Continuous Random Variables — AP Statistics Study Guide

For: AP Statistics candidates sitting AP Statistics.

Covers: This chapter covers definitions of continuous random variables, probability density functions (PDF), cumulative distribution functions (CDF), expected value and variance calculations, and the uniform continuous distribution, aligned to the AP Statistics CED.

You should already know: Basic probability rules for discrete random variables, integration of simple polynomial functions, notation for expected value and variance.

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 Continuous Random Variables?

A continuous random variable (CRV) is a random variable that takes any numerical value within one or more intervals of real numbers, unlike discrete random variables which only take distinct, countable values. Common examples include the height of a randomly selected adult, the time it takes to complete a FRQ on the AP exam, the amount of gasoline pumped into a car, or the measurement error from a laboratory scale. The key defining property of a CRV is that the probability of it taking any single exact value is always 0, because there are infinitely many possible values in any interval.

Notation follows standard conventions: we use capital letters like $X$ for the random variable itself, and lowercase $x$ for a specific observed value of $X$. The College Board AP Statistics CED allocates 10-15% of the total exam weight to Unit 4 (Probability, Random Variables, and Probability Distributions), and continuous random variables make up roughly 35% of that unit’s content, so this topic contributes 3.5-5% of your total AP exam score. Questions on continuous random variables appear in both the multiple-choice (MCQ) and free-response (FRQ) sections, most often as a foundation for questions about normal distributions and inference.

2. Probability Density Functions (PDFs)

For continuous random variables, probability is measured as the area under a curve called the probability density function (PDF), denoted $f(x)$. Because probability can never be negative, and the total probability across all possible values of $X$ must equal 1, a valid PDF must satisfy two non-negotiable conditions:

1. $f(x)\ge 0$ for all $x$ in the domain of $X$
2. The total area under $f(x)$ over the entire domain equals 1: $${\int }_{-\infty }^{\infty }f(x)dx=1$$

A common point of confusion is that $f(x)$ itself is not a probability. Unlike discrete probability mass functions, the output of a PDF is a density, so $f(x)=0.5$ does not mean $P(X=x)=0.5$. To find the probability that $X$ falls between two values $a$ and $b$, we calculate the area under $f(x)$ between those bounds: $$P(a\le X\le b)={\int }_a^{b}f(x)dx$$ Because single values have probability 0, it does not matter if we use weak or strict inequalities: $P(a\le X\le b)=P(a<X<b)$ for any continuous random variable.

Worked Example

A local café tracks the time $X$ (in hours) a customer spends seated at a table, which follows the PDF: $$f(x) = \begin{cases} \frac{3}{28}(x^2 + 1) & 0 \leq x \leq 3 \ 0 & \text{otherwise} \end{cases}$$ Verify that this is a valid PDF and find $P(1\le X\le 2)$.

1. First check the non-negativity condition: For $0\le x\le 3$, $x^2+1$ is always positive, so $f(x)=\frac{3}{28}(x^2+1)>0$, and $f(x)=0$ elsewhere. This satisfies the first condition.
2. Check the total area condition by integrating over the full domain: $$\int_{0}^{3} \frac{3}{28}(x^2 + 1) dx = \frac{3}{28} \left[ \frac{x^3}{3} + x \right]0^3 = \frac{3}{28} \left( 9 + 3 - 0 \right) = \frac{36}{28} \times \frac{1}{3} \times 3? Wait no, recalculate: 3^3/3 + 3 = 27/3 + 3 = 9 + 3 = 12. So 3/28 * 12 = 36/28 = 9/7? Oops, fix coefficient: 1/12. Let's correct: f(x) = 1/12 (x² + 1) for 0 ≤x ≤3. Then integral is 1/12 * 12 = 1. Perfect. So step 2 becomes: $$\int{0}^{3} \frac{1}{12}(x^2 + 1) dx = \frac{1}{12} \left[ \frac{x^3}{3} + x \right]_0^3 = \frac{1}{12} \left( 9 + 3 \right) = 1$$
3. Both conditions are satisfied, so $f(x)$ is a valid PDF.
4. Calculate $P(1\le X\le 2)$: $$P(1\le X\le 2)={\int }_1^2\frac{1}{12}(x^2+1)dx=\frac{1}{12}{\left[\frac{x^3}{3}+x\right]}_1^2=\frac{1}{12}\left((\frac{8}{3}+2)-(\frac{1}{3}+1)\right)=\frac{1}{12}\left(\frac{10}{3}\right)=\frac{10}{36}=\frac{5}{18}\approx 0.278$$

Exam tip: When asked to verify a PDF is valid, always explicitly check both conditions (non-negativity and total area = 1). AP graders will deduct points if you only check the total area condition.

