Example 1: Why 1.96 gives a 95% interval
Using A(1.96) = 0.4750, find P(−1.96 < Z < 1.96).
- The curve is symmetric about 0
P(−1.96 < Z < 1.96) = 2A(1.96)= 2 × 0.4750 = 0.95
Answer: 0.95
This page lists the formulas used in DSE M1, grouped by area. In the examination a standard normal distribution table is provided; plan to know the other formulas by heart. The table gives A(z) = P(0 ≤ Z ≤ z).
(a + b)ⁿ = Σ C(n, r) aⁿ⁻ʳ bʳeˣ = 1 + x + x²/2! + x³/3! + …(xⁿ)′ = nxⁿ⁻¹, (eˣ)′ = eˣ, (ln x)′ = 1/x, (aˣ)′ = aˣ ln a(uv)′ = u′v + uv′, (u/v)′ = (u′v − uv′)/v²dy/dx = (dy/du)(du/dx)∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C, ∫ (1/x) dx = ln|x| + C, ∫ ekx dx = ekx/k + C∫ab f(x) dx = F(b) − F(a)(h/2)[y₀ + 2(y₁ + … + yn−1) + yn], h = (b − a)/nP(A | B) = P(A ∩ B) / P(B)P(B | D) = P(D | B)P(B) / [P(D | A)P(A) + P(D | B)P(B)]E(X) = Σ x P(X = x)Var(X) = E(X²) − [E(X)]²E(aX + b) = aE(X) + b, Var(aX + b) = a²Var(X)B(n, p): P(X = k) = C(n, k) pᵏ(1 − p)ⁿ⁻ᵏ, mean np, variance np(1 − p)Po(λ): P(X = k) = e−λλᵏ / k!, mean λ, variance λZ = (X − μ)/σ, where X ~ N(μ, σ²)A(z) = P(0 ≤ Z ≤ z); P(Z > z) = 0.5 − A(z) for z > 0X̄ ~ N(μ, σ²/n)x̄ ± 1.96 σ/√nx̄ ± 2.575 σ/√nThe values 1.96 and 2.575 come from the table: A(1.96) = 0.4750 and A(2.575) = 0.4950.
Using A(1.96) = 0.4750, find P(−1.96 < Z < 1.96).
P(−1.96 < Z < 1.96) = 2A(1.96)= 2 × 0.4750 = 0.95Answer: 0.95