Math Core

Averages are more stable than individual measurements. One apple might be unusually heavy, but the average weight of a bag of apples rarely strays far from the typical weight. This lesson makes that intuition exact by describing the center, spread and shape of the sampling distribution of the sample mean $\bar{x}$ .

Center and spread of $\bar{x}$

Suppose a quantitative variable has population mean $\mu$ and population standard deviation $\sigma$ . Take a random sample of size $n$ and compute $\bar{x}$ . Each observation $X_i$ has mean $\mu$ and standard deviation $\sigma$ , and

\bar{x} = \frac{X_1 + X_2 + \cdots + X_n}{n}.

From Unit 4, the mean of a sum is the sum of the means, so the sum has mean $n\mu$ . If the observations are independent, variances add, so the sum has variance $n\sigma^2$ . Dividing by $n$ gives

\mu_{\bar{x}} = \frac{n\mu}{n} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sqrt{n\sigma^2}}{n} = \frac{\sigma}{\sqrt{n}}.

Sampling distribution of a sample mean

For a random sample of size $n$ from a population with mean $\mu$ and standard deviation $\sigma$ :

Center: $\mu_{\bar{x}} = \mu$ . The sample mean is an unbiased estimator of $\mu$ .
Spread: $\sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}}$ , as long as the 10% condition holds ( $n \le 0.10N$ ) when sampling without replacement.
Shape: if the population is normal, the sampling distribution of $\bar{x}$ is exactly normal for any sample size.

The standard deviation $\sigma_{\bar{x}}$ is sometimes called the standard deviation of the sample mean. It tells you how far $\bar{x}$ typically falls from $\mu$ . Notice the square root again: averaging 4 observations cuts the spread in half, and averaging 100 observations cuts it to one tenth.

When the population is normal

This lesson focuses on populations that are normally distributed. In that case, any average of observations is also normally distributed, no matter how small the sample. So you can standardize with

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

and use the standard normal table. (When the population is not normal, you need the central limit theorem, which is the next lesson.)

Here is a table excerpt with the values used in this lesson.

$z$	$-1.20$	$-0.90$	$0.40$	$0.50$	$1.50$	$2.00$
Area to the left	$0.1151$	$0.1841$	$0.6554$	$0.6915$	$0.9332$	$0.9772$

Worked example: One apple versus the average of nine

The weights of apples from an orchard are approximately normal with mean $\mu = 150$ grams and standard deviation $\sigma = 12$ grams. A bag contains a random sample of 9 apples.

(a) Find the probability that one randomly selected apple weighs more than 156 grams.

z = \frac{156 - 150}{12} = 0.50, \qquad P(X > 156) = 1 - 0.6915 = 0.3085.

(b) Find the probability that the mean weight of the 9 apples is more than 156 grams.

The orchard produces far more than 90 apples, so the 10% condition holds. The population is normal, so $\bar{x}$ is normal with

\mu_{\bar{x}} = 150, \qquad \sigma_{\bar{x}} = \frac{12}{\sqrt{9}} = 4.

z = \frac{156 - 150}{4} = 1.50, \qquad P(\bar{x} > 156) = 1 - 0.9332 = 0.0668.

A single apple over 156 grams happens about 31% of the time, but a bag of 9 averaging over 156 grams happens less than 7% of the time. The averages are much less spread out.

Dashed: weights of individual apples (SD 12). Solid: sample means for n = 9 (SD 4). The shaded area above 156 is about 0.0668.Open in grapher →

Working backward to a sample size

Because $\sigma_{\bar{x}} = \sigma / \sqrt{n}$ , you can solve for the sample size that gives a desired precision.

Worked example: Choosing n

A lab scale produces measurements that are normally distributed around the true weight, with standard deviation $\sigma = 0.6$ milligrams. How many repeated measurements should be averaged so that the standard deviation of the average is 0.15 milligrams?

\frac{0.6}{\sqrt{n}} = 0.15 \quad\Longrightarrow\quad \sqrt{n} = \frac{0.6}{0.15} = 4 \quad\Longrightarrow\quad n = 16.

Averaging 16 measurements gives a standard deviation one fourth as large as a single measurement.

Probabilities for an interval

Worked example: Mean between two values

Scores on a reading test are normally distributed with $\mu = 500$ and $\sigma = 100$ . A teacher's class is treated as a random sample of 25 students from a very large population of test takers. Find the probability that the class mean is between 480 and 540.

The 10% condition holds, and the population is normal, so $\bar{x}$ is normal with mean 500 and standard deviation $\dfrac{100}{\sqrt{25}} = 20$ .

z_1 = \frac{480 - 500}{20} = -1.00, \qquad z_2 = \frac{540 - 500}{20} = 2.00.

P(480 < \bar{x} < 540) = 0.9772 - 0.1587 = 0.8185.

Technology gives $0.8186$ . The tiny difference comes from rounding in the table.

Common mistake

The most common error is using $\sigma$ instead of $\sigma/\sqrt{n}$ when the question is about a sample mean. Read carefully: "one randomly selected apple" uses $\sigma$ ; "the mean of 9 apples" uses $\sigma/\sqrt{n}$ .

Tip

On free-response questions, state the full sampling distribution before calculating: " $\bar{x}$ is approximately normal with mean 150 and standard deviation $12/\sqrt{9} = 4$ because the population is normal." Then show the $z$ -score or the calculator command with labeled inputs, such as normalcdf(lower: 156, upper: 1000, mean: 150, SD: 4).

Practice

Use the table excerpt above where needed. Give probabilities to four decimal places.

Practice 1

The amount of time customers spend in a large grocery store has standard deviation $\sigma = 18$ minutes. For random samples of 36 customers, what is the standard deviation of the sampling distribution of $\bar{x}$ , in minutes?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Resting pulse rates of adults are approximately normal with mean 72 beats per minute and standard deviation 10. A random sample of 5 adults is selected. Which describes the sampling distribution of $\bar{x}$ ?

Practice 3

IQ scores are normally distributed with $\mu = 100$ and $\sigma = 15$ . A random sample of 25 people is selected from a large population. Find the probability that the sample mean IQ is less than 97.3.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

IQ scores are normally distributed with $\mu = 100$ and $\sigma = 15$ . Which event is more likely: (I) one randomly selected person has an IQ above 106, or (II) the mean IQ of a random sample of 25 people is above 106?

Practice 5

A machine fills cereal boxes with weights that have standard deviation $\sigma = 10$ grams. How many boxes must be in a random sample so that the standard deviation of the sample mean weight is 2 grams?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

A teacher wants to use $\sigma_{\bar{x}} = \sigma/\sqrt{n}$ for the mean height of a random sample of students selected without replacement from her school of 800 students. What is the largest sample size for which the 10% condition is satisfied?

Practice 7

Heights of adult women in a large population are approximately normal with mean 64 inches and standard deviation 2.5 inches. A random sample of 4 women is selected. Find the probability that their mean height is between 62.5 and 66.5 inches.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Center and spread of xˉ\bar{x}xˉ

When the population is normal

Working backward to a sample size

Probabilities for an interval

Practice

Center and spread of $\bar{x}$