Finding Probabilities for Sample Means

From a Sample-Mean Model to a Probability

A probability about a sample mean describes how often a statistic would fall in a specified range across repeated samples. For example, a clinic might ask how likely it is that the average wait time among 40 independently selected patients exceeds 30 minutes. The relevant variation is not the variation among individual patients; it is the variation of the sample mean across samples of 40 patients.

When the population is normal, the sample mean has an exact normal distribution. When the population is not normal, the Central Limit Theorem (CLT) can often justify a normal approximation for the sample mean, provided its conditions are met and the sample size is adequate for the population shape. In either case, the calculation follows the same path: specify the model, find its standard error, standardize the cutoff, and find the corresponding normal area.

Sample-mean model: For a random sample of size \(n\) from a population with mean \(\mu\) and finite standard deviation \(\sigma\), the sample mean \(\bar{x}\) has mean \(\mu\) and standard error \(\sigma/\sqrt{n}\). If the population is normal, then \(\bar{x}\sim N(\mu,\sigma^2/n)\) exactly. If the population is not normal, the CLT says that the distribution of \(\bar{x}\) is approximately normal for a sufficiently large sample, subject to its conditions.

The model assumes observations are independent and drawn from the same population, with a finite population standard deviation. For sampling without replacement, the usual model requires the sampling fraction to be negligible; if that is not true, the finite population correction from the earlier material on sampling without replacement may be needed. The CLT approximation also needs a sample size adequate for the population’s shape. There is no single sample-size cutoff that guarantees accuracy for every population.

The standard error describes the spread of sample means, not the spread of individual observations. It decreases as \(n\) increases, because averaging more observations generally makes the mean less variable. When the population standard deviation is known, the standard error used here is \(\sigma/\sqrt{n}\), not \(s/\sqrt{n}\).

Standardizing a Sample-Mean Cutoff

Let \(Z\) denote a standard normal random variable, with mean 0 and standard deviation 1. Write \(\Phi(z)=P(Z\le z)\) for the standard normal cumulative probability at \(z\). For a sample mean, a cutoff \(c\) is standardized by subtracting its model mean and dividing by its standard error:

z=\frac{c-\mu}{\sigma/\sqrt{n}}

This \(z\)-score says how many standard errors the cutoff is above or below \(\mu\). Under an exact normal model, the transformation gives exact normal probabilities. Under a CLT model, the resulting probabilities are approximate. For example, a lower-tail probability is calculated as \(P(\bar{x}\le c)\approx\Phi(z)\) when a CLT approximation is being used. An upper-tail probability is \(P(\bar{x}>c)\approx1-\Phi(z)\). For an interval, subtract the lower cumulative area from the upper one.

Probability interpretation: A probability for \(\bar{x}\) refers to the proportion of repeated samples of the stated size whose means fall in the specified range, according to the model. It does not describe the proportion of individual observations in that range.

Worked Example: An Exact Normal-Model Probability

Suppose a hypothetical population of packages has normally distributed weights with mean \(\mu=72\) grams and standard deviation \(\sigma=12\) grams. A random sample of \(n=36\) packages is taken independently. Find the probability that the sample mean weight exceeds 75 grams.

Check the model. The observations are assumed independent and drawn from the same population. Because the population itself is normal, the sampling distribution of \(\bar{x}\) is exactly normal for this sample size; a large-sample CLT approximation is not needed.

Find the standard error. The sample mean is measured in grams, so its standard error is also in grams:

SE=\frac{\sigma}{\sqrt{n}}=\frac{12}{\sqrt{36}}=\frac{12}{6}=2\text{ grams}

Standardize the cutoff. The cutoff of 75 grams is 3 grams above the population mean, or \(3/2=1.5\) standard errors above it:

z=\frac{75-72}{2}=1.50

Find and interpret the area. The upper-tail area is \(1-\Phi(1.50)\), approximately \(1-0.9332=0.0668\). Thus, under this model, the probability that the mean weight in a random sample of 36 packages exceeds 75 grams is about 0.0668, or 6.68%. This is a probability about sample means, not about a single package.

Using the CLT for a Non-Normal Population

For a non-normal population, first check whether a normal approximation is reasonable rather than assuming that every sample mean is normal. The observations should be independent and from the same population, the population should have a finite standard deviation, and the sample size should be large enough in light of the population’s shape. Earlier material on population shape and sample size explains why a more challenging population shape can require more observations. If these conditions are not credible, a calculated normal area may not be a reliable probability.

Worked Example: A Lower-Tail CLT Probability

Suppose a hypothetical population of delivery times has mean \(\mu=500\) seconds and standard deviation \(\sigma=80\) seconds. Its shape is known to be reasonably well behaved, and a random sample of \(n=64\) deliveries is selected independently. Use the CLT to approximate the probability that the sample mean is below 480 seconds.

Check the approximation. The sample is assumed to consist of independent observations from the same population, and the population standard deviation is finite and given. With 64 observations and a population shape judged suitable for this sample size, the CLT provides a reasonable approximate normal model. This is an approximation, not an exact claim that the population is normal.

