Parameters of Random Variables — AP® Statistics

Part 1 · Hook

Why this matters

A food truck plans tomorrow's supplies. Some orders are 1 taco, some are 5. How many tacos is an average order? Not the middle of 1 to 5: small orders are far more common. Weight each order size by how often it happens and you get the mean of the random variable, the number every planner wants.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. The mean of the data 2, 4, 9 is

5
4
15
9

Show the answer

$\frac{2 + 4 + 9}{3} = 5$ .

Correct: 5:
4:
15:
9:

2. For X with $P(X = 1) = 0.5$ , $P(X = 2) = 0.3$ , $P(X = 3) = 0.2$ , what is $P(X \ge 2)$ ?

0.5
0.3
0.2
0.8

Show the answer

$0.3 + 0.2 = 0.5$ .

Correct: 0.5:
0.3:
0.2:
0.8:

Part 4 · See it

See it first

Probability histogram of X, the number of tacos in an order: 0.20, 0.35, 0.25, 0.15 and 0.05 for 1 to 5 tacos. A triangle under the axis marks the mean, 2.5 tacos, where the histogram would balance; arrows show one standard deviation, about 1.12 tacos, on each side. — The mean, 2.5 tacos, is where the probability histogram balances; the standard deviation, about 1.12 tacos, is a typical distance from it. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Some values of X are more likely than othersa plain average of the values is wrong; each value must count by its probability
Multiply each value by its probability and addthe mean (expected value) $\mu_X = \sum x_i p_i$ , the long-run average
Weight the squared distances from the mean the same waythe variance; its square root is the standard deviation $\sigma_X$
Interpret both in contextthe mean is the average over many repetitions; the SD is how far values typically land from it

Part 6 · Key ideas

Key ideas

The mean or expected value of a discrete random variable is $\mu_X = E(X) = \sum x_i p_i$ : each value times its probability, added up. It is the long-run average and need not be a possible value.
The standard deviation is $\sigma_X$ $= \sqrt{\sum (x_i - \mu_X)^2 p_i}$ : a typical distance of the values from the mean, in the units of X.
Show the expression, the substitution and the answer: $1(0.20) + 2(0.35) + \dots$ $= 2.5$ .
Interpret in context: "Over many orders, the average number of tacos per order is about 2.5."

Part 7 · Misconception

A common mistake

The wrong idea: The mean of X is the average of its possible values: $\frac{1 + 2 + 3 + 4 + 5}{5}$ $= 3$ tacos.

What actually happens: Each value counts by its probability. Small orders are more likely, so $\mu_X = \sum x_i p_i = 2.5$ tacos.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Tacos per order at a food truck

A food truck tracks how many tacos each order contains. Let X = the number of tacos in a randomly chosen order (invented data).

Probability distribution of X, tacos per order
$x$	1	2	3	4	5
$P(X = x)$	0.20	0.35	0.25	0.15	0.05

1. What is the mean (expected value) of X?

3 tacos
2 tacos
2.5 tacos
0.2 tacos

Show the answer

$\mu_X$ $= \sum x_i p_i$ $= 1(0.20)$ $+ 2(0.35)$ $+ 3(0.25)$ $+ 4(0.15)$ $+ 5(0.05)$ $= 0.20$ $+ 0.70$ $+ 0.75$ $+ 0.60$ $+ 0.25$ $= 2.5$ tacos.

3 tacos: That averages 1 to 5 without the probabilities.
2 tacos: That is the most likely value, not the mean.
Correct: 2.5 tacos: Right: each value weighted by its probability.
0.2 tacos: That averages the probabilities, which always gives 1 divided by the number of values.

2. What is the standard deviation of X, to two decimal places?

1.25 tacos
1.58 tacos
0.25 tacos
1.12 tacos

Show the answer

$\sigma_X$ $= \sqrt{\sum (x_i - \mu_X)^2 p_i}$ $= \sqrt{(1 - 2.5)^2(0.20) + (2 - 2.5)^2(0.35) + (3 - 2.5)^2(0.25) + (4 - 2.5)^2(0.15) + (5 - 2.5)^2(0.05)}$ $= \sqrt{1.25}$ $\approx 1.12$ tacos.

1.25 tacos: That is the variance. Take its square root.
1.58 tacos: That treats 1 to 5 as a data set with no probabilities.
0.25 tacos: That is the squared deviation of 2 or 3 from the mean, not the whole sum.
Correct: 1.12 tacos: Right: the square root of the weighted squared deviations, 1.25.

3. Which is the best interpretation of the mean of X in context?

Over many orders, the average number of tacos per order is about 2.5
Most orders contain 2.5 tacos, the most common order size
Half of the orders have more than 2.5 tacos and half have fewer
The next order will contain about 2 or 3 tacos

Show the answer

An expected value is a long-run average: over many randomly chosen orders, the average number of tacos per order is about 2.5.

Correct: Over many orders, the average number of tacos per order is about 2.5: Right: a long-run average over many orders.
Most orders contain 2.5 tacos, the most common order size: No order contains 2.5 tacos; the most common is 2.
Half of the orders have more than 2.5 tacos and half have fewer: That would describe a median; here 55% of orders have fewer than 2.5.
The next order will contain about 2 or 3 tacos: The mean does not predict one order; it describes the long run.

Table

A fundraiser game

A school fundraiser game costs $2 to play. A player wins $10 with probability 0.10 and $5 with probability 0.15, and otherwise wins nothing. Let W = a player's net gain in dollars (prize minus the $2 cost) on one play.

Probability distribution of W, net gain on one play
$w$ (dollars)	8	3	−2
$P(W = w)$	0.10	0.15	0.75

4. Write the expression for the expected value of W.

Type the expression as on the exam, with the values substituted.

Show the answer

$\mu_W$ $= 8(0.10)$ $+ 3(0.15)$ $+ (-2)$ $(0.75)$ $= 0.80$ $+ 0.45 - 1.50$ $= -0.25$ dollars.

Answer: $8 \cdot 0.1 + 3 \cdot 0.15 + (-2) \cdot 0.75$

5. The expected value of W is −$0.25. Build an interpretation in context.

Write the conclusion in context.

Show the answer

An expected value is a long-run average. A negative mean net gain means players lose money on average, which is how the school raises money.

A full-credit answer: If many people play the game, players lose an average of about $0.25 per play, so in the long run the school raises about 25 cents for each play.

6. For a randomly chosen family in a town, the expected number of children is 1.8. A student says this is impossible, since no family has 1.8 children. How should you respond?

The student is right: it should be rounded to 2 children
The expected value is probably a mistake, since it is not a whole number of children
It is an average over many families, so it need not be a possible value
It means most families in the town have 1 or 2 children

Show the answer

An expected value is a long-run average over many families. It need not be one of the possible values.

The student is right: it should be rounded to 2 children: Rounding throws away information; the average is 1.8.
The expected value is probably a mistake, since it is not a whole number of children: Means of counts are often not whole numbers.
Correct: It is an average over many families, so it need not be a possible value: Right: the long-run mean can fall between whole numbers.
It means most families in the town have 1 or 2 children: The mean alone does not say what most families have.

Part 9 · Summary

Summary

The mean of a discrete random variable weights each value by its probability:

\mu_X = \sum x_i p_i

. It is the average over many repetitions. The standard deviation

\sigma_X

= \sqrt{\sum (x_i - \mu_X)^2 p_i}

is a typical distance from that mean. Write the expression with the numbers substituted, then interpret both in context.

Part 10 · Up next

What comes next

Part 11 · Connections