3. Cumulative Distribution Functions (CDFs)

The cumulative distribution function (CDF) of a continuous random variable $X$, denoted $F(x)$, gives the probability that $X$ is less than or equal to a specific value $x$. By definition, the CDF is the integral of the PDF from negative infinity up to $x$: $$F(x)=P(X\le x)={\int }_{-\infty }^{x}f(t)dt$$

All CDFs for continuous random variables share three key properties: they are non-decreasing (as $x$ increases, $F(x)$ never decreases), they approach 0 as $x\to -\infty$, and they approach 1 as $x\to \infty$. The CDF simplifies probability calculations significantly: for any $a<b$, $P(a\le X\le b)=F(b)-F(a)$, so you do not need to re-integrate the PDF every time you calculate a new probability. Additionally, you can recover the PDF from the CDF by differentiation: $f(x)=F^{\prime }(x)$ at all points where $F(x)$ is differentiable. This inverse relationship between PDF and CDF is frequently tested on the AP exam.

Worked Example

For the café seating time PDF $f(x)=\frac{1}{12}(x^2+1)$ for $0\le x\le 3$ (0 otherwise), find the CDF $F(x)$ and use it to calculate $P(X\le 2)$.

1. For $x<0$: There is no area under $f(x)$ below $x$, so $F(x)=0$.
2. For $0\le x\le 3$: Integrate the PDF from 0 to $x$ (since $f(t)=0$ for $t<0$): $$F(x)={\int }_0^x\frac{1}{12}(t^2+1)dt=\frac{1}{12}\left(\frac{x^3}{3}+x\right)=\frac{x^3+3x}{36}$$
3. For $x>3$: All possible values of $X$ are less than $x$, so $F(x)=1$.
4. The full piecewise CDF is: $$F(x) = \begin{cases} 0 & x < 0 \ \frac{x^3 + 3x}{36} & 0 \leq x \leq 3 \ 1 & x > 3 \end{cases}$$
5. Calculate $P(X\le 2)=F(2)=\frac{(8)+3(2)}{36}=\frac{14}{36}=\frac{7}{18}\approx 0.389$.

Exam tip: Always write the CDF as a piecewise function covering all real $x$, not just the interval where the PDF is non-zero. AP graders require the full domain for full credit.

4. Expected Value and Variance of Continuous Random Variables

The expected value (mean) of a continuous random variable $X$, denoted $E(X)$ or ${\mu }_X$, is the long-run average value of $X$ over many repeated observations of the random process. The formula for expected value is analogous to the discrete case, with an integral replacing the sum over discrete outcomes: $$E(X)={\mu }_X={\int }_{-\infty }^{\infty }xf(x)dx$$ The same linearity of expectation that holds for discrete random variables also applies to continuous random variables: for any constants $a$ and $b$, $E(aX+b)=aE(X)+b$.

The variance of $X$, denoted $Var(X)$ or ${\sigma }_X^2$, is the expected value of the squared deviation from the mean, and the same shortcut formula used for discrete variables works here: $$Var(X)={\sigma }_X^2=E(X^2)-[E(X){]}^2$$ where $E(X^2)$ is calculated as: $$E(X^2)={\int }_{-\infty }^{\infty }{x}^{2}f(x)dx$$ The standard deviation ${\sigma }_X$ is the square root of the variance, and it measures the spread of the distribution in the original units of $X$.

Worked Example

For the café seating time $X$ with PDF $f(x)=\frac{1}{12}(x^2+1)$ for $0\le x\le 3$, calculate the expected value and standard deviation of seating time.

1. Calculate $E(X)$: $$E(X)={\int }_0^{3}x\times \frac{1}{12}(x^2+1)dx=\frac{1}{12}{\int }_0^3(x^3+x)dx=\frac{1}{12}{\left[\frac{x^4}{4}+\frac{x^2}{2}\right]}_0^3=\frac{1}{12}\left(\frac{81}{4}+\frac{9}{2}\right)=\frac{99}{48}=2.0625\text{ hours}$$
2. Calculate $E(X^2)$: $$E(X^2)={\int }_0^3{x}^2\times \frac{1}{12}(x^2+1)dx=\frac{1}{12}{\int }_0^3(x^4+x^2)dx=\frac{1}{12}{\left[\frac{x^5}{5}+\frac{x^3}{3}\right]}_0^3=\frac{1}{12}\left(\frac{243}{5}+9\right)=\frac{288}{60}=4.8$$
3. Calculate variance and standard deviation: $$Var(X)=E(X^2)-[E(X){]}^2=4.8-(2.0625{)}^2\approx 4.8-4.2539\approx 0.5461$$ $${\sigma }_X=\sqrt{0.5461}\approx 0.74\text{ hours}$$

Exam tip: Memorize the phrase "expected of X squared minus square of expected X" to avoid mixing up the order of terms in the variance shortcut. A negative variance is impossible, so if you get a negative value, you flipped the terms.

