Week 10 · Chapter

Chapter 10 — Sampling Distributions & the Central Limit Theorem

Introduction to Statistics Free sample edition

The module's primary reading · ~45–55 minutes · Objective 1 (distinguish populations from samples and parameters from statistics, classify variables by type and level of measurement, and identify appropriate sampling methods and study designs, including what each can and cannot conclude), Objective 5 (use the normal model and sampling distributions, including the Central Limit Theorem, to reason about variability in individual values and in sample statistics) · Read it first this week; the lecture, the tutorial, the quiz, and the Data Lab all build on it.

In plain language

When you take a random sample and compute an average or a percentage, a second sample would give a slightly different number. This week is about that wobble, which statisticians call sampling variability (the change in a sample's summary from one sample to the next). You will learn how big the wobble typically is and how it shrinks as samples grow. You will also meet the Central Limit Theorem (the reason averages of many values tend to pile into a bell shape), which is why a well-drawn poll of about a thousand people can tell you something about millions.

Why you already care about this

Roll a fair die 30 times and average the faces. You might get 3.4. Roll it 30 more times and you might get 3.7. Ask 1,000 randomly chosen adults whether they favor a new park, and a second random 1,000 will not give exactly the same share. Weigh 30 penguins picked at random from the 342 measured birds in the field log, and a second 30 will have a different average weight. Nothing went wrong in any of these. A sample is only part of the population, and a different part gives a different summary.

That raises a practical question. If every sample gives a different answer, how can a pollster, a clinic manager, or a field biologist trust any one of them? Here is the good news: the wobble itself follows a pattern you can describe. It has a center, a typical size, and a shape. Once you know that pattern, you can say how far one sample's average is likely to land from the truth, and you can spot an average that is too far off to be ordinary luck.

By the end of this chapter you will be able to:

explain why a statistic such as a sample mean changes from sample to sample while the parameter it estimates stays fixed;
describe the sampling distribution of a sample mean, with its mean and its standard deviation σ ⁄ √n;
tell the standard deviation of single values apart from the standard error of a sample mean;
state the Central Limit Theorem and apply the course's rule of thumb for a large enough sample;
judge whether one sample mean would be surprising;
describe the sampling distribution of a sample proportion and check the large-counts rule of thumb before using a normal model for it.

Carry this sentence with you: single values scatter — averages of many values huddle near the mean.

Where we are

Already covered: Week 1 separated a population from a sample and a parameter from a statistic. Week 3 gave you the mean and the standard deviation. Weeks 5 to 7 built probability, random variables, and the binomial. Week 8 added z-scores, and Week 9 gave you the normal model and the empirical rule.
This week: the sampling distribution of the sample mean and of the sample proportion, the standard error, and the Central Limit Theorem.
How they connect: a sample mean is a random variable, so it has a mean and a standard deviation, as in Week 6; the Central Limit Theorem says its distribution is close to Week 9's normal model.
Where we're going next: Week 11 turns this week's standard error into a confidence interval for a mean, and Week 12 does the same for a proportion.

Each earlier week feeds this one. In Week 1 you learned that μ (the population mean, read "mu") is a fixed number about the whole population, while x̄ (the sample mean, read "x bar") is computed from one sample. In Week 6 you learned that a random variable has a mean and a standard deviation. In Week 9 you learned that a normal model with mean μ and standard deviation σ (read "sigma") puts about 68% of its values within one standard deviation of the mean and about 95% within two. This chapter puts those three ideas together: x̄ is a random variable, and its distribution is often close to a normal model.

1. Sampling variability: why a second sample gives a different answer

Picture a food truck owner who wants to know the average order size. One morning she pulls 5 receipts at random and gets an average of $11. The next morning she pulls 5 more at random and gets $13. The truck did not change overnight. The two samples caught different customers. That difference is not a mistake. It is what happens every time you summarize part of a group instead of all of it.

