Exploring the Normal Approximation to the Binomial Distribution, this section will delve into its conditions, application, and practical problem-solving.

## Normal Approximation to Binomial Distribution

### 1. Conditions for Normal Approximation:

- np > 5
**:**Ensures sufficient number of successes. - n(1-p) > 5
**:**Ensures sufficient number of failures. **Purpose:**To ensure a symmetric distribution suitable for normal approximation.

### 2. Applying Continuity Correction Factor:

**For**$P(X \leq x)$**:**Use $P(X \leq x + 0.5)$.**For**$P(X \geq x)$**:**Use P(X > x - 0.5).**For**$P(X = x)$**:**Use P(x - 0.5 < X < x + 0.5).**Purpose:**To align the discrete binomial distribution with the continuous normal distribution for more accurate results.

## Examples

### Example 1: $P(X \geq 10)$ for Binomial Distribution $n = 30, p = 0.2$

**Conditions Check:**$np = 6 , n(1-p) = 24$ (Both > 5, condition met).**Continuity Correction:**Adjust to P(X > 9.5).**Z-Score Calculation:**$Z = \frac{9.5 - 6}{\sqrt{4.8}} \approx 1.60$.**Probability:**$P(X \geq 10) \approx 0.0551$ (5.51% chance for 10+ successes)**Graph:**

### Example 2: $P(3 \leq X \leq 8)$ for Binomial Distribution $n = 50, p = 0.1$

**Conditions Check:**$np = 5 , n(1-p) = 45$ (Both > 5, condition met).**Continuity Correction:**Adjust to $P(2.5 \leq X \leq 8.5)$.**Z-Scores:**$Z_1 = -1.18$ (for X = 2.5), $Z_2 = 1.65$ (for X = 8.5).**Probability:**$P(3 \leq X \leq 8) \approx 0.8312$ (83.12% chance for 3-8 successes).**Graph:**

Written by: Dr Rahil Sachak-Patwa

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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.