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The normal distribution – The normal approximation to the binomial distribution

📘 Notes

Normal Approximation to the Binomial Distribution (AS Statistics)

When the number of trials is large, a binomial distribution can be approximated by a normal distribution.

1. When Can We Use the Normal Approximation?

If \(X \sim B(n,p)\), we may approximate using a normal distribution when:

\[ np \ge 5 \quad \text{and} \quad n(1-p) \ge 5 \]

Then:

\[ X \sim N(\mu, \sigma^2) \] where \[ \mu = np, \qquad \sigma^2 = np(1-p) \]

The standard deviation is:

\[ \sigma = \sqrt{np(1-p)} \]

2. Continuity Correction

The binomial distribution is discrete, but the normal distribution is continuous.

So we apply a continuity correction.

Binomial Probability Normal Approximation
\(P(X = k)\) \(P(k - 0.5 < X < k + 0.5)\)
\(P(X \le k)\) \(P(X < k + 0.5)\)
\(P(X < k)\) \(P(X < k - 0.5)\)
\(P(X \ge k)\) \(P(X > k - 0.5)\)
\(P(X > k)\) \(P(X > k + 0.5)\)

3. Worked Example

A biased coin has probability \(p = 0.6\) of landing heads.

The coin is tossed \(n = 100\) times.

Find \(P(X \ge 65)\) using a normal approximation.


Step 1: Check conditions

\[ np = 100(0.6) = 60 \] \[ n(1-p) = 100(0.4) = 40 \]

Both are greater than 5, so the approximation is valid.


Step 2: Mean and variance

\[ \mu = np = 60 \] \[ \sigma^2 = np(1-p) = 100(0.6)(0.4) = 24 \] \[ \sigma = \sqrt{24} \approx 4.90 \]

Step 3: Continuity correction

\[ P(X \ge 65) \approx P(X > 64.5) \]

Step 4: Standardise

\[ Z = \frac{64.5 - 60}{4.90} \] \[ Z \approx 0.92 \]

Step 5: Use normal tables

\[ P(Z > 0.92) = 0.1788 \]
Final answer: \[ P(X \ge 65) \approx 0.179 \]

4. Exam Tips

  • Always check the conditions \(np \ge 5\) and \(n(1-p) \ge 5\).
  • State the approximating distribution clearly.
  • Always apply the continuity correction.
  • Standardise using \(Z = \frac{x-\mu}{\sigma}\).
  • Use correct probability notation when writing answers.