Unit 2 · Topic 2.9 · about 20 minutes

Parameters of Random Variables

Calculate the mean and standard deviation of a discrete random variable from its distribution, and interpret both as long-run statements in context.

Predict first

At a coffee shop's drive-through, the number of drinks X in an order is 1 with probability 0.50, 2 with probability 0.25, 3 with probability 0.15 and 4 with probability 0.10. Over many orders, what is the average number of drinks per order?

Parameters of a distribution

A probability distribution has a center and a spread, just as a data set does. A number that measures a characteristic of a probability distribution, or of a population, is a parameter. A parameter is a single, fixed value: the distribution of X has exactly one mean and one standard deviation. That is the difference from a statistic, which changes from sample to sample, as in 1.2 Variables.

The mean, or expected value

The expected value of a discrete random variable X, also called its mean, is written $E (X)$ or $μ_{X}$ . Multiply each possible value by its probability and add:

μ_{X} = E (X) = \sum x_{i} \cdot P (x_{i})

The expected value is the long-run average of X. If you could watch thousands of drive-through orders, the mean number of drinks per order would settle near 1.85. It is not the value you should expect on any one order, and it does not have to be a value X can take.

The standard deviation

The standard deviation of X, written $SD (X)$ or $σ_{X}$ , measures how far the values of X typically fall from the mean over the long run:

σ_{X} = \sqrt{\sum (x_{i} - μ_{X})^{2} \cdot P (x_{i})}

Square each value's distance from the mean, weight it by the value's probability, add, and take the square root. The quantity under the square root is the variance, written $V (X)$ or $σ_{X}^{2}$ . The variance is in squared units (orders squared, dollars squared), which is why the standard deviation is the number you interpret.

Worked exampleWedding cakes at a bakery

A bakery tracks X, the number of wedding cake orders it receives in a week. From years of records, X is 0 with probability 0.15, 1 with probability 0.30, 2 with probability 0.30, 3 with probability 0.15 and 4 with probability 0.10. Find and interpret the mean and the standard deviation of X.

Check that it is a distribution. $0.15 + 0.30 + 0.30 + 0.15 + 0.10 = 1$ .
Mean. $μ_{X} = 0 (0.15) + 1 (0.30) + 2 (0.30) + 3 (0.15) + 4 (0.10) = 1.75$ .
Variance. Weight each squared distance from 1.75 by its probability, as in the table below: $σ_{X}^{2} = 0.459375 + 0.16875 + 0.01875 + 0.234375 + 0.50625 = 1.3875$ .
Standard deviation. $σ_{X} = \sqrt{1.3875} \approx 1.18$ orders.
Interpret both in context. Over many weeks, this bakery averages 1.75 wedding cake orders per week. The number of orders in a week typically differs from that mean of 1.75 by about 1.18 orders.

Answer.

$μ_{X} = 1.75$ orders and $σ_{X} \approx 1.18$ orders, both describing this bakery's weeks over the long run.

The bakery calculation, one row per value of X
$x$	$P (x)$	$x \cdot P (x)$	$(x - 1.75)^{2} \cdot P (x)$
0	0.15	0	0.459375
1	0.30	0.30	0.16875
2	0.30	0.60	0.01875
3	0.15	0.45	0.234375
4	0.10	0.40	0.50625
Sum	1	$μ_{X} = 1.75$	$σ_{X}^{2} = 1.3875$

Distribution of weekly wedding cake orders

The mean, 1.75, is the balance point of the bars. It falls between two possible values, since no week has 1.75 orders.

Expected value and money

Raffles and insurance companies both run on expected value. A school raffle sells 1,000 tickets for 10 dollars each, with one prize of 2,000 dollars and five prizes of 100 dollars. Let X be one ticket's net gain, the prize minus the 10 dollars paid. X is 1,990 with probability 0.001, 90 with probability 0.005 and $- 10$ with probability 0.994, so

E (X) = 1990 (0.001) + 90 (0.005) + (- 10) (0.994) = - 7.50

Over many tickets, buyers lose an average of 7.50 dollars per ticket, and that is the money the raffle raises. Any one buyer either loses 10 dollars or wins a prize; nobody loses exactly 7.50.

Check your understanding

Let X be the number of pets in a randomly chosen household in a town, with $P (X = 0) = 0.35$ , $P (X = 1) = 0.30$ , $P (X = 2) = 0.20$ , $P (X = 3) = 0.10$ and $P (X = 4) = 0.05$ . Find $E (X)$ .

Your answer

Let X be the number of pets in a randomly chosen household in a town, with $P (X = 0) = 0.35$ , $P (X = 1) = 0.30$ , $P (X = 2) = 0.20$ , $P (X = 3) = 0.10$ and $P (X = 4) = 0.05$ . Then $E (X) = 1.2$ . Which is the best interpretation of $E (X)$ ?

For Driver A, the number of late deliveries X in a day is 0 or 4, each with probability 0.5. For Driver B, the number Y is 1 with probability 0.25, 2 with probability 0.5 and 3 with probability 0.25. Which statement is true?

A bakery counts its wedding cake orders each week. The number of orders in a week, X, has $μ_{X} = 1.75$ and $σ_{X} \approx 1.18$ . Which is the best interpretation of the standard deviation?

A carnival game costs 2 dollars to play. You roll a fair die, and if it shows a 6 you win 10 dollars; otherwise you win nothing. Let X be your net gain on one play, so X is 8 with probability $\frac{1}{6}$ and $- 2$ with probability $\frac{5}{6}$ . Which statement is correct?

Practice

Practice until it is automatic

New numbers every time. Each one is checked the moment you answer, with the full working shown.

Discrete random variables: mean and standard deviation practice page

Course alignment, for teachers

AP Statistics topic 2.9, Unit 2: Probability, Random Variables, and Probability Distributions.

Skill 3.B: Calculate summary statistics, relative positions of points within a distribution, and predicted responses.
Skill 4.D: Interpret statistical calculations and results to assess meaning or a claim.

Parameters of a distribution

The mean, or expected value

The standard deviation

Worked exampleWedding cakes at a bakery

Expected value and money

Check your understanding

Practice until it is automatic

Labs that use this topic