Using the CLT for Sample Totals

Why Use the CLT for a Total?

Many practical questions concern an aggregate: the combined weight of several packages, the total time spent on a set of tasks, or the total amount of material used in a production run. If each observation comes from the same population, the central limit theorem (CLT) can help approximate probabilities for that aggregate. The key is to work out the total’s centre and spread correctly before standardising.

A total and a sample mean contain the same information when they are based on the same $n$ observations: the total is $n$ times the mean. This relationship lets us use results about sample means to answer questions about totals. The standard deviation, however, must also be converted; it is not multiplied by $n$.

Definitions and the Normal Approximation

Definition: Let $X_1,\ldots,X_n$ be independent observations from a population with mean $\mu$ and finite variance $\sigma^2$. The sample total is $S_n=X_1+\cdots+X_n$, and the sample mean is $\bar{x}=S_n/n$. The total is measured in the same units as an individual observation; the mean is also in those units.

The formulae for the mean and standard deviation of a total follow from independence. The expected value of a sum is the sum of the expected values, so $E(S_n)=n\mu$. For independent observations, the variance of a sum is the sum of the variances, so $\operatorname{Var}(S_n)=n\sigma^2$. Therefore, the standard deviation of the total is $\sigma\sqrt{n}$.

Theorem: If $X_1,\ldots,X_n$ are independent and identically distributed with mean $\mu$ and finite variance $\sigma^2$, then $$ E(S_n)=n\mu,\qquad SD(S_n)=\sigma\sqrt{n}. $$ For a sufficiently large $n$, the CLT approximates the distribution of $S_n$ by a normal distribution with this mean and standard deviation. Equivalently, the standardised total is approximately standard normal: $$ \frac{S_n-n\mu}{\sigma\sqrt{n}}\ \approx\ N(0,1). $$

The approximation is exact for every $n$ if the population itself is normal. Otherwise, its accuracy depends on the population distribution and sample size, as discussed in the earlier material on skewness and CLT accuracy. Independence and finite variance are essential conditions for the standard CLT result used here; the earlier material on independence and finite-variance conditions gives further detail. In applications, the observations should represent the same population distribution and should not be dependent in a way that invalidates the independence assumption.

Since $S_n=n\bar{x}$, a total cutoff $c$ corresponds to a mean cutoff $c/n$. Both approaches give the same standardised value: $$ \frac{c-n\mu}{\sigma\sqrt{n}} = \frac{c/n-\mu}{\sigma/\sqrt{n}}. $$ The standard deviation of the sample mean is $\sigma/\sqrt{n}$; multiplying it by $n$ gives the total’s standard deviation, $\sigma\sqrt{n}$.

A Reliable Method

For a probability about a total, first identify the individual observations and the event in context. Then derive the total’s mean and standard deviation from the population parameters. Standardise the relevant cutoff or cutoffs using those total-scale quantities, and use the standard normal distribution to approximate the probability. Keep the scale clear throughout: a cutoff for a total is not a cutoff for one observation.

State the model.
Identify $X_i$, $n$, $\mu$, and $\sigma$, and check that the observations can reasonably be treated as independent and identically distributed with finite variance.

Find the total’s centre and spread.
Use $E(S_n)=n\mu$ and $SD(S_n)=\sigma\sqrt{n}$. State both in the units of the total.

Standardise the event.
For a total cutoff $c$, calculate $z=(c-n\mu)/(\sigma\sqrt{n})$. For an interval, calculate a standardised value for each endpoint.

Find and interpret the probability.
Use the standard normal distribution to approximate the probability, then state what that probability means for the aggregate in the original context.

Worked Example: A Total Below a Cutoff

Suppose the times required for 64 independent tasks are modelled as observations from one population with mean 12 minutes and standard deviation 4 minutes. Assume that the population has finite variance and is not extremely skewed. Approximate the probability that the combined time for the 64 tasks is less than 800 minutes.

Check the model and conditions. The individual observations are task times, measured in minutes. We assume they are independent and share the stated population distribution. The population has finite variance, and with $n=64$, the CLT provides a reasonable normal approximation under the stated shape assumption.

Let $S_{64}$ be the combined time. Its mean and standard deviation are $$ E(S_{64})=64(12)=768\text{ minutes}, \qquad SD(S_{64})=4\sqrt{64}=32\text{ minutes}. $$

Standardise the cutoff of 800 minutes: $$ z=\frac{800-768}{32}=1. $$ Thus, $$ P(S_{64}<800)\approx P(Z<1)=0.8413, $$ where $Z\sim N(0,1)$. The approximate probability that the 64 tasks take less than 800 minutes in total is 0.8413, or about 84.13%.

