Comparing Sample-Mean Distributions Across Sample Sizes

Why Compare Sample-Mean Distributions?

A sample mean is random because the observations in its sample are random. Changing the sample size changes the set of values the mean can take and how probability is allocated among them. For Bernoulli observations, these features can be studied exactly using a binomial count. This gives a direct comparison across sample sizes without approximating the distribution by a continuous curve.

The comparison has three basic parts: identify each distribution’s support, locate its center, and calculate its variance. Exact event probabilities require one additional step: translate the event for the sample mean into the corresponding integer-count event. This last step is important because the sample mean can take only particular discrete values.

Exact Distribution at a Given Sample Size

Definition: Let \(X_1,\ldots,X_n\) be independent and identically distributed Bernoulli random variables with \(P(X_i=1)=p\) and \(P(X_i=0)=1-p\), where \(0<p<1\). Define the sample mean by \(\bar{X}_n=(X_1+\cdots+X_n)/n\). If \(K_n=X_1+\cdots+X_n\), then \(K_n\sim\operatorname{Bin}(n,p)\) and \(\bar{X}_n=K_n/n\). The support is \(\{0,1/n,2/n,\ldots,1\}\), and \(P(\bar{X}_n=k/n)=\binom{n}{k}p^k(1-p)^{n-k}\) for \(k=0,\ldots,n\).

The model assumes a fixed success probability \(p\), exactly two possible values for each observation, and independence across observations. Here “success” simply means the observation equals 1; it need not mean a favorable outcome in a real process. If \(p=0\) or \(p=1\), the sample mean is constant rather than having all \(n+1\) support points. The examples below use hypothetical processes, not observed company or industry data. Interest and mortality assumptions do not apply to these Bernoulli models.

For a fixed \(p\), changing \(n\) changes the spacing between the attainable sample means. With \(n=4\), the spacing is \(1/4\); with \(n=10\), it is \(1/10\). Both supports include 0 and 1, but they need not have the same interior points. The binomial probabilities also change: the same value, if attainable for both sample sizes, need not have the same probability under both distributions.

Center and Variance Across Sample Sizes

Theorem: Under the independent, identically distributed Bernoulli model with success probability \(p\), the sample mean has \(E[\bar{X}_n]=p\) and \(\operatorname{Var}(\bar{X}_n)=p(1-p)/n\). Thus its center does not depend on \(n\), while its variance is inversely proportional to \(n\).

The reason follows from the mean and variance of a Bernoulli observation. Each \(X_i\) has \(E[X_i]=p\) and \(\operatorname{Var}(X_i)=p(1-p)\). Linearity of expectation gives the sample-mean expectation. Independence makes the variance of the sum equal to the sum of the variances, after which division by \(n\) scales the variance by \(1/n^2\):

E[\bar{X}_n] =\frac{1}{n}\sum_{i=1}^{n}E[X_i] =\frac{np}{n} =p, \qquad \operatorname{Var}(\bar{X}_n) =\frac{1}{n^2}\sum_{i=1}^{n}\operatorname{Var}(X_i) =\frac{np(1-p)}{n^2} =\frac{p(1-p)}{n}.

This derivation depends on independence when adding variances. If the observations are dependent, the variance of their sum also includes covariance terms, so the displayed variance formula need not hold. When the model assumptions do hold, increasing the sample size reduces the variance, but it does not change the expected value \(p\).

A smaller variance describes a reduction in overall squared spread around the center; it does not mean that every particular tail probability must decrease when \(n\) changes. The support is discrete, and an event boundary can include a different set of count values at different sample sizes. Exact probabilities should be calculated from the relevant binomial distribution rather than inferred from variance alone.

Worked Comparisons

Worked Example: Comparing an Upper-Tail Event at Two Sample Sizes

A hypothetical inspection process accepts a unit with probability \(p=0.30\). Each inspected unit is modeled as an independent Bernoulli observation, with 1 for accepted and 0 for not accepted. Compare the sample-mean distributions for \(n=2\) and \(n=5\), then calculate \(P(\bar{X}_n\geq 0.60)\) for each. There is no interest or mortality assumption in this probability model.

Method. Use \(K_n=n\bar{X}_n\sim\operatorname{Bin}(n,0.30)\), compare support, expectation, and variance, and translate the event into a count condition. This avoids treating the discrete sample mean as continuous.

