4. Further Probability & Statistics

If a rod has linear density ρ(x), its mass on [a,b] is ∫ₐᵇρ(x)dx. A piecewise density uses the appropriate expression on each subinterval, and masses or centres require separate integrals.

Check continuity or a jump at the boundary, split the integral exactly there, then form x̄=(1/M)∫xρ(x)dx. Units of density must match the coordinate units.

If ρ=2 on 0≤x<1 and ρ=4 on 1≤x≤3, total mass is 2(1)+4(2)=10; the centre uses the corresponding weighted first moments.

A density value is not a mass, and using one formula across a discontinuity silently changes the physical model.

For a discrete random variable X, E(X)=ΣxP(X=x). It is the long-run mean of repeated trials and may be a value that no single outcome can take.

Check that probabilities sum to one before calculating. For a function g(X), use E[g(X)]=Σg(x)P(X=x), not generally g(E(X)).

If X is 0 or 2 with probabilities 0.7 and 0.3, E(X)=0.6; no individual trial has outcome 0.6. But over many trials the average approaches 0.6.

Expected value is not the most likely value and is not a promise about the next observation.

For a continuous variable, the probability density function f(x) satisfies P(a≤X≤b)=∫ₐᵇf(x)dx, while the cumulative distribution function F(x)=P(X≤x)=∫_{−∞}ˣf(t)dt.

A density value is not itself a probability; probabilities are areas. The CDF is non-decreasing, approaches 0 and 1 at the ends, and where differentiable F′(x)=f(x).

If f(x)=2x on 0≤x≤1, then P(X≤0.5)=∫₀^.5 2x dx=0.25 and F(0.5)=0.25.

For a continuous variable P(X=x)=0, even though f(x) may be positive; use an interval or tail probability.

Once F is known, P(a<X≤b)=F(b)−F(a), P(X>b)=1−F(b), and quantiles are found by solving F(x)=p. Endpoints do not matter for continuous distributions.

State the support before evaluating a CDF. If a piecewise formula is given, use the correct branch and check continuity at the join.

If F(x)=x² on 0≤x≤1, then P(0.2<X≤0.8)=0.64−0.04=0.60; the density is not needed for this subtraction.

F(b)−F(a) is not F(b−a), and a CDF value is cumulative probability, not a density height.

Set a null hypothesis H₀ and an alternative H₁ before calculating. The test statistic or p-value measures how surprising the sample would be if H₀ were true.

Choose a significance level α, identify the rejection region or compare p with α, and state the decision in context. A one- or two-tailed alternative changes the region.

At α=0.05, p=0.03 leads to rejection of H₀; it does not prove H₁, only that the observed result is sufficiently inconsistent with the null model.

Failing to reject H₀ is not accepting it, and statistical significance does not measure the size or practical importance of an effect.

When samples estimate a common population mean, a pooled estimate weights each sample mean by its sample size: x̄=(n₁x̄₁+n₂x̄₂)/(n₁+n₂). Variance pooling requires its own assumptions.

Pool only when the samples target the same parameter and independence and comparable modelling assumptions are reasonable. Do not average sample means equally unless sample sizes match.

Means 12 from n=20 and 15 from n=10 give pooled mean (20·12+10·15)/30=13, not 13.5.

Pooling cannot repair biased sampling, and a pooled mean does not imply the two populations have identical distributions.

A hypothesis-test statistic compares the observed estimate with the null value after scaling by its standard error. Under H₀, its sampling distribution determines critical values or p-values.

Use the correct distribution and degrees of freedom, match the tail to H₁, and distinguish a statistic calculated from data from the random variable describing its repeated-sample behaviour.

A two-sided test with statistic z=2.1 has p=2P(Z≥2.1), not just one tail; whether this rejects H₀ depends on α.

A 2.1 standard-error difference is not automatically “significant”; the alternative, tail count and chosen α still matter.

A confidence interval has the form estimate ± critical value × standard error. It gives a range produced by a method that captures the fixed population parameter in a stated proportion of repeated samples.

Choose the correct normal or t-based model and use the sample size and variability assumptions. A 95% interval is about the method’s long-run coverage, not a 95% probability attached to the already fixed mean.

If x̄=20 and the margin is 1.5, report (18.5,21.5) in the measurement units and explain the confidence level.

A wider interval is not automatically worse: it may reflect higher confidence or a smaller sample. The interval does not contain 95% of individual observations.

For two independent means, an interval for μ₁−μ₂ is the sample difference plus or minus a critical value times its standard error. The order of subtraction fixes the sign.

Check independence, variance assumptions and whether a paired design should instead be analysed through within-pair differences. If zero lies in the interval, a two-sided 5% test would not reject equality.

An interval for μA−μB of (1.2,4.8) supports a positive difference; an interval (−0.6,2.1) does not establish a direction at that confidence level.

An interval crossing zero is not proof of no effect, and separate intervals for μA and μB are not equivalent to an interval for their difference.

Fit a probability model by estimating its parameters from data, then compare expected behaviour with observed frequencies or summary features. A fitted model is an approximation, not an explanation by itself.

