6.1 The Central Limit Theorem for Sample Means (Averages)

Aligned outcomes:

Learning Objectives

By the end of this section, you will be able to:

In this section, you will learn to:

state what the central limit theorem says about the distribution of sample means, and name the two conditions it needs;
write the distribution of $\overline{X}$ for a given population mean, population standard deviation, and sample size;
calculate the standard error of the mean and explain what it measures;
find the probability that a sample mean falls in a given range;
find a percentile of the sampling distribution and state what it means in the original units.

Everything so far in this course has asked about one value at a time. What is the probability a single adult is taller than 180 cm? What is the probability one exam score lands above 90? This section changes the question. Instead of asking about one value drawn from a population, we ask about the average of a whole sample of values drawn from it.

That turns out to be a completely different question with a much friendlier answer. The average of a sample behaves better than any single value does — it varies less, and, remarkably, it settles into a normal distribution no matter what shape the original population had. That last fact is the central limit theorem, and it is the reason the normal curve shows up everywhere in the rest of this course.

6.1.1 The Central Limit Theorem for Sample Means

Definition 6.1.1: Sampling Distribution of the Mean

Let $X$ be a random variable whose distribution may be known or unknown — it can be any distribution at all: skewed, uniform, exponential, bumpy. Using a subscript that matches the variable, write $\mu_X$ for the mean of $X$ and $\sigma_X$ for the standard deviation of $X$. If you draw random samples of size $n$ from that population, the distribution of the random variable $\overline{X}$, which consists of the sample means, is called the sampling distribution of the mean. The lowercase $\overline{x}$ is the value of $\overline{X}$ from one particular sample.

Definition 6.1.1 - Each sample gives one average, and the averages pile up into a distribution of their own.

Picture how that distribution gets built. Draw a random sample of size $n$ and compute its mean. Then draw another sample of size $n$ and compute its mean. Keep going. Each sample gives a different average, so those averages have a distribution of their own — a distribution built out of sample means rather than out of individual values.

The word sampling is doing real work in that name, so it is worth pausing on. There are two different distributions in play now and they are easy to confuse. One is the distribution of the population itself — every individual value, however it happens to be shaped. The other is the distribution of the averages you would get by sampling that population over and over. They have the same center, but they are not the same curve and they do not have the same spread. Almost every mistake students make in this section comes from computing with one when the problem asked about the other, so whenever you read a problem, decide first which of the two it is asking about. A question about one person, one part, one gas station is about $X$. A question about the average of 25 people, 49 parts, 16 gas stations is about $\overline{X}$.

Definition 6.1.2: The Central Limit Theorem for Sample Means

If you draw random samples of size $n$ from a population with mean $\mu_X$ and standard deviation $\sigma_X$, then as $n$ increases the random variable $\overline{X}$ tends to be normally distributed:

$$ \overline{X} \sim N\!\left(\mu_X, \frac{\sigma_X}{\sqrt{n}}\right) $$

Extremes cancel out

One person's commute can be wildly long. The average commute of 50 people almost never is, because for every unusually slow trip there is usually a fast one to offset it. That cancelling is why averages vary less than individuals do.

Definition 6.1.2 - Whatever the population looks like, the sample mean turns normal and narrows as n grows.

The sampling distribution of the mean approaches a normal distribution as the sample size $n$ increases, whatever the shape of the original population.

Read the two parameters separately, because they say different things. The mean of the sampling distribution is $\mu_X$ — the same center as the population. Averaging does not push the answer up or down; it just makes the answer more reliable. The standard deviation of the sampling distribution is $\sigma_X$ divided by $\sqrt{n}$, which is smaller than $\sigma_X$ whenever $n > 1$. That division is the whole point. Sample means cluster tighter around the center than individual values do, and the bigger the sample, the tighter they cluster.

One warning about the arithmetic. The theorem is stated in terms of variance first: the variance of the sampling distribution equals the population variance divided by the sample size, $\sigma_X^2/n$. Standard deviation is the square root of variance, so taking the square root of both sides gives $\sigma_X/\sqrt{n}$ — the $n$ comes out of the square root as $\sqrt{n}$, not as $n$. Dividing by $n$ instead of $\sqrt{n}$ is the single most common slip in this section, and it shrinks the spread far too aggressively. Note also what $n$ actually counts: it is the number of values averaged together in one sample, not the number of times you repeat the experiment. If you roll ten dice and average them, $n = 10$ — even if you do that a thousand times.

Definition 6.1.3: Standard Error of the Mean

The standard deviation of the sampling distribution of the mean is called the standard error of the mean:

$$ \sigma_{\overline{X}} = \frac{\sigma_X}{\sqrt{n}} $$

Definition 6.1.3 - The standard error is the population's own spread, divided by the square root of the sample size.

It describes how far, on average, a sample mean will fall from the population mean in repeated random samples of size $n$.

Because $\overline{X}$ has its own center and its own spread, it also has its own z-score, and that z-score is not the same as the one you would compute for a single value. The numerator still measures distance from the center, but the denominator is now the standard error rather than the population standard deviation:

$$ z = \frac{\overline{x} - \mu_X}{\left(\dfrac{\sigma_X}{\sqrt{n}}\right)} $$

Everything you already know about z-scores still applies here — sign gives direction, size gives distance in standard units — as long as you remember to divide by $\sigma_X/\sqrt{n}$ and not by $\sigma_X$. Using the wrong denominator produces a z-score that is too small by a factor of $\sqrt{n}$, which quietly turns a very unusual sample mean into an ordinary-looking one.

