Normal Approximation — Edexcel A-level Mathematics (Applied)

SPEC 4.3 • STATISTICAL DISTRIBUTIONS

Normal Approximation to Binomial

Using the normal distribution to estimate binomial probabilities for large n

Why Approximate?

When n is large, binomial calculations become impractical:

The Problem

If

X \sim B (200, 0.4)

, to find

P (X \geq 85)

we would need:

P (X = 85) + P (X = 86) + ... + P (X = 200)

That's 116 separate terms to calculate!

The Solution

When certain conditions are met, we can use the normal distribution as an approximation to the binomial. This gives us accurate estimates with much simpler calculations.

Historical Context

This approximation was discovered by Abraham de Moivre in 1733, long before calculators existed. It was essential for practical probability calculations.

Conditions for Valid Approximation

The normal approximation to the binomial is valid when:

Validity Conditions

n is large AND p is close to 0.5Equivalently, check that:

n p > 5

AND

n (1 - p) > 5

Why These Conditions?

The normal distribution is symmetrical

Binomial is only symmetric when p is close to 0.5

If p is too close to 0 or 1, binomial is skewed

Large n ensures smooth, bell-shaped distribution

Hint

The approximation is only valid when p is close to 0.5 because the normal distribution is symmetrical. When p is far from 0.5, the binomial is skewed and the symmetric normal curve won't fit well.

Matching Parameters

To approximate a binomial with a normal distribution, we match the mean and variance:

Approximation Formula

If

X \sim B (n, p)

, then we approximate with

Y \sim N (μ, σ^{2})

where:

μ = n p

σ^{2} = n p (1 - p)

σ = n p (1 - p)

ParameterBinomial B(n, p)Normal Approximation

Mean

E (X) = n p

μ = n p

Variance

Va r (X) = n p (1 - p)

σ^{2} = n p (1 - p)

Standard Deviation

n p (1 - p)

σ = n p (1 - p)

Remember

The normal distribution uses the same mean and variance as the original binomial distribution.

Continuity Correction

There's a crucial problem: binomial is discrete, but normal is continuous.

Key Rule

If you are using a normal approximation to a binomial distribution, you need to apply a continuity correction when calculating probabilities.

The diagrams show X ~ B(14, 0.5) being approximated by Y ~ N(7, 1.87²):

P(X > 5) ≈ P(Y > 5.5)

P(X = 9) ≈ P(8.5 < Y < 9.5)

P(X ≤ 3) ≈ P(Y < 3.5)

Key Idea

Each integer k in the binomial corresponds to the interval (k − 0.5, k + 0.5) in the normal. The bars have width 1, centred on each integer value.

Continuity Correction Rules

Apply these corrections when converting binomial to normal:

Binomial ProbabilityNormal ApproximationExplanation

P (X \leq k)

P (Y < k + 0.5)

Include k, so go up to k + 0.5

P (X < k)

P (Y < k - 0.5)

Exclude k, so stop at k − 0.5

P (X \geq k)

P (Y > k - 0.5)

Include k, so start from k − 0.5

P (X > k)

P (Y > k + 0.5)

Exclude k, so start from k + 0.5

P (X = k)

P (k - 0.5 < Y < k + 0.5)

Interval of width 1 around k

Memory Aid

"Include means expand" — If the inequality includes the value (≤ or ≥), expand the boundary outward by 0.5. If it excludes (< or >), shrink inward by 0.5.

Worked Example01

A fair coin is tossed 100 times. Use the normal approximation to find the probability of obtaining at most 45 heads.

1

Identify parameters:

X \sim B (100, 0.5)

2

Check conditions:

n p = 100 \times 0.5 = 50 > 5

✓ and

n (1 - p) = 100 \times 0.5 = 50 > 5

✓

3

Calculate normal parameters:

μ = n p = 50

,

σ = n p (1 - p) = 25 = 5

4

Apply continuity correction:

P (X \leq 45) \approx P (Y < 45.5)

5

Standardise:

z = \frac{45.5 - 50}{5} = \frac{- 4.5}{5} = - 0.9

6

Find probability:

P (Z < - 0.9) = 0.1841

Answer: P(X ≤ 45) ≈ 0.184 (3 s.f.)