Keep class intervals consistent, calculate expected counts from model probabilities, and check that expected counts are large enough for the chosen test. Parameters estimated from the same data affect degrees of freedom.

Fit a Poisson model using the sample mean λ̂, then calculate each class probability from λ̂ before forming expected counts.

Matching the mean does not guarantee a good fit; dispersion, tail behaviour and the test statistic still matter.

The chi-squared goodness-of-fit statistic is Σ(O−E)²/E. Under H₀ the proposed distribution is adequate, subject to model and expected-count conditions.

Combine tail classes when expected counts are too small, subtract parameters estimated from the data when determining degrees of freedom, and state the conclusion in context.

A large statistic relative to the critical value gives evidence against the fitted distribution; it does not identify which class caused the mismatch without inspecting contributions.

Rejecting H₀ does not prove every observation is wrong, and a small statistic cannot prove the model true.

For a contingency table, H₀ states that the two categorical variables are independent. Expected count in a cell is row total × column total ÷ grand total, then use the chi-squared statistic.

Use counts rather than percentages, check expected-count conditions, and describe any rejection as evidence of association—not proof of causation.

If a row total is 40, a column total is 30 and the grand total is 100, the independent expected count is 12. Compare the observed cell with 12.

Independence is not mutual exclusivity, and a significant association can be produced by confounding or sampling bias.

A non-parametric test uses limited assumptions, often the signs or ranks of observations rather than a normal model. It is useful when outliers, skew or ordinal data make a parametric test unsuitable.

State the null and alternative, preserve pairing or independence, and use the exact sampling distribution or critical region specified for the statistic.

For paired measurements, analyse the signs of within-pair differences rather than treating the two columns as independent samples.

“Non-parametric” does not mean assumption-free: independence, symmetry or a meaningful ordering may still be required.

For paired or one-sample data, the sign test counts positive and negative differences after removing zeros. Under a null median or no directional effect, the count follows a binomial model with p=0.5.

Choose the direction before counting, discard exact ties as specified, and use the smaller tail probability for a two-sided alternative only after defining the alternative.

With 9 non-zero paired differences, 8 positive results are assessed using X~Bin(9,0.5); the observed count is evidence about direction, not magnitude.

The sign test ignores how large differences are, and a tie is not automatically positive or negative.

To test a population median m₀, compare each observation with m₀ and count signs. Under H₀: median=m₀, the positive count has a Bin(n,0.5) distribution after ties are removed.

For H₁: median>m₀ use an upper tail; for median<m₀ use a lower tail; for a two-sided alternative use both tails. State the conclusion in the original units.

If 12 non-tied measurements contain 10 above m₀, calculate P(X≥10) for a one-sided test, then compare with α.

A rejection says the data are inconsistent with the stated median, not that every observation lies on one side or that the effect is practically large.

For matched observations, calculate a difference within each pair, then apply a sign or signed-rank procedure to those differences. Pairing removes between-subject variation only when the pairing is meaningful.

Keep the order of subtraction consistent, remove zero differences as required, and do not analyse the two samples as independent if the design is paired.

Before-and-after measurements on the same 10 people produce 10 differences; the test asks whether the typical change is zero, not whether the two raw columns have equal means.

Pairing is not merely putting two samples side by side; a random pairing can add noise and invalidate the intended interpretation.

For a non-negative integer-valued X, the probability generating function is G_X(s)=E(s^X)=ΣP(X=r)s^r. The coefficient of s^r is P(X=r).

Check G(1)=1. Derivatives at s=1 give factorial moments: G′(1)=E(X), while higher derivatives combine to produce variance.

For X~Bin(n,p), G_X(s)=(1−p+ps)^n, because each independent trial contributes 1−p+ps and products combine the trials.

A PGF is not an ordinary moment-generating function; its variable is a formal probability marker and the support must be non-negative integers.

For a PGF G, E(X)=G′(1). Also E[X(X−1)]=G″(1), so Var(X)=G″(1)+G′(1)−[G′(1)]².

Differentiate before substituting s=1, simplify carefully, and use the non-negative variance check. The second derivative is not itself E(X²); add the first factorial moment.

For a Bernoulli PGF G=1−p+ps, G′(1)=p and G″(1)=0, giving Var(X)=p−p²=p(1−p).

Confusing G″(1) with E(X²) drops the E(X) term and gives the wrong variance.

If X and Y are independent non-negative integer-valued variables, G_{X+Y}(s)=G_X(s)G_Y(s). Independence makes the generating function factorise because E(s^{X+Y})=E(s^X)E(s^Y).

Multiply the functions, then read coefficients or differentiate the product to obtain the sum’s distribution or moments. Different distributions can be combined when their supports are compatible.

The sum of independent Bin(n₁,p) and Bin(n₂,p) variables has PGF (1−p+ps)^{n₁+n₂}, hence Bin(n₁+n₂,p).

The product rule fails without independence; adding means is always valid under finite expectations, but multiplying PGFs is not.