Probabilities for sample means are calculated with technology. The command is the same normalcdf you have been using for any normal distribution — the only change is what you type for the standard deviation. Instead of the population standard deviation, you enter the standard error, so the syntax becomes normalcdf(lower value, upper value, mean, standard deviation ÷ √(sample size)). The mean and standard deviation in that list are the ones from the original distribution, and the sample size $n$ is what converts the second of them into a standard error. Getting that one substitution right is most of the work in this subsection; the rest is the ordinary business of deciding which area a question is asking for. Sketch the curve, mark the boundaries the problem names, shade the region you want, and then read off whether you need a left tail, a right tail, or a middle strip. A picture takes ten seconds and prevents the two errors that actually cost points here — shading the complement of what was asked, and forgetting to divide by $\sqrt{n}$ at all.

Try It Now 6.1.1

An unknown distribution has a mean of 45 and a standard deviation of eight. Samples of size $n = 30$ are drawn randomly from the population. Find the probability that the sample mean is between 42 and 50.

Solution

Step 1 — Write the distribution of the sample mean: $\mu_X = 45$, $\sigma_X = 8$, and $n = 30$, so by the central limit theorem

$$ \overline{X} \sim N\!\left(45, \frac{8}{\sqrt{30}}\right) $$

Step 2 — Compute the standard error:

$$ \frac{\sigma_X}{\sqrt{n}} = \frac{8}{\sqrt{30}} \approx 1.4606 $$

Step 3 — Find the area between the two boundaries: we want $P(42 < \overline{x} < 50)$, which is normalcdf(42, 50, 45, 8/√30).

$$ P(42 < \overline{x} < 50) = 0.9797 $$

Answer: $P(42 < \overline{x} < 50) = 0.9797$. Almost 98% of samples of size 30 will have a mean between 42 and 50, because the standard error of 1.46 is small enough that both boundaries sit well out in the tails.

Example 6.1.1: An Unknown Distribution with Mean 90

An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size $n = 25$ are drawn randomly from the population.

a. Find the probability that the sample mean is between 85 and 92.

b. Find the value that is two standard deviations above the expected value, 90, of the sample mean.

Solution

Part a — Step 1 — Name the variables: let $X$ = one value from the original unknown population, and let $\overline{x}$ = the mean of a sample of size 25. The question asks for a probability about the sample mean, so we work with $\overline{X}$, not $X$.

Part a — Step 2 — Write the distribution. Since $\mu_X = 90$, $\sigma_X = 15$, and $n = 25$,

$$ \overline{X} \sim N\!\left(90, \frac{15}{\sqrt{25}}\right) $$

The standard error is $\dfrac{15}{\sqrt{25}} = \dfrac{15}{5} = 3$. Notice how much tighter that is than the population's spread of 15.

Part a — Step 3 — Find $P(85 < \overline{x} < 92)$ and draw a graph.

$$ P(85 < \overline{x} < 92) = 0.6997 $$

Figure 6.1.1 - The shaded strip between 85 and 92 under the sampling distribution of the sample mean.

On a TI-83/83+/84/84+: the syntax is normalcdf(lower value, upper value, mean, standard error of the mean), which spells out as normalcdf(lower value, upper value, μ, σ/√n). For this problem, normalcdf(85, 92, 90, 15/√25) = 0.6997.

The probability that the sample mean is between 85 and 92 is 0.6997.

Part b — Step 1 — Set up the formula. "Two standard deviations above the expected value" means two standard errors above 90, because the expected value in question belongs to $\overline{X}$:

$$ \text{value} = \mu_X + (\text{number of standard deviations})\left(\frac{\sigma_X}{\sqrt{n}}\right) $$

Part b — Step 2 — Substitute:

$$ \text{value} = 90 + 2\left(\frac{15}{\sqrt{25}}\right) = 90 + 2(3) = 96 $$

Answer: (a) $P(85 < \overline{x} < 92) = 0.6997$. (b) The value two standard deviations above the expected value is 96. The standard error here is $\frac{15}{\sqrt{25}} = 3$, which is a description of how far, on average, the sample mean will land from the population mean in repeated random samples of size 25.

Try it in rāSHio

The only new move here is what you type for the spread. In rāSHio choose Distributions → Normal, set mean 90 and standard deviation 3 — that is $15/\sqrt{25}$, the standard error, not the population's 15 — then pick Between with bounds 85 and 92 to get the 0.6997. Feed it 15 instead and it returns 0.1836, which answers a question about one value rather than an average.

Figure 6.1.2 — Reading a normal area in rāSHio: Distributions → Normal on the panel's Left tab. The Between, Right and Inverse modes the notes use are the tabs beside it; for a sample mean, type the standard error where it asks for the standard deviation.

Try It Now 6.1.2

Mei-Ling Chen is a school counselor who proctors the SAT. She knows the length of time taken on the SAT for a group of students is normally distributed with a mean of 2.5 hours and a standard deviation of 0.25 hours. She draws a sample of size $n = 60$ randomly from the population. Find the probability that the sample mean is between two hours and three hours.

