Imagine you want to know how much time students at a university spend studying each week. There is a perfectly straightforward way to find out: ask every single student, verify every answer, and calculate the result.

There are two problems. First, this would be expensive. Second, the answers might not be entirely accurate. What counts as studying? Does watching a lecture count? What about having the textbook open while watching television?

Welcome to statistics. The mathematics is often the easy part.

We rarely see the whole picture

The population is the complete group about which we want to know something. If our question concerns all students currently enrolled at a particular university, they constitute the population.

A sample is the subset we actually observe. We might survey 200 students and estimate something about the university’s 20,000 students.

A number summarizing the population—such as its true average study time—is a parameter. A number calculated from our sample is a statistic. Parameters are generally unknown. Statistics are what we have in our hands.

A statistic can be useful without being an exact representation of the parameter. The question is how much uncertainty accompanies it.

Two different jobs

Descriptive statistics summarize observed information. A histogram of the 200 students’ reported study hours describes that sample. So do its mean and median.

Inferential statistics use observed data to make claims beyond the observations. If we use the 200 students to estimate the average for all 20,000, we are doing inference.

That second job is harder than it looks. Which students were asked? Did they volunteer? Did everyone respond? Could the questionnaire have nudged respondents toward particular answers?

A beautifully precise average from a hopelessly biased sample is still a hopelessly biased answer.

Variation is not an inconvenience

Suppose the five students we asked report 2, 4, 6, 8, and 10 hours. Their mean is 6 hours. That does not mean anyone must study exactly 6 hours, or that the next student will.

Statistics begins by accepting that observations vary. Rather than pretending variation does not exist, we describe it and ask whether our results are stable in the presence of it.

This becomes especially important when comparing groups. A difference of two points on a test could be meaningful, irrelevant, or entirely unsurprising given the variability in scores. The difference itself cannot tell us which.

What this course will do

We will start by learning how to describe data. Then we will study randomness and sampling. Eventually we will build toward confidence intervals, hypothesis tests, regression models, and the rather remarkable problem of distinguishing patterns from coincidence.

The goal is not to memorize a table of formulas. The goal is to know which question each formula answers, what assumptions it requires, and where it can lead us astray.

Check your understanding

1. A city surveys 400 residents about transit. Identify the population, the sample, a possible parameter, and a possible statistic.
2. A news report says, “The survey found that 61% support the proposal.” Does that number necessarily describe the population?
3. Give one way that a large sample could still produce a misleading estimate.

A design problem, before any formulas

Here is a challenge you will return to in the Week 1 investigation. A university wants to know whether a statistics-preparation workshop causes students to perform better. Someone surveys the students who attended, compares their marks with those who did not, and claims that the difference is the benefit of attending.

What did the researcher actually learn?

For the people measured, the researcher may have described an association between attendance and scores. But students who attend voluntarily may differ from those who decline in prior knowledge, time, motivation, or access. We do not know what their scores would have been without attending. And even a carefully randomized experiment involving willing volunteers would not necessarily represent all university students.

These are three separate limitations:

• Description: do we correctly summarize the observations we collected?
• Causal inference: is the observed difference a credible estimate of the effect of an intervention?
• Generalization: which population could this result describe?

The point of the first week is not to learn a formula for each problem. It is to stop one kind of claim from quietly turning into another. Go to the interactive study-design workshop and predict what happens when participation and treatment selection change.

Specify the claim before collecting its denominator

A research question should identify the target population, measurement, time window, and quantity to estimate. “Do students study enough?” is underspecified. “What fraction of currently enrolled undergraduates report at least ten hours of independent study during the previous seven days?” names a measurable proportion, although the quality of self-report still needs investigation.

Suppose a university has 20,000 undergraduates and selects 200 by simple random sampling. Of 160 respondents, 96 report at least ten hours. The observed respondent proportion is $96/160=0.60$. It is not $96/200=0.48$ unless nonresponse has explicitly been coded as nonsupport for a justified target, which would answer a different question. The response rate is $160/200=0.80$. Distinguish the denominator selected, the denominator responding, and the population to which an estimate is intended to apply.

An inferential report connects question, design, evidence, estimate, uncertainty, and limitations. Each link can fail independently. More elaborate computation cannot repair an undefined outcome, a missing population, or a self-selected comparison silently described as random assignment.

Further investigation

1. Write a fully specified question about travel time to campus, including its population and time window.
2. In the example above, how many selected students did not respond, and what is the response rate?
3. Give one limitation that would remain even if every selected student responded.

We are going to do something statisticians are surprisingly prone to neglect: ask what the numbers mean before doing mathematics with them.

Suppose a hospital records a patient’s temperature as 39. Is the patient feverish? We cannot answer without knowing whether that number means degrees Celsius or Fahrenheit. And that is a relatively simple measurement problem.

A dataset is a collection of observations about units: people, events, places, machines, or anything else we can observe. A variable is an attribute measured for each unit. One person can contribute multiple variables, such as age, reported study hours, and preferred transportation.

Names are not numbers

A categorical variable puts observations into groups. A student’s major or preferred musical genre is categorical. We can count categories and calculate their proportions. Averaging majors, however, makes no sense.

A quantitative variable records numerical amounts or counts. Distance traveled, number of books read, and exam scores are quantitative.

Quantitative variables are often described as discrete (counts such as number of siblings) or continuous (measurements such as temperature, which can in principle take values across an interval). The distinction depends partly on the measurement process: a thermometer might report tenths of a degree even though temperature itself varies continuously.

Measurement scales constrain conclusions

Four names are often useful:

• Nominal: categories without an inherent ordering, such as blood type.
• Ordinal: ordered categories, such as poor, fair, good, and excellent.
• Interval: numerical differences are meaningful, but zero is not an absence of the quantity, such as temperature in Celsius.
• Ratio: differences and ratios are meaningful and zero represents an absence, such as mass.

If a respondent gives a service a rating of 4 rather than 2, we can say the rating is higher. We cannot automatically conclude the respondent liked it twice as much. The numbers might label an ordinal scale rather than measure a ratio.

Where did these data come from?

Suppose a school’s website asks visitors whether students should have longer holidays. Ten thousand people respond. That is a lot of observations, but it is not necessarily a useful estimate of what the school’s students think.

The visitors were not randomly selected from the students. They might not even be students.

This is called selection bias: the mechanism that puts people in our dataset is related to what we want to measure.

Other problems include missing responses, ambiguous questions, miscalibrated instruments, and variables that are convenient to collect but poorly represent the underlying concept.

An easy mistake: treating codes as quantities

Consider the table:

The mean study time is $17/3 \approx 5.67$ hours. The mean major code, 2, is arithmetically calculable but conceptually meaningless. Those codes are labels.

A valid operation on numbers is not necessarily a valid operation on data.

Check your understanding

1. Is a ZIP code categorical or quantitative, even though it contains digits?
2. What additional information would you need to interpret a recorded temperature of 39?
3. Why might a voluntary online poll of 100,000 people be less informative than a carefully selected survey of 1,000?

Turning a question into a defensible measurement

Suppose we ask whether tutoring improves statistics performance. Before surveying anyone, we have to choose what that means. Is it a final exam score, an independently marked common assessment, a course grade, passing the module, or confidence answering questions? Those are related but different constructs. Asking students how much they feel they learned does not directly measure their demonstrated proficiency.

Write a miniature data dictionary before collecting data:

Notice that offered and attended are different variables. If someone receives an invitation but does not attend, including them according to their randomized offer answers a different question from analyzing only people who chose to participate.

The data dictionary is not administrative decoration. It determines whether the analysis can answer the question. You will use these definitions in the Week 1 study-design studio.

The unit of observation controls the denominator

Imagine a clinic file with one row per visit. A person attending five times supplies five rows. The mean waiting time across visits targets a visit-level average; the mean of each person’s average targets a person-level quantity. If frequent visitors have different waits, the two summaries can differ even when every row is accurate. Define whether the unit is a person, visit, household, or measurement before computing a sample size.

Derived variables need operational definitions. Body mass index uses mass in kilograms divided by squared height in metres. Supplying height in centimetres without converting changes the numerical result by a factor of 10,000. A data dictionary should record units, coding, valid ranges, timing, missing-value codes, and transformations. Plausibility checks require subject-matter limits, not just a column’s software type.

Ordinal codes preserve order but not necessarily equal spacing. A shift from category 1 to 2 on a satisfaction scale need not equal a shift from 4 to 5. Means of such codes can be used under an explicitly defended measurement model, but the numeric labels alone do not establish that model. Report category distributions when the spacing assumption is uncertain.

Further practice

1. A file has 500 visits from 120 people. Which number is the count of independent people?
2. Is a postal code a quantitative variable because software stores it as an integer?
3. Specify a data-dictionary entry for travel time that makes its units and missingness unambiguous.

Someone discovers that people who drink tea tend to live longer. Should we issue everyone a kettle? Maybe. But the observation itself does not tell us whether tea made anyone live longer.

Before touching a statistical test, we need to understand how the data came into existence.

