Probability, random variables, and distributions

1. THE TASK VERB, AND THE THREE LINES EVERY CALCULATION SHOWS

2. TWO CATEGORICAL VARIABLES: THE TWO-WAY TABLE AND THE THREE RELATIVE FREQUENCIES

A two-way table (contingency table) counts observational units by two categorical variables at once. Everything in Topics 2.1 and 2.2 is one question — which total do I divide by? — asked three ways:

• joint relative frequency: a cell over the grand total — the share of everyone who is both things;
• marginal relative frequency: a row or column total over the grand total — the share of everyone who is one thing, ignoring the other variable;
• conditional relative frequency: a cell over its row total or its column total — the share among the people in that row or column.

The conditional is the one that answers questions about association: if the conditional distribution of one variable differs across the levels of the other — 45% of device owners read daily against 35% of non-owners — the variables are associated. If the conditional distributions are the same at every level, they are not. Association is a claim about the sample; whether it travels to the population is Unit 5’s chi-square question.

The displays. A side-by-side bar chart puts the conditional distributions next to each other; a segmented bar chart stacks each conditional distribution to 100%, so association reads as segments of different heights; a mosaic plot is a segmented bar chart whose bar widths are the marginal proportions.

3. PROBABILITY IS A LONG-RUN RELATIVE FREQUENCY — AND A SIMULATION IS HOW YOU WATCH IT

A random process produces results determined by chance; one run of it is a trial, the result of a trial is an outcome, and a collection of outcomes is an event. The probability of an event is its long-run relative frequency — the fraction of trials in which it occurs, as the number of trials grows without bound. That is a definition, not a metaphor, and the law of large numbers is the theorem behind it: for independent trials, the relative frequency settles closer and closer to one value.

This is one simulated sequence, shown at selected checkpoints: 5 heads in the first 10 tosses and 530 in the first 1,000. Equal horizontal steps represent equal multiplicative changes in the toss count. A larger number of trials usually makes the relative frequency more stable; it need not move closer to $0.5$ at every step. A run of tails does not make heads “due”: each independent toss still has probability $0.5$ of heads.

A simulation, described so a grader can run it. Five parts, and the exam scores each:

1.

assign digits (or cards, or a spinner) to outcomes so the probabilities match — “let 0–4 represent a girl and 5–9 a boy”;

2.

one trial: what is read and how much — “read three digits; that is one family of three”;

3.

record: the statistic from the trial — “count the girls; note whether the count is at least 2”;

4.

repeat: many trials — “repeat for 20 families,” and on the exam “many times” means a stated large number;

5.

estimate: the relative frequency — “the proportion of the 20 families with at least two girls estimates the probability.”

A simulation estimates; more independent trials reduce the typical sampling error, although an individual estimate can move farther away before moving closer. Problem 2 runs one against a digit table and compares it with the exact answer.

4. THE RULES OF PROBABILITY — AND THE PICTURE THAT MAKES THEM OBVIOUS

The sample space is the set of all possible non-overlapping outcomes; its probability is $1$. When the outcomes are equally likely,

and every probability is a number in $\left[\mathrm{0,1}\right]$. The complement of $E$ — written $E^{\prime }$, $Ē$ or $E^C$, “not $E$” — has $P\left(E^{\prime }\right)=1-P\left(E\right)$, which is the fastest route to every “at least one” question in the course.

Two events at once. The joint probability $P(A\cap B)$ is the chance both occur. Events are mutually exclusive (disjoint) when they cannot occur together: $P(A\cap B)=0$. The conditional probability of $A$ given $B$ is

the share of $B$’s probability in which $A$ also happens — and it is read straight off a two-way table as a cell over its row or column total. Rearranged, it is the general multiplication rule:

Independence: $A$ and $B$ are independent when knowing one happened does not change the probability of the other — $P(A\mid B)=P\left(A\right)$, equivalently $P(B\mid A)=P\left(B\right)$, equivalently $P(A\cap B)=P\left(A\right)P\left(B\right)$. Any one of the three, checked with numbers, is the justification; the third is the multiplication rule for the independent case. And the addition rule for the union, “$A$ or $B$ or both”:

the overlap subtracted once because adding $P\left(A\right)$ and $P\left(B\right)$ counted it twice. For mutually exclusive events the overlap is $0$ and the rule is a plain sum.

The distinction the exam tests most. Mutually exclusive is a statement about overlap (none); independent is a statement about information (none). Two events with positive probability that are mutually exclusive are dependent: if $A$ happened, $B$ certainly did not, so $P(B\mid A)=0\ne P\left(B\right)$. Do not let the words decide either one — compute.

5. THE FORMULA PICKER: THE WORD IN THE QUESTION NAMES THE RULE

The reverse conditional. “Given a positive test, what is the probability of the disease?” is $P(D\mid +)$, and the problem hands you $P(+\mid D)$ — the other direction. The route is a tree: branches for $D$ and $D^{\prime }$ with their probabilities, then $+$ and $-$ on each branch with the conditionals given; multiply along paths for the joint probabilities; then

No formula has to be memorized under the name Bayes; it is the conditional-probability definition with the denominator built from the tree. Problem 4 draws it.

The check that catches a wrong rule. A probability above $1$, or a conditional larger than the joint it came from divided by something less than $1$ — stop. And the two-way table is the universal verification: every rule in this card can be read as counts in cells, and if your formula’s answer disagrees with the counts, the formula was applied wrong.

6. DISCRETE RANDOM VARIABLES: THE DISTRIBUTION, ITS MEAN, ITS STANDARD DEVIATION

A random variable is a variable whose numerical value is the outcome of a random process — the number of girls in a family, the dollars a spin pays. A discrete one takes countable values, and its probability distribution lists every value with its probability: as a table, a graph, or a formula. Two checks before anything else: every probability is between $0$ and $1$, and they sum to $1$. A cumulative distribution gives $P(X\le x)$ for each value — a running total of the table.