Solution

Step 1 — Write the distribution of the sample mean: $\mu_X = 2.5$, $\sigma_X = 0.25$, and $n = 60$, so

$$ \overline{X} \sim N\!\left(2.5, \frac{0.25}{\sqrt{60}}\right) $$

Step 2 — Compute the standard error:

$$ \frac{0.25}{\sqrt{60}} \approx 0.0323 $$

Step 3 — Find the area: normalcdf(2, 3, 2.5, 0.25/√60).

$$ P(2 < \overline{x} < 3) = 1.0000 $$

Answer: $P(2 < \overline{x} < 3) \approx 1$. Both boundaries sit more than 15 standard errors from the center, so essentially every sample of 60 students has a mean time in that range. The individual students vary far more than their average does.

Example 6.1.2: The Length of a Soccer Match

Rajesh Menon runs an "over 40" recreational soccer league and keeps records of how long each match lasts. He finds the length of time, in hours, it takes his group to play one soccer match is normally distributed with a mean of two hours and a standard deviation of 0.5 hours. He draws a sample of size $n = 50$ randomly from the population. Find the probability that the sample mean is between 1.8 hours and 2.3 hours.

Solution

Step 1 — Name the variables: let $X$ = the time, in hours, it takes to play one soccer match, and let $\overline{x}$ = the mean time for a sample of 50 matches. The question asks about the sample mean.

Step 2 — Fill in the blanks of the central limit theorem. If $\mu_X = \_\_\_\_$, $\sigma_X = \_\_\_\_$, and $n = \_\_\_\_$, then $\overline{X} \sim N(\_\_\_\_, \_\_\_\_)$ by the central limit theorem for means. Here:

$$ \mu_X = 2, \quad \sigma_X = 0.5, \quad n = 50, \quad\text{and}\quad \overline{X} \sim N\!\left(2, \frac{0.5}{\sqrt{50}}\right) $$

Step 3 — Find $P(1.8 < \overline{x} < 2.3)$ and draw a graph. The standard error is $\frac{0.5}{\sqrt{50}} \approx 0.0707$, so 1.8 sits about 2.8 standard errors below the center and 2.3 sits about 4.2 above it. Shade the strip between them.

$$ P(1.8 < \overline{x} < 2.3) = 0.9977 $$

On a TI-83/83+/84/84+: normalcdf(1.8, 2.3, 2, 0.5/√50) = 0.9977.

Answer: the probability that the mean time is between 1.8 hours and 2.3 hours is 0.9977.

Try it in rāSHio

Same two-bound setup, much tighter curve. In rāSHio choose Distributions → Normal, set mean 2 and standard deviation 0.0707 — $0.5/\sqrt{50}$ — then pick Between with bounds 1.8 and 2.3. The 0.9977 it returns is nearly the whole curve, which is what a standard error one seventh the population's spread buys you.

6.1.2 Percentiles for Sample Means

The questions so far have all run in one direction: you were handed two values on the axis and asked for the area between them. Percentile questions run the other way. You are handed an area and asked for the value on the axis that cuts it off. The command flips too — invNorm instead of normalcdf — and, as before, the only adjustment for sample means is that the standard deviation you type in is the standard error.

The syntax is k = invNorm(area to the left of k, mean, standard deviation ÷ √(sample size)), where $k$ is the $k^{\text{th}}$ percentile, the mean and standard deviation come from the original distribution, and $n$ is the sample size. The area you supply always has to be the area to the left of the value you want, which is worth checking twice, since problems often describe the region from the other side. "The top 10%" means an area of 0.90 to the left. "The bottom quartile" means an area of 0.25 to the left. Translate the phrase into a left area first, then type it in.

Which command does this problem want?

If the answer should come out as a probability between 0 and 1, use normalcdf. If the answer should come out carrying a unit — years, minutes, dollars — use invNorm. The unit on the answer tells you which direction the question runs.

There is one more habit worth building before the examples, because percentile answers are easy to compute and easy to misread. A percentile of the sampling distribution is a statement about averages, not about individuals. When you find that the $95^{\text{th}}$ percentile of the sample mean age is 36.5 years, you have not said that 95% of users are younger than 36.5. You have said that 95% of samples of that size would produce an average age below 36.5 — a much narrower claim, because averages hug the center far more tightly than individuals do. The two numbers can be dramatically different: with a population standard deviation of 15 and a sample of 100, the individual $95^{\text{th}}$ percentile sits about 25 years above the mean while the sample-mean version sits about 2.5 years above it. Whenever a problem asks you to interpret a percentile in a complete sentence, name which of the two populations you are describing. That one word — "average" — is usually the difference between a correct interpretation and a plausible-sounding wrong one.

Try It Now 6.1.3

In an article on Flurry Blog, a gaming marketing gap for men between the ages of 30 and 40 is identified. Amara Okafor is researching a startup game targeted at the 35-year-old demographic. Her idea is to develop a strategy game that can be played by men from their late 20s through their late 30s. Based on the article's data, industry research shows that the average strategy player is 28 years old with a standard deviation of 4.8 years. She takes a sample of 100 randomly selected gamers. If her target market is 29- to 35-year-olds, should she continue with her development strategy?

Solution

Step 1 — Write the distribution of the sample mean: $\mu_X = 28$, $\sigma_X = 4.8$, $n = 100$, so the standard error is $\frac{4.8}{\sqrt{100}} = 0.48$ and

$$ \overline{X} \sim N(28, 0.48) $$

Step 2 — Find the probability that a sample average lands in the target range: normalcdf(29, 35, 28, 0.48).

$$ P(29 < \overline{x} < 35) = 0.0186 $$

Step 3 — Interpret it. Only about 1.86% of samples of 100 gamers would have an average age between 29 and 35. The center of the market is 28, and averages of 100 players cluster within about half a year of that.