Sampling variability is the natural change in a statistic, such as a sample mean or a sample proportion, from one random sample to the next, even when every sample comes from the same population by the same method.

Keep the two roles from Week 1 straight. The population mean μ is a parameter: one fixed number, even when nobody knows it. The sample mean x̄ is a statistic: it changes with the sample. Sampling variability is the reason a statistic is an estimate and not the parameter itself.

Worked example — the fair die as a population. A fair die is a population you can list in full.

The values. One roll of a fair die takes the values 1, 2, 3, 4, 5, 6, each with chance 1 ⁄ 6, so its histogram is flat.
The parameters. You know its mean, the parameter μ = 3.5, and its population standard deviation, the parameter σ = √(35 ⁄ 12) ≈ 1.7078 points.
The statistic. Now roll the die 30 times and average the faces. That average is x̄, and a second set of 30 rolls gives a different x̄.

In words: μ stays at 3.5 forever, while x̄ wobbles around it from one set of 30 rolls to the next. A game designer who tests a die with 30 rolls should expect the average to miss 3.5 by a little every time.

Misconception check — "a different answer means someone made an error." Students often think two honest random samples should agree, so a second sample that disagrees must be wrong. Both samples can be done perfectly and still differ, because each one catches a different part of the population. The cure: expect the statistic to move, and ask how much it typically moves. That is the question the rest of this chapter answers.

Memory hook for the week: the parameter stands still; the statistic wobbles.

2. A sampling distribution you can build by hand: averages of two rolls

To study the wobble, you need every possible sample mean and its chance. That list has a name.

A sampling distribution is the distribution of the values a statistic takes across all possible random samples of the same size from the same population; it shows how much the statistic varies from sample to sample and where it centers.
The sampling distribution of the sample mean is the distribution of x̄ across all possible random samples of size n; its mean is the population mean μ and its standard deviation is σ ⁄ √n.

For most real populations you cannot list every possible sample. For the smallest case, though, you already have the list. In Week 5 you built the table of the sums of two fair dice. An average of two rolls is that sum divided by 2.

Worked example — every possible average of two rolls. Use the two-dice sums from Week 5.

The possible averages. When you average two rolls, you divide their sum by 2, so the average can be 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, or 6, eleven possible values.
The lowest average. An average of 1 needs a sum of 2, which has chance 1 ⁄ 36.
The middle average. An average of 3.5 needs a sum of 7, which has chance 6 ⁄ 36, about 16.7%.
The pattern. The chances rise from 1 ⁄ 36 to 6 ⁄ 36 and fall back to 1 ⁄ 36, a symmetric triangle.

Sampling distribution of the average of two rolls of a fair die: each possible average and its chance out of 36
Average of two rolls	1	1.5	2	2.5	3	3.5	4	4.5	5	5.5	6
Chance (out of 36)	1	2	3	4	5	6	5	4	3	2	1

Picture the two histograms side by side. One roll gives six bars of equal height, a flat shape. The average of two rolls gives a triangle with its peak at 3.5. Averaging only two values has already pulled the extreme results, like an average of 1 or 6, toward the middle. In words: a two-roll average lands on exactly 3.5 on about 1 try in 6, while it lands on 1 only about once in 36 tries.

Misconception check — "the sampling distribution is the sample's histogram." Students often think the sampling distribution is a graph of the values inside one sample. It is not. One sample of two rolls is only two numbers. The sampling distribution is the pattern of the averages across all possible samples. The cure: before you draw anything, ask "what is one dot in this picture: one roll, or one sample's average?"

3. Center and spread: the sampling distribution has mean μ and standard deviation σ ⁄ √n

Two rules describe any sampling distribution of x̄. Its center is the population mean μ, so x̄ does not systematically aim too high or too low. Its standard deviation is σ ⁄ √n, the population standard deviation divided by the square root of the sample size n (how many values go into one sample). Because you divide by √n, bigger samples give averages that stay closer to μ.