Observing is not intervening

In an observational study, researchers record exposures, characteristics, and outcomes without assigning the exposure of interest. A survey of commuting habits is observational. So is a study comparing medical outcomes among patients who independently chose different treatments.

A randomized experiment uses a chance mechanism to assign an intervention, such as treatment A or treatment B. Under a suitable design, randomization prevents investigators and participants from systematically choosing treatment groups based on pre-treatment characteristics.

Random assignment does not mean that two realized groups will have precisely the same age, income, or health. Chance still produces imbalances. Rather, those imbalances arise from a known mechanism we can use in inference.

Random sampling is a different machine

Imagine recruiting 100 volunteers and randomly assigning 50 to each intervention. The experiment may support causal inference about the volunteer population under its assumptions, but it does not automatically represent everyone in a city.

Random sampling concerns who enters the study and the population to which we can generalize. Random assignment concerns why groups differ on treatment. They serve distinct purposes.

Controls, blinding, and ethics

A control group supplies a meaningful comparator: usual practice, placebo, or another intervention. A placebo may help distinguish treatment effects from expectations, but it is not ethically acceptable to deny established effective care merely to make a cleaner experiment.

Blinding can reduce measurement or behavior changes related to knowledge of treatment assignments. When blinding is impossible, clearly measured endpoints, pre-registered outcomes, and independent assessment can still improve the study.

In medical and social research, informed consent, safety monitoring, privacy, and ethical oversight are parts of the design—not paperwork to bolt on after the results are interesting.

A small example

A school compares 20 students who enrolled in an optional tutoring program with 20 who did not. The tutoring group gains four more points on average. Maybe tutoring helped. Maybe students seeking tutoring were more motivated or had different prior preparation. A difference between means does not resolve this.

If the school randomly offered access to tutoring, that can improve causal identification of the offer’s effect. It may not identify the effect of actually attending tutoring when attendance differs from the assignment.

Check yourself

1. A random telephone survey asks about voting intentions. Is treatment randomly assigned?
2. A randomized trial recruits only university students. What population-generalization concern remains?
3. Why might measuring outcomes without knowing treatment assignment improve an experiment?

Further reading

See the experimental-design chapter of OpenStax Introductory Statistics 2e.

A model in which sampling and assignment can be separated

Consider a hypothetical university of 1,000 students. Half enter with more prior preparation, and half have less. Suppose a workshop would help both groups, but its benefit is larger for those with less preparation.

Imagine that 90% of the more-prepared students volunteer for a study but only 20% of the less-prepared students do. The volunteers are not a miniature copy of the university. Because effects differ between groups, the average treatment benefit among volunteers need not equal the benefit in the university population.

Now suppose the researchers randomize workshop access among those volunteers. This improves our ability to learn an effect among the selected participants. It does not turn the volunteers into a representative sample.

Finally, imagine that researchers instead allow students to choose whether to attend. More-prepared students may select into treatment at a different rate. The raw difference between those who attended and those who did not can combine a real treatment effect with differences in prior preparation.

These are distinct mechanisms. Random sampling addresses who we can generalize to. Random assignment addresses a causal comparison. Neither automatically supplies the other.

Work through each mechanism in the interactive study-design studio. The model reports expected counts rather than actual experimental data, so a perfectly balanced expectation is not a promise of perfectly balanced real treatment groups.

Assignment, receipt, and analysis population

Random assignment specifies the offered intervention, but participants may not receive or adhere to it. An intention-to-treat comparison follows assigned groups and estimates the effect of the assignment strategy under its design. Comparing only those who complied can reintroduce selection: people who adhere may differ in prognosis or resources. A treatment-received analysis needs additional assumptions to identify a causal effect.

Suppose 50 volunteers are assigned to each arm of a hypothetical trial. Ten assigned to treatment never attend, and five controls obtain similar services elsewhere. Discarding these 15 people destroys the original randomized comparison unless the missingness or compliance mechanism is addressed. Retaining assignment may dilute the difference relative to perfect adherence, but it preserves a clearly defined policy question about offering the treatment.

Blocking randomizes within groups such as site or baseline severity to improve balance or precision. Blinding reduces some behavior and measurement changes associated with knowledge of assignment. Neither replaces ethical consent or outcome validity. State who was randomized, the allocation unit, the implemented mechanism, and the planned analysis population before interpreting a treatment coefficient.

Further practice

1. Distinguish an allocation unit from an outcome-measurement unit in a school-randomized study.
2. Why can analyzing only adherent participants defeat randomization?
3. Does an observational study become randomized when a computer randomly orders its rows?

The spreadsheet says a patient’s temperature is 101. In Fahrenheit, that suggests a fever. In Celsius, it suggests the patient was briefly inside an industrial process.

The computer will calculate the average either way.

A mathematical operation on a column is not evidence that the column represents a coherent quantity. We have to understand the records first.

Start with the unit of observation

What does one row describe: one person, one appointment, one hospital admission, or one measurement? If a patient has five visits and another patient has one, treating six rows as six independent people changes the scientific question.

A data dictionary records what variables mean, allowed values, units, collection dates, and whether a value is a category, count, or measurement.

Missing does not mean zero

Suppose recorded incomes are 40, 50, 60, and blank. Replacing the blank with zero produces a mean of 37.5. Omitting the blank produces 50. Neither is guaranteed to estimate the intended population mean; the correct treatment depends on why the value is missing.

Unknown is not the same thing as none. Use explicit missing-value markers rather than inventing a number.

Also inspect duplicates, impossible dates, changing codes, mismatched units, and whether identifiers inadvertently reveal personal information.

A reproducible cleaning log

Keep the original source read-only. Record transformations as scripts or documented operations: which rows were excluded, which categories were recoded, which measurement conversions were made, and why.

For example, converting pounds to kilograms is reproducible if the original units and conversion are documented. Deleting “suspicious” high observations without stating the criterion is not.

A tidy-looking result can still be inaccurate if someone has quietly deleted the events that made the dataset interesting.

Example audit in R

R’s airquality example includes historical observations and missing measurements:

data(airquality, package = "datasets")
dim(airquality)
colSums(is.na(airquality))
summary(airquality)

We can inspect missingness before fitting a model. If we use na.omit(), report how many rows were dropped and which variables caused the exclusion. Our air-quality field notebook demonstrates this.

Check yourself

1. Why does replacing a missing measurement with zero alter the research claim?
2. An observation appears twice. When should it be removed, and when might it be legitimate?
3. What metadata would you require before sharing an analysis-ready dataset?

Further reading

The R documentation for the airquality dataset demonstrates why source details matter.

Demonstrate a transformation without destroying the evidence

Suppose five synthetic records report minutes as $[10,12,-99,15,1200]$, and the documented codebook says $-99$ means missing. The raw mean is 227.6, a valid arithmetic calculation of an invalidly mixed code and measurement. Converting only the documented code to missing leaves observed values $[10,12,15,1200]$ and mean 309.25. That surprising result does not authorize deleting 1200. It requires checking units, provenance, and whether the long duration is genuine.

If source verification establishes that 1200 was recorded in seconds while the other entries were minutes, a documented conversion gives $[10,12,15,20]$, mean 14.25. Preserve the raw entry, record the corrected units and reason, and keep the missing record as missing. This sequence illustrates why a cleaning log must distinguish confirmed correction from a guess made to make results look reasonable.

Duplicate detection also needs a key. Two people can share a name and outcome; one person can legitimately have repeated visits. Dropping identical-looking rows without defining record identity can erase valid observations. A reproducible pipeline should assert expected keys and retain counts at every filtering stage.

Further practice

1. Why is converting every zero to missing unsafe?
2. Which file should remain unchanged when a verified unit error is corrected?
3. What should be reported if the origin of the value 1200 cannot be resolved?

One of my favorite statistical questions is also one of the least useful without context: What is the average?

Consider five annual incomes, measured in thousands of dollars:

The arithmetic mean adds the observations and divides by their number:

The mean is the balancing point of the values. It uses every observation, which is both its strength and its vulnerability.

The median tells another story

The median is the middle value after sorting the observations. Here it is also 50.

But imagine the fifth person wins a tremendous prize. Replace 60 with 600. Our data become:

The mean is now 158. The median remains 50. Neither calculation is wrong, but they answer slightly different questions.

The mean describes a mathematical balance point and is strongly affected by extreme observations. The median describes the middle of the ordered observations and is generally more resistant to outliers.

If there are an even number of observations, the median is conventionally the average of the two middle values.

Spread: how different are the observations?

A center without a description of variation is often misleading.

The simplest measure of spread is the range:

A more sophisticated measure is the sample standard deviation, $s$:

The square root returns the answer to the original units. Dividing by $n-1$, rather than $n$, produces the familiar sample variance estimator used to estimate population variance under standard assumptions.

Standard deviation is not the average distance from the mean. It is the square root of the average squared deviation with that sample correction.

Standard deviation, like the mean, can be strongly affected by extreme values. For heavily skewed data, the median and interquartile range may give a more representative description of a typical observation.

Experiment with an outlier

Below, change the final observation. The mean, median, and sample standard deviation will update immediately.

