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Welcome, Detective.

Top Edu Prep presents Unit 5 — a case file by Mr. Kim Min-ho

You've just been assigned to the Sampling Detective Agency. Your job is simple to describe and tricky to master: you can never question an entire population — there are too many people, too many bottles, too many voters. All you ever get is a sample, a small stack of clues. From that sample, you must make a fair and honest guess about the truth of the whole population.

Every time you pull a new sample and calculate something from it — a mean, a proportion — you get a slightly different number. If you did this over and over, all those numbers would form their own pattern. That pattern is called a sampling distribution, and understanding its shape is the single most important skill for everything that comes after this unit.

Four case files are waiting on your desk. Open them in order — each one builds on the last.

01 · Sample Means

What happens when you average a sample, over and over?

02 · Sample Proportions

Same idea, but for yes/no, pass/fail, votes for/against.

03 · Central Limit Theorem

The rule that makes almost everything in statistics possible.

04 · Standard Error

How confident can you really be in a sample's answer?

Why this unit matters: Units 6–9 (confidence intervals, hypothesis testing) are built entirely on the ideas here. If the sampling distribution logic isn't solid now, everything after it will feel like memorized steps instead of understanding. Take your time on this file.

CASE 01

The Distribution of Sample Means

Every sample tells a slightly different story. This is the pattern behind those stories.

01The Concept, Plainly

Imagine a population with mean $\mu$ and standard deviation $\sigma$. If you repeatedly pull a random sample of size $n$ and compute its mean $\bar{x}$ each time, those thousands of $\bar{x}$ values form a new distribution — the sampling distribution of the sample mean.

Here's the reassuring part: those sample means aren't wild or biased. On average, they land right on the true population mean. They're just a little spread out around it — and how spread out depends only on $n$.

Core Formulas $ \mu_{\bar{x}} = \mu \qquad \qquad \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} $

Plain-language translation: the "average of the averages" equals the population's true average. The "spread of the averages" shrinks as your sample gets bigger — bigger samples give more reliable, less jumpy estimates.

2Worked Evidence

EVIDENCE A · BASIC SETUP

A bottling machine fills bottles with a mean of $\mu = 500$ mL and a standard deviation of $\sigma = 20$ mL. For random samples of $n = 25$ bottles, find the mean and standard error of $\bar{x}$.

SOLVED

Mean of the sampling distribution: $ \mu_{\bar{x}} = \mu = 500 $ mL.

Standard error: $ \sigma_{\bar{x}} = \dfrac{20}{\sqrt{25}} = \dfrac{20}{5} = 4 $ mL.

Read it like a detective: if you kept grabbing 25-bottle samples forever, their means would cluster around 500 mL, typically within about 4 mL of it.

EVIDENCE B · PROBABILITY

Adult male height has $\mu = 175$ cm and $\sigma = 10$ cm. For a random sample of $n = 100$ men, find $P(\bar{x} > 176.5)$.

SOLVED

Standard error: $ \sigma_{\bar{x}} = \dfrac{10}{\sqrt{100}} = \dfrac{10}{10} = 1 $ cm.

Convert to a z-score: $ z = \dfrac{176.5 - 175}{1} = 1.5 $.

From the standard normal table: $ P(Z > 1.5) = 1 - 0.9332 = 0.0668 $.

Verdict: about a 6.7% chance a sample of 100 men averages above 176.5 cm.

EVIDENCE C · COMPARING SAMPLE SIZES

Test scores have $\mu = 70$, $\sigma = 12$. Compare the standard error for $n = 16$ versus $n = 36$.

SOLVED

$ n = 16: \ \sigma_{\bar{x}} = \dfrac{12}{\sqrt{16}} = \dfrac{12}{4} = 3 $

$ n = 36: \ \sigma_{\bar{x}} = \dfrac{12}{\sqrt{36}} = \dfrac{12}{6} = 2 $

Verdict: more than double the sample size, but the standard error only dropped from 3 to 2 — the shrinkage follows a square-root pattern, not a straight line. Case 04 explains exactly why.

CASE 02

The Distribution of Sample Proportions

Same detective logic — now for yes/no evidence instead of numbers.

01The Concept, Plainly

Not everything is a measurement. Sometimes the question is a yes-or-no: did the customer buy it, did the coin land heads, did the voter say yes. If the true population proportion is $p$, a sample of size $n$ gives you a sample proportion $\hat{p} = \dfrac{x}{n}$, where $x$ is the count of "yes" results.

Just like sample means, repeated sample proportions form their own sampling distribution — centered on the truth, with a predictable spread.