Worked example — checking both rules on two rolls. Use the triangle from Section 2.

The center. The sum of two dice has expected value 7, so the average has mean 7 ⁄ 2 = 3.5 points, the same as μ.
The spread from the sums. The sum of two dice has a standard deviation equal to the given value 2.4152, so the average has the given spread 2.4152 ⁄ 2, about 1.21 points, rounded to two decimals.
The spread from the rule. You get the same answer from the rule σ ⁄ √n, using the parameter σ = 1.7078: 1.7078 ⁄ √2, about 1.21 points.

In words: both routes agree. A two-roll average typically lands about 1.21 points from 3.5, while one roll typically lands about 1.71 points away.

Worked example — averages of 30 rolls. Now let each sample be n = 30 rolls.

The center. The sampling distribution of x̄ for n = 30 has mean μ = 3.5 points.
The spread. σ ⁄ √n, with the parameter σ = 1.7078 and n = 30, is 1.7078 ⁄ √30 = 1.7078 ⁄ 5.4772 ≈ 0.3118 points, rounded to four decimals as pinned.
The comparison. 1.7078 ⁄ 0.3118 ≈ 5.48, so a 30-roll average wobbles about 5.5 times less than a single roll, because √30 ≈ 5.48.

In words: one roll typically misses 3.5 by about 1.71 points, and a 30-roll average misses by only about 0.31 points. That is a 5.5-fold tightening for a 30-roll average versus a single roll. A board-game tester who averages 30 rolls gets a far steadier number than one who looks at a single roll.

Misconception check — "the average of 30 rolls has the same spread as one roll." Students often reuse σ for an average, as if averaging changed nothing. Averaging cancels some highs against some lows, so the average varies less. The cure: whenever the quantity is a mean of n values, divide σ by √n.

4. Standard deviation versus standard error: two different wobbles

The last section used two different spreads, and they answer two different questions. σ answers "how much do single values vary?" σ ⁄ √n answers "how much does the sample mean vary from sample to sample?" The second one has its own name.

The standard error is the standard deviation of a statistic's sampling distribution: how much the statistic typically wobbles from sample to sample. For a sample mean it is σ ⁄ √n, and it shrinks as the sample size n grows.

We write SE for the standard error. When σ is unknown, which is most of the time in real data, you estimate the standard error with s ⁄ √n, using the sample standard deviation s from Week 3. That version takes center stage in Week 11.

Worked example — one commute versus the average of several. Under Week 9's commute model, you suppose single trip times follow a normal model with μ = 32 minutes and σ = 6 minutes.

One trip. A single trip varies with standard deviation σ = 6 minutes.
Four trips. If you average n = 4 trips, that average has standard error σ ⁄ √n = 6 ⁄ √4 = 6 ⁄ 2 = 3 minutes.
Thirty-six trips. If you average n = 36 trips, you get σ ⁄ √n = 6 ⁄ 6 = 1 minute, a third of the four-trip standard error, because 36 is 9 times 4 and √9 = 3.

In words: going from 1 trip to 4 trips cuts the wobble in half, from 6 minutes to 3. A commuter who tracks the average of 36 trips can expect it to sit within a minute or two of 32, even though single trips swing much more.

That pattern is worth memorizing. To cut the standard error in half, you need four times the sample. To cut it to a third, you need nine times the sample. Precision gets expensive.

Worked example — a clinic's waiting room. Suppose a clinic's single patient wait times have standard deviation σ = 12 minutes. The average wait of 16 randomly chosen patients has standard error 12 ⁄ √16 = 12 ⁄ 4 = 3 minutes. The average of 64 patients has standard error 12 ⁄ √64 = 12 ⁄ 8 = 1.5 minutes. In words: a manager who checks 64 patients instead of 16 does four times the work to halve the wobble.

