Random variables, binomial and normal distributions
Random variables and distributions close the syllabus: model a discrete or continuous variable, compute expectation and variance, and use the binomial and normal models, including standardisation and the normal approximation to the binomial.
Core rules
- Discrete X: P(X = x_i) = p_i with Σp_i = 1; E[X] = Σx_i p_i; Var(X) = Σx_i² p_i − (E[X])²; σ = √Var.
- Continuous X with density f: P(a ≤ X ≤ b) = ∫_a^b f(x)dx, the total area under f is 1, and E[X] = ∫x f(x)dx.
- Linear transformations: E[aX + b] = aE[X] + b but Var(aX + b) = a²Var(X); the standard deviation, not the variance, carries the units of X.
- Binomial B(n, p): P(X = k) = C(n,k)p^k(1−p)^(n−k), with E[X] = np and Var(X) = np(1−p).
- Normal N(μ, σ): standardise with z = (x − μ)/σ, use the table of P(Z ≤ z), and exploit symmetry: P(Z ≤ −z) = 1 − P(Z ≤ z), P(−z ≤ Z ≤ z) = 2P(Z ≤ z) − 1. Inverse problems read the table backwards and undo the standardisation.
- Approximation: if np > 5 and n(1 − p) > 5, B(n,p) ≈ N(np, √(np(1−p))) with continuity correction, so P(X ≤ k) becomes P(Z ≤ (k + 0.5 − np)/σ).
Worked example
A variable is N(500, 10), mean 500 with standard deviation 10. Then P(X > 515) = P(Z > (515 − 500)/10) = P(Z > 1.5) = 1 − 0.9332 = 0.0668, about 6.7%. Likewise P(485 < X < 515) = P(−1.5 < Z < 1.5) = 2(0.9332) − 1 = 0.8664. Note the second parameter is the standard deviation; writing σ² = 100 in the wrong slot would change every answer.
Traps that cost marks
- Mixing up variance and standard deviation, especially in N(μ, σ) notation.
- Using Var(aX + b) = aVar(X) + b instead of a²Var(X).
- Forgetting the continuity correction when approximating a binomial.
- Applying the normal approximation when np or n(1−p) is small.
- Reading the table as P(Z > z), or omitting the 1 − for a right-hand tail.