Try setting the final value to 60, then to 600. Ask yourself which summary best represents a typical member of this tiny group, and why.

Check your understanding

1. For $2,4,6,8$, calculate the mean and median.
2. If the largest observation increases by 100, what happens to the mean of a five-observation dataset?
3. Can two datasets share the same mean but have dramatically different variability?

Why the mean and median solve different problems

For any proposed center $c$, expand $x_i-c=(x_i-\bar x)+(\bar x-c)$. Summing squared deviations gives

because centered deviations sum to zero. The mean uniquely minimizes squared error. A median instead minimizes $\sum_i|x_i-c|$; moving a proposed center across an ordered observation changes how many distances increase and decrease. For an even sample, any point between the two middle values minimizes absolute error; the midpoint is the conventional reported median. Neither center is universally correct without specifying what kind of loss or description matters.

For the original five incomes, squared deviations from 50 are $100,25,0,25,100$, summing to 250. The sample variance is $250/4=62.5$ and the standard deviation is $\sqrt{62.5}\approx7.9057$ thousand dollars. The variance has squared-income units; the standard deviation returns to income units. For a descriptive population consisting of exactly these five people, dividing by five instead gives population variance 50. The choice of denominator reflects the intended quantity, not a universal preference for one button.

The $n-1$ sample variance is unbiased for population variance under independent identically distributed finite-variance sampling. Its square root is not generally an unbiased estimator of population standard deviation. That distinction matters when interpreting the word “unbiased.”

Additional practice

1. Calculate the mean, median, sample variance, and range for $[1,2,3,10]$.
2. Add 100 to every observation. What changes? Multiply every value by $-2$ instead; how does the standard deviation change?
3. Explain which summary you would report for typical income and which for total income per person.

A good graph can make a complicated dataset understandable in seconds. A bad graph can make a perfectly ordinary dataset look terrifying.

The numbers don’t get to object. That is our job.

Count categories with a bar chart

Imagine asking 100 commuters how they get to work. Forty say car, 30 bus, 20 train, and 10 bicycle. These are categorical observations. A bar chart is appropriate: each category gets its own bar, and the height represents a count or proportion.

Bars for distinct categories typically have gaps between them. Rearranging unordered categories does not change the underlying data.

For a quantitative variable such as commute time, a categorical bar chart is usually not the right choice. We care about numerical order, distance, and the distribution across intervals.

A histogram is not simply a bar chart with touching bars

A histogram partitions quantitative values into intervals called bins. Each observation falls into one bin. For equal-width bins, we can draw bar heights proportional to counts. For unequal-width bins, the area of each bar, not merely its height, must represent frequency.

The interval boundaries matter. Are scores of exactly 20 placed in the 10–20 bin or the 20–30 bin? A graph should use a consistent convention.

Most importantly, the choice of bin width can change the story a reader sees. Too few bins can hide structure; too many can emphasize random variation.

Describe what you actually see

When reading a distribution, look for four things:

• Shape: symmetric, right-skewed, left-skewed, single-peaked, or multimodal.
• Center: where a typical observation seems to fall.
• Spread: how widely values are dispersed.
• Unusual observations: gaps, clusters, or apparent outliers that might merit investigation.

Right-skew means a longer tail toward higher values, not necessarily that most observations are high. Income and waiting-time distributions often provide intuitive examples, though specific datasets must be examined rather than assumed to follow a particular shape.

The same data, many pictures

The laboratory uses a synthetic dataset, generated for instruction. Move the number of bins without changing a single observation. Watch which details become prominent. Then examine the frequency table, which includes the numerical counts that the visual representation might obscure.

Check your understanding

1. Would you usually use a bar chart or histogram for students’ favorite subjects?
2. Why is it risky to compare raw counts from groups of very different sizes?
3. Can changing the number of histogram bins change the sample mean?

Unequal bins require density, not raw height

A histogram represents frequencies by rectangle areas. For equal-width bins, counts and heights differ only by a common scale. With unequal widths, raw count heights distort the comparison. A frequency-density height is count divided by bin width; a relative-frequency-density height additionally divides by total count, so all rectangle areas sum to one.

Suppose 20 synthetic waits contain 10 values in $[0,5)$ and 10 in $[5,15]$. Counts are equal, but the first interval is half as wide. Relative density heights are $10/(20\cdot5)=0.10$ and $10/(20\cdot10)=0.05$ per minute. Both areas are $0.5$. Equal count heights would imply twice as much area in the second interval and tell the wrong distributional story.

Specify endpoint conventions to prevent double counting. A common scheme uses left-closed, right-open intervals except the final bin includes its right endpoint. Also retain the original observations or an empirical cumulative distribution when possible: a histogram is a lossy summary whose appearance depends on width and origin.

Additional practice

1. A bin of width 4 contains 8 of 40 observations. Find its relative-frequency-density height.
2. Can two histograms from the same data have different apparent peaks?
3. Why should a time plot retain chronological ordering rather than sort observations by value?

The mean and standard deviation are useful, but they can be dramatically affected by unusual values. Sometimes it is more useful to summarize where the middle of the observations sits.

To do that, we sort the data.

Percentiles and quartiles

The $p$th percentile roughly describes the point at or below which $p$ percent of the data lie. The median is the 50th percentile.

The first quartile, $Q_1$, is the 25th percentile; the third quartile, $Q_3$, is the 75th percentile. The interquartile range is

It describes the spread of the central portion of the data. We commonly report a median and IQR together for distributions that are skewed or contain extreme observations.

There is a subtlety: for a finite sample, software packages use several different rules for determining percentiles. They can disagree, particularly for small datasets. This is not necessarily a software bug. It is a choice of definition.

One explicitly stated convention

For the laboratories on this site, we use a common linear interpolation rule (sometimes called type 7). For $n$ sorted values $x_{(1)} \le \cdots \le x_{(n)}$, estimate a quantile at probability $p$ by interpolating at zero-based position

If $h$ falls between two integers, interpolate between the corresponding adjacent observations.

For example, take

For $p=.25$, we have $h=4(.25)=1$, so $Q_1=4$. Similarly, $Q_3=8$, the median is 6, and $\mathrm{IQR}=4$.

This definition is easy to calculate consistently. Other perfectly legitimate textbook conventions may give different quartiles.

Is it an outlier?

A common exploratory rule calls observations outside the inner fences potential outliers:

For our five observations, the fences are $4-6=-2$ and $8+6=14$. No value lies beyond them.

Importantly, this is a screening convention, not a scientific definition of an error. A high observation might be completely legitimate. Deleting it automatically can introduce bias and remove the very phenomenon we hope to understand.

A boxplot often shows the median, quartiles, whiskers, and observations outside the chosen fence. Different plotting packages can also use different rules for whiskers.

What not to infer

The interquartile range does not describe every observation. A small IQR does not rule out extreme tails. And quartiles alone cannot tell us whether a distribution has one peak or two.

No single statistic describes everything about a dataset. That is why we use several summaries together.

Check your understanding

1. For the sorted data $2,4,6,8,10$, what are the median, IQR, and upper 1.5-IQR fence using this lesson’s rule?
2. Does an observation beyond the upper fence have to be a measurement error?
3. Why should we state which method produced the sample quartiles when reporting small datasets?

Interpolation and boxplot conventions

For sorted $[1,2,4,8]$, the type-7 first-quartile position is $(4-1)(0.25)=0.75$ in zero-based indexing. Interpolate 75% of the way from 1 to 2, giving $Q_1=1.75$. The third-quartile position is 2.25, yielding $Q_3=5$. Thus IQR is $3.25$, with fences $-3.125$ and $9.875$. The maximum 8 is not flagged.

A boxplot’s whiskers commonly end at the most extreme observed values within the fences, rather than at the numerical fences themselves. In this example whiskers end at 1 and 8. R’s boxplot hinges can differ from type-7 quartiles, so an independently calculated quantile table and a package boxplot can legitimately disagree unless conventions are aligned.

Quantiles are resistant to changing one extreme value because they depend mainly on order positions, but they are not immune to contamination. The median’s resistance is often summarized by an approximately 50% breakdown point in suitable large-sample settings: enough altered observations can move the center arbitrarily far. The phrase “robust” names a specific behavior under perturbation, not a guarantee that the statistic answers the right scientific question.

Additional practice

1. Calculate type-7 quartiles for $[0,4,8,12]$.
2. If the largest observation 12 becomes 120, which quartile changes under this interpolation rule?
3. Why should an outlier analysis preserve an audit of any exclusion?

A politician presents two bars. One is twice as tall as the other. The figures represented are 51% and 49%. The graphic is not necessarily numerically false. But if the vertical axis starts at 48%, it could easily leave an impression that the data do not justify.

That is the fundamental difficulty with statistical graphics: the picture is also an argument.

First, find the denominator

Suppose one town reports 400 crashes and another reports 200. The first has more recorded crashes. But suppose the first town has ten times the residents and considerably more road travel. Counts and rates answer different questions.

Before comparing groups, identify the units: crashes per resident, per mile driven, per licensed driver, or simply total events. None of these denominators is intrinsically correct for every investigation.