Worked Example02

A factory produces components with a 15% defect rate. In a batch of 200 components, find the probability that at least 35 are defective.

1

Identify parameters:

X \sim B (200, 0.15)

2

Check conditions:

n p = 200 \times 0.15 = 30 > 5

✓ and

n (1 - p) = 200 \times 0.85 = 170 > 5

✓

3

Calculate normal parameters:

μ = 30

,

σ = 200 \times 0.15 \times 0.85 = 25.5 = 5.050

4

Apply continuity correction:

P (X \geq 35) \approx P (Y > 34.5)

5

Standardise:

z = \frac{34.5 - 30}{5.050} = \frac{4.5}{5.050} = 0.891

6

Find probability:

P (Z > 0.891) = 1 - P (Z < 0.891) = 1 - 0.8136 = 0.1864

Answer: P(X ≥ 35) ≈ 0.186 (3 s.f.)

Worked Example03

If X ~ B(150, 0.4), use the normal approximation to find P(X = 60).

1

Check conditions:

n p = 150 \times 0.4 = 60 > 5

✓ and

n (1 - p) = 150 \times 0.6 = 90 > 5

✓

2

Calculate normal parameters:

μ = 60

,

σ = 150 \times 0.4 \times 0.6 = 36 = 6

3

Apply continuity correction:

P (X = 60) \approx P (59.5 < Y < 60.5)

4

Standardise both bounds:

z_{1} = \frac{59.5 - 60}{6} = - 0.0833

,

z_{2} = \frac{60.5 - 60}{6} = 0.0833

5

Find probability:

P (- 0.0833 < Z < 0.0833) = P (Z < 0.0833) - P (Z < - 0.0833)

6

Calculate:

= 0.5332 - 0.4668 = 0.0664

Answer: P(X = 60) ≈ 0.0664 (3 s.f.)

Worked Example04

A survey shows 35% of people prefer Brand A. In a random sample of 80 people, find the probability that between 25 and 32 (inclusive) prefer Brand A.

1

Identify parameters:

X \sim B (80, 0.35)

2

Check conditions:

n p = 80 \times 0.35 = 28 > 5

✓ and

n (1 - p) = 80 \times 0.65 = 52 > 5

✓

3

Calculate normal parameters:

μ = 28

,

σ = 80 \times 0.35 \times 0.65 = 18.2 = 4.266

4

Apply continuity correction:

P (25 \leq X \leq 32) \approx P (24.5 < Y < 32.5)

5

Standardise:

z_{1} = \frac{24.5 - 28}{4.266} = - 0.821

,

z_{2} = \frac{32.5 - 28}{4.266} = 1.055

6

Find probability:

P (Z < 1.055) - P (Z < - 0.821) = 0.8543 - 0.2058 = 0.6485

Answer: P(25 ≤ X ≤ 32) ≈ 0.649 (3 s.f.)

Worked Example05

Determine whether the normal approximation is appropriate for X ~ B(40, 0.08). If not, explain why.

1

Calculate

n p = 40 \times 0.08 = 3.2

2

Calculate

n (1 - p) = 40 \times 0.92 = 36.8

3

Check conditions:

n p = 3.2 < 5

✗

4

The first condition is NOT satisfied.

Answer: The normal approximation is NOT appropriate because np = 3.2 < 5. The distribution would be heavily skewed to the right (small p value), so the symmetric normal distribution would not provide an accurate approximation.

Worked Example06

For X ~ B(50, 0.5), compare P(X ≤ 22) using: (a) exact binomial, (b) normal approximation. Comment on the accuracy.

1

(a) Exact binomial: Using calculator,

P (X \leq 22) = 0.2399

2

(b) Normal approximation:

μ = 50 \times 0.5 = 25

,

σ = 50 \times 0.5 \times 0.5 = 3.536

3

With continuity correction:

P (X \leq 22) \approx P (Y < 22.5)

4

Standardise:

z = \frac{22.5 - 25}{3.536} = - 0.707

5

P (Z < - 0.707) = 0.2398

Answer: Exact = 0.2399, Approximation = 0.2398. The difference is only 0.0001, showing excellent accuracy when conditions are well satisfied (np = n(1-p) = 25).

Worked Example07

A quality control process accepts a batch if fewer than 8 items in a sample of 120 are defective. If the defect rate is 5%, find the probability that a batch is accepted.