Misconception check — confusing σ with σ ⁄ √n. Students often call σ ⁄ √n "the standard deviation of the data", or use σ when they mean the standard error. They describe different things. σ describes one commute; σ ⁄ √n describes the average of n commutes. The cure: before you divide by √n, ask "am I describing one value or an average?"

Pause and reflect — choose one. Pick either of the 2 prompts; they count the same.

(a) Think of a number you track in your own life, such as your weekly grocery bill or your phone's daily screen time. Explain in your own words why the average of several weeks would wobble less than any single week.
(b) A friend says, "More data always gives a more accurate answer, so a survey of 4,000 is four times as precise as a survey of 1,000." Explain what quadrupling the sample really does to the standard error.

5. The Central Limit Theorem: averages pile into a bell

Look back at the die. One roll is flat. Two-roll averages make a triangle. If you kept going to 30-roll averages, the picture would round off into a smooth, symmetric bell centered at 3.5. That change of shape is not special to dice. It is one of the most useful results in statistics.

The Central Limit Theorem is the result that, for random samples of size n from a population with mean μ and standard deviation σ, the sampling distribution of x̄ is approximately normal when n is large, whatever the population's shape; it is the distribution of the mean, not the data, that becomes approximately normal.

How large is "large"? There is no single magic number, so the course uses a working rule.

The CLT rule of thumb is the working rule this course uses for when n is large enough for the Central Limit Theorem: about 30 for a moderately skewed population, fewer for a roughly symmetric one, and any n when the population is assumed to follow a normal model.

It is a rule of thumb, not a law. A very skewed population may need more than 30, and some courses state the rule a little differently. No graded question in this course depends on the exact cutoff.

Worked example — three populations, three verdicts. Apply the rule out loud.

The die. The flat die population is symmetric, so averages look close to a bell well before n = 30; by n = 30 the sampling distribution of x̄ is approximately normal.
The commute model. Because the commute model is normal, the average of any number of trips, even 4, follows a normal model exactly; no rule of thumb is needed.
A skewed population. Suppose the amounts customers spend at a bookstore are right-skewed, with most small purchases and a few large ones. The average of 5 purchases would still lean right, while the average of 40 purchases meets the rule of thumb, so a normal model for that average is reasonable.

In words: the shape of the population matters for small samples, and it fades for large ones. A store analyst studying averages of 40 purchases can use a normal model even though single purchases are far from a bell.

Misconception check — "the CLT says the data become normal." Students often think that collecting 30 or more values makes the data themselves bell-shaped. It does not. If you sample 100 bookstore purchases, the histogram of those 100 purchases still leans right, like the population. What becomes approximately normal is the distribution of the average across many samples. The cure: say "the sampling distribution of x̄ is approximately normal", never "the data are normal now".

Caveat: the Central Limit Theorem needs random samples. A huge convenience sample, as you saw in Week 1, can be biased, and no sample size fixes a biased method.

6. Use it: judge whether one sample mean is surprising

Now the payoff. Once you know the center, the standard error, and the shape, Week 9's tools apply to a sample mean. The z-score uses the standard error in place of σ: z = (x̄ − μ) ⁄ (σ ⁄ √n).

Worked example — a 30-roll average. Use μ = 3.5 and the standard error 0.3118 from Section 3.

Two standard errors below. 3.5 − 2 × 0.3118 = 2.88, rounded to two decimals.
Two standard errors above. 3.5 + 2 × 0.3118 = 4.12, rounded to two decimals.
The verdict. By the empirical rule about 95% of 30-roll averages land between 2.88 and 4.12, so an average outside roughly 2.9 to 4.1 would be surprising for a fair die.

In words: if a friend rolls a die 30 times and averages 4.6, that is far outside the usual range, and you have good reason to suspect the die is not fair. An average of 3.8 is ordinary luck.

Worked example — one long commute versus a long average. Use the commute model again.