Answer: no — you should not continue with this strategy as stated. The evidence says the typical strategy player is younger than your target window, and the sampling distribution is tight enough that a sample average landing in the 29-to-35 range would be unusual.

Example 6.1.3: The Mean Age of Tablet Users

In a recent study, it was reported that the mean age of tablet users is 34 years. Suppose the standard deviation is 15 years. Take a sample of size $n = 100$.

a. What are the mean and standard deviation for the sample mean ages of tablet users?

b. What does the distribution look like?

c. Find the probability that the sample mean age is more than 30 years.

d. Find the $95^{\text{th}}$ percentile for the sample mean age (to one decimal place).

Solution

Part a — Step 1 — The mean of the sampling distribution. Since the sample mean tends to target the population mean, $\mu_{\overline{X}} = \mu = 34$.

Part a — Step 2 — The standard deviation of the sampling distribution is the standard error:

$$ \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{100}} = \frac{15}{10} = 1.5 $$

Part b — The shape. The central limit theorem says that for large sample sizes $n$, the sampling distribution will be approximately normal. With $n = 100$, the curve is a normal distribution centered at 34 with a standard error of 1.5.

Part c — A right-tail area. We want $P(\overline{X} > 30)$, so the lower bound is 30 and the upper bound stands in for infinity:

$$ P(\overline{X} > 30) = \text{normalcdf}(30,\ E99,\ 34,\ 1.5) = 0.9962 $$

Part d — A percentile. Let $k$ = the $95^{\text{th}}$ percentile. The area to the left of $k$ is 0.95:

$$ k = \text{invNorm}\!\left(0.95,\ 34,\ \frac{15}{\sqrt{100}}\right) = 36.5 $$

Answer: (a) mean 34, standard error 1.5. (b) approximately normal, $\overline{X} \sim N(34, 1.5)$. (c) 0.9962. (d) 36.5 years — 95% of samples of 100 users would have an average age below 36.5 years.

Try it in rāSHio

This example runs the tool in both directions. In rāSHio choose Distributions → Normal, set mean 34 and standard deviation 1.5 — $15/\sqrt{100}$ — then pick Right with cutoff 30 for part c's 0.9962, and switch to Inverse (Left) with probability 0.95 for part d's 36.5 years.

Try It Now 6.1.4

Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are $n = 34$ and $\overline{x} = 16.01$ ounces. If the cans are filled so that $\mu = 16.00$ ounces (as labeled) and $\sigma = 0.143$ ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?

Solution

Step 1 — Write the distribution of the sample mean: $\mu_X = 16.00$, $\sigma_X = 0.143$, $n = 34$, so the standard error is

$$ \frac{0.143}{\sqrt{34}} \approx 0.0245 $$

Step 2 — Find the right-tail area: normalcdf(16.01, E99, 16, 0.143/√34).

$$ P(\overline{x} > 16.01) = 0.3417 $$

Step 3 — Interpret it. An average of 16.01 ounces or more happens in about 34% of samples of 34 cans even when the machine is filling to exactly 16 ounces. That is a completely ordinary result, not a surprising one.

Answer: $P(\overline{x} > 16.01) = 0.3417$. No — the results do not suggest the cans are overfilled. The observed average sits well inside the range you would expect from ordinary sampling variation.

Example 6.1.4: App Engagement Time

The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.

a. What are the mean and standard deviation for the sample mean number of minutes of app engagement?

b. What is the standard error of the mean?

c. Find the $90^{\text{th}}$ percentile for the sample mean time of minutes for app engagement. Interpret this value in a complete sentence.

d. Find the probability that the sample mean is between eight minutes and 8.5 minutes.

Solution

Part a — The two parameters:

$$ \mu_{\overline{x}} = \mu = 8.2 \qquad\text{and}\qquad \sigma_{\overline{x}} = \frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt{60}} \approx 0.13 $$

Part b — The standard error is that second number, $\frac{1}{\sqrt{60}} \approx 0.13$ minutes. It lets us calculate the probability that a sample mean lands a given distance from the center, in repeated samples of size 60.

Part c — A percentile. Let $k$ = the $90^{\text{th}}$ percentile, so the area to the left of $k$ is 0.90:

$$ k = \text{invNorm}\!\left(0.90,\ 8.2,\ \frac{1}{\sqrt{60}}\right) = 8.37 $$

In a complete sentence: 90% of samples of 60 tablet users would have an average engagement time below 8.37 minutes.

Part d — A middle strip.

$$ P(8 < \overline{x} < 8.5) = \text{normalcdf}\!\left(8,\ 8.5,\ 8.2,\ \frac{1}{\sqrt{60}}\right) = 0.9293 $$

Answer: (a) mean 8.2 minutes, standard error about 0.13 minutes. (b) 0.13 minutes. (c) 8.37 minutes — 90% of samples of 60 users average less than 8.37 minutes of engagement. (d) 0.9293.

Try it in rāSHio

Read the answer's units to pick the mode. In rāSHio choose Distributions → Normal, set mean 8.2 and standard deviation 0.1291 — $1/\sqrt{60}$ — then use Inverse (Left) at 0.90 for part c's 8.37 minutes, an answer in minutes, and Between with bounds 8 and 8.5 for part d's 0.9293, an answer that is a probability.