Scales change the impression

In a bar chart, bar length normally represents an amount. Starting the baseline above zero can exaggerate small absolute differences. In a line chart, however, a truncated axis may legitimately reveal a small change over time, provided the scale is explicit and the interpretation does not falsely imply a dramatic absolute difference.

Logarithmic axes are useful for multiplicative change: a move from 10 to 100 and from 100 to 1,000 occupies the same vertical distance on a base-ten logarithmic scale. Do not use such an axis without explaining it, and do not interpret equal visual increments as equal arithmetic differences.

Two-dimensional symbols can be even worse. Doubling both a circle’s diameter and its height quadruples its area; a visually enlarged illustration can overstate the magnitude of the data.

Context the chart cannot supply

Every useful graphic should answer four questions. What was measured? How was it collected? Which comparisons does the graphic support? And which uncertainties remain?

If it displays estimates, show uncertainty where relevant. A graph that compares two estimates without acknowledging their precision invites overconfidence. If observations are incomplete, show the missingness or state what was excluded.

The data should remain inspectable. A table and downloadable source often complement the picture.

Worked comparison

A chart compares 98 and 102 on a baseline of 95. The displayed heights are 3 and 7, making one bar appear over twice the other. But the values themselves differ by only 4/98 ≈ 4.1% relative to the smaller value. For an ordinary quantity bar chart, restoring a zero baseline removes that visual amplification.

This does not tell us whether the 4-unit difference matters scientifically. That requires measurement, study design, context, and perhaps uncertainty intervals.

Check yourself

1. Which denominator would help compare traffic injuries between two cities with different populations? Could you justify a different one?
2. Why is a zero baseline especially important for ordinary bar heights?
3. A time-series graph uses a logarithmic scale. What must the caption explain?

Further reading

ASA GAISE College recommendations emphasize context, critical communication, and real data.

Audit the mapping from numbers to ink

For a bar chart, length from the baseline encodes magnitude, so a truncated baseline can magnify a small difference. A line chart can use a narrower vertical range to reveal variation, but should label that range and avoid presenting the visual slope as a unit-free effect. Axis choice is a claim about emphasis that readers should be able to inspect.

If a circular symbol’s radius is doubled to represent a doubling in count, its area quadruples. To make area proportional to count, radius must scale with the square root of count. For values 25 and 100, radii in ratio 1:2 give areas in ratio 1:4. State what a symbol encodes rather than allowing width, area, and volume to compete silently.

An uncertainty display must name its interval: sample standard deviation, standard error, confidence interval, and prediction interval convey different quantities. Two overlapping 95% intervals for separate means do not by themselves establish that a difference is nonsignificant; the uncertainty of the difference depends on both variances and their dependence. Plot the relevant contrast and its own interval where that is the question.

Further practice

1. If area represents count, what radius ratio represents a count ratio of nine?
2. A chart reports a rise from 98 to 102. Calculate absolute and relative changes.
3. Why should a caption identify the denominator and missing records?

Suppose you measure how many hours students studied and their examination scores. If higher study time tends to accompany higher scores, the variables are associated.

Now for the critical part: that does not automatically mean the extra study time caused higher scores.

Students differ in prior preparation, difficulty of coursework, sleep, access to resources, and many other ways. Statistics can help us identify patterns. Inferring causes requires more thought.

A scatterplot preserves the pairs

A dataset containing $n$ pairs of observations $(x_i,y_i)$ can be drawn as a scatterplot. Each point represents a case.

A plot can suggest a positive association, negative association, curved relationship, clusters, or no obvious pattern. Before calculating a single coefficient, inspect the graph.

If the relationship is curved, a measure of linear association may summarize it poorly.

Pearson’s correlation

The sample Pearson correlation coefficient is

For nonconstant paired data, $r$ lies between $-1$ and $1$.

• Near $+1$ indicates a strong positive linear association.
• Near $-1$ indicates a strong negative linear association.
• Near $0$ indicates weak linear association; there might still be a pronounced nonlinear relationship.

Correlation is unchanged if we add a constant to either variable or multiply both by positive constants. It is not immune to outliers.

If either variable has zero variability, Pearson correlation is undefined. There is simply no direction of linear covariation to standardize.

The case of one surprising observation

Imagine seven observations following a clear upward trend. Now add an eighth observation very far above or below the trend.

That one observation can change the correlation dramatically. A statistic summarizing the entire sample may look convincing even when one data point exerts disproportionate influence.

Try the laboratory below. Move the last observation vertically, keeping the first seven fixed. Watch the graph and the correlation value together.

In a real analysis, this would be a reason to investigate data collection, fit diagnostics, and the scientific context—not to discard a point because the result is inconvenient.

A final warning

A correlation between ice cream sales and swimming-related emergencies might be associated with season or temperature rather than ice cream causing emergencies. An observational association alone cannot identify the correct causal explanation.

We will return to causal interpretation, confounding, and regression in the later modeling unit. For now, the most important habit is simple: look at the paired data before telling a story about the coefficient.

Check your understanding

1. If one variable is constant, what happens to Pearson’s $r$?
2. Does $r=0$ necessarily imply the variables are independent?
3. If moving one point greatly changes $r$, what should the analyst do next?

Centering, scaling, and the limits of a single number

Define $S_{xx}=\sum(x_i-\bar x)^2$, $S_{yy}=\sum(y_i-\bar y)^2$, and $S_{xy}=\sum(x_i-\bar x)(y_i-\bar y)$. Pearson’s correlation is $r=S_{xy}/\sqrt{S_{xx}S_{yy}}$, provided both sums of squares are positive. Cauchy–Schwarz gives $|r|\leq1$. If either variable is constant, the denominator is zero and correlation is undefined, not zero.

For pairs $(-1,1),(0,0),(1,1)$, the response is exactly $y=x^2$. Yet $S_{xy}=0$, so $r=0$. Zero linear correlation does not imply absence of a relationship. A scatterplot should precede or accompany the coefficient.

Adding a constant to a variable leaves correlation unchanged. Multiplying one variable by a positive constant leaves it unchanged; multiplying by a negative constant reverses its sign. Thus changing centimetres to metres does not change $r$, while reversing a scale’s orientation can. The regression slope changes magnitude under unit changes because it has response-per-predictor units; correlation is dimensionless.

Additional practice

1. Compute $r$ for $(1,2),(2,4),(3,6)$.
2. What is $r$ if every response in those pairs is negated?
3. Why can a single influential point increase or decrease $r$ dramatically in a small sample?

Imagine a town with 100,000 residents. We want to know what fraction supports a new park. We cannot ask everyone, so we select a thousand people.

An intelligent question is not just how many people did we ask? It is how did those people end up in our sample?

Random sampling and the big idea

In a simple random sample of size $n$, every possible group of $n$ members of the population has an equal chance of being selected.

This does not guarantee that our sample is perfectly representative. By chance alone, it could contain more supporters than opponents. But it gives us a defined random mechanism with which to reason about uncertainty.

If we repeatedly draw random samples, their statistics will differ. This is sampling variability. It remains even when the selection process is impeccable.

Bias is a different problem

Now imagine taking our survey exclusively outside a meeting of people campaigning for the park. Even if we survey ten thousand attendees, we should not expect their answers to represent the town.

That is selection bias. It is not cured simply by collecting more of the wrong observations.

Another concern is nonresponse bias. Even if we select residents randomly, people who do not respond may differ systematically from those who do.

Bias and random error are different concepts. Random sampling error tends to diminish with larger samples, under appropriate conditions. Systematic bias need not.

Try the accompanying response-selection laboratory. The town’s opinions stay fixed. Only the relative probability that two groups respond changes, shifting the expected survey percentage without changing the population percentage. The display reports mathematical expectations, not a simulated or real opinion poll.

Why the square root keeps appearing

For independent, identically distributed observations with finite population variance $\sigma^2$, the sample mean has standard error

Doubling a sample does not halve its standard error. To halve the standard error, we generally need four times as many observations.

That relationship is crucial to understanding surveys, clinical studies, and other evidence-based claims. There can be enormous costs associated with modest improvements in precision.

Real survey designs often require adjustments for clustering, weighting, finite populations, and other departures from the ideal model.

Correlation is not randomization

Random sampling helps us learn about a population. Random assignment in an experiment helps us compare interventions by breaking systematic relationships between treatment assignment and potential outcomes.

These ideas are related, but they solve different problems. We can have an experiment with random assignment and a convenience sample, or a representative survey with no experimental intervention.

Check your understanding

1. What happens to the standard error of a sample mean if $n$ increases from 100 to 400?
2. Can an enormous convenience sample have a very small calculated standard error and still be seriously biased?
3. What is the difference between randomly selecting study participants and randomly assigning treatments?

Separate variance, bias, and mean squared error

For an estimator $T$ of a fixed parameter $\theta$, its bias is $E[T]-\theta$ under a specified repetition model. Expanding the square around $E[T]$ gives

Larger independent samples can reduce variance while leaving a recruitment bias untouched. A precise estimate of the wrong target can therefore have substantial mean squared error. “Unbiased” is a repeated-procedure property, not a statement that every realized sample estimate is correct.

