Normal Distribution
The Normal Distribution
The normal distribution is the most important continuous probability distribution in statistics. It models many natural phenomena and provides the foundation for much of statistical inference.
Properties of the Normal Distribution
A continuous random variable X follows a normal distribution, written X ~ N(μ, σ²), if its probability density function forms the classic bell-shaped curve.
Key properties:
- The curve is symmetric about the mean μ
- The mean, median and mode are all equal (at μ)
- The total area under the curve equals 1
- The curve extends infinitely in both directions but approaches zero
- Approximately 68% of values lie within 1σ of the mean
- Approximately 95% lie within 2σ
- Approximately 99.7% lie within 3σ
The Standard Normal Distribution
The standard normal distribution Z ~ N(0, 1) has mean 0 and standard deviation 1.
Any normal variable X can be converted to Z using the standardisation formula:
Z = (X - μ) / σ
This allows you to use standard normal tables (which give P(Z ≤ z)) for any normal distribution.
Worked Example: X ~ N(50, 16) (so μ = 50, σ = 4). Find P(X > 56).
Standardise: Z = (56 - 50)/4 = 1.5
P(X > 56) = P(Z > 1.5) = 1 - P(Z ≤ 1.5) = 1 - 0.9332 = 0.0668
Finding Probabilities
Use the symmetry of the distribution and the relationship P(Z > z) = 1 - P(Z ≤ z):
| Probability | Method |
|---|---|
| P(X < a) | Standardise, read from tables |
| P(X > a) | = 1 - P(X < a) |
| P(a < X < b) | = P(X < b) - P(X < a) |
| P(X < -z) | = P(X > z) = 1 - P(X < z) (symmetry) |
Worked Example: X ~ N(100, 225). Find P(88 < X < 112).
σ = √225 = 15.
P(X < 112) = P(Z < (112-100)/15) = P(Z < 0.8) = 0.7881
P(X < 88) = P(Z < (88-100)/15) = P(Z < -0.8) = 1 - 0.7881 = 0.2119
P(88 < X < 112) = 0.7881 - 0.2119 = 0.5762
Inverse Normal (Finding Values)
Given a probability, find the corresponding value of X. This involves reading the tables in reverse (or using the inverse normal function on a calculator).
Worked Example: X ~ N(70, 25). Find the value of a such that P(X < a) = 0.9.
From tables: P(Z < 1.2816) = 0.9, so z = 1.2816.
Using X = μ + zσ: a = 70 + 1.2816 × 5 = 70 + 6.408 = 76.41
Finding μ and σ
If you know probabilities and need to find the parameters:
Worked Example: X ~ N(μ, σ²). P(X > 80) = 0.1 and P(X < 60) = 0.2. Find μ and σ.
From tables: P(Z > 1.2816) = 0.1, so (80 - μ)/σ = 1.2816 ... (i)
P(Z < -0.8416) = 0.2, so (60 - μ)/σ = -0.8416 ... (ii)
From (i): 80 - μ = 1.2816σ
From (ii): 60 - μ = -0.8416σ
Subtracting: 20 = 2.1232σ, so σ = 9.42 (to 3 s.f.)
From (i): μ = 80 - 1.2816(9.42) = 80 - 12.07 = 67.9 (to 3 s.f.)
Normal Approximation to the Binomial
When X ~ B(n, p) with n large, np > 5 and nq > 5, the binomial can be approximated by:
X ≈ N(np, npq)
A continuity correction is needed because we are approximating a discrete distribution with a continuous one:
- P(X ≤ k) ≈ P(Y ≤ k + 0.5)
- P(X ≥ k) ≈ P(Y ≥ k - 0.5)
- P(X = k) ≈ P(k - 0.5 < Y < k + 0.5)
Worked Example: X ~ B(100, 0.4). Approximate P(X ≤ 35).
μ = 100 × 0.4 = 40, σ² = 100 × 0.4 × 0.6 = 24, σ = 4.899
With continuity correction: P(X ≤ 35) ≈ P(Y ≤ 35.5)
Z = (35.5 - 40)/4.899 = -0.918
P(Z ≤ -0.918) = 1 - P(Z ≤ 0.918) ≈ 1 - 0.8207 = 0.1793
Exam Tips
- Always define your variable and state the distribution before calculating.
- When standardising, ensure you use the standard deviation σ, not the variance σ². The notation N(μ, σ²) uses the variance as the second parameter.
- For inverse normal problems, sketch the curve and shade the relevant area — this helps you determine the correct sign of z.
- Never forget the continuity correction when approximating a binomial with a normal distribution. State it explicitly.
- If the question says "using a suitable approximation", you must justify why the approximation is valid (check np > 5 and nq > 5).