Problem Set 6.1

Use the following information to answer the next six exercises: Cruz Delgado, who uses they/them pronouns, is a personnel manager in a large corporation. Each month they must review 16 of the employees. From past experience, they have found that the reviews take them approximately four hours each to do with a population standard deviation of 1.2 hours. Let $X$ be the random variable representing the time it takes them to complete one review. Assume $X$ is normally distributed. Let $\overline{x}$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

Problem 1. What is the mean, standard deviation, and sample size?

Solution

Step 1 — Read the givens off the setup. Each review takes about four hours on average, with a population standard deviation of 1.2 hours, and Cruz reviews 16 employees a month.

Answer: the mean is $\mu_X = 4$ hours, the standard deviation is $\sigma_X = 1.2$ hours, and the sample size is $n = 16$.

Problem 2. Complete the distributions.

a) $X \sim$ ____ (____, ____)

b) $\overline{X} \sim$ ____ (____, ____)

Solution

Step 1 — The distribution of one review. $X$ is given as normally distributed with mean 4 and standard deviation 1.2.

$$ X \sim N(4,\ 1.2) $$

Step 2 — The distribution of the sample mean. By the central limit theorem, $\overline{X}$ has the same center but a standard error of $\sigma_X/\sqrt{n}$:

$$ \frac{1.2}{\sqrt{16}} = \frac{1.2}{4} = 0.3 $$ $$ \overline{X} \sim N(4,\ 0.3) $$

Answer: (a) $X \sim N(4, 1.2)$. (b) $\overline{X} \sim N(4, 0.3)$.

Problem 3. Find the probability that one review will take Cruz from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

a) Sketch and shade the region on the curve below.

Exercise Figure - Sketch and shade the probability for one review.

b) $P(\_\_\_\_ < x < \_\_\_\_) = \_\_\_\_$

Solution

Step 1 — Decide which variable the question is about. "One review" means a single value, so this is a question about $X \sim N(4, 1.2)$ — not about the sample mean.

Step 2 — Sketch and shade. Draw the normal curve centered at 4 with the horizontal axis labelled in hours, mark 3.5 and 4.25, and shade the strip between them.

Step 3 — Compute the area: normalcdf(3.5, 4.25, 4, 1.2).

$$ P(3.5 < x < 4.25) = 0.2441 $$

Answer: $P(3.5 < x < 4.25) = 0.2441$. Only about 24% of individual reviews land in that window, because a single review varies a lot.

Problem 4. Find the probability that the mean of a month's reviews will take Cruz from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

a) Sketch and shade the region on the curve below.

Exercise Figure - Sketch and shade the probability for the mean of a month's reviews.

b) $P(\_\_\_\_ < \overline{x} < \_\_\_\_) = \_\_\_\_$

Solution

Step 1 — Decide which variable the question is about. "The mean of a month's reviews" is the average of 16 values, so this is a question about $\overline{X} \sim N(4, 0.3)$.

Step 2 — Sketch and shade. Same picture, but the axis is now labelled $\overline{x}$ and the curve is much narrower — its standard error is 0.3 instead of 1.2. Shade between 3.5 and 4.25.

Step 3 — Compute the area: normalcdf(3.5, 4.25, 4, 1.2/√16).

$$ P(3.5 < \overline{x} < 4.25) = 0.7499 $$

Answer: $P(3.5 < \overline{x} < 4.25) = 0.7499$.

Problem 5. What causes the probabilities in Problem 6.1.3 and Problem 6.1.4 to be different?

Solution

Step 1 — Compare the two distributions. Problem 6.1.3 used $X \sim N(4, 1.2)$; Problem 6.1.4 used $\overline{X} \sim N(4, 0.3)$. Same center, very different spread.

Step 2 — Say what the spread does. The interval from 3.5 to 4.25 is a fixed width. Measured against a standard deviation of 1.2 it is narrow, so it catches little area. Measured against a standard error of 0.3 it is wide — it reaches more than 1.6 standard errors below the mean and more than 0.8 above — so it catches much more.

Answer: the two probabilities are different because they come from two different distributions. Averaging 16 reviews divides the spread by $\sqrt{16} = 4$, so the sampling distribution is four times narrower than the population and the same interval captures far more of it.

Problem 6. Find the $95^{\text{th}}$ percentile for the mean time to complete one month's reviews. Sketch the graph.

a) Sketch and shade the region on the curve below.

Exercise Figure - Sketch the 95th percentile for the mean time to complete one month's reviews.

b) The $95^{\text{th}}$ percentile = ____

Solution

Step 1 — Identify the direction. The question hands you a percentage (95%) and wants a time in hours, so this is an invNorm problem on the sampling distribution $\overline{X} \sim N(4, 0.3)$.

Step 2 — Sketch and shade. Draw the curve for $\overline{x}$ centered at 4 and shade everything to the left of an unknown boundary $k$, so that the shaded area is 0.95.

Step 3 — Compute $k$: invNorm(0.95, 4, 1.2/√16).

$$ k = 4 + 1.645(0.3) \approx 4.49 $$

Answer: the $95^{\text{th}}$ percentile is about 4.49 hours. In 95% of months, the average review time for the 16 reviews will be below 4.49 hours.

Problem 7. Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

a) In words, $X$ = ____

b) $X \sim$ ____ (____, ____)

c) In words, $\overline{X}$ = ____

d) $\overline{X} \sim$ ____ (____, ____)

e) Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.

f) Find the probability that the average of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.

g) Explain why there is a difference in part e and part f.

