Discrete Random Variables — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: Definition of discrete random variables, probability mass functions, expected value (mean), variance, linear transformations, and combining independent discrete random variables, aligned to AP Statistics CED learning objectives.

You should already know: Basic probability rules for sample spaces. Summation notation for finite calculations. Basic properties of mean and standard deviation for data.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Statistics style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Discrete Random Variables?

A discrete random variable (DRV) is a variable that takes on a countable number of distinct values, where each value has an associated probability of occurring. Unlike continuous random variables (which take uncountably infinite values over an interval, like time spent waiting in line), DRVs can be listed one by one: common examples include the number of heads in 10 coin flips, the number of customers arriving at a store in an hour, or the number of defective items in a production batch.

Notation convention per AP Statistics uses capital letters (e.g., $X$, $Y$) for the random variable itself, and lowercase letters (e.g., $x$, $y$) for specific observed values of the variable. According to the AP Statistics CED, this topic and its related subtopics make up approximately 10-15% of Unit 4’s exam weight, and DRV questions appear on both the multiple-choice (MCQ) and free-response (FRQ) sections of the exam. DRVs are the foundation for all common probability distributions tested on the AP exam, including binomial and geometric distributions, which are special cases of discrete random variables. Any question involving counting distinct outcomes will rely on the rules and properties covered here.

2. Probability Distributions for Discrete Random Variables

A probability distribution (specifically called a probability mass function, or PMF, for discrete variables) describes all possible values of a DRV and the probability that the variable takes each value. For a DRV $X$, the PMF is written $p(x)=P(X=x)$, and it must satisfy two non-negotiable requirements: (1) $0\le p(x)\le 1$ for all possible $x$, and (2) the sum of all $p(x)$ over all possible $x$ equals 1, or ${\sum }_{\text{all }x}p(x)=1$.

We can represent a DRV distribution as a table, a histogram, or a list of outcomes and probabilities. A cumulative distribution function (CDF) gives the probability that $X$ is less than or equal to a specific value: $F(x)=P(X\le x)={\sum }_{k\le x}p(k)$. The CDF simplifies calculating probabilities for ranges of outcomes: $P(a<X\le b)=F(b)-F(a)$.

Worked Example

A student organization sells up to 5 snack bags per person at a campus event. Let $X$ = the number of snack bags bought by a randomly selected person. The partial probability distribution is given below:

(a) Find the missing probability for $P(X=4)$; (b) Calculate $P(2\le X<5)$.

1. By the fundamental rule of probability distributions, the sum of all probabilities must equal 1. Let $c=P(X=4)$: $$0.21+0.28+0.19+c+0.08=1$$
2. Sum the known probabilities: $0.21+0.28+0.19+0.08=0.76$. This gives $c=1-0.76=0.24$, so $P(X=4)=0.24$.
3. For part (b), $P(2\le X<5)$ includes the values $x=2,3,4$ (since 5 is not less than 5).
4. Sum the probabilities for these values: $0.28+0.19+0.24=0.71$, so $P(2\le X<5)=0.71$.

Exam tip: Always circle the inequality sign in the problem before calculating interval probabilities. AP questions frequently test the distinction between $P(X<a)$ (does not include $X=a$) and $P(X\le a)$ (does include $X=a$).

3. Expected Value and Variance of a Discrete Random Variable

The expected value (or mean) of a discrete random variable $X$, written ${\mu }_X=E(X)$, is the long-run average value of $X$ we would expect to observe if we repeated the random process infinitely many times. The formula for expected value is: $$E(X)={\mu }_X=\sum _{\text{all }x}x\cdot P(X=x)$$

Variance, written ${\sigma }_X^2=Var(X)$, measures the spread of the distribution, or the average squared deviation of $X$ from its mean. The definition formula is ${\sigma }_X^2=\sum (x-{\mu }_X{)}^{2}P(X=x)$, but a more calculation-friendly equivalent formula (used on nearly all AP exam problems) is: $${\sigma }_X^2=E(X^2)-[E(X){]}^2=\left(\sum _{\text{all }x}{x}^{2}P(X=x)\right)-{\mu }_X^2$$

The standard deviation ${\sigma }_X$ is the square root of the variance, and it measures spread in the original units of $X$, unlike variance which is in squared units.

Worked Example

Using the snack bag distribution from the previous example ($p(1)=0.21,p(2)=0.28,p(3)=0.19,p(4)=0.24,p(5)=0.08$), calculate the expected value, variance, and standard deviation of $X$.

1. Calculate expected value: $$E(X)=(1)(0.21)+(2)(0.28)+(3)(0.19)+(4)(0.24)+(5)(0.08)=2.7$$
2. Calculate $E(X^2)$ for the variance formula: $$E(X^2)=(1^2)(0.21)+(2^2)(0.28)+(3^2)(0.19)+(4^2)(0.24)+(5^2)(0.08)=9.68$$
3. Calculate variance: $$Var(X)=E(X^2)-[E(X){]}^2=9.68-(2.7{)}^2=9.68-7.29=2.39$$
4. Calculate standard deviation: ${\sigma }_X=\sqrt{2.39}\approx 1.55$.