Core Formulas $ \mu_{\hat{p}} = p \qquad \qquad \sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} $

Condition to check: this normal-shaped pattern only kicks in reliably when the sample is large enough — commonly checked with $np \geq 10$ and $n(1-p) \geq 10$. Small samples with rare events can behave oddly.

2Worked Evidence

EVIDENCE A · BASIC SETUP

60% of students prefer online classes ($p = 0.6$). For samples of $n = 100$ students, find the mean and standard error of $\hat{p}$.

SOLVED

Mean: $ \mu_{\hat{p}} = p = 0.6 $

Standard error: $ \sigma_{\hat{p}} = \sqrt{\dfrac{0.6 \times 0.4}{100}} = \sqrt{\dfrac{0.24}{100}} = \sqrt{0.0024} \approx 0.049 $

Verdict: sample proportions from repeated 100-student surveys would typically land within about 4.9 percentage points of 60%.

EVIDENCE B · PROBABILITY

A coin is claimed to be fair ($p = 0.5$). In 100 flips, find $P(\hat{p} > 0.55)$ — the chance of seeing 55%+ heads just by random luck.

SOLVED

Standard error: $ \sigma_{\hat{p}} = \sqrt{\dfrac{0.5 \times 0.5}{100}} = \sqrt{0.0025} = 0.05 $

z-score: $ z = \dfrac{0.55 - 0.5}{0.05} = 1.0 $

$ P(Z > 1.0) = 1 - 0.8413 = 0.1587 $

Verdict: about a 15.9% chance — not rare at all. Seeing 55 heads out of 100 is not strong evidence the coin is rigged.

EVIDENCE C · SAMPLE SIZE EFFECT

With $p = 0.3$, compare the standard error for $n = 50$ versus $n = 200$.

SOLVED

$ n = 50: \ \sigma_{\hat{p}} = \sqrt{\dfrac{0.3 \times 0.7}{50}} = \sqrt{0.0042} \approx 0.065 $

$ n = 200: \ \sigma_{\hat{p}} = \sqrt{\dfrac{0.3 \times 0.7}{200}} = \sqrt{0.00105} \approx 0.032 $

Verdict: quadrupling the sample size ($50 \to 200$) cut the standard error roughly in half — because $\sqrt{4} = 2$. Same square-root pattern as Case 01.

CASE 03

The Central Limit Theorem

The single rule that makes all of statistical inference possible.

01The Concept, Plainly

Here's the twist in the story: the population itself can look like anything — skewed, lumpy, flat, weird — and it doesn't matter. The Central Limit Theorem (CLT) guarantees that as long as your sample size $n$ is reasonably large (a common rule of thumb is $n \geq 30$), the sampling distribution of $\bar{x}$ will look approximately normal — a smooth, symmetric bell curve — regardless of the original population's shape.

skewed population data → averaged into samples → bell-shaped pattern emerges

Why detectives love this rule: it means you never need to know the population's true shape to use normal-distribution tools on $\bar{x}$ or $\hat{p}$ — you just need a big enough sample. This is the reason confidence intervals and hypothesis tests (Units 6–9) work at all.

2Worked Evidence

EVIDENCE A · SKEWED POPULATION

Household income is heavily right-skewed with $\mu = \$65{,}000$, $\sigma = \$40{,}000$. For $n = 64$ households, describe the shape of the sampling distribution of $\bar{x}$ and find its standard error.

SOLVED

Even though individual incomes are strongly skewed, $n = 64 \geq 30$, so by the CLT, the sampling distribution of $\bar{x}$ is approximately normal.

Standard error: $ \sigma_{\bar{x}} = \dfrac{40000}{\sqrt{64}} = \dfrac{40000}{8} = 5000 $

Verdict: the raw data is lopsided, but the sample means behave like a tidy bell curve centered at $65,000.

EVIDENCE B · UNIFORM (NOT NORMAL AT ALL)

A fair die roll is uniform across 1–6 ($\mu = 3.5$, $\sigma \approx 1.71$) — flat, not bell-shaped at all. For samples of $n = 40$ rolls, describe $\bar{x}$'s distribution.

SOLVED

Standard error: $ \sigma_{\bar{x}} = \dfrac{1.71}{\sqrt{40}} \approx \dfrac{1.71}{6.32} \approx 0.27 $

Even though a single die roll is perfectly flat (uniform), $n = 40 \geq 30$ is large enough that the CLT still applies: $\bar{x}$ is approximately normal, centered at 3.5.

Verdict: this is the CLT at its most dramatic — turning a flat population into a bell curve of averages.

EVIDENCE C · WHEN YOU DON'T EVEN NEED THE CLT

If a population is already normal (like adult heights), does $\bar{x}$ need a large $n$ to be normal?

