6. Probability & Statistics 2

If events occur independently at constant mean rate λ per interval, X~Po(λ) and P(X=r)=e^{−λ}λ^r/r!.

Match λ to the interval length, check that events are countable and rare enough for the model, and use complements for “at least one” questions.

If a call centre averages 3 calls per minute, P(2 calls in one minute)=e^{−3}3²/2.

Changing the interval changes λ proportionally; it is not a universal parameter for every time window.

For X~Po(λ), E(X)=λ and Var(X)=λ. For a time or area interval scaled by k, the mean becomes kλ under a constant-rate model.

Use the equality as a model check, not as a statement that every sample has equal mean and variance. Estimate λ from appropriate exposure.

If the observed average is 4 events per hour, a two-hour interval has Po(8), not Po(4).

Sample variance need not equal sample mean exactly; the equality describes the distributional parameter.

Independent Poisson variables add to a Poisson variable: if X~Po(λ₁) and Y~Po(λ₂), then X+Y~Po(λ₁+λ₂).

Check independence and compatible exposure definitions before combining. For a difference or conditional count, the simple sum rule does not apply.

Counts from two independent sensors with means 2 and 3 per hour combine to Po(5) per hour.

Adding observed counts is always possible, but adding Poisson distributions requires independence and parameter interpretation.

Bin(n,p) has a fixed number n of independent trials with success probability p; Po(λ) models a count in an interval with a rate. They answer different sampling questions.

Use binomial when trials are explicit and bounded; use Poisson when exposure and event rate are natural. A Poisson approximation to binomial needs large n, small p and λ=np.

Ten quality checks with defect probability 0.02 are binomial; defects over a long production interval may be modelled Poisson with the measured rate.

A small p alone does not justify Poisson approximation; n and the product np also matter.

A binomial or Poisson count may be approximated by N(μ,σ²) when its distribution is sufficiently spread. Convert integer events to intervals using continuity correction, such as P(X≤k)≈P(Y<k+0.5).

State μ and σ² from the original model, standardise the corrected boundary and compare the approximation with an exact calculation when accuracy matters.

P(X≤10) becomes P(Y<10.5), not P(Y≤10), under a continuous normal approximation.

The normal variable is continuous, so forgetting the half-unit correction can materially shift a tail probability.

For X~N(μ,σ²), z=(x−μ)/σ converts a value to the standard normal scale. Interval and tail probabilities follow from the standard CDF.

Sketch the region, standardise both bounds and use symmetry or complements consistently. Keep the distinction between variance σ² and standard deviation σ.

If μ=50 and σ=8, P(42<X<58)=P(−1<Z<1)≈0.6827.

Using σ² in the denominator produces a dimensionally wrong z-score.

For a continuous X, probabilities are areas over intervals: P(a≤X≤b)=∫ₐᵇf(x)dx, and P(X=x)=0 for every single point.

State the support and use a density or CDF that is non-negative and normalised. Endpoint inclusion does not change a continuous probability.

If X is uniform on [0,4], P(1<X<3)=2/4=0.5.

A density height is not a probability; only an area over a range is.

A PDF f(x) satisfies f(x)≥0 and ∫f(x)dx over its support=1. Interval probabilities are integrals of f, and the CDF is their accumulated area.

Find an unknown constant by normalising, then use the correct support for probabilities and moments. A density may exceed 1 when its units are inverse-length.

If f(x)=kx on 0≤x≤2, normalisation gives k=1/2; then P(X>1)=∫₁²x/2 dx=3/8.

A PDF value of 1.5 is not impossible; probabilities are areas and the total area, not the peak height, is constrained.

A population contains all units of interest; a sample is selected to estimate population characteristics. A parameter describes the population, while a statistic is calculated from the sample.

Define the target population before sampling and consider coverage, non-response and selection bias. Larger samples reduce random error but do not automatically remove systematic bias.

A survey of 500 randomly chosen voters is a sample; the proportion supporting a policy in all eligible voters is a population parameter.

A sample statistic is not the parameter itself, and a large biased sample can still mislead.

Simple random, systematic, stratified, quota and cluster samples select units differently. Stratification preserves subgroup representation; cluster sampling selects groups and may increase dependence.

Match the method to the population list and objective, and state how units are chosen. Convenience or voluntary-response samples often under-represent reluctant groups.

To estimate a school’s results across year groups, stratify by year and sample within each stratum rather than sampling only one class.

Random selection within a biased sampling frame cannot represent people missing from the frame.

For independent observations Xᵢ with mean μ and variance σ², the sample mean X̄ has E(X̄)=μ and Var(X̄)=σ²/n. Its standard error is σ/√n (or estimated with s/√n).