1

Identify parameters:

X \sim B (120, 0.05)

, find

P (X < 8)

2

Check conditions:

n p = 120 \times 0.05 = 6 > 5

✓ and

n (1 - p) = 120 \times 0.95 = 114 > 5

✓

3

Calculate normal parameters:

μ = 6

,

σ = 120 \times 0.05 \times 0.95 = 5.7 = 2.387

4

Apply continuity correction:

P (X < 8) \approx P (Y < 7.5)

(exclude 8, so use 8 − 0.5)

5

Standardise:

z = \frac{7.5 - 6}{2.387} = 0.628

6

Find probability:

P (Z < 0.628) = 0.7350

Answer: P(batch accepted) ≈ 0.735 (3 s.f.)

Common Mistakes to Avoid

Mistake 1: Forgetting Continuity Correction

Without the correction, answers can be significantly wrong. Always adjust by ±0.5 when approximating.

Mistake 2: Wrong Direction of Correction

Remember: ≤ and ≥ expand outward (include the value), < and > shrink inward (exclude the value).

Mistake 3: Not Checking Conditions

Always verify both np > 5 AND n(1−p) > 5 before using the approximation.

Mistake 4: Using Wrong Variance Formula

Remember:

σ^{2} = n p (1 - p)

, NOT

σ^{2} = n p

or

σ^{2} = n pq

with q ≠ 1−p.

Practice

Question 1: A biased die has probability 0.2 of showing a six. The die is rolled 90 times. Use the normal approximation to find the probability of obtaining more than 20 sixes.

1

X \sim B (90, 0.2)

. Check: np = 18 > 5 ✓, n(1−p) = 72 > 5 ✓

2

μ = 18

,

σ = 90 \times 0.2 \times 0.8 = 14.4 = 3.795

3

P (X > 20) \approx P (Y > 20.5)

(exclude 20, so add 0.5)

4

z = \frac{20.5 - 18}{3.795} = 0.659

5

P (Z > 0.659) = 1 - 0.7451 = 0.2549

Answer: P(X > 20) ≈ 0.255 (3 s.f.)

Practice

Question 2: In a large population, 45% support a new policy. A random sample of 150 people is taken. Find the probability that exactly 70 people in the sample support the policy.

1

X \sim B (150, 0.45)

. Check: np = 67.5 > 5 ✓, n(1−p) = 82.5 > 5 ✓

2

μ = 67.5

,

σ = 150 \times 0.45 \times 0.55 = 37.125 = 6.093

3

P (X = 70) \approx P (69.5 < Y < 70.5)

4

z_{1} = \frac{69.5 - 67.5}{6.093} = 0.328

,

z_{2} = \frac{70.5 - 67.5}{6.093} = 0.492

5

P (0.328 < Z < 0.492) = 0.6886 - 0.6286 = 0.0600

Answer: P(X = 70) ≈ 0.0600 (3 s.f.)

Practice

Question 3: A medical trial has a 30% success rate. In a trial of 200 patients, find the probability that between 55 and 65 patients (inclusive) respond successfully to the treatment.

1

X \sim B (200, 0.3)

. Check: np = 60 > 5 ✓, n(1−p) = 140 > 5 ✓

2

μ = 60

,

σ = 200 \times 0.3 \times 0.7 = 42 = 6.481

3

P (55 \leq X \leq 65) \approx P (54.5 < Y < 65.5)

4

z_{1} = \frac{54.5 - 60}{6.481} = - 0.849

,

z_{2} = \frac{65.5 - 60}{6.481} = 0.849

5

P (- 0.849 < Z < 0.849) = 0.8021 - 0.1979 = 0.6042

Answer: P(55 ≤ X ≤ 65) ≈ 0.604 (3 s.f.)

Summary

Key Concepts

ConceptDetails

When to UseWhen np > 5 AND n(1−p) > 5

Parameters

μ = n p

,

σ = n p (1 - p)

Continuity CorrectionAdjust boundaries by ±0.5

Key RuleInclude → expand; Exclude → shrink

Exam Skills

Always check conditions before using the approximation

State your continuity correction clearly

Show the standardisation step with the z-value

Use correct inequality symbols throughout