For simple random sampling without replacement from a finite population of size $N$, an estimated mean’s variance includes a finite-population correction. Using population variance defined with denominator $N-1$, it is $(1-n/N)S^2/n$. The corresponding standard-error factor is $\sqrt{1-n/N}$. Sampling nearly everyone leaves little sampling uncertainty, although measurement error and nonresponse can remain. The familiar independent-with-replacement expression is a useful approximation when the sampling fraction is small.

Additional practice

1. Estimator A has variance 4 and bias 3; estimator B has variance 10 and zero bias. Compare mean squared errors.
2. Under an independent mean model, how much must sample size increase to reduce standard error by a factor of five?
3. If half a finite population is sampled without replacement, what is the standard-error correction relative to the corresponding with-replacement formula using $S^2$?

Suppose we want to survey a school district with forty schools. It would be convenient to visit two schools, interview everyone there, and call the result a district survey.

That might work for some questions. But the students in a school often resemble one another more than they resemble students at another school. Convenience is not a sampling design.

Simple random and stratified samples

In a simple random sample of size $n$, every set of $n$ population units has the same chance of selection.

In a stratified sample, we divide the population into defined groups (strata) and draw probability samples separately from each. If the strata are meaningful—say, elementary versus high schools—this can ensure each is represented and may improve precision.

However, oversampling one stratum means its raw sample percentage cannot necessarily be used as the overall population percentage. Appropriate design weights may be needed.

Cluster sampling

A cluster sample selects groups, such as schools, then studies all or some units inside the selected groups. This can be much cheaper than visiting hundreds of dispersed locations.

But outcomes within one cluster may be correlated. A sample of 500 students from two schools does not necessarily supply as much independent information as 500 students sampled across the district.

Under common approximations for equally sized clusters, the design effect can be related to cluster size $m$ and intracluster correlation $\rho$:

For $m=25$ and $\rho=.10$, this approximation gives $1+24(.10)=3.4$. The standard error can be roughly $\sqrt{3.4}$ times an independent-sample calculation under the corresponding conditions.

This is a teaching approximation, not a universal survey-analysis formula.

Systematic selection and nonresponse

Selecting every tenth name from an ordered list can be useful when the starting point is randomized. But if the list has periodic structure, systematic sampling can produce unexpected bias.

Even a beautifully designed probability sample faces nonresponse. People may answer at different rates, and missing respondents may differ in the outcome itself. We explored that in the selection bias laboratory.

What a useful survey report must say

Describe the target population, frame, sample design, selection probabilities, response rate, weighting, missingness, and uncertainty method. “Surveyed 5,000 people” does not tell us whether those people are representative.

Check yourself

1. What is one advantage of stratifying by school level?
2. Why might two schools with 500 respondents require a different variance calculation than a simple random sample of 500 students?
3. What is the approximate design effect when $m=11$ and $\rho=.05$ under the simplified formula?

Weights change the target of a sample mean

Suppose a target population is 80% group A and 20% group B, but a stratified sample deliberately selects 50 from each. If observed group means are 10 and 30, the unweighted mean is 20. The population-composition estimate is $0.8(10)+0.2(30)=14$. Both summarize numbers correctly, but they target different mixtures. Report the population weights, their source, and the sampling design.

For independent simple random samples within strata, an estimated variance of $\hat\mu=\sum_hW_h\bar x_h$ is $\sum_hW_h^2(1-n_h/N_h)s_h^2/n_h$, where $W_h=N_h/N$ and the finite-population correction reflects sampling without replacement. Neglecting the correction may be reasonable when fractions are small, but ignoring weights and design altogether is a different and often consequential approximation.

For roughly equal cluster size $m$ and intracluster correlation $\rho$, the approximation $DEFF=1+(m-1)\rho$ compares variance with an independent-sample reference. If 100 observations have design effect 2.5, a rough effective sample size is 40 and the standard-error inflation is $\sqrt{2.5}\approx1.581$. This is a planning approximation, not a universal formula for arbitrary weights, unequal clusters, or all statistics.

Inverse-probability weights require meaningful inclusion probabilities and an appropriate sampling frame. Calibration to known margins cannot guarantee correction for every unmeasured nonresponse mechanism. Extreme weights can increase variance, so a weighted estimate needs an uncertainty calculation that respects its construction.

Further problems

1. A population is 30% A and 70% B, with estimated stratum means 5 and 15. Find the composition-weighted mean.
2. Under the equal-cluster approximation, find design effect for $m=10$, $\rho=0.2$.
3. Why are many pupils from two classrooms not equivalent to the same number from many independently selected classrooms?

Suppose we toss a coin and it lands heads. What was the probability of that outcome?

You might be tempted to say 50 percent. But that answer rests on an assumption: the coin is fair, it was tossed in a particular way, and our model treats the toss as random. A heavily weighted coin would not have the same probabilities.

Probability is a mathematical description of uncertainty under specified assumptions. It is not a guarantee that reality will behave exactly as expected over a handful of observations.

What counts as an outcome?

A sample space, often written $\Omega$, lists the possible outcomes of an experiment. For a single coin toss, $\Omega=\{H,T\}$. An event is a collection of outcomes. In a toss of two coins, the event “at least one head” consists of $HH$, $HT$, and $TH$.

For an event $A$, we assign a probability $P(A)$ satisfying

For mutually exclusive events, the probability of their union is the sum of their probabilities.

Under our fair-coin model, all four two-toss outcomes are equally likely, so the probability of at least one head is $3/4$.

But notice what made the calculation possible: the model assigning equal probability to the four outcomes.

Complements are often easier

The complement of event $A$, written $A^c$, is everything in the sample space that is not in $A$.

For example, if each of five independent tosses has a one-half chance of tails, the probability of at least one head is easiest to obtain by subtracting the probability of all tails:

This does not mean that five tosses are guaranteed to include a head. It means the all-tails outcome has probability $1/32$ under the model.

Addition and overlap

For arbitrary events $A$ and $B$,

That subtraction prevents us from counting the intersection twice.

If 40 percent of a group likes tea, 35 percent likes coffee, and 15 percent likes both, then 60 percent likes tea or coffee (or both):

Independent is not mutually exclusive

Two events are independent if knowing that one occurred does not change the probability of the other:

Two events are mutually exclusive if they cannot both happen:

These are very different ideas. Drawing a heart and drawing a spade on a single card draw are mutually exclusive, not independent. If one happened, the other definitely did not.

Repeated independent fair-coin tosses are independent by the model’s assumptions, but outcomes on different tosses can both be heads.

Exercise

1. In three independent fair-coin tosses, what is the probability of at least one tail?
2. Events $A$ and $B$ have probabilities .6 and .3, with overlap .2. Find $P(A\cup B)$.
3. Can two events with positive probability be both independent and mutually exclusive?

Read further

The free OpenStax probability chapter provides additional examples and exercises.

State the probability model before using its rules

A sample space lists possible outcomes; events are subsets of that space; a probability assignment is nonnegative, totals one, and adds over disjoint events. For a finite equally likely sample space, probability is favorable-count divided by total-count. Equal likelihood is an assumption, not something established by naming outcomes.

For synthetic events A and B with probabilities 0.4 and 0.5 and intersection probability 0.1, inclusion–exclusion gives $P(A\cup B)=0.8$. The chance neither occurs is $0.2$. The conditional probability $P(A\mid B)=0.1/0.5=0.2$, so these events are not independent. Simply multiplying 0.4 by 0.5 would impose a different intersection model.

For independent repetitions of an event with probability $p$, the chance of at least one occurrence in $n$ trials is $1-(1-p)^n$. With $p=0.1$ and $n=3$, it is $0.271$, not $0.3$. Adding the single-event probabilities double-counts outcomes containing multiple occurrences. Without independence, the complement formula still holds but the product for the no-occurrence probability may fail.

Additional practice

1. If $P(A)=0.6$, $P(B)=0.5$, and $P(A\cap B)=0.3$, calculate union probability and test independence.
2. Can two positive-probability mutually exclusive events be independent?
3. Give a model assumption needed to interpret a coin probability as a long-run experimental frequency.

Twenty coin tosses have a million or so possible sequences. Yet when we ask how many contain exactly ten heads, counting each sequence by hand becomes an irritating way to spend an afternoon.

Fortunately, mathematicians noticed this problem before we did.

Multiply when decisions happen in stages

If there are three shirt choices and four independent trouser choices, there are $3\times4=12$ outfits. The multiplication principle counts a sequence of decisions when each step has a specified number of options.

If repetition is allowed, selecting a four-digit PIN with digits 0–9 gives $10^4=10,000$ possible strings. If digits must be distinct, the count is $10\times9\times8\times7=5,040$.

These count ordered selections. The sequence 1234 is not the same PIN as 4321.

Order sometimes does not matter

A committee containing Alice and Bob is the same committee regardless of which name we write first. Selecting $k$ people from $n$ without caring about order is counted by a combination:

A factorial $n!$ is the product of positive integers through $n$, with $0!=1$.

