AP Statistics Unit 2 review: Probability, Random Variables, and Probability Distributions
Unit 2 connects categorical tables, probability rules, random variables and sampling distributions. In the revised course it is 15–25% of multiple choice. Expect to choose a probability model, show its parameters and bounds, and interpret an expected value or standard deviation in context.
Unit 2 on the May 2027 exam
| 2027 exam | |
|---|---|
| Weight | 15–25% of the multiple-choice section |
| Exam date | Tuesday, May 11, 2027, afternoon session |
| Multiple choice | 42 questions in 90 minutes, worth 50% of the exam score |
| Free response | 4 questions in 90 minutes, worth 50%. All answers are entered in Bluebook: two multi-focus questions, one inference question and a final multi-focus question. |
| Calculator | A graphing calculator with statistical capabilities is expected on both sections. Bluebook includes Desmos graphing. Nongraphing handheld calculators are not allowed in 2027. |
| Class time | About 24 class periods in College Board's pacing guide |
| 2026 results | 17% of students scored a 5 on the whole exam, and about 62% scored 3 or higher |
Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.
What Unit 2 covers
College Board's course description splits the unit into 12 topics. The second column is what you should be able to do for each one.
| Topic | What you should be able to do |
|---|---|
| 2.1 Tabular and Graphical Representations for the Distributions of Two Categorical Variables | Construct and interpret two-way tables and displays for two categorical variables. |
| 2.2 Summary Statistics for Two Categorical Variables | Calculate marginal, joint and conditional relative frequencies and compare groups. |
| 2.3 Estimating Probabilities Using Simulation | Design a simulation with a specified random mechanism and estimate event probabilities. |
| 2.4 Introduction to Probability | Use the probability scale, complement rule and long-run interpretation. |
| 2.5 Mutually Exclusive Events | Recognize disjoint events and calculate probabilities for their union. |
| 2.6 Conditional Probability | Calculate conditional probabilities using the conditioned group as the denominator. |
| 2.7 Independent Events and Unions of Events | Apply independence and the general addition rule without double-counting intersections. |
| 2.8 Introduction to Random Variables and Probability Distributions | Define a random variable and verify that a probability distribution is valid. |
| 2.9 Parameters of Random Variables | Calculate and interpret expected value and standard deviation for a random variable. |
| 2.10 The Binomial Distribution | Check binomial conditions and calculate counts, probabilities, mean and standard deviation. |
| 2.11 The Normal Distribution | Standardize normal values and find probabilities or percentiles with appropriate bounds. |
| 2.12 Sampling Distributions and the Central Limit Theorem | Distinguish data distributions from sampling distributions and explain the central limit theorem. |
Translate the wording before choosing a probability rule
Write the event in symbols and identify the reference group. Independence is a condition to check, not a synonym for disjointness.
| Wording or model | Calculation | Condition or interpretation |
|---|---|---|
| A or B | P(A) + P(B) − P(A and B) | Subtract the overlap once. |
| A given B | P(A and B)/P(B) | Restrict the denominator to B; P(B) must be positive. |
| Independent A and B | P(A and B) = P(A)P(B) | Knowing one event does not change the other’s probability. |
| Exactly k binomial successes | C(n,k)pᵏ(1 − p)ⁿ⁻ᵏ | Fixed n, independent trials, two outcomes and constant p. |
| Binomial count | Mean np; SD √(np(1 − p)) | These describe counts, not proportions. |
| Normal variable | z = (x − μ)/σ | Identify and shade the requested tail or interval. |
| Sampling mean | Mean μ; SD σ/√n for independent observations | The CLT concerns the distribution of sample means. |
Common mistakes
Each of these loses points on questions you otherwise understand.
- Giving a calculator output without the distribution or bounds. Name the random variable, parameters and event before the result. CED printed p. 58.
- Reporting an expected value without context or units. Explain what quantity averages to that value over repetitions. CED pp. 57–58.
- Treating disjoint nonzero-probability events as independent. If A occurs, a disjoint B cannot occur, so its conditional probability changes. CED topics 2.5–2.7.
- Saying the CLT makes raw observations normal. It concerns the sampling distribution of a mean as sample size increases. CED topic 2.12.
6 practice questions with answers
Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.
Question 1 · 2.7
P(A) = 0.35, P(B) = 0.50 and P(A and B) = 0.20. What is P(A or B)?
- A0.85
- B0.175
- C0.65
- D0.20
Show the answer
C. The addition rule gives 0.35 + 0.50 − 0.20 = 0.65. Adding without subtracting would count the overlap twice.
Question 2 · 2.6
Among 80 commuters, 30 cycle and 18 of those cyclists wear reflective jackets. What is P(jacket | cyclist)?
- A0.60
- B0.225
- C0.375
- D0.167
Show the answer
A. Conditioning on being a cyclist restricts the reference group to 30 people. The desired probability is 18/30 = 0.60, not 18/80.
Question 3 · 2.10
Each of 6 independent packages has a 0.20 probability of arriving late. What is the probability that exactly 2 arrive late?
- A0.0400
- B0.2621
- C0.0164
- D0.2458
Show the answer
D. For X ~ Binomial(6, 0.20), P(X = 2) = C(6,2)(0.20)²(0.80)⁴ = 15(0.04)(0.4096) = 0.24576. The combination counts the possible pairs of late packages.
Question 4 · 2.9
A random reward X is $0 with probability 0.50, $4 with probability 0.30 and $10 with probability 0.20. What is E(X)?
- A$4.67
- B$3.20
- C$7.00
- D$2.00
Show the answer
B. E(X) = 0(0.50) + 4(0.30) + 10(0.20) = $3.20. This is a long-run mean reward, even though $3.20 is not a possible single reward.
Question 5 · 2.11
A normal distribution has mean 70 and standard deviation 8. What is the z-score for 82?
- A−1.50
- B0.667
- C1.50
- D12.0
Show the answer
C. z = (82 − 70)/8 = 1.50. The observation is 1.5 standard deviations above the mean. A z-score is not itself a probability.
Question 6 · 2.12
Independent observations come from a population with mean 40 and standard deviation 12. For samples of size 36, what is the SD of the sample mean?
- A2
- B6
- C12
- D72
Show the answer
A. The sampling standard deviation is σ/√n = 12/√36 = 2. The population SD of individual observations remains 12; averaging reduces the variability of the statistic.
How to study Unit 2
Translate each phrase into an event before calculating: at least means a right tail, at most means a left tail, and given changes the reference group. Check that the result lies between 0 and 1 and compare it with a rough estimate from a table or sketch.
For binomial questions, state n and p and verify independence and a fixed success probability. For normal questions, draw and shade the region first. For sampling distributions, label the horizontal axis with the statistic, such as a sample mean, so it is not confused with individual observations.
Ashore's Stats course has a lesson and practice questions for every topic in this unit. The free starting check takes about 15 minutes and shows which ones are costing you points.
Find the Stats topics costing you points
The starting check takes about 15 minutes. Free, no card.
Starting check
SAT Math · 1 of 5
Calculator allowed
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