > 78) = 0.15, so P(X < 78) = 0.85

•

z-value for 0.85 = 1.0364

•

(78 - μ)/12 = 1.0364

•

μ = 78 - 12(1.0364) = 65.6 (3 s.f.)

b)

Need x such that P(X < x) = 0.30

•

z-value for 0.30 = -0.5244

•

x = 65.6 + 12(-0.5244) = 59.3 (3 s.f.)

Summary

Key Formulas

Distribution: X ~ N(μ, σ²)

Standard Normal: Z ~ N(0, 1)

Standardisation: Z = (X - μ)/σ

Properties

Symmetric about μ

68% within μ ± σ

95% within μ ± 2σ

P(X = exact) = 0

Calculator Functions

TaskFunction

Find P(a < X < b)Normal CD

Find x given P(X < x) = pInverse Normal

Exam Tips

Always check if given σ or σ² in the notation

For "or" probabilities, use complement method: P(X < a or X > b) = 1 - P(a < X < b)

For finding μ and σ together, set up simultaneous equations

Normal + Binomial: find p from normal, then use B(n, p)