For eight volunteers selecting a four-person experimental group,

That number was the entire reference distribution for our exact randomization laboratory.

Return to the coin

Each sequence of 20 independent fair tosses has probability $2^{-20}$. For exactly ten heads, the heads can occupy any choice of ten of the twenty positions. Therefore,

The probability calculation is valid under the fair independent-toss model. It would be a mistake to use equal sequence probabilities for a biased coin without modification.

Counting isn’t always the best tool

When events overlap, naive addition double-counts. With sampling without replacement, probabilities change after each draw. Sometimes conditional probability or simulation is more direct than heroic combinatorics.

Counting rules serve a model; they do not excuse us from specifying the model.

Check yourself

1. How many three-letter codes can be made from A–Z if repetition is allowed?
2. In how many ways can three people be chosen from seven when order doesn’t matter?
3. Why is $\binom{10}{3}$ relevant to the probability of exactly three heads in ten fair tosses?

Derive combinations and distinguish sampling models

For $k$ distinct selections from $n$ objects, ordered selection without replacement gives $n(n-1)\cdots(n-k+1)=n!/(n-k)!$ possibilities. Each unordered subset appears in $k!$ different orders. Dividing gives $\binom nk$. The reasoning, not the factorial button, determines when the formula applies.

Suppose a fictional box contains 5 red and 7 blue counters. Draw 4 without replacement. Every four-counter subset is equally likely under simple random drawing, and exactly two red counters can be chosen in $\binom52\binom72$ ways. Therefore

This is a hypergeometric calculation. A binomial model with success probability $5/12$ would instead assume replacement or independent trials with constant success probability. Without replacement, the chance of a red counter on the next draw changes with what has already been removed. Independence is a probability assumption; the ability to count outcomes does not create it.

For a binomial count, $\binom nk$ counts the possible positions of $k$ successes in $n$ trials. Each particular sequence has probability $p^k(1-p)^{n-k}$ under independent constant-probability trials. Multiplying gives the binomial mass function. “There are two outcomes” does not imply they are equally probable, nor does it establish independence.

Further problems

1. How many ordered three-letter strings can be formed from five distinct letters without repetition? How many unordered three-letter subsets?
2. In the counter example, calculate the probability all four are blue.
3. Why does shuffling an existing sample not create new independent participants?

A survey finds that 60% of residents use public transit. Of transit users, 30% commute before 7 a.m.; of nonusers, only 10% do. What fraction of all residents commute that early?

The answer is not 30% and not 10%. We must weight both subgroups.

Conditional probabilities live inside denominators

If $T$ denotes transit use and $E$ denotes early commuting, then

It means among transit users, what fraction commute early? Changing what we condition on changes the denominator and therefore the question.

Weighted branches

Partition the population by transit use:

Using the fictional percentages gives

So 22% commute early. A probability tree makes the same calculation visible: multiply along branches, then add disjoint branches ending in the event we care about.

This is called the law of total probability.

Check independence

Two events $A$ and $B$ are independent when

when conditioning is defined. Equivalently, $P(A\cap B)=P(A)P(B)$.

In our example, $P(E\mid T)=.30$ but $P(E)=.22$. The events are not independent. Knowing transit use changes the probability of early commuting.

Compare this with mutually exclusive events, which cannot happen together. A person cannot be simultaneously classified in two nonoverlapping transit categories, but the category and an early commute can certainly co-occur.

Why this matters for Bayesian reasoning

Bayes’ theorem reverses a conditional question by using the probability of the evidence across possible explanations. The base-rate laboratory is a probability tree in disguise: disease/no disease branches, followed by positive/negative test branches.

Always label which conditional probability each percentage represents. Sensitivity is not positive predictive value.

Check yourself

1. Find the probability that a resident both uses transit and commutes early.
2. If everyone commuted early at the same rate regardless of transit, what equality would hold?
3. Why can you add $P(E\cap T)$ and $P(E\cap T^c)$ without subtracting an overlap?

Build the entire joint table

For 1,000 hypothetical residents, the stated percentages correspond to the following expected counts:

Thus $P(T\mid E)=180/220=9/11\approx0.81818$, whereas $P(E\mid T)=180/600=0.30$. Conditioning in the other direction changes the denominator. The joint probability $0.18$ is the same intersection whichever event is named first, but the conditional probabilities are not interchangeable.

For a partition $B_1,\ldots,B_k$ with positive probabilities, $P(A)=\sum_iP(A\mid B_i)P(B_i)$. This follows by splitting $A$ into disjoint intersections $A\cap B_i$ and applying the multiplication rule. Bayes then divides one of those intersections by their total: $P(B_j\mid A)=P(A\mid B_j)P(B_j)/P(A)$ when $P(A)>0$.

Pairwise independence is weaker than mutual independence for three or more events. Toss two fair independent coins. Let A mean the first is heads, B the second is heads, and C mean the two agree. Each pair is independent, yet $P(A\cap B\cap C)=1/4$, not $(1/2)^3$. Conditioning on two events can reveal a dependence invisible in pairwise checks.

Further problems

1. From the table, compute $P(T\mid E^c)$.
2. If $P(A)=0.4$, $P(B)=0.5$, and $P(A\cap B)=0.2$, are A and B independent? Mutually exclusive?
3. Why is conditioning on an event of probability zero not defined by the elementary ratio formula?

A medical screening test is described as “90 percent accurate.” Is a positive result a 90 percent chance that you have the condition?

Not necessarily. In fact, that inference can be wrong by a remarkable amount. The trouble is that there are two different conditional probabilities, and reversing them changes the question.

We will use a fictional test and hypothetical numbers rather than make any claims about a real disease.

Conditional probability changes the sample space

For events $A$ and $B$ with $P(B)>0$,

Read this as “the probability of $A$, given that $B$ occurred.”

A conditional probability does not simply tell us that two things are related. It says we are restricting attention to cases where the condition is known to hold.

For a diagnostic test, consider two distinct probabilities:

• Sensitivity: $P(+\mid D)$, the probability of a positive test given that the condition is present.
• Positive predictive value: $P(D\mid +)$, the probability of the condition given a positive test.

These are not the same fraction.

Enter Bayes

Bayes’ theorem gives us a way to reverse a conditional probability while accounting for how common the underlying condition is:

Here $P(D)$ is prevalence, and $P(+\mid D^c)$ is the false-positive rate, which equals $1-\text{specificity}$.

Suppose the hypothetical condition affects 1 percent of people. The fictional test has 90 percent sensitivity and 95 percent specificity.

Out of 10,000 screened people, we expect approximately 100 to have the condition. Around 90 of those would test positive. Of the 9,900 without the condition, about 495 would test positive anyway.

Therefore, among expected positive results,

Just 15.4 percent, not 90 percent.

The test hasn’t failed to achieve the assumed sensitivity or specificity. The low prevalence means that even a relatively small false-positive rate generates many positives among the much larger group without the condition.

Why the base rate matters

This is the base-rate problem. The same evidence can carry a different meaning in populations where the underlying condition is more or less common.

Change the prevalence in the laboratory below while leaving sensitivity and specificity untouched. Notice how the positive predictive value moves.

This is equally relevant to fraud detection, spam filtering, equipment fault monitoring, and other settings where we attempt to detect rare events.

What this example does not imply

The example does not recommend screening, diagnosis, or treatment for anyone. Real tests have different characteristics in different populations and settings. Sampling, verification, selection, and measurement also matter.

The lesson is narrower and more general: when interpreting evidence, remember how probable the underlying event was before the evidence arrived.

Exercise

1. For a hypothetical test with 80 percent sensitivity, 90 percent specificity, and 10 percent prevalence, find $P(D\mid +)$.
2. Is a 99 percent specificity equivalent to a 99 percent positive predictive value?
3. Why might the same positive test have different predictive values in two populations?

Read further

See the probability material in OpenStax Introductory Statistics 2e and the current ASA GAISE teaching recommendations.

Likelihood ratios update odds

For condition D and evidence E, Bayes can be written

The last factor is the likelihood ratio. For a positive screening result with sensitivity 0.90 and specificity 0.95, it is $0.90/0.05=18$. At prevalence 0.01, prior odds are $1/99$, so posterior odds are $18/99=2/11$ and posterior probability is $(2/11)/(1+2/11)=2/13\approx0.153846$.

A hypothetical population of 10,000 makes the denominators tangible: 100 have the condition, with 90 expected positive results. Among 9,900 without it, 495 expected results are positive. Thus 90 of 585 expected positives correspond to the condition. These are model-based expected counts, not guaranteed realized counts or a report about a particular medical test.

Repeated positive results cannot automatically be treated as independent evidence. Multiplying the same likelihood ratio twice assumes the second test is conditionally independent of the first given condition status, or otherwise has the specified conditional evidence ratio. Shared specimen contamination or correlated errors can violate that assumption. An apparently stronger posterior can therefore be a consequence of an unjustified dependence model.

Additional practice

1. With prevalence 0.10 and the same sensitivity and specificity, calculate positive predictive value.
2. If prior odds are $1/4$ and the evidence likelihood ratio is 3, calculate posterior probability.
3. Why is this worked probability model not individualized medical advice?