Solution

Step 1 — Name the variables. $X$ = the amount of change one daytime statistics student carries. $\overline{X}$ = the average amount of change carried by a sample of 25 daytime statistics students.

Step 2 — The distribution of $X$. An exponential distribution with mean 0.88 has decay rate $m = 1/0.88 \approx 1.1364$:

$$ X \sim \text{Exp}(1.1364) $$

Step 3 — The distribution of $\overline{X}$. By the central limit theorem, the standard error is $0.88/\sqrt{25} = 0.176$:

$$ \overline{X} \sim N(0.88,\ 0.176) $$

Step 4 — Part e, the individual. For an exponential distribution, $P(X < x) = 1 - e^{-x/0.88}$. Shade the strip from 0.80 to 1.00 under the decaying exponential curve.

$$ P(0.80 < X < 1.00) = \left(1 - e^{-1.1364}\right) - \left(1 - e^{-0.9091}\right) = 0.0819 $$

Step 5 — Part f, the average. Shade the same strip, but under the normal sampling distribution: normalcdf(0.80, 1.00, 0.88, 0.176).

$$ P(0.80 < \overline{x} < 1.00) = 0.4276 $$

Answer: (a) $X$ = the amount of change a student carries. (b) $X \sim \text{Exp}(1.1364)$. (c) $\overline{X}$ = the average amount of change carried by a sample of 25 students. (d) $\overline{X} \sim N(0.88, 0.176)$. (e) 0.0819. (f) 0.4276. (g) The distributions are different: part e asks about one student drawn from a strongly right-skewed exponential distribution, and part f asks about the average of 25 students, which the central limit theorem makes normal and much less spread out.

Problem 8. Suppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 fly balls.

a) If $\overline{x}$ = average distance in feet for 49 fly balls, then $\overline{X} \sim$ ____ (____, ____)

b) What is the probability that the 49 balls traveled an average of less than 240 feet? Sketch the graph. Scale the horizontal axis for $\overline{x}$. Shade the region corresponding to the probability. Find the probability.

c) Find the $80^{\text{th}}$ percentile of the distribution of the average of 49 fly balls.

Solution

Step 1 — The distribution of the sample mean. $\mu_X = 250$, $\sigma_X = 50$, $n = 49$, so the standard error is $50/\sqrt{49} = 50/7 \approx 7.14$:

$$ \overline{X} \sim N\!\left(250,\ \frac{50}{\sqrt{49}}\right) = N(250,\ 7.14) $$

Step 2 — Part b, a left-tail area. Sketch the curve with the axis scaled for $\overline{x}$, mark 240, and shade everything to its left. The z-score is

$$ z = \frac{240 - 250}{50/7} = -1.4 $$

normalcdf(-1E99, 240, 250, 50/7) gives

$$ P(\overline{x} < 240) = 0.0808 $$

Step 3 — Part c, the $80^{\text{th}}$ percentile: invNorm(0.80, 250, 50/7).

$$ k = 250 + 0.8416\left(\frac{50}{7}\right) \approx 256.01 $$

Answer: (a) $\overline{X} \sim N(250, 7.14)$. (b) 0.0808. (c) about 256.01 feet.

Problem 9. According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

a) In words, $X$ = ____

b) In words, $\overline{X}$ = ____

c) $\overline{X} \sim$ ____ (____, ____)

d) Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.

e) Would you be surprised if one taxpayer finished their Form 1040 in more than 12 hours? In a complete sentence, explain why.

Solution

Step 1 — Name the variables. $X$ = the length of time, in hours, for one individual to complete IRS Form 1040. $\overline{X}$ = the mean length of time, in hours, for a sample of 36 taxpayers to complete Form 1040.

Step 2 — The distribution of $\overline{X}$. The population distribution is unknown, but $n = 36$ is large enough for the central limit theorem. The standard error is $2/\sqrt{36} = 1/3$:

$$ \overline{X} \sim N\!\left(10.53,\ \frac{1}{3}\right) $$

Step 3 — Part d, the average of 36.

$$ z = \frac{12 - 10.53}{1/3} = 4.41 \qquad P(\overline{x} > 12) \approx 0.000005 $$

Step 4 — Part e, one taxpayer. Now use the population standard deviation of 2 hours.

$$ z = \frac{12 - 10.53}{2} = 0.735 \qquad P(x > 12) = 0.2312 $$

Answer: (a) the time for one individual to complete Form 1040, in hours. (b) the mean time for a sample of 36 taxpayers, in hours. (c) $\overline{X} \sim N(10.53, \frac{1}{3})$. (d) Yes, that would be surprising — the probability is almost zero, because averaging 36 taxpayers shrinks the spread to a third of an hour and 12 hours is more than four standard errors out. (e) No, that would not be surprising — about 23% of individual taxpayers take more than 12 hours.

Problem 10. Suppose that a category of world-class runners are known to run a marathon (26 miles) in an average of 145 minutes with a standard deviation of 14 minutes. Consider 49 of the races. Let $\overline{x}$ be the average of the 49 races.

a) $\overline{X} \sim$ ____ (____, ____)

b) Find the probability that the runner will average between 142 and 146 minutes in these 49 marathons.

c) Find the $80^{\text{th}}$ percentile for the average of these 49 marathons.

d) Find the median of the average running times.

