Worked Problems on Sample Means and Aggregate Sums

Choosing the Quantity to Approximate

Applied problems about many observations often ask about either their average or their <strongaggregate total. The first step is to identify which quantity the event describes. A threshold for an average is not standardized in the same way as a threshold for a total: their centers and standard deviations differ, even though they are built from the same observations.

The central limit theorem (CLT) provides a normal approximation for sums and averages of many independent observations under suitable conditions. It does not make the original observations normal. Instead, it describes the distribution of their sum or average approximately. The approximation is most dependable when no single observation dominates the total variability and the number of observations is sufficiently large; substantial skewness or heavy tails can require extra caution.

Definition: For observations $X_1,\ldots,X_n$, the sample average is $\bar{X}=(X_1+\cdots+X_n)/n$, and their aggregate sum is $S_n=X_1+\cdots+X_n$. If the observations are independent and identically distributed with finite mean $\mu$ and variance $\sigma^2$, then $E[\bar{X}]=\mu$, $\operatorname{Var}(\bar{X})=\sigma^2/n$, $E[S_n]=n\mu$, and $\operatorname{Var}(S_n)=n\sigma^2$.

Thus, the standard deviation of the average, also called its standard error, is $\sigma/\sqrt{n}$; the standard deviation of the sum is $\sigma\sqrt{n}$. If the observations are identically distributed but dependent, these variance formulas need not hold. Independence is part of the model, not a detail to discard after identifying the mean.

Standardizing an Average or a Sum

When the iid conditions apply and the CLT approximation is appropriate, standardize an average using its mean $\mu$ and standard error $\sigma/\sqrt{n}$. Standardize a sum using its mean $n\mu$ and standard deviation $\sigma\sqrt{n}$. The resulting standardized quantity is approximated by a standard normal variable, denoted $Z\sim N(0,1)$.

\bar{X}\ \approx\ N\!\left(\mu,\frac{\sigma^2}{n}\right), \qquad S_n\ \approx\ N(n\mu,n\sigma^2).

For independent observations that are not identically distributed, use the actual mean and variance of each observation. If $X_1,\ldots,X_n$ are independent with finite means $\mu_j$ and variances $\sigma_j^2$, then the sum has mean $\sum_{j=1}^{n}\mu_j$ and variance $\sum_{j=1}^{n}\sigma_j^2$. A normal approximation may be suitable when the terms are sufficiently numerous and no one term dominates the total variability, subject to the usual CLT regularity conditions. The corresponding average is the sum divided by $n$.

E[S_n]=\sum_{j=1}^{n}\mu_j,\qquad \operatorname{Var}(S_n)=\sum_{j=1}^{n}\sigma_j^2,\qquad E[\bar{X}]=\frac{1}{n}\sum_{j=1}^{n}\mu_j,\qquad \operatorname{Var}(\bar{X})=\frac{1}{n^2}\sum_{j=1}^{n}\sigma_j^2.

If an event is stated using an average but the data are easier to sum, multiply each average bound by $n$ to obtain the equivalent sum bound. Conversely, divide a sum bound by $n$ to express it as an average bound. Either route gives the same standardized value when the calculations are consistent.

Justification: For independent observations, the variance of a sum is the sum of the variances: the covariance terms are zero. In the iid case, $E[S_n]=n\mu$ and $\operatorname{Var}(S_n)=n\sigma^2$. Since $\bar{X}=S_n/n$, scaling gives $E[\bar{X}]=n\mu/n=\mu$ and $\operatorname{Var}(\bar{X})=n\sigma^2/n^2=\sigma^2/n$. The CLT then approximates the standardized sum, and therefore the standardized average, by $N(0,1)$ under its conditions.

Worked Applications

Worked Example: Probability for an Average Processing Time

In a hypothetical service center, the processing times for individual applications are independent and identically distributed with mean 12 minutes and standard deviation 4 minutes. For a random group of 64 applications, estimate the probability that the average processing time is between 11.5 and 13 minutes.