Find the standard error. The units remain seconds:

SE=\frac{80}{\sqrt{64}}=\frac{80}{8}=10\text{ seconds}

Standardize 480 seconds. This cutoff is 20 seconds below the population mean, which is 2 standard errors below it:

z=\frac{480-500}{10}=-2.00

Find and interpret the area. The standard normal lower-tail area is \(\Phi(-2.00)\approx0.0228\). Therefore, the probability that the mean time for 64 independently selected deliveries is below 480 seconds is approximately 2.28%, under the CLT model.

Probabilities for a Range of Sample Means

An interval probability uses two cutoffs. Standardize each endpoint using the same \(\mu\) and standard error, then subtract the lower normal area from the upper normal area. This subtraction works because the interval’s probability is the area to the left of the upper endpoint minus the area to the left of the lower endpoint.

P(a<\bar{x}<b)\approx \Phi\left(\frac{b-\mu}{\sigma/\sqrt{n}}\right) - \Phi\left(\frac{a-\mu}{\sigma/\sqrt{n}}\right)

The formula is exact if the population is normal and the assumptions for the sample-mean model hold. For a non-normal population, it gives an approximation when the CLT is applicable. Since the normal model is continuous, including or excluding an endpoint makes no difference to its calculated area. If the sample-mean distribution is not well approximated by a continuous normal curve, however, that model may not represent the endpoint probabilities accurately.

Worked Example: A Probability Between Two Cutoffs

Suppose a hypothetical population has mean \(\mu=100\) units and standard deviation \(\sigma=15\) units. A random sample of \(n=100\) observations is drawn independently from a population whose shape makes a CLT approximation reasonable at this sample size. Approximate the probability that the sample mean lies between 98 and 102 units.

Check the model and find the standard error. The observations are assumed independent and from the same population, and the population standard deviation is finite. Given the stated population shape and sample size, use the CLT approximation. The standard error of the sample mean is:

SE=\frac{15}{\sqrt{100}}=\frac{15}{10}=1.5\text{ units}

Standardize both endpoints. The lower endpoint is 2 units below the mean and the upper endpoint is 2 units above it:

z_{\text{lower}}=\frac{98-100}{1.5}\approx-1.33, \qquad z_{\text{upper}}=\frac{102-100}{1.5}\approx1.33

Subtract the cumulative areas. Using \(\Phi(1.33)\approx0.9082\) and \(\Phi(-1.33)\approx0.0918\):

P(98<\bar{x}<102)\approx \Phi(1.33)-\Phi(-1.33) =0.9082-0.0918 =0.8164

Thus, the approximate probability that the mean of 100 observations falls between 98 and 102 units is 0.816, or 81.6%. This calculation concerns the distribution of sample means across repeated samples of 100, not the fraction of individual observations between 98 and 102 units.

Why the Steps Matter

The arithmetic in these examples is short, but each step protects the interpretation. Using \(\sigma\) rather than a standard deviation measured on individual observations is not enough by itself: the quantity must be divided by \(\sqrt{n}\) to obtain the standard error of the mean. A cutoff must then be compared with \(\mu\), not with zero, and expressed in standard-error units before a standard normal table or calculator can be used.

A useful order of work is:

State the population information.
Record \(\mu\), \(\sigma\), and \(n\), including the units of the measurements.

Check the model.
Establish whether the population is normal for an exact model or whether the CLT conditions and sample size support an approximation.

Calculate the standard error.
Use \(SE=\sigma/\sqrt{n}\), in the same units as the sample mean.

Standardize and find the area.
Convert each cutoff to a \(z\)-score, then use the appropriate lower tail, upper tail, or difference of areas.

State the result in context.
Describe the probability as applying to sample means from repeated samples of the stated size, and identify whether it is exact or approximate.

Common Mistakes

Using \(\sigma\) as the standard deviation of \(\bar{x}\). The population standard deviation describes individual observations. The sample-mean standard error is \(\sigma/\sqrt{n}\).
Standardizing against zero. A sample-mean cutoff is standardized as \((c-\mu)/SE\), because the sampling distribution is centered at \(\mu\).
Using the wrong tail. For an upper-tail probability, subtract the lower cumulative area from 1. For a between-cutoffs probability, subtract the lower endpoint’s cumulative area from the upper endpoint’s.
Calling a CLT result exact. If the population is not normal, the normal-area result is an approximation whose reliability depends on the conditions and sample size.
Interpreting a mean probability as an individual probability. A result such as 6.68% describes repeated sample means of a specified size, not the percentage of individual observations above the cutoff.

Key takeaway: A sample-mean probability comes from the distribution centered at \(\mu\) with standard error \(\sigma/\sqrt{n}\). Check whether its normal model is exact or approximate, standardize the cutoff or endpoints, and interpret the area as a probability about repeated sample means.

Check Your Understanding

For each question, state whether a normal model is exact or approximate where relevant, and keep the interpretation focused on sample means.

A normal population has \(\mu=40\) minutes and \(\sigma=10\) minutes. For samples of \(n=25\), what is the standard error of \(\bar{x}\)?
For that same population and sample size, what \(z\)-score corresponds to a sample-mean cutoff of 44 minutes?
A non-normal population has a finite standard deviation. What additional features of the sample and population should be considered before using a CLT approximation for \(\bar{x}\)?
If the standardized cutoff for a lower-tail event is \(z=-1.00\), should the probability be read as \(\Phi(-1.00)\) or \(1-\Phi(-1.00)\)?
In words, what does an approximate probability of 0.10 that \(\bar{x}>c\) mean for repeated samples of size \(n\)?