You are offered a game. With probability one-half you win 10 dollars; otherwise you lose 8 dollars. Would you play?

The average outcome over many plays is positive. But that does not mean you’ll win one dollar on your next attempt.

Expected value is probability-weighted

For discrete outcomes $x_i$ with probabilities $p_i$,

For this game,

The expected gain is one dollar per play. In a single play the gain is either 10 dollars or a loss of 8 dollars; the expected value is not even a possible outcome.

Variance describes the spread

Expectation alone tells us little about risk. A game that always pays one dollar and a game that pays 1,001 dollars or loses 999 dollars with equal probabilities both have expectation one dollar. Their variability is rather different.

The population variance of a discrete random variable is

For the first game with mean $1$, deviations are $9$ and $-9$. Variance is $81$ and standard deviation is $9$.

A useful computational shortcut is $\operatorname{Var}(X)=E[X^2]-E[X]^2$.

Units and linear transformations

For constants $a$ and $b$,

Adding ten dollars changes the mean by ten but leaves the spread unchanged. Multiplying dollar amounts by two doubles the standard deviation and quadruples the variance.

If $X$ and $Y$ are independent, the variance of their sum is the sum of their variances. Without independence, a covariance term is required. Don’t silently assume independence because it makes a formula convenient.

A binomial connection

For $X\sim\mathrm{Binomial}(n,p)$, expectation is $np$ and variance is $np(1-p)$, derived from a sum of independent Bernoulli indicators.

Check yourself

1. If $X$ equals 0 with probability .75 and 8 with probability .25, find $E[X]$.
2. Find the variance and standard deviation of that variable.
3. If all values are multiplied by three, what happens to the variance?

Derive the variance identities

Expanding the centered square gives $E[(X-\mu)^2]=E[X^2]-2\mu E[X]+\mu^2=E[X^2]-\mu^2$, assuming the second moment is finite. This identity is algebraically convenient, but direct subtraction of two nearly equal large values can lose floating-point accuracy. Computing centered deviations is often numerically safer.

Linearity of expectation does not require independence: $E[X+Y]=E[X]+E[Y]$ whenever the expectations are defined appropriately. Variance does require attention to dependence:

For a difference the covariance term has a minus sign. If $Y=X$, the sum is $2X$ and its variance is $4\operatorname{Var}(X)$, not twice the variance. Independence is sufficient for zero covariance when moments exist, but zero covariance need not imply independence. For example, a symmetric variable $X$ and $Y=X^2$ can have zero covariance despite a deterministic relationship.

A Bernoulli indicator $I$ equals one for success and zero otherwise. Since $I^2=I$, $E[I]=p$ and $\operatorname{Var}(I)=p-p^2$. A binomial count is the sum of $n$ independent indicators with the same success probability, giving mean $np$ and variance $np(1-p)$. For the sample proportion $X/n$, the variance is $p(1-p)/n$. This derivation explains the square-root sample-size denominator in proportion standard errors.

Further problems and solutions

1. If $E[X]=3$ and $\operatorname{Var}(X)=4$, find the mean and variance of $10-2X$.
2. If $X,Y$ each have variance 4 and covariance 3, find the variance of their difference.
3. Why can expected profit alone be insufficient for comparing two investments or games?

Suppose we toss a fair coin twenty times and count the number of heads. There is no mystery about the procedure, but the count is not fixed in advance.

That count is a random variable: a numerical function of an uncertain outcome. The entire set of probabilities assigned to its possible values is its probability distribution.

Discrete random variables

Let $X$ be the number of heads in twenty independent, fair coin tosses. The possible values are $0,1,\ldots,20$.

A probability mass function lists $P(X=k)$ for each possible $k$. Every entry is nonnegative, and all the probabilities sum to one.

The count is not equally likely to be 0 or 10. There is exactly one way to get 20 heads, but many different sequences with exactly 10 heads.

The binomial distribution

A binomial model applies when we have:

1. A fixed number $n$ of trials.
2. Exactly two relevant outcomes per trial, often called success and failure.
3. A constant probability $p$ of success for each trial.
4. Independent trials.

Then $X$, the number of successes, follows a binomial distribution:

The term $\binom nk$ is the number of ways to arrange $k$ successes among $n$ trials.

For a fair coin, $p=.5$. With two tosses, the probability of exactly one head is

If you count the sequences $HT$ and $TH$, you get exactly the same answer.

Expected value isn’t necessarily a possible outcome

For a binomial random variable,

The expected number of heads in twenty fair tosses is $20(.5)=10$, and its standard deviation is $\sqrt{20(.5)(.5)}=\sqrt5\approx2.236$.

In one trial of a different experiment, the expected value might be 2.6 successes. That does not mean 2.6 successes can actually occur. An expectation is a long-run weighted average, not a promise about a single experiment.

Real data can violate the model

Imagine surveying ten people who all live in the same household. Their answers may be correlated. If we pretend their responses are independent Bernoulli trials, the simple binomial model may badly misrepresent the uncertainty.

Likewise, changing success probabilities from person to person violate the constant-$p$ assumption of the basic model.

Those are not small print. They are reasons to ask whether a probability model is suitable before applying it.

From probability to inference

Once we specify a binomial model, we can ask: If the coin really is fair, how surprising would a count of 18 heads in 20 tosses be?

That’s the doorway into hypothesis testing. The exact binomial test laboratory lets you explore that question. Unlike a normal approximation, it calculates the finite-sample probabilities directly.

Exercise

1. Give the four conditions for a binomial count.
2. What is the probability of exactly 2 heads in 3 independent fair tosses?
3. If $X\sim\operatorname{Binomial}(40,.25)$, what are its mean and variance?

Read further

See OpenStax: Discrete Random Variables.

Derive and compute the distribution

Let $I_i$ indicate success on trial $i$. Under independent identical Bernoulli trials, $X=\sum_iI_i$ and

For $n=5$ and $p=0.2$, exactly two successes has probability $10(0.2)^2(0.8)^3=0.2048$. At least one has probability $1-0.8^5=0.67232$. The expected count is 1 and standard deviation $\sqrt{0.8}\approx0.89443$. A count mean and a single count outcome are different objects, even when their numerical values happen to coincide.

If probabilities differ by trial while trials remain independent, the count generally follows a Poisson-binomial distribution, not this binomial formula. Sampling without replacement from a small finite population instead calls for a hypergeometric model. Dependence or changing probability must be examined before selecting a familiar mass function.

For large $n$, direct factorial arithmetic can overflow. Software works with log probabilities or stable recurrences. A numerical display of zero can mean underflow rather than an impossible event. In R, dbinom(k,n,p,log=TRUE) retains log probabilities, while pbinom computes cumulative probabilities without requiring manual summation of rounded masses.

Additional practice

1. For $X\sim\mathrm{Binomial}(4,0.5)$, calculate $P(X=2)$ and $P(X\geq3)$.
2. Calculate mean and variance of $X/4$ in that example.
3. Why does a basketball player’s number of successes not necessarily satisfy independent constant-probability assumptions?

A network operations desk averages two incident tickets an hour. What is the probability that the next hour brings no tickets?

The tempting answer is “not very likely.” Statistics demands something a little more specific.

A count model

A Poisson random variable with rate parameter $\lambda>0$ has probability mass function

Both its mean and variance are $\lambda$. For a one-hour observation window with an average of two events, $\lambda=2$.

The probability of zero events is

That is about 13.5%, under the stated model. The average of two events does not mean every hour contains two.

Exposure changes the expected count

If the event rate is two per hour and we assume it remains constant through time, a half-hour window has $\lambda=1$, while a three-hour window has $\lambda=6$.

This rate-per-exposure interpretation is especially important when comparing counts from different periods or populations. A hospital with twice as many patient-days may record twice as many incidents even if its incidence rate is unchanged.

Where the model can fail

A Poisson process assumes a particular kind of randomly occurring event structure, including a stable rate and independent increments in its simplest form.

Network incidents may cluster during outages. Injuries may share environmental causes. Data recorded in short or long windows may have variance much larger than their mean—a phenomenon called overdispersion relative to the Poisson model.

When a dataset’s observed variance greatly exceeds its observed mean, the basic Poisson model deserves scrutiny. Negative-binomial models or explicit clustered processes may be more realistic.

Poisson as a binomial limit

A binomial count with many trials and a small success probability approaches a Poisson distribution if $np\to\lambda$ appropriately. That makes the Poisson a useful approximation for rare, independent events across many opportunities—but only when the assumptions resemble reality.

Check yourself

1. If $X\sim\operatorname{Poisson}(3)$, calculate $P(X=0)$.
2. A process averages four calls per hour. What is $\lambda$ for 15 minutes under constant rate?
3. Why might a spate of correlated outages make a simple Poisson model poor?

Use base R dpois(0,3) to verify the first numerical result.

A process model and a count distribution are different claims

A homogeneous Poisson process assumes independent counts in disjoint time windows, a constant event rate, and appropriately rare single events over very short intervals. It produces a Poisson count in a window of length $t$ with parameter $\lambda=rt$. Observing one histogram resembling a Poisson mass function does not establish independent increments or a constant rate over time.

