Paper II · 4 Random Variables, Distributions & Inference · 4.1 Probability and probability distributions
4.1.3 Normal distribution: properties, standard normal, area rules and uses
The normal distribution N(μ, σ²) is the symmetric bell-shaped continuous distribution fixed by its mean μ and variance σ²; every normal question is solved by changing X to the standard normal Z = (X − μ)/σ and reading areas from the Z table.
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Core facts
- The shape, moments and β values of the normal curve are also in the skewness note (STA-DK-02). This note is about using the distribution.
- pdf: f(x) = [1/(σ√(2π))] e^(−(x − μ)²/(2σ²)), −∞ < x < ∞.
- Notation X ~ N(μ, σ²): the second parameter is the variance, not the SD.
- Found by De Moivre (1733) as a limit of the binomial; developed by Laplace and Gauss (hence "Gaussian").
- Mean = median = mode = μ. Symmetric about x = μ. Skewness β₁ = 0; kurtosis β₂ = 3 (mesokurtic).
- Points of inflexion: x = μ ± σ. The curve never touches the x-axis (asymptotic).
- Maximum ordinate at x = μ: 1/(σ√(2π)) ≈ 0.3989/σ.
- Quartiles: μ ± 0.6745σ. QD ≈ (2/3)σ, MD ≈ (4/5)σ (exactly σ√(2/π) ≈ 0.7979σ).
- So QD : MD : SD ≈ 10 : 12 : 15.
- Standard normal variate Z = (X − μ)/σ ~ N(0, 1); pdf φ(z) = (1/√(2π)) e^(−z²/2).
- Area rules: μ ± σ → 68.27%; μ ± 2σ → 95.45%; μ ± 3σ → 99.73%.
- μ ± 1.96σ → 95%; μ ± 2.58σ → 99%; μ ± 1.645σ → 90%.
- mgf: M_X(t) = e^(μt + σ²t²/2).
- Additive property: if X ~ N(μ₁, σ₁²) and Y ~ N(μ₂, σ₂²) are independent, then X ± Y ~ N(μ₁ ± μ₂, σ₁² + σ₂²).
- aX + b ~ N(aμ + b, a²σ²).
Explained
1. The three-step method.
- Convert: z = (x − μ)/σ.
- Read the table area from 0 to |z|.
- Combine with 0.5 (each half of the curve has area 0.5) and use symmetry.
Table values used below (area from 0 to z): 0.5 → 0.1915; 1 → 0.3413; 1.4 → 0.4192; 1.5 → 0.4332; 1.9 → 0.4713; 2 → 0.4772; 3 → 0.4987.
Rules of thumb.
- P(Z > z) = 0.5 − area(0 to z).
- P(Z < z) = 0.5 + area(0 to z), for z > 0.
- P(−a < Z < b) = area(0 to a) + area(0 to b).
- P(a < Z < b), both positive = area(0 to b) − area(0 to a).
- Using the cumulative form Φ(z) = P(Z ≤ z): Φ(−z) = 1 − Φ(z).
Example 1. X ~ N(50, 100). Find P(X > 70) and P(40 < X < 60). Solution. σ = √100 = 10. For 70: z = (70 − 50)/10 = 2. P(Z > 2) = 0.5 − 0.4772 = 0.0228. For 40 and 60: z = −1 and 1. Area = 2 × 0.3413 = 0.6826. Answer: 0.0228; 0.6826
Example 2. Marks of 1,000 students are normal with mean 50 and SD 10. How many score between 45 and 60? Solution. z = −0.5 and z = 1. Area = 0.1915 + 0.3413 = 0.5328. 0.5328 × 1,000 = 532.8. Answer: about 533 students
Example 3. X ~ N(30, 25). Find P(X < 20). Solution. σ = 5. z = (20 − 30)/5 = −2. P(Z < −2) = P(Z > 2) = 0.5 − 0.4772. Answer: 0.0228
2. Working backwards: finding μ and σ.
Example 4. In a normal distribution, 31% of the items are under 45 and 8% are over 64. Find μ and σ. (Area 0 to 0.5 = 0.19; area 0 to 1.4 = 0.42.) Solution.
- 31% below 45 means area 0.19 between 45 and μ, so 45 is z = −0.5: 45 = μ − 0.5σ.
- 8% above 64 means area 0.42 between μ and 64, so 64 is z = 1.4: 64 = μ + 1.4σ.
- Subtract: 19 = 1.9σ, so σ = 10 and μ = 50.
Answer: μ = 50, σ = 10
Deeper
1. Sums and differences of normal variables.
Example 5. X ~ N(10, 9) and Y ~ N(20, 16) are independent. Find the distribution of X + Y and of Y − X. Solution. X + Y ~ N(30, 9 + 16) = N(30, 25), SD 5. Y − X ~ N(10, 25), also SD 5: the variances add even for a difference. Answer: N(30, 25) and N(10, 25)
Consequence: the mean x̄ of n independent N(μ, σ²) values is exactly N(μ, σ²/n). Its SD, σ/√n, is called the standard error.