Worked Example: A Total Between Two Cutoffs

A hypothetical process produces individual units with weights that are independent observations from a population with mean 2.40 grams and standard deviation 0.50 grams. Assume the population has finite variance and is not extremely skewed. For a batch of 100 units, approximate the probability that the total weight is between 235 and 245 grams.

Check the model and conditions. Each observation is one unit’s weight. We assume the 100 weights are independent and identically distributed with the stated finite-variance population. The sample size is large, and the stated shape assumption makes using the CLT approximation reasonable.

Let $S_{100}$ be the batch’s total weight. Then $$ E(S_{100})=100(2.40)=240\text{ grams}, \qquad SD(S_{100})=0.50\sqrt{100}=5.00\text{ grams}. $$ The standardised endpoints are $$ z_{\text{lower}}=\frac{235-240}{5.00}=-1, \qquad z_{\text{upper}}=\frac{245-240}{5.00}=1. $$ Therefore, $$ P(235\leq S_{100}\leq245) \approx P(-1\leq Z\leq1) =0.8413447-0.1586553 =0.6826894 \approx 0.6827. $$ The approximate probability that the batch weighs between 235 and 245 grams is 0.6827, or about 68.27%.

Worked Example: An Aggregate Above a Cutoff

Suppose the service times for 36 independent cases are observations from a population with mean 72 minutes and standard deviation 15 minutes. For this hypothetical process, assume the population distribution is approximately symmetric, has finite variance, and produces independent observations. Approximate the probability that the total service time exceeds 2,700 minutes.

Check the model and conditions. Each observation is one case’s service time. The problem specifies a common population distribution and independence, and the finite-variance and approximate-symmetry assumptions support a CLT approximation for $n=36$.

For the total $S_{36}$, $$ E(S_{36})=36(72)=2592\text{ minutes}, \qquad SD(S_{36})=15\sqrt{36}=90\text{ minutes}. $$ The cutoff is $2,700$ minutes, so $$ z=\frac{2700-2592}{90}=\frac{108}{90}=1.20. $$ Hence, $$ P(S_{36}>2700)\approx P(Z>1.20) =1-0.8849 =0.1151. $$ The approximate probability that the total service time exceeds 2,700 minutes is 0.1151, or about 11.51%.

Why the Square Root Matters

The standard deviation of a total grows like $\sqrt{n}$, not like $n$. If $n$ independent observations each have standard deviation $\sigma$, their variances add to $n\sigma^2$, and taking the square root gives $\sigma\sqrt{n}$. By contrast, the mean of the total grows as $n\mu$. Confusing these two scaling rules can produce a substantially incorrect standardised cutoff and therefore an incorrect probability.

It is sometimes simpler to divide the event by $n$ and work with the sample mean. For example, $S_n>c$ is equivalent to $\bar{x}>c/n$. In that approach, the mean has standard deviation $\sigma/\sqrt{n}$. Either approach is valid, but use a total’s mean and standard deviation with total cutoffs, or a sample mean’s mean and standard deviation with mean cutoffs. Do not mix the scales.

Common mistake: Treating the total’s standard deviation as $n\sigma$, because the total is the sum of $n$ observations. This incorrectly adds standard deviations. Independence makes variances add, so $\operatorname{Var}(S_n)=n\sigma^2$ and $SD(S_n)=\sigma\sqrt{n}$. A second scale error is comparing a total cutoff directly with $\mu$; compare it with $n\mu$, or divide the cutoff by $n$ before comparing it with $\mu$.

A CLT probability is an approximation unless the population is normal. The result does not say that the individual observations become normally distributed; it says that the distribution of the standardised sum is approximated by a normal distribution under the stated conditions. In reporting a result, retain the aggregate’s context and units, and make clear that it is approximate.

Check Your Understanding

Use the total’s scale throughout, or convert every part of the event consistently to the sample-mean scale.

Independent observations have population mean 8 and population standard deviation 3. For $n=25$, find the mean and standard deviation of the total.
For the setting in Question 1, what standardised value corresponds to a total cutoff of 220?
In your own words, explain why the standard deviation of a sum of independent observations is $\sigma\sqrt{n}$, rather than $n\sigma$.
A total event is $S_n\leq c$. Write the equivalent event in terms of $\bar{x}$, and state the standard deviation used when standardising that mean.
Name two conditions that support using the standard CLT approximation for a sample total.

Using the CLT for Sample Totals

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