One trip. A trip of 38 minutes has z = (38 − 32) ⁄ 6 = 1, and P(Z < 1) = 0.8413, so about 100% − 84.1% = 15.9% of single trips take longer than 38 minutes.
Four trips. With standard error 3 minutes, an average of 38 minutes has z = (38 − 32) ⁄ 3 = 2, so by the empirical rule only about 2.5% of four-trip averages run longer than 38 minutes.
Compare. About 15.9% versus about 2.5%, so a long average is about 6 times rarer than one long trip.

In words: one slow morning happens about once in every six or seven trips, but a four-trip average that long is about a 6-fold rarer event. A commuter who sees a four-trip average of 38 minutes has a real reason to suspect the route has changed.

Misconception check — using σ in the z-score for an average. Students often compute z = (38 − 32) ⁄ 6 = 1 for the four-trip average. That treats the average like a single trip and makes a rare average look ordinary. The cure: for a sample mean, the denominator is the standard error, σ ⁄ √n.

7. Proportions: the sampling distribution of p̂

Many questions are about a share rather than an average: the share of voters who favor a proposal, or the share of patients who miss an appointment. The same ideas carry over.

The population proportion is the share of a whole population that has a trait of interest, written p; it is a parameter, usually unknown, that a sample proportion estimates.
The sample proportion is the share of a sample that has a trait of interest, written p̂ and read "p hat"; it is a statistic that estimates the population proportion p.
The sampling distribution of the sample proportion is the distribution of p̂ across all possible random samples of size n; its mean is the population proportion p, its standard deviation is √(p(1 − p) ⁄ n), and it is approximately normal when np and n(1 − p) are both at least 10.

The last condition is the large-counts rule of thumb from Week 7: the expected numbers of successes, np, and failures, n(1 − p), must both be at least 10.

Worked example — a poll of 1,000. Suppose 50% of a large population favors a proposal, so the population proportion p (the share in favor) is 0.5.

The center. The sample proportion p̂ from your random sample has mean p = 0.5.
The spread. √(p(1 − p) ⁄ n) = √(0.5 × 0.5 ⁄ 1,000) = √0.00025 ≈ 0.0158, rounded to four decimals.
The typical range. Two standard deviations is about 2 × 0.0158 ≈ 0.03, so about 95% of such polls land within about 3 percentage points of the true 50%.
The check. np = 1,000 × 0.5 = 500 and n(1 − p) = 500, both at least 10, so a normal model for p̂ is reasonable.

In words: a random poll of 1,000 people usually lands within about 3 points of the truth. Notice that the population size does not appear anywhere in √(p(1 − p) ⁄ n). That is why a well-drawn poll of about a thousand people can speak for millions: the wobble depends on the size of the sample, not the size of the country.

Worked example — bigger and smaller polls. Keep p = 0.5. A poll of n = 100 gives √(0.25 ⁄ 100) = 0.05, and n = 2,500 gives √(0.25 ⁄ 2,500) = 0.01; 25 times the sample cuts the spread to one-fifth, because √25 = 5. In words: a campaign that pays for 2,500 interviews instead of 100 buys a five-fold steadier estimate, not a 25-fold one.

Worked example — when the check fails. The proportion of sixes in 12 rolls has np = 12 × 1 ⁄ 6 = 2, below 10, so its sampling distribution is lumpy and right-skewed and a normal model is not appropriate. In words: with only about 2 expected sixes, most samples show 0, 1, 2, or 3 sixes, piled against zero, and a smooth bell would mislead you.

Misconception check — "a bigger population needs a bigger sample." Students often think a poll of a whole country must be much larger than a poll of one city. As long as the sample is random and a small part of the population, the standard deviation of p̂ depends on n and p, not on the population size. The cure: look at the formula and notice what is missing from it.

Limitation: this spread describes only the luck of random sampling. A poll with a biased sample, low response, or a slanted question can miss by much more, and Week 12 returns to that point.

Pause and reflect — choose one. Pick either of the 2 prompts; they count the same.