Solution

Step 1 — The distribution of the sample mean. $\mu_X = 145$, $\sigma_X = 14$, $n = 49$, so the standard error is $14/\sqrt{49} = 2$:

$$ \overline{X} \sim N(145,\ 2) $$

Step 2 — Part b, a middle strip: normalcdf(142, 146, 145, 2). The boundaries sit at $z = -1.5$ and $z = 0.5$.

$$ P(142 < \overline{x} < 146) = 0.6247 $$

Step 3 — Part c, the $80^{\text{th}}$ percentile: invNorm(0.80, 145, 2).

$$ k = 145 + 0.8416(2) \approx 146.68 $$

Step 4 — Part d, the median. The sampling distribution is normal, and a normal curve is symmetric about its mean, so the median equals the mean.

Answer: (a) $\overline{X} \sim N(145, 2)$. (b) 0.6247. (c) about 146.68 minutes. (d) 145 minutes.

Problem 11. The length of songs in a collector's music album collection is uniformly distributed from two to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.

a) In words, $X$ = ____

b) $X \sim$ ____

c) In words, $\overline{X}$ = ____

d) $\overline{X} \sim$ ____ (____, ____)

e) Find the first quartile for the average song length, $\overline{X}$.

f) The IQR (interquartile range) for the average song length, $\overline{X}$, is from ____ to ____.

Solution

Step 1 — Name the variables. $X$ = the length of one song, in minutes, in the collection. $\overline{X}$ = the average length, in minutes, of the songs from a sample of five albums.

Step 2 — The distribution of $X$. Song lengths run uniformly from two to 3.5 minutes:

$$ X \sim U(2,\ 3.5) $$

Step 3 — The parameters of the uniform. For $U(a, b)$, the mean is $\frac{a+b}{2}$ and the standard deviation is $\frac{b-a}{\sqrt{12}}$:

$$ \mu_X = \frac{2 + 3.5}{2} = 2.75 \qquad \sigma_X = \frac{1.5}{\sqrt{12}} \approx 0.4330 $$

Step 4 — The distribution of $\overline{X}$. There are 43 songs across the five albums, so $n = 43$ and the standard error is $0.4330/\sqrt{43} \approx 0.0660$:

$$ \overline{X} \sim N(2.75,\ 0.0660) $$

Step 5 — Part e, the first quartile: invNorm(0.25, 2.75, 0.0660).

$$ Q_1 = 2.75 - 0.6745(0.0660) \approx 2.71 \text{ minutes} $$

Step 6 — Part f, the IQR. The third quartile is the mirror image, $Q_3 = 2.75 + 0.0445 \approx 2.79$.

Answer: (a) the length of a song, in minutes, in the collection. (b) $X \sim U(2, 3.5)$. (c) the average length, in minutes, of the songs from a sample of five albums. (d) $\overline{X} \sim N(2.75, 0.0660)$. (e) about 2.71 minutes. (f) from about 2.71 to about 2.79 minutes, an IQR of roughly 0.09 minutes.

Problem 12. In 1940 the average size of a U.S. farm was 174 acres. Let's say that the standard deviation was 55 acres. Suppose we randomly survey 38 farmers from 1940.

a) In words, $X$ = ____

b) In words, $\overline{X}$ = ____

c) $\overline{X} \sim$ ____ (____, ____)

d) The IQR for $\overline{x}$ is from ____ acres to ____ acres.

Solution

Step 1 — Name the variables. $X$ = the size, in acres, of one U.S. farm in 1940. $\overline{X}$ = the average size, in acres, of a sample of 38 U.S. farms in 1940.

Step 2 — The distribution of $\overline{X}$. The standard error is $55/\sqrt{38} \approx 8.9221$:

$$ \overline{X} \sim N(174,\ 8.9221) $$

Step 3 — Part d, the IQR. The quartiles sit $0.6745$ standard errors either side of the center:

$$ 0.6745(8.9221) \approx 6.014 $$ $$ Q_1 = 174 - 6.014 \approx 167.99 \qquad Q_3 = 174 + 6.014 \approx 180.01 $$

Answer: (a) the size of one U.S. farm in 1940, in acres. (b) the average size of a sample of 38 U.S. farms in 1940, in acres. (c) $\overline{X} \sim N(174, 8.9221)$. (d) the IQR runs from about 167.99 acres to about 180.01 acres.

Problem 13. Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

a) When the sample size is large, the mean of $\overline{X}$ is approximately equal to the mean of $X$.

b) When the sample size is large, $\overline{X}$ is approximately normally distributed.

c) When the sample size is large, the standard deviation of $\overline{X}$ is approximately the same as the standard deviation of $X$.

Solution

Step 1 — Statement a. The central limit theorem gives $\overline{X}$ the same center as the population, $\mu_{\overline{X}} = \mu_X$, for any sample size.

Step 2 — Statement b. The theorem says the sampling distribution approaches a normal distribution as $n$ grows, whatever the shape of the population.

Step 3 — Statement c. The standard deviation of $\overline{X}$ is $\sigma_X/\sqrt{n}$, which shrinks as $n$ grows. It equals $\sigma_X$ only when $n = 1$.

Answer: (a) True — the mean of the sampling distribution is approximately the mean of the data distribution. (b) True — by the central limit theorem, the larger the sample, the closer the sampling distribution of the means comes to normal. (c) False — the standard deviation of the sampling distribution decreases as the sample size increases, so it does not equal the standard deviation of $X$.

Problem 14. The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of about ten. Suppose that 16 individuals are randomly chosen. Let $\overline{x}$ = average percent of fat calories.

a) $\overline{X} \sim$ ____ (____, ____)

b) For the group of 16, find the probability that the average percent of fat calories consumed is more than five. Graph the situation and shade in the area to be determined.

c) Find the first quartile for the average percent of fat calories.

