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Put $v_k=p_k(1-p_k)$ and assume $0<p_k<1$. For fixed $n$ and a fixed nonzero difference $\Delta =p_0-p_1$, the approximate statistical power of the Wald test increases as its noncentrality $|\Delta |/\sqrt{V}$ increases, whereDifferentiating with respect to the continuous allocation givesHence the unique minimum and its associated minimum variance areThis is Neyman allocation. Integer allocations use a feasible adjacent integer to the continuous optimum, comparing the resulting variances; $n_0,n_1\ge 1$ must still hold. It is a large-sample normal approximation argument, not an exact finite-sample power theorem for arbitrary response-adaptive randomization. Under the null $p_0=p_1$, the approximate power is the significance level for any allocation, although the same allocation still minimizes the stated variance.

The pivotal quantity $U=M/\theta$ lies in $[0,1]$ and satisfies $\mathbb{P}(U\le t)=t^n$. ThusRearranging these inequalities yields the exact confidence intervalFor every $\theta >0$, its coverage probability is exactly $1-\alpha$; the result uses the finite-sample probability distribution, with no normal approximation.

The independent uniform choices from the student list describe sampling with replacement. Under the stated independent normal distribution model for the heights, the population variance is unknown and the given $8.5$ cm is the sample standard deviation $s$. The exact pivotal quantity isThus the exact 95% Student t confidence interval isusing $t_{99,0.975}\approx 1.9842$. The supplied approximation $\Phi (2)=0.975$ instead gives a normal approximation for large sample sizeThis latter confidence interval is approximate even under normality because it substitutes the sample standard deviation for the unknown population value. There is no finite population correction under the specified independent selections. Interpreting the finite student list literally as a discrete population makes normality an idealized model; it does not justify changing the stated sampling design.

The number $R$ of successes has a binomial distribution. Each arrangement of $r$ successes and $n-r$ failures has probability $p^r{q}^{n-r}$ by independence, and the binomial coefficient counts the arrangements. ThereforeThe final equality is the binomial theorem. Writing $R$ as a sum of independent Bernoulli random variables gives its expectation and variance: $\mathbb{E}R=np$, $\mathrm{Var}(R)=npq$. Thus the matching normal distribution hasThe normal approximation is appropriate when both expected counts $np$ and $nq$ are sufficiently large; large $n$ alone is insufficient if $p$ is extremely close to zero or one.

For the guessing student, $n=60$, $p=1/4$, $q=3/4$. Therefore the expected number of correct answers is $\boxed{15}$, and ${\sigma }^2=45/4$, $\sigma =3\sqrt{5}/2$. The inclusive integer range $13,14,15,16,17$ becomes $12.5\le X\le 17.5$ after the continuity correction. Standardizing the normal distribution givesHere $(5/2)/(3\sqrt{5}/2)=\sqrt{5}/3$, and the last step uses the supplied bound. The approximate probability is $0.54394$. The normal approximation provides the requested conclusion; as a separate check, summing the five exact binomial probabilities gives approximately $0.54370$, also exceeding one half.

For $0<{\sigma }^2<\infty$, the central limit theorem givesIf $g$ is differentiable at $\mu$, the delta method applies to the first-order expansionwhere the remainder notation means that its product with $\sqrt{n}$ tends to zero in probability. HenceThe useful nondegenerate normal approximation requires $g^{\prime }(\mu )\ne 0$. When $g^{\prime }(\mu )=0$, the first-order limit is zero and a higher-order expansion may be needed.

For the independent statistical sample, write the likelihood function and log-likelihood function asThe expected Fisher information in the whole statistical sample isUnder the regularity conditions that permit differentiation under the integral sign, the score function has expected value zero andFor an interior statistical parameter, assuming identifiability and $0<I_1(\theta )<\infty$, the regular asymptotic normality of a maximum likelihood estimator statesThus the large-sample normal approximation is ${\hat{\theta }}_n\dot{\sim }N(\theta ,I_n(\theta {)}^{-1})$. The inverse information is the leading sampling variance approximation; the displayed convergence specifies its precise asymptotic meaning.

Using the raw moments from part (i), the variance of one observation isThe central limit theorem for the independent statistical sample givesFor $g(x)=x/(1-x)$, its derivative at the mean is $g^{\prime }(\mu )=(1-\mu {)}^{-2}=(\theta +1{)}^2$. The delta method therefore givesEquivalently, the large-sample normal approximation isHere $v$ denotes the leading asymptotic variance, rather than an assertion about an exact finite-sample variance.

Condition on the total of 400 complaints. Assume each complaint independently has the same probability $p$ of being received on Monday; in particular, repeated complaints from one incident must not create material dependence. The Monday count $S$ then has a binomial distribution with parameters $(400,p)$. Test $H_0:p=0.2$ against $H_1:p>0.2$, rejecting for large $S$ as suggested by the monotone likelihood ratio.

Under the null hypothesis, ${\mathbb{E}}_{0}S=80$ and ${\mathrm{Var}}_0(S)=400(0.2)(0.8)=64$. Both expected successes and expected failures are large, so the central limit theorem gives a suitable normal approximation:Using the paper's supplied rounded table, the upper-tail p-value is approximatelyA more accurate evaluation of the standard normal distribution cumulative distribution function gives about 0.0062. For example, choose significance level 0.05 before examining the data; the supplied upper 5% standard normal quantile is approximately 1.6, and $2.5>1.6$. Reject $H_0$ at 5%, and also at 1%, using this approximation. The observed binomial proportion is $100/400=0.25$, providing evidence that the Monday probability exceeds 0.2 under the sampling assumptions.

If applying a continuity correction, approximate ${\mathbb{P}}_0(S\ge 100)$ by $1-\Phi ((99.5-80)/8)=1-\Phi (2.4375)\approx 0.0074$; this gives the same 5% and 1% conclusions. The uncorrected value 0.005 is approximate, so a decision exactly at the 0.5% boundary is not warranted by the supplied rounded table.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$ and expected value $\mu =\mathbb{E}[X_1]$. The weak law of large numbers states that their sample mean has convergence in probability to $\mu$: for every fixed $\epsilon >0$,Thus a large sample is unlikely to give an average with a specified fixed-size error. The strong law of large numbers makes the stronger statement of almost sure convergence along an entire sequence. Under the displayed assumptions both laws apply; their names distinguish the conclusions, not two competing predictions.

When the variance ${\sigma }^2$ is finite, variance additivity for independent random variables gives $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. The Chebyshev inequality immediately supplies a useful quantitative form and a brief proof:For independent trials with a Bernoulli distribution of parameter $p=0.2$, the expected value is $0.2$ and the variance is $0.16$. At $n=1000$, the chance of an error of at least $0.1$ is therefore at most $0.016$. The exact binomial distribution gives a much smaller probability, as the figure under the strong law of large numbers shows. The bound is conservative, but it is valid for every sample size under these assumptions and does not require a normal approximation.

The weak law of large numbers addresses a fixed tolerance $\epsilon$. The central limit theorem instead resolves errors on the shrinking scale $\sigma /\sqrt{n}$ and gives their approximate normal distribution. In particular, the weak law of large numbers alone supplies neither a bell-shaped error law nor a universal rate of approach. For example, a Pareto distribution with minimum one and shape $3/2$ has expected value three but infinite variance: both the weak law of large numbers and strong law of large numbers apply, while the finite-variance statement of the central limit theorem does not.