Distinguish variability of individual observations from variability of means, and account for finite-population or dependence conditions when relevant.

If σ=12 and n=36, the standard error of X̄ is 2, even though individual values vary with standard deviation 12.

Increasing n reduces standard error by √n, not by n, and does not necessarily reduce measurement bias.

For normal data, X̄ is exactly normal; for large independent samples it may be approximately normal by the central limit effect. Its mean is μ and variance σ²/n.

Standardise with the standard error, state whether σ is known or estimated, and check independence and sample-size assumptions before using normal probabilities.

With μ=100, σ=15 and n=25, X̄~N(100,9), so a one-standard-error interval is 100±3.

The sample mean does not inherit the population variance unchanged; averaging reduces variance by n.

For independent observations with mean μ and variance σ², E(X̄)=μ and Var(X̄)=σ²/n. A linear transformation aX̄+b has mean aμ+b and variance a²σ²/n.

Separate the spread of individual observations from the standard error of the mean, and state independence assumptions before using the formula.

If σ=10 and n=25, the mean has standard deviation 2; quadrupling n halves the standard error.

A sample mean is not guaranteed to equal μ, and multiplying by a changes variance by a², not a.

An estimator T is unbiased for θ when E(T)=θ. Unbiasedness concerns long-run average, not whether one estimate is close to θ.

Check the expectation algebra and compare variance separately; an unbiased estimator can be noisy, while a biased estimator may have lower variance.

The sample mean is unbiased for μ under random sampling, while dividing by n rather than n−1 gives a biased estimate of population variance.

Unbiased does not mean error-free or best for every loss function.

A mean confidence interval is estimate ± critical value×standard error, using a normal or t model as specified. The confidence level describes repeated-method coverage.

Use the correct critical distribution, sample size and known/estimated standard deviation; interpret the result in context of the population mean.

A sample mean 42 with margin 3 gives (39,45) at the stated confidence level.

The fixed population mean is not random after the sample is taken; “95% confidence” is not a 95% probability that this interval contains it.

For a population proportion p, estimate with p̂=x/n and use p̂±critical value√[p̂(1−p̂)/n] when the normal approximation is justified.

Check that expected successes and failures are sufficiently large, keep the interval within [0,1] in interpretation, and state the confidence level.

If p̂=0.40 and n=100, the estimated standard error is √(0.24/100)≈0.049.

A proportion is not a mean of arbitrary measurements, and the normal interval can be poor for small counts or extreme p̂.

Set H₀ and H₁, choose a significance level, calculate a test statistic under H₀ and reject only when the result lies in the specified critical region or has p≤α.

Match one- or two-tailed wording, retain the parameter context and report the decision rather than claiming certainty.

A two-sided p-value of 0.04 rejects H₀ at 5% but not at 1%.

A p-value is not the probability that H₀ is true, and failure to reject is not proof of equality.

Approximate a binomial count by N(np,np(1−p)) or a Poisson count by N(λ,λ) when the distribution is sufficiently spread. Apply continuity correction to integer boundaries.

Translate “at most”, “at least” and exact counts into half-unit intervals before standardising, then compare with the exact model when the approximation is marginal.

P(X≥12) becomes P(Y>11.5) in the continuous approximation.

Using the uncorrected boundary can produce a visibly different tail probability.

Under H₀, the statistic has a known or approximated distribution. Critical values or p-values are calculated from that distribution and the direction of H₁.

Use the stated variance or standard error, preserve tails and degrees of freedom, and conclude in the original units.

A one-sided upper test uses P(T≥t_obs), whereas a two-sided test counts both tails of comparable extremity.

The same statistic value can lead to different decisions at different α or under different alternative hypotheses.

For testing μ=μ₀ with known σ, use Z=(X̄−μ₀)/(σ/√n). If σ is estimated and the syllabus model requires it, use the corresponding t statistic.

State H₀, calculate the standard error, choose the tail and compare with the correct critical value. The observed mean is evidence, not the null value itself.

With x̄=52, μ₀=50, σ=10, n=100, z=2; the decision depends on α and whether the test is one- or two-sided.

Using σ instead of σ/√n understates the evidence by ignoring the sample size.

For X~N(μ,σ²), standardise observations or sample means with z=(x−μ)/σ and use the standard normal tail. A two-sided alternative uses both tails.

Write the null parameter and variance clearly, draw the rejection region and convert back to the original variable when reporting a critical boundary.

Testing μ=100 with σ=15 and n=25 uses standard error 3, so x̄=106 gives z=2.

The test statistic’s denominator is the standard error for a mean, not the individual standard deviation σ.