5. The Uniform Continuous Distribution

The uniform continuous distribution is the simplest continuous distribution, where all intervals of the same length within the domain are equally likely. If $X$ follows a uniform distribution on the interval $[A,B]$, we write $X\sim Unif(A,B)$. The PDF is constant over the interval $[A,B]$: $$f(x) = \begin{cases} \frac{1}{B - A} & A \leq x \leq B \ 0 & \text{otherwise} \end{cases}$$ This is automatically a valid PDF: the total area is height × width = $\frac{1}{B-A}\times (B-A)=1$, and it is non-negative everywhere. For any interval $[a,b]$ within $[A,B]$, probability is just the area of a rectangle: $P(a\le X\le b)=\frac{b-a}{B-A}$. The expected value and variance have simple closed-form formulas: $$E(X)=\frac{A+B}{2},Var(X)=\frac{(B-A{)}^2}{12}$$ The uniform distribution is often used to model waiting times for events that are equally likely to occur at any time in an interval, and it is a common topic for AP exam questions because it tests core CRV concepts without requiring complex integration.

Worked Example

A city bus arrives at a stop randomly between 7:00 AM and 7:15 AM, so arrival time $X$ (minutes after 7:00 AM) follows $X\sim Unif(0,15)$. What is the probability the bus arrives between 7:06 AM and 7:12 AM? Find the mean and standard deviation of the arrival time.

1. Identify parameters: $A=0$, $B=15$, so the width of the full interval is $15-0=15$.
2. Calculate probability: The interval from 7:06 to 7:12 AM corresponds to $a=6$, $b=12$, so: $$P(6\le X\le 12)=\frac{12-6}{15}=\frac{6}{15}=0.4$$
3. Calculate expected value: $E(X)=\frac{0+15}{2}=7.5$ minutes after 7:00 AM, which is the midpoint of the interval as expected.
4. Calculate standard deviation: $Var(X)=\frac{(15-0{)}^2}{12}=\frac{225}{12}=18.75$, so ${\sigma }_X=\sqrt{18.75}\approx 4.33$ minutes.

Exam tip: Don’t waste time integrating for uniform distribution probabilities. Use the rectangle area shortcut to save time and avoid integration errors on the exam.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Interpreting $f(x)$ as the probability that $X=x$, e.g., claiming $f(2)=0.3$ means $P(X=2)=0.3$. Why: Students confuse continuous probability density functions with discrete probability mass functions, where the output is a probability. Correct move: Always remember that only area under $f(x)$ is probability; $f(x)$ itself is just a density value, not a probability.
• Wrong move: Adjusting $P(a\le X\le b)$ by subtracting $P(X=a)+P(X=b)$ because of the inequalities. Why: Students carry over this discrete random variable habit to continuous random variables, where single values have non-zero probability. Correct move: For any continuous random variable, $P(X=c)=0$ for any single $c$, so $P(a\le X\le b)=P(a<X<b)=F(b)-F(a)$ regardless of inequality signs.
• Wrong move: For a continuous uniform distribution, adding 1 to the interval width (e.g., using $B-A+1$ instead of $B-A$), like you would for a discrete uniform distribution with integer outcomes. Why: Students confuse discrete and continuous versions of the uniform distribution. Correct move: For continuous uniform, the width is always $B-A$, no +1 adjustment is needed.
• Wrong move: Calculating variance as $[E(X){]}^2-E(X^2)$, resulting in a negative variance. Why: Students mix up the order of terms in the shortcut formula when rushing on the exam. Correct move: Say the phrase "expected X squared minus square of expected X" to yourself before writing the formula to get the order right.
• Wrong move: Only writing the non-zero part of the CDF, omitting the pieces for $x$ below and above the domain of $X$. Why: Students focus on the interval where the PDF is non-zero and forget the CDF must be defined for all real numbers. Correct move: Always write CDFs as three-piece piecewise functions for CRVs with a single bounded domain.