Method and assumptions. Use the iid sample-average CLT approximation. The model gives $\mu=12$, $\sigma=4$, and $n=64$; individual processing times need not be normal, but the approximation is being used for their average.

Calculate the center and standard error. The average has approximate mean 12 minutes. Its standard error is

\frac{\sigma}{\sqrt{n}}=\frac{4}{\sqrt{64}}=\frac{4}{8}=0.5\text{ minutes}.

Standardize each endpoint. At 11.5 minutes, $z=(11.5-12)/0.5=-1$. At 13 minutes, $z=(13-12)/0.5=2$. Therefore the required probability is approximately

P(11.5<\bar{X}<13) \approx P(-1<Z<2) =\Phi(2)-\Phi(-1) =0.9772-0.1587 =0.8185.

Using more precise normal probabilities, the result rounds to 0.8186 to four decimal places. The approximation says that about 81.86% of groups of this size would have an average in the stated interval under the model.

Check and common mistake avoided. The standard error, 0.5 minutes, is smaller than the individual standard deviation of 4 minutes, as expected for an average of 64 independent observations. Using 4 as the standard error would instead give bounds $-0.125$ and $0.25$, and a probability of only about 0.1484. That calculation incorrectly treats the spread of one observation as the spread of the average.

Worked Example: Aggregate Total with Two Claim Groups

A hypothetical insurer models two groups of independent claims. There are 40 claims in Group A, each with mean payment $500 and standard deviation $400, and 30 claims in Group B, each with mean payment $800 and standard deviation $600. Assume payments are independent within and between groups, with finite variances. Estimate the probability that the aggregate payment $S$ is between $39,852.71 and $48,147.29.

Method and assumptions. Add the individual means and variances, then use a normal approximation for the independent aggregate. The group-specific counts are fixed; each claim amount has the stated group distribution. These are hypothetical model parameters, not observed industry data.

Calculate the aggregate mean. Group A contributes $40(500)=20{,}000$ dollars in expected payments. Group B contributes $30(800)=24{,}000$ dollars. Thus,

E[S]=20{,}000+24{,}000=44{,}000\text{ dollars}.

Calculate the aggregate variance and standard deviation. Group A's variance contribution is $40(400^2)=40(160{,}000)=6{,}400{,}000$ dollars squared. Group B's is $30(600^2)=30(360{,}000)=10{,}800{,}000$ dollars squared. Independence gives

\operatorname{Var}(S)=6{,}400{,}000+10{,}800{,}000 =17{,}200{,}000, \qquad \operatorname{SD}(S)=\sqrt{17{,}200{,}000}\approx 4{,}147.29\text{ dollars}.

Standardize the bounds. The lower bound is $4,147.29 below the mean, and the upper bound is $4,147.29 above it. Dividing each difference by the unrounded standard deviation gives standardized bounds approximately $-1.0000$ and $1.0000$. Hence

P(39{,}852.71<S<48{,}147.29) \approx P(-1<Z<1) =0.8413-0.1587 =0.6826.

Using more precise values of $\Phi(-1)$ and $\Phi(1)$, the estimate is 0.6827 to four decimal places. It represents the approximate probability that the total payment falls within one aggregate standard deviation of its mean.

Check and common mistake avoided. The total standard deviation is about $4,147.29. Standard deviations are not added across claims: adding 40 standard deviations of $400 and 30 standard deviations of $600 would not give the spread of the total. For independent amounts, add variances first and take one square root at the end.

Worked Example: Translating a Total Trigger into a Standardized Event

A hypothetical portfolio has 100 independent annual cost amounts with a common exponential distribution of mean $250. For an exponential variable with mean $\theta$, the variance is $\theta^2$; thus each amount has variance $250^2=62{,}500$ dollars squared. Estimate the probability that the portfolio's total cost exceeds $28,750.

