Random Variables

Unit 4 · Probability, Random Variables & Distributions

What AP Stats asks here

A random variable maps outcomes to numbers. Discrete RVs use a probability mass function whose values sum to 1; continuous RVs use a density curve. The expected value $E(X)$ is the long-run average; it is always a weighted average of outcomes with probabilities as weights — never a simple average of the distinct values.

Discrete RV summaries

Expected value

\mu_X = E(X) = \sum x_i p_i

Variance

\sigma_X^2 = \sum (x_i - \mu_X)^2 p_i

Standard deviation

\sigma_X = \sqrt{\sigma_X^2}

Linear transformation Y = aX + b

Mean

\mu_Y = a\mu_X + b

Standard deviation

\sigma_Y = |a|\sigma_X

Variance

\sigma_Y^2 = a^2 \sigma_X^2

AP Tip: E(X) is the long-run average per repetition. In a carnival game with E(X) = −$0.33, playing 100 games costs about $33 on average — even though any single game's result varies.

Caution: Adding a constant b shifts the distribution but does not change its spread. Scale the SD by |a|, never by |a| + b.

Type 1

Discrete vs continuous

Discrete = countable values (counts, integers). Continuous = any value in an interval (time, length, weight).

Example 1

Classify: (i) number of cars passing through an intersection in 1 hour; (ii) time until a phone battery dies; (iii) heads or tails on one fair flip (coded as 0/1). Which classification is correct? (A) All three are continuous. (B) (i) discrete; (ii) continuous; (iii) discrete (a Bernoulli RV). (C) (i) continuous; (ii) discrete; (iii) discrete. (D) All three are discrete.

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Type 2

Expected value as weighted average

$E(X)$ is a probability-weighted average of outcomes. In a game, it tells you the long-run average net outcome per play.

Example 2

A carnival game charges $2 to play. You roll a fair die and win $10 if you roll a 6, otherwise nothing. Let X = net winnings per play. What is

E(X)

? (A)

+

8.00 (B)

+

1.67 (C)

-

0.33 (D)

-

2.00

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Type 3

Linear transformation of expectation

$E(aX + b) = aE(X) + b$ . The SD scales by $|a|$ , never by $a + b$ .

Example 3

X

has

\mu_X = 50

and

\sigma_X = 8

. For

Y = 3X + 5

, which gives

(\mu_Y, \sigma_Y)

? (A)

(155, 24)

(B)

(155, 29)

(C)

(150, 24)

(D)

(155, 8)

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