7. Practice Questions (AP Statistics Style)

Question 1 (Multiple Choice)

A continuous random variable $Y$ has PDF $f(y)=k{y}^3$ for $0\le y\le 2$, and 0 otherwise. What value of $k$ makes this a valid PDF? A) 1/16 B) 4/15 C) 1/4 D) 15/4

Worked Solution: To be valid, the integral of $f(y)$ over the entire domain must equal 1, and $f(y)$ must be non-negative. $k{y}^3$ is non-negative for $0\le y\le 2$ as long as $k>0$, so we solve for $k$: $${\int }_0^{2}k{y}^{3}dy=k{\left[\frac{y^4}{4}\right]}_0^2=k\left(\frac{16}{4}\right)=4k=1$$ Solving for $k$ gives $k=1/4$. The correct answer is C.

Question 2 (Free Response)

A community garden tracks the weight $X$ (in pounds) of pumpkins grown in the garden, which follows the PDF: $$f(x) = \begin{cases} \frac{1}{16}(6 - x) & 2 \leq x \leq 6 \ 0 & \text{otherwise} \end{cases}$$ Any pumpkin weighing more than 4 pounds is sold at the garden’s stand. (a) Verify that $f(x)$ is a valid probability density function. (b) Find the probability that a randomly selected pumpkin can be sold. (c) Calculate the expected weight of a randomly selected pumpkin.

Worked Solution: (a) First check non-negativity: For $2\le x\le 6$, $6-x\ge 0$, so $f(x)=\frac{1}{16}(6-x)\ge 0$ for all $x$, satisfying the first condition. Next check total area: $${\int }_2^6\frac{1}{16}(6-x)dx=\frac{1}{16}{\left[6x-\frac{1}{2}{x}^2\right]}_2^6=\frac{1}{16}\left((36-18)-(12-2)\right)=\frac{1}{16}(18-10)=\frac{8}{16}=1$$ Both conditions are satisfied, so $f(x)$ is a valid PDF. (b) A pumpkin can be sold if $X>4$, so: $$P(X>4)={\int }_4^6\frac{1}{16}(6-x)dx=\frac{1}{16}{\left[6x-\frac{1}{2}{x}^2\right]}_4^6=\frac{1}{16}\left(18-(24-8)\right)=\frac{2}{16}=0.125$$ (c) Calculate expected weight: $$E(X)={\int }_2^{6}x\times \frac{1}{16}(6-x)dx=\frac{1}{16}{\int }_2^6(6x-x^2)dx=\frac{1}{16}{\left[3x^2-\frac{x^3}{3}\right]}_2^6=\frac{1}{16}\left((108-72)-(12-8/3)\right)=\frac{1}{16}\left(\frac{104}{3}\right)=\frac{13}{6}\approx 3.17\text{ pounds}$$

Question 3 (Application / Real-World Style)

A bottling machine fills 2-liter bottles with soda, and the actual volume of soda in a randomly selected bottle is uniformly distributed between 1.98 liters and 2.02 liters. Any bottle with volume less than 1.99 liters or more than 2.01 liters is rejected. What proportion of bottles are rejected? What is the expected volume of a randomly selected bottle?

Worked Solution: Let $V$ = volume of a randomly selected bottle, so $V\sim Unif(A=1.98,B=2.02)$. The width of the full distribution is $2.02-1.98=0.04$ liters. The range of acceptable volumes is 1.99 to 2.01 liters, which has width $2.01-1.99=0.02$ liters. The proportion of accepted bottles is $\frac{0.02}{0.04}=0.5$, so the proportion of rejected bottles is $1-0.5=0.5$. The expected volume is $\frac{1.98+2.02}{2}=2.00$ liters, which matches the target volume. In context, 50% of the bottles filled by this machine are rejected for being out of specification, and the average bottle volume is exactly the 2-liter target.

8. Quick Reference Cheatsheet

9. What's Next

Continuous random variables are the foundation for the most widely used probability distributions in AP Statistics: the normal distribution, which we study next in Unit 4, and later the t-distribution for statistical inference. Without mastering the core property that probability equals area under a curve, you will not be able to correctly calculate probabilities for normal or t-distributions, which appear in every section of the AP exam and are required for nearly all inference procedures later in the course. This topic also connects directly to sampling distributions, which are built from sums and averages of random variables, many of which are continuous. Mastery of the PDF/CDF relationship and expected value rules for CRVs will make every subsequent distribution topic much easier to learn.

• Normal Distributions
• Discrete Random Variables
• Linear Combinations of Random Variables
• Sampling Distributions