SOLVED

No — if the population is already normal, $\bar{x}$ is exactly normal for any sample size, even $n = 5$.

Verdict: the CLT's real power shows up precisely when the population is not normal — it's the safety net for messy, real-world data.

CASE 04

Standard Error

How much should you trust a single sample's answer?

01The Concept, Plainly

The standard error (SE) is just the standard deviation of a sampling distribution — how much sample statistics (means or proportions) bounce around from sample to sample. A small SE means samples are reliable and consistent; a large SE means samples are jumpy and untrustworthy.

Recap of Both Formulas $ SE_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} \qquad \qquad SE_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} $

The one relationship to memorize: SE is inversely proportional to $\sqrt{n}$. To cut the standard error in half, you don't need double the sample — you need 4 times the sample. To cut it to a third, you need 9 times the sample.

2Worked Evidence

EVIDENCE A · WORKING BACKWARD

A sample of $n = 25$ gives a standard error of 8. What sample size is needed to shrink the standard error to 4?

SOLVED

First find $\sigma$: $ 8 = \dfrac{\sigma}{\sqrt{25}} \Rightarrow \sigma = 8 \times 5 = 40 $

To halve the SE, quadruple $n$: new $n = 25 \times 4 = 100$

Check: $ \dfrac{40}{\sqrt{100}} = \dfrac{40}{10} = 4 $ ✓

EVIDENCE B · POLLING

A political poll with $n = 600$ (assume $p \approx 0.5$) has a standard error of about 0.02 (2 percentage points). How many people would be needed to bring that down to about 0.01 (1 point)?

SOLVED

Check the original: $ \sqrt{\dfrac{0.5 \times 0.5}{600}} = \sqrt{0.000417} \approx 0.0204 $ — matches.

Halving SE needs 4× the sample: $ 600 \times 4 = 2400 $

Check: $ \sqrt{\dfrac{0.25}{2400}} = \sqrt{0.000104} \approx 0.0102 $ ✓

Verdict: this is exactly why "doubling your poll size" barely tightens the margin of error — pollsters need much larger jumps for real precision gains.

EVIDENCE C · QUALITY CONTROL

A machine's output has $\sigma = 15$. As $n$ grows from 9 to 900 (a 100× increase), how does the standard error change?

SOLVED

$ n = 9: \ SE = \dfrac{15}{\sqrt{9}} = \dfrac{15}{3} = 5 $

$ n = 900: \ SE = \dfrac{15}{\sqrt{900}} = \dfrac{15}{30} = 0.5 $

Verdict: a 100× larger sample only buys a 10× smaller SE — because $\sqrt{100} = 10$. Precision is expensive, and it gets more expensive the more of it you want.

CASE CLOSED

Case Summary Board

Everything from Unit 5, pinned to one board.

✓Quick Reference

Case File	Formula	Key Idea
01 · Sample Means	$ \mu_{\bar{x}} = \mu, \ \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} $	Sample means center on the true mean; spread shrinks as $n$ grows.
02 · Sample Proportions	$ \mu_{\hat{p}} = p, \ \sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} $	Same logic, applied to yes/no data instead of measurements.
03 · Central Limit Theorem	$ n \geq 30 \Rightarrow $ approx. normal	Works no matter what the population's shape looks like.
04 · Standard Error	$ SE \propto \dfrac{1}{\sqrt{n}} $	4× the sample size is needed to cut SE in half.

Detective's Toolkit — one line per case

01 — A sample mean's own average is the population's average.

02 — A sample proportion's own average is the population's proportion.

03 — Big enough $n$ → bell curve, guaranteed, whatever the population looked like.

04 — Bigger samples are more trustworthy, but only by the square root of how much bigger.

Before you move to Unit 6: make sure you can, without looking, write both standard error formulas from memory and explain in one sentence why the Central Limit Theorem matters. If you can do that, this case is truly closed.

Case File	Formula	Key Idea
01 · Sample Means	\( \mu_{\bar{x}} = \mu, \ \sigma_{\bar{x}} = \dfrac{\sigma}{\sqrt{n}} \)	Sample means center on the true mean; spread shrinks as \(n\) grows.
02 · Sample Proportions	\( \mu_{\hat{p}} = p, \ \sigma_{\hat{p}} = \sqrt{\dfrac{p(1-p)}{n}} \)	Same logic, applied to yes/no data instead of measurements.
03 · Central Limit Theorem	\( n \geq 30 \Rightarrow \) approx. normal	Works no matter what the population's shape looks like.
04 · Standard Error	\( SE \propto \dfrac{1}{\sqrt{n}} \)	4× the sample size is needed to cut SE in half.