For \(n=2\), the support is \(\{0,0.5,1\}\). Its center is \(E[\bar{X}_2]=0.30\), and its variance is \(0.30(0.70)/2=0.105\).

The event \(\bar{X}_2\geq0.60\) is equivalent to \(K_2\geq 2(0.60)=1.2\). Since \(K_2\) is an integer, this means \(K_2\geq2\). Therefore,

P(\bar{X}_2\geq0.60) =P(K_2=2) =\binom{2}{2}(0.30)^2(0.70)^0 =0.0900.

For \(n=5\), the support is \(\{0,0.2,0.4,0.6,0.8,1\}\). Its center remains \(0.30\), and its variance is \(0.30(0.70)/5=0.042\).

Now \(\bar{X}_5\geq0.60\) is equivalent to \(K_5\geq3\). The three included counts have probabilities \(P(K_5=3)=\binom{5}{3}(0.30)^3(0.70)^2=10(0.027)(0.49)=0.1323\), \(P(K_5=4)=\binom{5}{4}(0.30)^4(0.70)=5(0.0081)(0.70)=0.02835\), and \(P(K_5=5)=(0.30)^5=0.00243\). Thus \(P(\bar{X}_5\geq0.60)=0.1323+0.02835+0.00243=0.16308\), or **0.1631** to four decimal places.

The larger sample has smaller variance, but this particular upper-tail probability is larger. The two discrete distributions allocate probability differently around the threshold; variance alone does not determine this event probability. The method also avoids rounding the cutoff \(1.2\) down: \(K_2\) must be at least 2, not at least 1.

Worked Example: Applying a Ceiling to an Exact Threshold

A hypothetical sensor records whether each trial meets a specified condition. The outcomes are independent Bernoulli observations with success probability \(p=0.40\). Compare sample sizes \(n=2\) and \(n=5\), and calculate the probability that the sample mean is at least \(0.80\). No interest rate or mortality assumption is relevant.

Method. Apply the binomial model and convert the inequality to an integer threshold using the ceiling. This avoids omitting a count that meets the sample-mean condition.

For \(n=2\), the support is \(\{0,0.5,1\}\), the expectation is \(0.40\), and the variance is \(0.40(0.60)/2=0.12\). The condition \(\bar{X}_2\geq0.80\) requires \(K_2\geq1.6\), hence \(K_2\geq2\). Therefore,

P(\bar{X}_2\geq0.80) =P(K_2=2) =(0.40)^2 =0.1600.

For \(n=5\), the support is \(\{0,0.2,0.4,0.6,0.8,1\}\), the expectation is still \(0.40\), and the variance is \(0.40(0.60)/5=0.048\). The event requires \(K_5\geq5(0.80)=4\), so the included cases are \(K_5=4\) and \(K_5=5\):

\begin{aligned} P(\bar{X}_5\geq0.80) &=P(K_5=4)+P(K_5=5)\\ &=\binom{5}{4}(0.40)^4(0.60)+(0.40)^5\\ &=5(0.0256)(0.60)+0.01024\\ &=0.0768+0.01024\\ &=0.08704 =0.0870\text{ to four decimal places.} \end{aligned}

The variance falls from \(0.12\) to \(0.048\), the center stays at \(0.40\), and the exact upper-tail probability falls from \(0.1600\) to \(0.0870\). The calculation includes both counts 4 and 5; using only the probability of five successes would incorrectly omit the case in which exactly four of the five trials succeed.

Worked Example: Comparing Probability Within an Interval

A hypothetical system produces a flagged outcome with probability \(p=0.50\) on each independent trial. Compare \(n=3\) and \(n=4\), including the probability that the sample mean is between \(0.25\) and \(0.75\), inclusive. The Bernoulli parameter is \(p=0.50\); interest and mortality assumptions do not apply.

Method. Find which integer counts place the mean inside the interval, then add their binomial probabilities. This avoids assuming that the endpoints are attainable for both sample sizes.

For \(n=3\), the support is \(\{0,1/3,2/3,1\}\), the expectation is \(0.50\), and the variance is \(0.25/3=1/12\). The event \(0.25\leq\bar{X}_3\leq0.75\) becomes \(0.75\leq K_3\leq2.25\), so the possible counts are \(K_3=1,2\). Each has probability \(\binom{3}{1}(0.5)^3=3/8\) or \(\binom{3}{2}(0.5)^3=3/8\). The interval probability is \(3/8+3/8=6/8=0.7500\).

