Continuous random variables and the bell curve
A continuous random variable can take any value in a range, like height or time. We cannot list the probability of one exact value (it is 0). Instead we use a curve, the probability density function, and a probability is the area under the curve between two values. The total area is 1.
Properties of the normal curve
- Bell-shaped and symmetric about the mean μ.
- Mean = median = mode, all at the peak.
- The spread is set by the standard deviation σ; the points of inflection are at μ ± σ.
- The tails get closer to the axis but never touch it.
We write X ~ N(μ, σ²). Note the second number is the variance.
The 68–95–99.7 rule
For every normal distribution:
- P(μ − σ < X < μ + σ) ≈ 0.68
- P(μ − 2σ < X < μ + 2σ) ≈ 0.95
- P(μ − 3σ < X < μ + 3σ) ≈ 0.997
Because the curve is symmetric, each half holds 50%. So P(X > μ + σ) ≈ (1 − 0.68) ÷ 2 = 0.16.
A value more than 3σ from the mean is very unusual; this is often used to spot outliers.
Standard normal and z-scores
The standard normal Z has μ = 0 and σ = 1: Z ~ N(0, 1). Any normal X is changed to Z with
z = (x − μ) ÷ σ
The table (or calculator) gives Φ(z) = P(Z < z). Useful facts:
- P(Z > z) = 1 − Φ(z)
- Φ(−z) = 1 − Φ(z) (symmetry)
- P(a < Z < b) = Φ(b) − Φ(a)
z-scores also let you compare values from different groups: 80 marks in a test with μ = 70, σ = 5 (z = 2) is better than 85 in a test with μ = 75, σ = 10 (z = 1).
When σ is unknown and the sample is small, statisticians use the t-distribution: it looks like the normal curve but with fatter tails, and it gets closer to the normal as the sample grows.
Inverse normal, unknown μ or σ, and the binomial link
Inverse normal
If you know the probability and want the value: find z from the table (or invNorm), then x = μ + zσ. Top 5% cut-off: z = 1.645.
Finding μ or σ
Turn a given probability into a z-value, then solve z = (x − μ)/σ. With two conditions you get two equations for μ and σ.
Normal approximation to the binomial
If X ~ B(n, p) with n large and np > 5 and n(1 − p) > 5, then X is roughly N(np, np(1 − p)). Add or subtract 0.5 (continuity correction) because a whole-number count becomes a continuous value: P(X ≤ 55) ≈ P(Y < 55.5).
Key formulas and definitions
- X ~ N(μ, σ²)
- z = (x − μ) ÷ σ
- P(X < x) = Φ(z)
- P(X > x) = 1 − Φ(z)
- Φ(−z) = 1 − Φ(z)
- P(a < X < b) = Φ(z_b) − Φ(z_a)
- x = μ + zσ (inverse normal)
- B(n, p) ≈ N(np, np(1 − p)), with continuity correction ± 0.5
Worked examples
1. X ~ N(50, 16). Find the z-score of x = 58.
σ = √16 = 4. z = (58 − 50) ÷ 4 = 2.
2. Marks are normal with μ = 60, σ = 10. Using the empirical rule, what percent score between 40 and 80?
40 and 80 are μ ± 2σ, so about 95%.
3. Heights: μ = 170 cm, σ = 8 cm. Find P(height > 182 cm).
z = 12 ÷ 8 = 1.5. Φ(1.5) = 0.9332. P(X > 182) = 1 − 0.9332 = 0.0668.
4. Marks: μ = 60, σ = 10. Find P(55 < X < 70).
z = −0.5 and 1. Φ(1) − Φ(−0.5) = 0.8413 − 0.3085 = 0.5328.
5. Rice packets: μ = 500 g, σ = 4 g. The top 5% heaviest packets are checked. Above what mass?
Top 5% → z = 1.645. x = 500 + 1.645 × 4 = 506.58 g.
6. A fair coin is tossed 100 times. Use a normal approximation to estimate P(at most 55 heads).
np = 50, σ = √(100 × 0.5 × 0.5) = 5. With continuity correction: z = (55.5 − 50) ÷ 5 = 1.1. Φ(1.1) ≈ 0.8643.
Common mistakes
- Using the variance instead of the SD: in N(50, 16), σ = 4, not 16.
- Forgetting 1 − Φ(z) for "greater than" questions.
- Thinking P(X = 170) is a number bigger than 0 for a continuous variable. It is 0; use intervals.
- Leaving out the continuity correction when approximating a binomial count.