Method and assumptions. The total is a sum of 100 iid, finite-variance amounts. The exponential distribution is right-skewed, so a normal approximation is not exact; with 100 independent terms, use the CLT as an approximation and retain that caveat when interpreting the result.

Find the sum's mean and standard deviation. The expected total is $100(250)=25{,}000$ dollars. The total variance is $100(62{,}500)=6{,}250{,}000$, so the standard deviation is $\sqrt{6{,}250{,}000}=2{,}500$ dollars.

Standardize the trigger. The threshold is $28{,}750-25{,}000=3{,}750$ dollars above the mean. Dividing by the aggregate standard deviation gives $z=3{,}750/2{,}500=1.5$. Therefore

P(S>28{,}750)\approx P(Z>1.5)=1-\Phi(1.5)=1-0.9332=0.0668.

The estimated probability is 0.0668 to four decimal places, or about 6.68%. Equivalently, the average cost must exceed $28{,}750/100=\$287.50$; its standard error is $2{,}500/100=\$25$, giving $(287.50-250)/25=1.5$, the same standardized value.

Check and common mistake avoided. Converting the total threshold into an average threshold leaves the standardized result unchanged, provided both the mean and standard deviation are divided by 100. Dividing only the threshold, or using $250 as the standard deviation of the total, produces an inconsistent calculation.

Interpreting the Approximation and Avoiding Traps

A computed probability is an approximation to a model-based probability, not a guarantee about a particular portfolio or sample. Before reporting it, check that the model describes independent observations, that the means and variances are finite, and that the standard deviation belongs to the quantity in the event. For nonidentical observations, also check that the total variance is not dominated by a single term. When these conditions are doubtful, a normal approximation may be poor, especially in a far tail.

Continuity corrections are relevant when approximating probabilities for integer-valued sums, because a discrete event includes whole integer outcomes while a normal variable is continuous. They are not used for the continuous processing times or claim amounts in the examples above. For an integer-valued aggregate, choose a correction that matches the event's strict or inclusive boundary before standardizing; do not apply a correction automatically to every sum.

Identify the target.
Decide whether the event concerns an average, a sum, or a rescaled version of one of these.

Assemble the model quantities.
Find the relevant means and variances. For independent terms, add variances, not standard deviations.

Check whether a normal approximation is reasonable.
Consider sample size, skewness, finite variance, independence, and whether any one term dominates.

Standardize the event.
Subtract the target quantity's mean and divide by its standard deviation. Use $\sigma/\sqrt{n}$ for an iid average and $\sigma\sqrt{n}$ for an iid sum.

Interpret and check.
Translate the standard normal probability back into the original average or aggregate context, and verify that the spread and direction make sense.

A frequent exam trap is to start standardizing before deciding whether the question concerns $\bar{X}$ or $S_n$. A useful check is that, for $n>1$, the average of iid observations has less variability than one observation, while the sum has more variability than one observation. This topic is aligned with the SOA VEE Mathematical Statistics syllabus.

Check Your Understanding

For each problem, state whether the target is an average or a sum before calculating. Probabilities are approximate normal-approximation results.

A hypothetical set of 49 independent, identically distributed inspection times has mean 18 minutes and standard deviation 7 minutes. Estimate the probability that the average time is at most 20 minutes. (Answer: 0.9772)
A hypothetical total consists of 20 independent payments with mean $120 and variance $900 each, and 30 independent payments with mean $80 and variance $400 each. Estimate the probability that the total exceeds $5,100. (Answer: 0.0416)
For 64 independent, identically distributed measurements with mean 5.2 units and standard deviation 1.6 units, estimate the probability that the average is between 5.0 and 5.6 units. (Answer: 0.8186)
A hypothetical aggregate is the sum of 100 independent, identically distributed contributions, each with mean $30 and variance $25. Estimate the probability that the total exceeds $3,125. (Answer: 0.0062)
For 36 independent, identically distributed costs with mean $100 and standard deviation $24, estimate the probability that the total is between $3,456 and $3,816. (Answer: 0.7745)

What You'll Learn