2. Normal approximation to the binomial and Poisson.
- Binomial B(n, p) ≈ N(np, npq) when n is large and p is not near 0 or 1 (a working rule: np and nq both at least 5).
- Poisson P(λ) ≈ N(λ, λ) when λ is large.
- Continuity correction: a discrete count is spread over an interval of width 1. So P(X ≥ 60) uses 59.5; P(X ≤ 40) uses 40.5; P(X = 50) uses 49.5 to 50.5.
Example 6. A fair coin is tossed 100 times. Find (approximately) P(at least 60 heads). (Area 0 to 1.9 = 0.4713.) Solution. μ = np = 50, σ = √(npq) = √25 = 5. With continuity correction, z = (59.5 − 50)/5 = 1.9. P = 0.5 − 0.4713 = 0.0287. (The exact binomial value is about 0.0284.) Answer: about 0.029
3. Central limit theorem (CLT). For a large sample from any population with mean μ and finite variance σ², the sample mean x̄ is approximately N(μ, σ²/n). This is why large-sample (z) tests use the normal table, whatever the shape of the population.
4. Uses of the normal distribution.
- Many natural measurements (heights, errors of measurement) are close to normal. Gauss used it for errors, so it is also the normal law of errors.
- It approximates the binomial, Poisson, chi-square and t distributions for large n.
- Sampling distributions of means and proportions are normal for large samples (CLT), so it is the basis of large-sample tests and confidence limits (x̄ ± 1.96 SE for 95%).
- Quality control: control limits at μ ± 3σ; a point outside means a cause to investigate (only 0.27% fall outside by chance).
- Standardising and grading scores (Z scores).
5. Other properties.
- The total area is 1, and the curve is symmetric, so P(Z > 0) = 0.5.
- All odd central moments are 0; μ₂ = σ², μ₄ = 3σ⁴. In general μ₂ᵣ = 1 · 3 · 5 … (2r − 1) σ²ʳ.
- A linear function of independent normal variables is normal.
- Changing μ shifts the curve sideways. A larger σ makes the curve flatter and wider, with a lower peak.
| Fact | Value |
|---|---|
| Area μ ± 0.6745σ | 50% (quartiles) |
| Area μ ± 1.645σ | 90% |
| Area μ ± 1.96σ | 95% |
| Area μ ± 2.58σ | 99% |
Common traps
- ✗ "N(50, 100) has SD 100." ✓ The second parameter is the variance; SD = 10.
- ✗ "P(X = 60) for a normal X is the height of the curve at 60." ✓ For any continuous variable P(X = 60) = 0.
- ✗ "P(Z > 2) = 0.4772." ✓ The table gives area 0 to 2; P(Z > 2) = 0.5 − 0.4772 = 0.0228.
- ✗ "Points of inflexion at μ ± 2σ." ✓ At μ ± σ.
- ✗ "For independent normals, V(X − Y) = σ₁² − σ₂²." ✓ σ₁² + σ₂².
- ✗ "95% of the area lies within μ ± 2σ exactly." ✓ μ ± 2σ holds 95.45%; 95% needs μ ± 1.96σ.
- ✗ "QD ≈ 4σ/5 and MD ≈ 2σ/3." ✓ QD ≈ 2σ/3, MD ≈ 4σ/5.
- ✗ "Normal approximation for P(X ≥ 60) uses 60.5." ✓ Continuity correction moves outward: use 59.5.
- ✗ "The CLT says the population becomes normal." ✓ It is the sample mean that becomes approximately normal.
Quick revision
- X ~ N(μ, σ²); second parameter = variance.
- Z = (X − μ)/σ ~ N(0, 1); Φ(−z) = 1 − Φ(z).
- Mean = median = mode; inflexion at μ ± σ; asymptotic.
- 68.27%, 95.45%, 99.73% for 1σ, 2σ, 3σ.
- 1.645σ → 90%, 1.96σ → 95%, 2.58σ → 99%.
- QD ≈ 2σ/3, MD ≈ 4σ/5; QD : MD : SD ≈ 10 : 12 : 15.
- Independent normals: X ± Y ~ N(μ₁ ± μ₂, σ₁² + σ₂²).
- x̄ ~ N(μ, σ²/n); SE = σ/√n.
- B(n, p) ≈ N(np, npq); P(λ) ≈ N(λ, λ); use ±0.5 correction.
- Uses: large-sample tests, CLT, control limits μ ± 3σ.
Practise this topic. Questions on Normal distribution & its uses, with answers explained.
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