When the number of trials is large, a binomial distribution can be approximated by a normal distribution.
If \(X \sim B(n,p)\), we may approximate using a normal distribution when:
Then:
The standard deviation is:
\[ \sigma = \sqrt{np(1-p)} \]The binomial distribution is discrete, but the normal distribution is continuous.
So we apply a continuity correction.
| Binomial Probability | Normal Approximation |
|---|---|
| \(P(X = k)\) | \(P(k - 0.5 < X < k + 0.5)\) |
| \(P(X \le k)\) | \(P(X < k + 0.5)\) |
| \(P(X < k)\) | \(P(X < k - 0.5)\) |
| \(P(X \ge k)\) | \(P(X > k - 0.5)\) |
| \(P(X > k)\) | \(P(X > k + 0.5)\) |
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
\[ P(Z > 0.92) = 0.1788 \]