Solution

Step 1 — The distribution of the sample mean. $\mu_X = 36$, $\sigma_X = 10$, $n = 16$, so the standard error is $10/\sqrt{16} = 2.5$:

$$ \overline{X} \sim N(36,\ 2.5) $$

Step 2 — Part b, a right-tail area. Sketch the curve centered at 36 and shade everything to the right of 5. That boundary is

$$ z = \frac{5 - 36}{2.5} = -12.4 $$

standard errors below the center — so far into the left tail that essentially the entire curve lies to its right. normalcdf(5, 1E99, 36, 2.5) returns 1.

Step 3 — Part c, the first quartile: invNorm(0.25, 36, 2.5).

$$ Q_1 = 36 - 0.6745(2.5) \approx 34.31 $$

Answer: (a) $\overline{X} \sim N(36, 2.5)$. (b) approximately 1 — a group average below 5% fat calories is effectively impossible when the population averages 36%. (c) about 34.31%.

Problem 15. The distribution of income in some developing countries is considered wedge shaped (many low income people, very few middle income people, and even fewer high income people). Suppose we pick a country with a wedge shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.

a) In words, $X$ = ____

b) In words, $\overline{X}$ = ____

c) $\overline{X} \sim$ ____ (____, ____)

d) How is it possible for the standard deviation to be greater than the average?

e) Why is it more likely that the average of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?

Solution

Step 1 — Name the variables. $X$ = the yearly income of one resident of the developing country. $\overline{X}$ = the average yearly income of a sample of 1,000 residents.

Step 2 — The distribution of $\overline{X}$. The population is wedge shaped, not normal, but $n = 1{,}000$ is more than large enough for the central limit theorem. The standard error is $8{,}000/\sqrt{1{,}000} \approx 252.98$:

$$ \overline{X} \sim N\!\left(2{,}000,\ \frac{8{,}000}{\sqrt{1{,}000}}\right) $$

Step 3 — Part d, why the standard deviation can exceed the mean. Income cannot be negative, so the distribution is squeezed against zero on the left and stretched far out on the right by a small number of very high earners. Very wide differences among data values can produce a standard deviation larger than the average.

Step 4 — Part e, comparing two intervals. The sampling distribution is centered at $2,000, and a normal curve is tallest at its center. The interval $2,000 to $2,100 sits right under the peak; the interval $2,100 to $2,200 sits further out where the curve is already falling away.

$$ P(2{,}000 < \overline{x} < 2{,}100) = 0.1537 $$ $$ P(2{,}100 < \overline{x} < 2{,}200) = 0.1317 $$

Answer: (a) the yearly income of someone in a developing country. (b) the average salary from samples of 1,000 residents. (c) $\overline{X} \sim N\!\left(2{,}000, \frac{8{,}000}{\sqrt{1{,}000}}\right)$. (d) Very wide differences in data values can give a distribution an average smaller than its standard deviation. (e) The sampling distribution puts higher probability closer to the population mean, so the interval next to $2,000 captures more area than the one further out.

Problem 16. Which of the following is NOT TRUE about the distribution for averages?

a) The mean, median, and mode are equal.

b) The area under the curve is one.

c) The curve never touches the x-axis.

d) The curve is skewed to the right.

Solution

Step 1 — Check each statement against the central limit theorem. The distribution for averages is normal, so it is symmetric about its mean; its mean, median, and mode all coincide; the total area under it is one; and its tails approach the horizontal axis without ever touching it.

Step 2 — Find the one that breaks. Symmetry rules out any skew, so "skewed to the right" cannot describe a sampling distribution of means.

Answer: d. The curve is not skewed to the right — the distribution for averages is normal, and a normal curve is symmetric.

Problem 17. The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:

a) $\overline{X} \sim N(4.59, 0.10)$

b) $\overline{X} \sim N\left(4.59, \frac{0.10}{\sqrt{16}}\right)$

c) $\overline{X} \sim N\left(4.59, \frac{16}{0.10}\right)$

d) $\overline{X} \sim N\left(4.59, \frac{\sqrt{16}}{0.10}\right)$

Solution

Step 1 — Identify the parameters. $\mu_X = 4.59$, $\sigma_X = 0.10$, and $n = 16$ gas stations.

Step 2 — Apply the central limit theorem. The center stays at 4.59 and the standard deviation becomes the standard error, $\sigma_X/\sqrt{n}$:

$$ \overline{X} \sim N\!\left(4.59,\ \frac{0.10}{\sqrt{16}}\right) $$

Choice a keeps the population standard deviation instead of the standard error, and choices c and d invert the fraction.

Answer: b.

Key Terms

sampling distribution of the mean — the distribution of the random variable $\overline{X}$, built from the means of all possible samples of size $n$ drawn from a population.

central limit theorem for sample means — the result that $\overline{X} \sim N\!\left(\mu_X, \frac{\sigma_X}{\sqrt{n}}\right)$ as $n$ grows, whatever the shape of the original population.

standard error of the mean — the standard deviation of the sampling distribution, $\sigma_{\overline{X}} = \frac{\sigma_X}{\sqrt{n}}$; how far a sample mean typically falls from the population mean.

sample mean — written $\overline{x}$, the average of the values in one particular sample; a single observed value of the random variable $\overline{X}$.