For \(n=4\), the support is \(\{0,0.25,0.50,0.75,1\}\), the expectation is again \(0.50\), and the variance is \(0.25/4=0.0625\). The interval becomes \(1\leq K_4\leq3\), so the included counts are 1, 2, and 3. Their probabilities are \(4/16\), \(6/16\), and \(4/16\), respectively. Thus the interval probability is \((4+6+4)/16=14/16=0.8750\).

The \(n=4\) support includes both interval endpoints, while the \(n=3\) support does not. The interval probabilities are exact and differ even though both distributions have the same center. The method avoids assuming that the two supports contain identical points.

Exact-Probability Checklist and Exam Traps

For a threshold event, the integer conversion is often the most error-prone step. Since \(\bar{X}_n=K_n/n\), for any real number \(c\),

\bar{X}_n\geq c\iff K_n\geq\lceil nc\rceil, \qquad \bar{X}_n\leq c\iff K_n\leq\lfloor nc\rfloor.

For a closed interval \(a\leq\bar{X}_n\leq b\), the count must satisfy \(\lceil na\rceil\leq K_n\leq\lfloor nb\rfloor\). For strict inequalities, check the integer boundary directly: for example, \(\bar{X}_n>c\) means \(K_n>nc\), so if \(nc\) is an integer, the next integer count is required. After translating, sum the binomial probabilities for all and only the included counts.

State the model.
Confirm that observations are independent Bernoulli variables with the same success probability \(p\).

Describe the distribution.
Set \(K_n=\sum X_i\sim\operatorname{Bin}(n,p)\) and \(\bar{X}_n=K_n/n\); list the support if comparing sample sizes.

Compare center and spread.
Use \(E[\bar{X}_n]=p\) and \(\operatorname{Var}(\bar{X}_n)=p(1-p)/n\).

Convert the event and calculate.
Translate the sample-mean condition into integer counts, evaluate each binomial probability, and add the required terms.

A common mistake is to use the variance comparison as if it determined every tail probability. A smaller variance means the variance around the common center is smaller under the model; it does not replace an exact count calculation for a specified event. Another trap is rounding a count threshold in the wrong direction. For example, \(K_n\geq1.6\) means \(K_n\geq2\), not \(K_n\geq1\), because \(K_n\) can only be an integer.

Key takeaway: At each sample size, a Bernoulli sample mean has a binomially determined discrete support, mean \(p\), and variance \(p(1-p)/n\). Compare exact event probabilities by translating the event into integer counts and summing the corresponding binomial masses; do not use a normal approximation or infer a particular tail probability from the variance alone.

Check Your Understanding

For each problem, use the exact binomial distribution rather than a normal approximation. Probabilities are rounded to four decimal places where needed.

A hypothetical quality check has independent Bernoulli outcomes with success probability \(p=0.20\). For \(n=3\) and \(n=6\), find the variance of each sample mean and calculate \(P(\bar{X}_n\geq0.50)\) for each sample size. (Answer: Variances \(0.0533\) and \(0.0267\); probabilities \(0.1040\) and \(0.0989\), respectively.)
A hypothetical trial succeeds independently with probability \(p=0.35\). For a sample of \(n=4\), calculate the probability that the sample mean is exactly \(0.50\), and state whether \(0.50\) is an attainable value. (Answer: It is attainable, and the probability is \(0.3105\).)
A hypothetical event occurs independently on each trial with probability \(p=0.60\). For \(n=5\), calculate \(P(\bar{X}_5>0.60)\) and give the variance of the sample mean. (Answer: \(P(\bar{X}_5>0.60)=0.3370\); variance \(0.048\).)
A hypothetical binary outcome has probability \(p=0.50\) on each independent trial. Compare \(P(1/3\leq\bar{X}_n\leq2/3)\) for \(n=2\) and \(n=6\). (Answer: The probabilities are \(0.5000\) for \(n=2\) and \(0.7813\) for \(n=6\).)
A hypothetical screening process has independent Bernoulli outcomes with \(p=0.30\). For \(n=4\) and \(n=10\), find each sample mean’s variance and calculate the probability that the sample mean is exactly \(0.50\). (Answer: Variances \(0.0525\) and \(0.021\); probabilities \(0.2646\) and \(0.1029\), respectively.)

What You'll Learn