(a) Think of a share that matters in your major or job, such as the share of customers who return a product. Describe how you would draw a random sample to estimate it, and how large n would need to be for np and n(1 − p) to reach 10.
(b) Explain to a friend why a poll of 1,000 adults can describe a whole country, using the words "sampling variability" and "standard deviation of p̂" in your answer.

8. A real population to sample from: the penguins

The die and the poll are tidy. Real data are messier, so the week's Data Lab tries the same ideas on the penguins you have met all term. You will treat the 342 measured body masses as if they were the whole population and draw many random samples from it.

Worked example — the penguin population. Treat the 342 measured body masses as the whole population.

The center. Their mean is 4,201.75 g, and dividing by N gives σ = 801.9545 × √(341 ⁄ 342) ≈ 800.78 g, a little below the sample standard deviation of 801.95 g.
The standard error. If you draw samples of n = 30 masses, σ ⁄ √n = 800.78 ⁄ √30 ≈ 146.20 g.
The shape. The population is mildly right-skewed, so by the rule of thumb the means of samples of 30 should pile up in a roughly normal shape centered near 4,201.75 g.

In words: single penguins differ from the mean by about 800 g, while averages of 30 penguins typically differ by about 146 g. A field team that weighs 30 birds gets an average it can trust far more than any one weighing.

Here N is the number of values in a population you can list in full. Dividing by N instead of n − 1 is Week 3's population formula. With 342 values the two versions differ by only about 1 g, so either one gives nearly the same standard error.

In practice (your tools). In Google Sheets or Excel, put the 342 masses in column A. =AVERAGE(A2:A343) returns the population mean, and =STDEV.P(A2:A343) returns σ, dividing by N. To draw one random sample of 30, put =RAND() in column B beside every mass, sort both columns by column B, and average the first 30 masses with =AVERAGE(A2:A31). Each new sort is a fresh random sample. The Data Lab walks you through repeating this many times.

Misconception check — "my sample of 30 should look like a bell." Students often expect the 30 masses in one sample to form a bell, because n is 30. One sample's histogram looks like the population: mildly right-skewed. The bell appears only when you collect the averages of many samples. The cure: in the Data Lab, make two histograms, one of a single sample's masses and one of many sample means, and compare their shapes.

Honesty note: treating 342 penguins as a "population" is a teaching move. In the field study they are themselves a sample from much larger colonies, and the Central Limit Theorem is what lets researchers reason from such samples.

Key terms

sampling variability — The natural change in a statistic, such as a sample mean or a sample proportion, from one random sample to the next, even when every sample comes from the same population by the same method. (first used: 1. Sampling variability: why a second sample gives a different answer)
sampling distribution — The distribution of the values a statistic takes across all possible random samples of the same size from the same population; it shows how much the statistic varies from sample to sample and where it centers. (first used: 2. A sampling distribution you can build by hand: averages of two rolls)
sampling distribution of the sample mean — The distribution of x̄ across all possible random samples of size n; its mean is the population mean μ and its standard deviation is σ ⁄ √n. (first used: 2. A sampling distribution you can build by hand: averages of two rolls)
standard error — The standard deviation of a statistic's sampling distribution: how much the statistic typically wobbles from sample to sample. For a sample mean it is σ ⁄ √n, and it shrinks as the sample size n grows. (first used: 4. Standard deviation versus standard error: two different wobbles)
Central Limit Theorem — The result that, for random samples of size n from a population with mean μ and standard deviation σ, the sampling distribution of x̄ is approximately normal when n is large, whatever the population's shape; it is the distribution of the mean, not the data, that becomes approximately normal. (first used: 5. The Central Limit Theorem: averages pile into a bell)
CLT rule of thumb — The working rule this course uses for when n is large enough for the Central Limit Theorem: about 30 for a moderately skewed population, fewer for a roughly symmetric one, and any n when the population is assumed to follow a normal model. (first used: 5. The Central Limit Theorem: averages pile into a bell)
population proportion — The share of a whole population that has a trait of interest, written p; it is a parameter, usually unknown, that a sample proportion estimates. (first used: 7. Proportions: the sampling distribution of p̂)
sample proportion — The share of a sample that has a trait of interest, written p̂ and read "p hat"; it is a statistic that estimates the population proportion p. (first used: 7. Proportions: the sampling distribution of p̂)
sampling distribution of the sample proportion — The distribution of p̂ across all possible random samples of size n; its mean is the population proportion p, its standard deviation is √(p(1 − p) ⁄ n), and it is approximately normal when np and n(1 − p) are both at least 10. (first used: 7. Proportions: the sampling distribution of p̂)

