Statistics 1: Normal Distribution

Statistics 1: Normal Distribution — CIE A Level Mathematics 9709

1. Introduction to the Normal Distribution

The Normal distribution (bell curve) is the most important continuous probability distribution. It describes many natural phenomena: heights, weights, exam scores, measurement errors.

Notation: X ~ N(μ, σ²) where μ = mean, σ² = variance (σ = standard deviation).

Properties:

• Symmetric about the mean μ
• Bell-shaped curve; extends from −∞ to +∞
• Mean = Median = Mode = μ
• Total area under curve = 1 (probability)
• 68-95-99.7 rule: 68% within 1σ of mean; 95% within 2σ; 99.7% within 3σ

2. Standardising: The Z-Score

To find probabilities, convert to the standard Normal Z ~ N(0, 1):

$$Z = rac{X - mu}{sigma}$$

Standard Normal tables give Φ(z) = P(Z ≤ z) for z > 0.

For negative z: Φ(−z) = 1 − Φ(z) (by symmetry)

3. Probability Calculations

Finding P(X < a): Standardise → look up Z table.

Finding P(X > a): P(X > a) = 1 − P(X < a)

Finding P(a < X < b): P(a < X < b) = P(X < b) − P(X < a) = Φ(z₂) − Φ(z₁)

Worked Example: X ~ N(100, 64). Find P(X < 112).

σ = √64 = 8. Z = (112 − 100)/8 = 1.5.

P(X < 112) = Φ(1.5) = 0.9332

Example: Find P(88 < X < 116) for X ~ N(100, 64).

Z₁ = (88−100)/8 = −1.5; Z₂ = (116−100)/8 = 2.

P = Φ(2) − Φ(−1.5) = 0.9772 − (1−0.9332) = 0.9772 − 0.0668 = 0.9104

4. Finding Values from Probabilities

'Inverse Normal': Given P(X < a) = p, find a.

Method: Look up p in body of Z-table → find z → x = μ + zσ.

Example: X ~ N(50, 25). Find a such that P(X < a) = 0.9.

σ = 5. From table: Φ(1.282) ≈ 0.9. So z = 1.282.

a = 50 + 1.282 × 5 = 50 + 6.41 = 56.4

Given probabilities on both sides: Set up two equations.

If P(X < a) = 0.9 and P(X > b) = 0.05 with X ~ N(μ, σ²):

(a − μ)/σ = 1.282 and (b − μ)/σ = 1.645 → two equations, solve for μ and σ.

5. Normal Approximation to the Binomial

When n is large and p is not too close to 0 or 1:

B(n, p) ≈ N(np, np(1−p))

Conditions for Normal approximation: np > 5 AND n(1−p) > 5.

Continuity correction: Since Binomial is discrete but Normal is continuous:

• P(X = k) becomes P(k−0.5 < Y < k+0.5)
• P(X ≤ k) becomes P(Y < k+0.5)
• P(X ≥ k) becomes P(Y > k−0.5)
• P(X < k) becomes P(Y < k−0.5)

Example: X ~ B(200, 0.4). Find P(X ≤ 75) using Normal approximation.

μ = 200×0.4 = 80, σ² = 200×0.4×0.6 = 48, σ = √48 = 6.928.

With continuity correction: P(Y < 75.5) where Y ~ N(80, 48).

Z = (75.5 − 80)/6.928 = −4.5/6.928 = −0.6495.

P = Φ(−0.65) = 1 − Φ(0.65) = 1 − 0.7422 = 0.2578

6. Finding μ and σ

Given two probability statements → two equations in μ and σ.

Example: X ~ N(μ, σ²). P(X < 20) = 0.25 and P(X < 30) = 0.75.

By symmetry: P(X<20) = 0.25 and P(X<30) = 0.75 → since 0.25 + 0.75 = 1, mean = (20+30)/2 = μ = 25.

P(X < 30) = 0.75 → Z = 0.674 → (30−25)/σ = 0.674 → σ = 7.42.