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Year 12 Maths

Binomial to Normal Approximation

Learn when and how to approximate the binomial distribution with a normal distribution, including the continuity correction, in HSC Advanced Mathematics.

When to Use the Normal Approximation

For large n, the binomial distribution B(n, p) can be approximated by a normal distribution. This avoids computing large binomial coefficients. The approximation is valid when:

np ≥ 5   and   n(1 − p) ≥ 5

Both conditions must be satisfied for the approximation to be reliable.

If X ~ B(n, p), we approximate with X ~ N(μ, σ2) where:

μ = np,   σ2 = np(1 − p)

Example: X ~ B(100, 0.4). Check: np = 40 ≥ 5, n(1−p) = 60 ≥ 5. Valid!

Approximate: X ~ N(40, 24), since μ = 40 and σ2 = 100 × 0.4 × 0.6 = 24.

Continuity Correction

Since we are approximating a discrete distribution with a continuous one, we apply a continuity correction of ±0.5 to improve accuracy.

P(X ≤ k) becomes P(X ≤ k + 0.5) in the normal approximation.

P(X ≥ k) becomes P(X ≥ k − 0.5).

P(X = k) becomes P(k − 0.5 ≤ X ≤ k + 0.5).

P(a ≤ X ≤ b) becomes P(a − 0.5 ≤ X ≤ b + 0.5).

Why? In the binomial, P(X = 5) covers just the integer 5. In the normal, we represent this as the area from 4.5 to 5.5 — giving the discrete value a "width" of 1.

The continuity correction makes the normal approximation more accurate, especially for moderate n.

Applications

The normal approximation is particularly useful in quality control, polling, and any situation where large numbers of independent trials are conducted.

Quality control: A factory produces items with a 3% defect rate. In a batch of 500, what is the probability of more than 20 defects?

X ~ B(500, 0.03). μ = 15, σ = √(500 × 0.03 × 0.97) = √14.55 ≈ 3.81.

P(X > 20) ≈ P(Z > (20.5 − 15)/3.81) = P(Z > 1.44) ≈ 0.0749 or about 7.5%.

Polling: If 60% of voters support a candidate, what is the probability that a random sample of 200 voters has fewer than 110 supporters?

X ~ B(200, 0.6). μ = 120, σ = √48 ≈ 6.93. P(X < 110) ≈ P(Z < (109.5 − 120)/6.93) = P(Z < −1.52) ≈ 0.0643.

Key Vocabulary

Normal Approximation

Using a normal distribution to estimate binomial probabilities when n is large and np ≥ 5, n(1−p) ≥ 5.

Continuity Correction

An adjustment of ±0.5 applied when approximating a discrete distribution with a continuous one.

Binomial Distribution

A discrete distribution B(n, p) counting successes in n independent trials, each with probability p.

Z-score

A standardised value z = (x − μ)/σ measuring how many standard deviations x is from the mean.

Worked Examples

1

X ~ B(80, 0.5). Use the normal approximation to find P(X ≤ 45).

Step 1: Check: np = 40 ≥ 5, n(1−p) = 40 ≥ 5. Valid.

Step 2: μ = 40, σ = √(80 × 0.5 × 0.5) = √20 ≈ 4.47.

Step 3: With continuity correction: P(X ≤ 45.5). z = (45.5 − 40)/4.47 ≈ 1.23.

Answer: P(X ≤ 45) ≈ P(Z ≤ 1.23) ≈ 0.8907.

2

X ~ B(200, 0.3). Find P(X = 60) using the normal approximation.

Step 1: μ = 60, σ = √(200 × 0.3 × 0.7) = √42 ≈ 6.48.

Step 2: P(X = 60) ≈ P(59.5 ≤ X ≤ 60.5). z1 = (59.5 − 60)/6.48 ≈ −0.077, z2 = (60.5 − 60)/6.48 ≈ 0.077.

Answer: P(X = 60) ≈ P(−0.077 ≤ Z ≤ 0.077) ≈ 0.0614.

3

A coin is tossed 150 times. Find the probability of getting between 70 and 85 heads (inclusive).

Step 1: X ~ B(150, 0.5). μ = 75, σ = √(37.5) ≈ 6.12.

Step 2: P(70 ≤ X ≤ 85) ≈ P(69.5 ≤ X ≤ 85.5). z1 = (69.5 − 75)/6.12 ≈ −0.90, z2 = (85.5 − 75)/6.12 ≈ 1.72.

Answer: P(70 ≤ X ≤ 85) ≈ P(Z ≤ 1.72) − P(Z ≤ −0.90) ≈ 0.9573 − 0.1841 = 0.7732.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

For X ~ B(n, p), the normal approximation is valid when:

Question 2

When approximating P(X ≥ 30) for a binomial, what continuity correction should be applied?

Question 3

X ~ B(100, 0.4). What are μ and σ for the normal approximation?

Question 4

Can the normal approximation be used for X ~ B(20, 0.1)?

Question 5

Why is the continuity correction needed?

Key Concepts Summary

Continuous Distributions