Exam tip: On FRQs, you will often need to interpret expected value or standard deviation in context. Always frame it as a long-run average, not a prediction for a single outcome: for this example, "Over many randomly selected people, the average number of snack bags bought is about 2.7."

4. Linear Transformations of Discrete Random Variables

A linear transformation changes a random variable $X$ into a new random variable $Y=aX+b$, where $a$ and $b$ are fixed constants. This occurs constantly in real-world problems: common examples include unit conversion (e.g., changing from dollars to cents, where $a=100,b=0$) or adding a fixed fee to a variable payout.

The rules for expected value and variance after linear transformation always hold, regardless of whether $X$ is discrete or continuous:

• Expected value: $E(aX+b)=aE(X)+b$
• Variance: $Var(aX+b)=a^{2}Var(X)$

The constant $b$ does not affect variance because adding $b$ shifts all values of $X$ by the same amount, so the spread of the distribution does not change. Only scaling by $a$ changes spread, and since variance is in squared units, we square $a$.

Worked Example

The student organization makes $3profitforeverysnackbagsold,plusafixed$0.50 donation from every person who buys at least one snack bag. Let $Y$ be the total profit the organization gets from a randomly selected person. Using the previous distribution of $X$ (${\mu }_X=2.7$, ${\sigma }_X^2=2.39$), find the expected value and standard deviation of $Y$.

1. Write the transformation relating $Y$ to $X$: Total profit = $3persnack+$0.50 fixed donation, so $Y=3X+0.5$.
2. Calculate expected value using the linear transformation rule: $$E(Y)=3{\mu }_X+0.5=3(2.7)+0.5=8.6$$
3. Calculate variance, remembering to square the coefficient $a=3$: $$Var(Y)=3^{2}Var(X)=9(2.39)=21.51$$
4. Calculate standard deviation: ${\sigma }_Y=\sqrt{21.51}\approx 4.64$.

Exam tip: The most common mistake here is forgetting to square $a$ when calculating variance. Write the formula $Var(aX+b)=a^{2}Var(X)$ on your paper before plugging in values to avoid this error.

5. Combining Independent Discrete Random Variables

We often want to find the expected value and variance of the sum or difference of two discrete random variables. Two random variables $X$ and $Y$ are independent if knowing the value of one does not change the probability distribution of the other.

The rules for combining random variables are:

• For any constants $a,b$, $E(aX+bY)=aE(X)+bE(Y)$. This rule always holds, even if $X$ and $Y$ are dependent.
• For independent $X$ and $Y$, $Var(aX+bY)=a^{2}Var(X)+b^{2}Var(Y)$. This rule only works for independent variables, and variances always add, even when we take the difference of two variables. For $X-Y$, this gives $Var(X-Y)=(1{)}^{2}Var(X)+(-1{)}^{2}Var(Y)=Var(X)+Var(Y)$.

Worked Example

At the campus event, there are two separate check-in lines. Let $X_1$ be the number of snack bags bought per person in the first line, and $X_2$ be the number per person in the second line. $X_1$ and $X_2$ are independent, and both have the same distribution as before (${\mu }_X=2.7$, ${\sigma }_X^2=2.39$). Let $D=X_1-X_2$, the difference in mean snack bags bought between the two lines. Find the expected value and variance of $D$.

1. Expected value of a difference is the difference of expected values: $$E(D)=E(X_1)-E(X_2)=2.7-2.7=0$$ This makes sense: there is no systematic difference between the two lines, so the expected difference is 0.
2. Since $X_1$ and $X_2$ are independent, we add the variances even for a difference: $$Var(D)=Var(X_1)+(-1{)}^{2}Var(X_2)=2.39+2.39=4.78$$
3. The standard deviation of $D$ is $\sqrt{4.78}\approx 2.19$.

Exam tip: Never subtract variances when calculating $Var(X-Y)$. The square of $-1$ is always 1, so variances add regardless of whether you add or subtract the random variables.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Calculating $Var(X-Y)$ as $Var(X)-Var(Y)$. Why: Students incorrectly extend the expected value rule (difference of expectations is expectation of differences) to variance. Correct move: Always square the coefficient for variance; for $X-Y$, coefficients are 1 and -1, both square to 1, so add variances.
• Wrong move: Forgetting to square $a$ when calculating $Var(aX+b)$. Why: Expected value uses $a$ to the first power, so students carry this over to variance by mistake. Correct move: Write the variance formula explicitly before plugging in numbers to confirm you squared $a$.
• Wrong move: Including $X=a$ in $P(X<a)$ when calculating interval probabilities. Why: Students rush and do not read inequality signs carefully. Correct move: Circle the inequality in the problem, then explicitly list all included $x$ values before summing probabilities.
• Wrong move: Interpreting expected value as "the most likely value of $X$". Why: Students confuse the long-run average (expected value) with the mode (most common outcome). Correct move: Always interpret expected value as the long-run average of many repeated observations of the random variable.
• Wrong move: Applying the variance addition rule to dependent random variables. Why: Students assume independence automatically if it is not explicitly stated. Correct move: Never add variances unless the problem explicitly states the random variables are independent.
• Wrong move: Guessing a missing probability instead of using the total probability rule. Why: Students forget the fundamental requirement that all probabilities sum to 1. Correct move: Any time you have a missing probability in a discrete PMF, use $\sum p(x)=1$ to solve for it.

7. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

Let $X$ be a discrete random variable with $E(X)=5$ and $Var(X)=4$. Let $Y=2X-3$. What are $E(Y)$ and $Var(Y)$? A) $E(Y)=7$, $Var(Y)=5$ B) $E(Y)=7$, $Var(Y)=16$ C) $E(Y)=10$, $Var(Y)=5$ D) $E(Y)=10$, $Var(Y)=16$

Worked Solution: We use the linear transformation rules for discrete random variables. First, for expected value: $E(aX+b)=aE(X)+b$. Here, $a=2$, $b=-3$, and $E(X)=5$, so $E(Y)=2(5)-3=10-3=7$. For variance: $Var(aX+b)=a^{2}Var(X)$, so $Var(Y)=2^2(4)=4\times 4=16$. This matches option B. Correct answer: $\boxed{B}$.

Question 2 (Free Response)

A coffee shop sells 1, 2, 3, or 4 lattes to a drive-thru customer. Let $X$ be the number of lattes bought by a randomly selected customer. The probability distribution of $X$ is given below:

(a) Find $P(X>2)$. (b) Calculate the expected value and standard deviation of $X$. (c) The coffee shop charges $4perlatte,plusafixed$1 service fee for any drive-thru order. Let $C$ be the total cost for a randomly selected customer's order. Find the expected value and standard deviation of $C$.

Worked Solution: (a) $X>2$ includes $x=3$ and $x=4$, so $P(X>2)=P(X=3)+P(X=4)=0.2+0.1=0.3$.

(b) Calculate expected value: $$E(X)=(1)(0.4)+(2)(0.3)+(3)(0.2)+(4)(0.1)=2.0$$ Calculate $E(X^2)=(1^2)(0.4)+(2^2)(0.3)+(3^2)(0.2)+(4^2)(0.1)=5.0$ Variance: $Var(X)=E(X^2)-[E(X){]}^2=5-2^2=1$ Standard deviation: ${\sigma }_X=\sqrt{1}=1.0$

(c) Total cost follows the transformation $C=4X+1$. Expected value: $E(C)=4E(X)+1=4(2)+1=9$ dollars. Variance: $Var(C)=4^{2}Var(X)=16(1)=16$ Standard deviation: ${\sigma }_C=\sqrt{16}=4$ dollars.

Question 3 (Application / Real-World Style)

A ecologist studying pollinators counts the number of monarch butterflies observed in a 10-minute survey of a milkweed patch. Let $X$ be the number of monarchs observed in a random 10-minute survey, with the following probability distribution:

Two independent 10-minute surveys are conducted on different days. Let $T$ be the total number of monarchs observed across the two surveys. Find the expected value and standard deviation of $T$, and interpret the expected value in context.

Worked Solution: First, calculate the expected value and variance of $X$ for one survey: $$E(X)=(0)(0.52)+(1)(0.28)+(2)(0.14)+(3)(0.06)=0.74$$ $$E(X^2)=0^2(0.52)+1^2(0.28)+2^2(0.14)+3^2(0.06)=1.38$$ $$Var(X)=1.38-(0.74{)}^2=0.8324$$ Let $X_1$ = number of monarchs in first survey, $X_2$ = number in second survey, so $T=X_1+X_2$. For independent surveys: $$E(T)=E(X_1)+E(X_2)=0.74+0.74=1.48$$ $$Var(T)=Var(X_1)+Var(X_2)=0.8324+0.8324=1.6648$$ $${\sigma }_T=\sqrt{1.6648}\approx 1.29$$ Interpretation: Over many pairs of independent 10-minute surveys, the average total number of monarchs observed per pair of surveys is approximately 1.48.

8. Quick Reference Cheatsheet

9. What's Next

Discrete random variables are the foundation for the specific discrete probability distributions you will study next in Unit 4: binomial random variables and geometric random variables, both of which are special cases of discrete random variables with their own simplified rules for expected value and variance. Without mastering the general rules for expected value, variance, linear transformations, and combining independent random variables covered here, you will not be able to correctly apply the rules for these specific distributions, which are heavily tested on the AP exam. Beyond Unit 4, the rules for combining random variables also extend to sampling distributions, which are the foundation for statistical inference (confidence intervals and hypothesis testing) in Units 5–8. This means a strong understanding of discrete random variables is critical for almost half of the AP Statistics exam content.

Binomial Random Variables Geometric Random Variables Sampling Distributions