The mean, or expected value, is a probability-weighted average of the values:

interpreted as the long-run average outcome per trial. The standard deviation is the typical distance of an outcome from that mean, over the long run:

and its square is the variance. These are parameters — fixed numbers describing the process — not statistics from a sample; the calculator’s one-variable statistics on a list of values with their probabilities as frequencies returns exactly them.

Interpreting them is the scored part. “$\mu =1.75$” is a number; “over many plays, the game pays an average of $1.75 per play” is the interpretation, and “$\sigma =4.44$” becomes “a single play’s payout typically differs from that average by about $4.44.” Context and units, as the CED says, or the parameter has not been interpreted.

7. THE BINOMIAL DISTRIBUTION: FOUR CONDITIONS, ONE FORMULA, TWO CALCULATOR FUNCTIONS

A binomial random variable counts successes in $n$ repeated trials when four things hold, and the exam asks you to check all four:

• binary — each trial is a success or a failure;
• independent — one trial’s outcome does not change another’s probability;
• fixed number of trials, $n$, decided before you start;
• same probability of success, $p$, on every trial.

Write $X\sim B(n,p)$. The probability of exactly $x$ successes is

where $\left(\genfrac{}{}{0.0pt}{}{n}{x}\right)$ counts the arrangements of $x$ successes among $n$ trials and $p^x{(1-p)}^{n-x}$ is the probability of any one arrangement. The mean and standard deviation need no table:

The boundary is an inequality, written first. “At least 10” is $P(X\ge 10)$; “more than 10” is $P(X\ge 11)$; “at most 7” is $P(X\le 7)$; “fewer than 7” is $P(X\le 6)$. Write it in $X$, then choose the tool: binompdf$(n,p,x)$ gives one value, binomcdf$(n,p,x)$ gives $P(X\le x)$, and everything else is a complement — $P(X\ge 10)=1-P(X\le 9)$. The scored line is the inequality with the distribution named, not the keystrokes.

When it is not binomial. Drawing without replacement from a small population breaks independence (each draw changes the next $p$); counting trials until the first success has no fixed $n$; a probability that drifts across trials breaks the same-$p$ condition. The CED’s verb is justify why a random variable is or is not binomial: name the condition that fails.

8. THE NORMAL DISTRIBUTION: SKETCH, SHADE, LABEL, STANDARDIZE

A continuous random variable takes any value in an interval, and probability is area: $P(a<X<b)$ is the area under the density curve between $a$ and $b$, the total area is $1$, and the probability of any single exact value is $0$ — so $P(X<68)$ and $P(X\le 68)$ are the same number, unlike the binomial. A normal distribution, $N(\mu ,\sigma )$, is the continuous, unimodal, symmetric, bell-shaped curve fixed by its mean and standard deviation; a smaller $\sigma$ makes it taller and narrower. The standard normal is $N\left(\mathrm{0,1}\right)$, and $z=(x-\mu )∕\sigma$ moves any normal question onto it.

The empirical rule (68–95–99.7) estimates the area within one, two and three standard deviations; the table or the calculator gives the exact area for any boundary. The CED names four shapes of question, and each is a picture before it is a calculation:

The routine, in the order the rubric wants it. Name the model, $X\sim N\left(\mathrm{64.5,2.5}\right)$. Sketch the curve, mark $\mu$ and the boundary in the problem’s units, shade the region the words describe. Standardize: $z=(68-64.5)∕2.5=1.4$. Read the area — from the table, $P(Z<1.4)=0.9192$, so $P(Z>1.4)=0.0808$; or from the calculator, normalcdf with the boundaries. Then the sentence. A percentile runs the routine backwards: the area is given, invert to $z$ (invNorm), then $x=\mu +\mathrm{z\sigma }$.

Comparing positions across two distributions. Two values from two normal models are compared by their $z$-scores or their percentiles, never by their raw values — Unit 1’s z-score card, now with a curve behind it. The larger $z$ is the higher relative position. Extremeness in either direction is measured by $\left|z\right|$; a value far below the mean can be more extreme than one above it.

9. SAMPLING DISTRIBUTIONS, RANDOMIZATION DISTRIBUTIONS, AND THE CENTRAL LIMIT THEOREM

Every statistic — a sample mean, a sample proportion — would come out differently in a different sample. The sampling distribution of a statistic is the distribution of its values across all possible samples of size $n$ from the population. It is a distribution of statistics, not of data, and it has the same three properties as any distribution: a center, a spread, a shape.

For the sample mean of $n$ independent observations from a population with mean $\mu$ and finite standard deviation $\sigma$:

and the central limit theorem says its shape is approximately normal — better as $n$ grows — whatever the population’s shape. If the population is itself normal, the sampling distribution is exactly normal at every $n$. The $\sqrt{n}$ is the whole reason averaging works: a mean of 36 observations varies a sixth as much as one observation.

By simulation. Draw many random samples of size $n$ from a population with the parameter set to a known value, compute the statistic each time, and pile up the values: the pile approximates the sampling distribution. A randomization distribution does the same for an experiment: reassign the observed response values to the treatment groups at random, many times, computing the statistic each time — the distribution of a difference that chance alone would produce. Both are how Units 3 and 4 will decide whether an observed result is unusual.

Three sentences before a calculation. Center: ${\mu }_{\overline{x}}=210$ minutes. Spread: ${\sigma }_{\overline{x}}=60∕\sqrt{36}=10$ minutes. Shape: approximately normal, by the central limit theorem, because $n=36$ is large. Then the normal routine of card 8, on $\overline{x}$.