Estimation, Poisson, hypothesis tests and Markov chains
HL statistics for Maths AI: linear combinations of random variables, unbiased estimates, the central limit theorem, confidence intervals for a mean, the Poisson distribution, hypothesis tests with binomial, Poisson and normal distributions, Type I and Type II errors, and Markov chains with transition matrices and steady states.
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In 30 seconds
- Combining variables: E(aX + bY) = aE(X) + bE(Y); for independent X and Y, Var(aX + bY) = a2Var(X) + b2Var(Y). Variances add even when you subtract.
- Central limit theorem: for large n (n ≥ 30 is the usual guide), the sample mean X̄ is approximately N(μ, σ2n), whatever the shape of the population.
- Confidence interval for μ: x̄ ± zσ√n when σ is known, or x̄ ± tsn−1√n when it isn't.
- Poisson Po(m): counts of random events in a fixed interval, with mean = variance = m.
- Errors: a Type I error rejects a true H0 (its probability is the significance level); a Type II error fails to reject a false H0. A Markov chain moves between states with a transition matrix T; the steady state s satisfies Ts = s.
By the end you can
- Use E(aX + b), Var(aX + b) and linear combinations of independent random variables. HL
- Find unbiased estimates of the mean and variance. HL
- Use the central limit theorem and find confidence intervals for a mean. HL
- Use the Poisson distribution, including sums of independent Poisson variables. HL
- Carry out hypothesis tests with binomial, Poisson and normal distributions, find critical regions, and Type I and II error probabilities. HL
- Use transition matrices for Markov chains and find steady states. HL
1Combining random variables HL
X ~ N(50, 32) and Y ~ N(30, 42) are independent. Find the distribution of X − Y and P(X − Y > 25).
- E(X − Y) = 20, Var(X − Y) = 9 + 16 = 25 M1A1
- X − Y ~ N(20, 52) A1
- P(X − Y > 25) = 0.159 A1
- Unbiased estimates: x̄ estimates μ; sn−12 = nn − 1sn2 estimates σ2. For 10, 12, 9, 11, 13: x̄ = 11 and sn−12 = 104 = 2.5.
2The central limit theorem and confidence intervals HL
A population has mean 70 and SD 12. A sample of 36 is taken. Find P(X̄ > 73).
- X̄ ~ N(70, 12236) = N(70, 22) (CLT) M1A1
- P(X̄ > 73) = 0.0668 A1
A sample of 25 has x̄ = 48.2 and sn−1 = 5. Find a 95% confidence interval for μ.
- σ unknown, so use t with 24 df: t = 2.064 M1
- 48.2 ± 2.064 × 55 M1
- 46.1 < μ < 50.3 A1
- Interpreting: "we are 95% confident μ is between 46.1 and 50.3". A wider interval comes from a higher confidence level or a smaller sample.
3The Poisson distribution HL
- Conditions: events happen singly, independently, at a constant average rate. If X ~ Po(m) and Y ~ Po(n) are independent, X + Y ~ Po(m + n).
A helpline gets 4 calls an hour on average. Find P(exactly 2 calls in an hour), P(at least 6 calls in an hour), and P(no calls in 30 minutes).
- X ~ Po(4): P(X = 2) = 0.147 A1
- P(X ≥ 6) = 1 − P(X ≤ 5) = 0.215 A1
- in 30 minutes Y ~ Po(2): P(Y = 0) = e−2 = 0.135 M1A1
4Hypothesis tests HL
A seed company says 30% of its seeds fail. A gardener suspects more fail. In a sample of 20, 10 fail. Test at the 5% level.
- X ~ B(20, p); H0: p = 0.3, H1: p > 0.3 A1
- P(X ≥ 10 ∣ p = 0.3) = 1 − P(X ≤ 9) = 0.0480 M1A1
- 0.0480 < 0.05, so reject H0: there is evidence that more than 30% fail R1
- Critical region: the set of results that lead to rejecting H0. Here P(X ≥ 10) = 0.048 ≤ 0.05 but P(X ≥ 9) = 0.113 > 0.05, so the critical region is X ≥ 10.
For this test, find the probability of a Type I error, and of a Type II error if really p = 0.5.
- Type I: P(X ≥ 10 ∣ p = 0.3) = 0.0480 A1
- Type II: P(X ≤ 9 ∣ p = 0.5) = 0.412 M1A1
- Other tests: a normal mean test uses X̄ ~ N(μ0, σ2/n) (z-test) or a t-test when σ is unknown; a Poisson mean test works like the binomial one; a correlation test checks H0: ρ = 0 using r and the GDC's linear regression t-test. All three are removed for 2029.
5Markov chains HL
Sunny days are followed by sunny days with probability 0.8, rainy days by sunny days with probability 0.4. Today is sunny. Find the probability that it is sunny in 2 days' time, and the long-term proportion of sunny days.
- T = (0.8 0.4; 0.2 0.6), s0 = (1; 0) A1
- T2s0 = (0.72; 0.28), so P(sunny) = 0.72 M1A1
- steady state: 0.8s + 0.4(1 − s) = s ⇒ s = 23, so ⅔ of days are sunny in the long run M1A1
- Eigenvector link: the steady state is the eigenvector of T with eigenvalue 1, scaled so its entries add to 1.
Test yourself — cover the answers first
Quick check
- HL Var(X) = 4. Find Var(3X − 2).
- HL X and Y are independent with variances 5 and 7. Find Var(X − Y).
- HL What does the CLT say about X̄ for large n?
- HL X ~ Po(3). Find Var(X).
- HL What is a Type I error?
- HL What must each column of a transition matrix add up to?
Exam-style
- HL Cars pass a point at an average of 2 per minute and lorries at 0.5 per minute, independently. Find the probability that exactly 3 vehicles pass in one minute.
- HL Find a 90% confidence interval for μ from a sample of 50 with x̄ = 120 when σ = 15 is known.
Show answers
- 36
- 12
- approximately N(μ, σ²/n)
- 3
- rejecting H₀ when it is true
- 1
- total V ~ Po(2.5)
P(V = 3) = 2.53e−2.53!
= 0.214 - z = 1.645
120 ± 1.645 × 15√50 = 120 ± 3.49
116.5 < μ < 123.5
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