AP® Statistics review sheet from Aim for Five (aimforfive.com/stats/units/2/2-9)
Unit 2 · Topic 2.9
2.9 Parameters of Random Variables
The mean (expected value) and standard deviation summarize a discrete random variable's distribution. The mean is the long-run average outcome, the standard deviation is the typical distance from that mean, and both need to be interpreted in context.
Key terms
- expected value
- mean of a random variable (μX)
- standard deviation of a random variable (σX)
- parameter
Expected value
The mean of a discrete random variable X, also called its expected value, is μX = E(X) = Σ xᵢ · P(xᵢ). Multiply each value by its probability and add.
It's a weighted average: values that are more likely count more. It's the average you'd get over a very large number of repetitions, so it doesn't have to be a value X can actually take. A home can't have 1.1 pets, but 1.1 can still be the expected number of pets (see the example below).
Standard deviation
The standard deviation is σX = √[Σ (xᵢ − μX)² · P(xᵢ)]. For each value: subtract the mean, square, multiply by the probability. Add those up (that's the variance, σX²) and take the square root.
It measures how much the outcomes typically vary from the mean over many repetitions. Both formulas are on the formula sheet. On a calculator, you can enter the values in one list and probabilities in another and run 1-Var Stats with the probability list as the frequency.
Why the standard deviation matters
Two random variables can have the same mean but very different risk. Game A pays 1 dollar every time. Game B pays 0 dollars with probability 0.9 and 10 dollars with probability 0.1. Both have an expected payout of 1 dollar. But Game A's standard deviation is 0, while Game B's is √[(0 − 1)²(0.9) + (10 − 1)²(0.1)] = √9 = 3 dollars. The standard deviation shows how much more unpredictable Game B is.
Insurance companies, casinos and businesses use expected value to set prices and the standard deviation to judge risk.
These are parameters
μX and σX describe the probability model, so they're parameters with single fixed values. They aren't computed from a sample. That's why they get Greek letters, unlike x̄ and s.
Interpreting in context
Mean: "If we randomly selected many homes, the average number of pets per home would be about 1.1 in the long run."
Standard deviation: "Over many randomly selected homes, the number of pets typically varies by about 0.94 from the mean of 1.1."
You won't be tested on combining or transforming random variables (such as finding the mean of X + Y). That was removed from the course.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Mean and standard deviation from a table
X = number of pets in a randomly chosen home. P(0) = 0.3, P(1) = 0.4, P(2) = 0.2, P(3) = 0.1. Find and interpret μX and σX.
Show the solutionHide the solution
- Step 1: μX = 0(0.3) + 1(0.4) + 2(0.2) + 3(0.1) = 0 + 0.4 + 0.4 + 0.3 = 1.1.
- Step 2: Squared deviations times probabilities: (0 − 1.1)²(0.3) = 0.363; (1 − 1.1)²(0.4) = 0.004; (2 − 1.1)²(0.2) = 0.162; (3 − 1.1)²(0.1) = 0.361.
- Step 3: Variance: 0.363 + 0.004 + 0.162 + 0.361 = 0.89.
- Step 4: σX = √0.89 ≈ 0.943.
Answer: μX = 1.1 pets: over many randomly chosen homes, the average is about 1.1 pets per home. σX ≈ 0.94: over many randomly chosen homes, the number of pets typically differs from 1.1 by about 0.94.
- Example 2Calculator allowed
Is the game worth playing?
A carnival game costs 3 dollars. You win 10 dollars with probability 0.15, 2 dollars with probability 0.35 and nothing otherwise. What are your expected net winnings per game?
Show the solutionHide the solution
- Step 1: Expected prize: 10(0.15) + 2(0.35) + 0(0.50) = 1.50 + 0.70 = 2.20 dollars.
- Step 2: Subtract the 3-dollar cost: 2.20 − 3 = −0.80 dollars.
- Step 3: Alternatively, use net values of 7, −1 and −3 dollars with the same probabilities: 7(0.15) − 1(0.35) − 3(0.50) = 1.05 − 0.35 − 1.50 = −0.80.
Answer: −0.80 dollars. Over many games, you'd lose about 80 cents per game on average.
- Example 3Calculator allowed
Trap: dividing by the number of values
A student finds the mean of the pet distribution as (0 + 1 + 2 + 3)/4 = 1.5. What went wrong?
Show the solutionHide the solution
- Step 1: That treats all four values as equally likely.
- Step 2: The probabilities aren't equal: 0 and 1 pets are much more likely than 3.
- Step 3: The weighted mean Σ x · P(x) = 1.1 is correct.
Answer: The mean is 1.1, not 1.5. Each value must be weighted by its probability.
Common mistakes
- Averaging the values without weighting them by probability.
- Forgetting the square root, reporting the variance as the standard deviation.
- Interpreting the expected value as the most likely single outcome. It's a long-run average.
- Leaving out "over many trials" or "in the long run" in an interpretation.
On the exam
- Interpretations should include the long-run idea and context. A bare number usually doesn't earn the interpretation point.
- Expected value questions often involve money and costs. Decide whether you're computing the prize or the net gain.
Connected topics
Videos
Check yourself
4 questions on 2.9 Parameters of Random Variables. Pick an answer to see if you got it, and why.
A school fundraiser game costs $2 to play. A player wins $10 with probability 0.1, wins $3 with probability 0.2, and wins nothing otherwise. Let X be the player's net gain (winnings minus the $2 cost). What is E(X), and what does it mean?
A random variable Y takes the values 1, 2 and 3 with probabilities 0.2, 0.5 and 0.3. What is the standard deviation of Y?
| Number of pets, x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| P(X = x) | 0.30 | 0.35 | 0.20 | ? | 0.05 |
Invented data: X is the number of pets in a randomly selected household in a town
What is the expected value of X?
Which is the best interpretation of the standard deviation of X, which is about 1.13?
0 of 4 answered