Discrete Random Variables: Mean and Standard Deviation
Define random variables and construct probability distributions. Compute and interpret μₓ = ∑x·P(x) and σₓ = √(∑(x−μ)²P(x)). Apply rules for linear transformations: μ(a+bX) = a + bμₓ, σ(a+bX) = |b|σₓ.
A lesson of AP Statistics, in Probability, about 10 minutes.
Discrete Random Variables
By the end of this lesson, you will be able to compute and interpret the mean (expected value), variance, and standard deviation of a discrete random variable, and apply rules for combining random variables.
Random Variables
A random variable is a variable whose value is a numerical outcome of a random phenomenon.
Discrete random variable: takes a countable number of distinct values (usually integers). Associated with counting.
Continuous random variable: can take any value in an interval. Associated with measuring. (Covered later with the normal distribution.)
Probability Distribution
The probability distribution of a discrete random variable X lists all possible values and their probabilities.
Requirements:
- Every probability $P(X = x)$ must satisfy $0 \leq P(X = x) \leq 1$
- The sum of all probabilities must equal 1: $\sum P(X = x) = 1$
Example: Roll a fair 6-sided die. X = number showing.
| x | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| P(X=x) | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 | 1/6 |
Expected Value (Mean)
The expected value is the long-run average value of the random variable over many repetitions.
$$\mu_X = E(X) = \sum x \cdot P(X = x)$$
Example: Profit from a game: win $5 with P = 0.4, lose $3 with P = 0.6.
$\mu_X = (5)(0.4) + (-3)(0.6) = 2 - 1.8 = $0.20$
On average, you expect to win $0.20 per play. This doesn't mean you will win exactly $0.20 on any single play.
Variance and Standard Deviation
$$\sigma_X^2 = Var(X) = \sum (x - \mu_X)^2 \cdot P(X = x)$$
$$\sigma_X = \sqrt{\sigma_X^2}$$
The standard deviation $\sigma_X$ has the same units as X and measures the typical spread of the distribution around the mean.
Example: Using the game above ($\mu_X = 0.20$):