Summary

A statistic changes from sample to sample; the parameter it estimates does not.
A sampling distribution shows every value a statistic can take across all samples of one size.
The sample mean is centered on the population mean, and its spread is sigma divided by the square root of n.
The standard deviation describes single values; the standard error describes the sample mean.
Four times the sample halves the standard error.
The Central Limit Theorem says averages from large samples follow an approximately normal model, whatever the population's shape.
The course's rule of thumb for large is about 30 for a moderately skewed population.
A sample proportion is centered on p, and a normal model fits it when the expected successes and failures are both at least 10.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

Suppose the weights of apples from one orchard have mean 150 g and standard deviation 20 g. Find the mean and the standard error of the average weight of 16 randomly chosen apples, and then of 100 apples.
A classmate says, "I collected 50 values, so by the Central Limit Theorem my data are normal now." Explain what is wrong with that sentence and how to fix it.
Suppose 20% of the students at a large campus bike to class. A random sample of 400 students is surveyed. Find the mean and standard deviation of the sample proportion who bike, and check whether a normal model is reasonable.
Suppose the battery life of one phone model follows a normal model with mean 20 hours and standard deviation 2 hours. About what share of single phones last less than 18 hours? About what share of averages of 4 phones fall below 18 hours?

Answers (worked reasoning — read after attempting)

The mean of the average is the population mean, 150 g, for any sample size. For 16 apples, the standard error is 20 ⁄ √16 = 20 ⁄ 4 = 5 g. For 100 apples, it is 20 ⁄ √100 = 20 ⁄ 10 = 2 g. In words: the average of 100 apples typically lands within about 2 g of 150 g, much closer than a single apple, which typically misses by about 20 g.
The Central Limit Theorem is about the sampling distribution of the mean, not the data. The 50 values keep the shape of the population they came from, skewed or not. A correct sentence is: "With n = 50, the sampling distribution of x̄ is approximately normal, so I can use a normal model for the sample mean." This is the "the data become normal" misconception.
The mean of p̂ is p = 0.2. The standard deviation is √(0.2 × 0.8 ⁄ 400) = √(0.16 ⁄ 400) = 0.4 ⁄ 20 = 0.02. Check the rule: np = 400 × 0.2 = 80 and n(1 − p) = 400 × 0.8 = 320, both at least 10, so a normal model is reasonable. In words: most such surveys land within about 2 × 0.02 = 0.04, or 4 percentage points, of the true 20%.
For one phone, z = (18 − 20) ⁄ 2 = −1. By symmetry, the share below z = −1 equals the share above z = 1. Since P(Z < 1) = 0.8413, that share is 1 − 0.8413, about 15.9%, so about 15.9% of single phones last less than 18 hours. For an average of 4 phones, the standard error is 2 ⁄ √4 = 1 hour, so z = (18 − 20) ⁄ 1 = −2. By the empirical rule about 5% of values fall outside two standard deviations, split evenly, so only about 2.5% of four-phone averages fall below 18 hours. If you used 2 hours instead of the standard error, you would wrongly get 15.9% for the average too.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, and the quiz — and next week the standard error becomes the heart of a confidence interval for a mean.

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