For $r=4$ calls per hour and a 15-minute window, $\lambda=1$. The probability of at least two calls is

For a full hour it is $1-e^{-4}(1+4)\approx0.90842$. Changing the exposure changes the entire probability distribution, not just the units printed beside the parameter.

If independent Poisson counts have parameters $\lambda_1$ and $\lambda_2$, their sum is Poisson with parameter $\lambda_1+\lambda_2$. A direct convolution proof expands $P(X+Y=k)$ and applies the binomial theorem to $\sum_j\lambda_1^j\lambda_2^{k-j}/[j!(k-j)!]$. Dependence invalidates this simple sum result.

For independent equal-exposure counts, a sample variance substantially greater than the sample mean can signal overdispersion relative to a simple Poisson model. Clustering, changing rates, heterogeneous exposure, and excess zeros can all contribute. A diagnostic signal prompts model investigation; it is not an automatic proof of any one alternative.

Further problems

1. A constant-rate model averages 3 events per hour. Find the expected count over 20 minutes and the chance of no event.
2. If two independent Poisson streams have rates 2 and 5 per hour, what is the one-hour combined distribution?
3. Why should days with different observation durations not be compared by raw event counts alone?

A probability density can be taller than one. This surprises people who were just told that probabilities lie between zero and one.

There is no contradiction: a density is not a probability. It is a function whose area over a range gives a probability.

Probability lives in an area

For a continuous random variable with density $f$,

The total area is one. The probability of one exact point is zero under an absolutely continuous distribution, even at the point where the density is tallest.

For a uniform variable between 0 and 2, $f(x)=1/2$ inside that interval. The probability of falling between .5 and 1.5 is the width 1 multiplied by density .5, giving .5.

A perfectly flat model

For $X\sim U(a,b)$, each equal-width interval inside the range has equal probability. The density is $1/(b-a)$.

Expected value is $(a+b)/2$ and variance is $(b-a)^2/12$. Uniform randomness is useful as a model and as a simulation ingredient, but many real physical measurements are anything but uniform.

Waiting for an event

In a simple Poisson process with rate $\lambda$ events per unit time, the time $T$ until the next event has an exponential distribution:

If the event rate is two per hour, the chance of waiting more than half an hour is $e^{-1}\approx.3679$.

The exponential model is memoryless: conditional on waiting already, the remaining waiting-time distribution has the same form. This is a strong modeling assumption, not a universal law of waiting rooms.

Normality is one model among many

A normal distribution also has a density, but unlike a uniform or exponential density, it has a symmetric bell shape across the real line.

We should decide which model to use from the science and observed patterns, not because one curve is easier to integrate or has appeared on more classroom slides.

Check yourself

1. What is $P(.2<X<.7)$ when $X\sim U(0,1)$?
2. Why is $P(X=1)=0$ for a continuous model not saying that the value can never be observed?
3. For an exponential waiting time with rate 3/hour, calculate $P(T>1\text{ hour})$.

A density is not a point probability

A probability density is nonnegative and integrates to one over its domain. Its units are inverse units of the variable, so multiplying by a small interval width produces a dimensionless approximate probability. A density can exceed one: the uniform density on $[0,0.2]$ is 5, yet its total area is exactly one. For a continuous variable, $P(X=x)=0$ at each single point. Zero point probabilities do not imply zero probability for an interval containing uncountably many points.

For $X\sim U(a,b)$, $E[X]=(a+b)/2$ and $\operatorname{Var}(X)=(b-a)^2/12$. These follow by integrating $xf(x)$ and $(x-E[X])^2f(x)$. For a uniform variable on $[2,8]$, the mean is 5 and variance 3. The chance of lying in $[3,6]$ is its length divided by total length, $3/6=1/2$.

For exponential waiting time with rate $r>0$, the survival function is $P(T>t)=e^{-rt}$ for $t\geq0$. Consequently

This is the memoryless property. It is a model implication, not a universal description of waiting. Aging equipment and scheduled service intervals usually require different assumptions. The mean is $1/r$ and variance $1/r^2$; doubling the rate halves expected waiting time.

The cumulative distribution function $F(x)=P(X\leq x)$ gives interval probabilities by $F(b)-F(a)$ for $a<b$ in the continuous case. For a differentiable CDF, its derivative is the density. CDFs are bounded by zero and one; densities need not be.

Further problems

1. For a uniform variable on $[10,14]$, find mean, variance, and $P(X>13)$.
2. For an exponential rate of 2 per hour, find the chance of waiting more than another hour after already waiting two hours.
3. Explain why a probability density of 4 at a point does not violate probability axioms.

You have undoubtedly seen a bell curve. It appears in discussions of exam scores, measurement error, biological traits, and rather more dubious claims about human behavior.

The curve is not automatically the right model merely because it is familiar. But when a normal distribution is appropriate, its mathematical structure makes many calculations elegant.

Where a bell curve comes from

A normally distributed random variable $X$ has two parameters: its mean $\mu$ and standard deviation $\sigma>0$.

Its probability density function is

The mean determines the center. The standard deviation determines how wide or narrow the curve is.

The height of a density curve is not itself a probability. Probability corresponds to area under the curve. The entire area equals 1.

For a continuous random variable, the probability of any single exact value is zero. We calculate the probability of an interval instead.

Standardization

If $X$ is normal with mean $\mu$ and standard deviation $\sigma$, the standardized variable is

$Z$ has a standard normal distribution: mean 0 and standard deviation 1. A $z$-score of 2 means an observation is two standard deviations above the mean.

For example, if exam scores are approximately normal with mean 70 and standard deviation 10, a score of 85 has

We can use the standard normal distribution to interpret the associated proportion of observations.

The 68–95–99.7 rule

For an exactly normal distribution, about 68% of the probability lies within one standard deviation of the mean, about 95% within two, and about 99.7% within three.

These are approximations. The commonly used interval $-1.96<Z<1.96$ captures approximately 95% of the standard normal probability.

That does not mean every dataset contains exactly 95% of its values within 1.96 sample standard deviations. The result is a statement about an idealized probability model.

Explore the probability area

Move the slider in the laboratory to change the symmetric interval $[-z,z]$. The shaded area and its approximate probability update together.

This is a model probability, not automatically a confidence interval for an unknown parameter. Later, we will examine how sampling distributions help construct confidence intervals and what their coverage really means.

Check your understanding

1. If $\mu=100$ and $\sigma=15$, what is the $z$-score of 130?
2. What does the total area under a probability density equal?
3. Is a tall density curve at a particular point evidence that the point itself has positive probability?

Translate an applied interval to a probability statement

For $X\sim N(100,15^2)$, the chance of a value between 85 and 130 is

The standard deviation is 15, while the second distribution parameter as written is the variance $225$. Books and software use different conventions; define which one is supplied. Standardization subtracts the mean and divides by the standard deviation, producing a dimensionless variable.

An upper 5% cutoff is $100+15z_{.95}\approx124.6728$, with $z_{.95}\approx1.644854$. A central 95% interval uses $z_{.975}\approx1.959964$ instead because 5% is split between both tails. Confusing a one-tail percentile with a two-tail interval critical value changes the answer.

The empirical 68–95–99.7 pattern is a normal-model approximation. It is not a property of every distribution with a mean and standard deviation. A normal model also assigns positive probability to every real interval; if a variable is bounded or strongly skewed, examine whether its tail predictions make sense in the application.

Additional practice

1. Standardize 70 for a normal variable with mean 50 and standard deviation 10.
2. Find its approximate upper-tail probability using $\Phi(2)\approx0.97725$.
3. Explain the difference between a 95th percentile of individual values and a 95% confidence interval for an unknown mean.

Here is something that sounds suspiciously like a mathematical magic trick. Start with a population that does not resemble a bell curve at all. Select random samples, calculate their averages, and repeat. The distribution of those averages will often begin to resemble a bell curve.

That is the central limit theorem. And it is one of the reasons statistical inference is possible.

First, distinguish three things

1. The population distribution describes the individual values in the population.
2. A sample is a set of values drawn from that population.
3. The sampling distribution of the sample mean describes the means produced by repeatedly drawing samples of the same size.

The sampling distribution is not a histogram of all the individual observations. It is a histogram of a statistic computed once per sample.

This distinction matters. If a distribution of individual incomes is skewed, the distribution of averages across many independently selected groups can be far more symmetric.

The theorem (a useful version)

Suppose $X_1,\ldots,X_n$ are independent and identically distributed with mean $\mu$ and finite, positive variance $\sigma^2$. Then, as $n$ grows,

In words: the standardized sample mean approaches the standard normal distribution.

It does not say that the original population becomes normal. It does not promise that a sample size of 30 is always sufficient. And it does not automatically apply to distributions without finite variance or to observations with unsuitable dependence.

The required sample size depends, among other things, on how skewed or heavy-tailed the population is.

Two discoveries at once

The theorem gives us a shape: approximate normality of the standardized sample mean for sufficiently large samples.

And the standard error,