HSS.MD.A.2Common CoreMathStatistics and ProbabilityGrades 9-12
HSS.MD.A.2: Expected Value as the Mean of a Probability Distribution
In plain English: HSS.MD.A.2 is an advanced (+) Common Core statistics and probability standard that asks students to calculate the expected value of a random variable by multiplying each value by its probability and adding. Students interpret the result as the mean of the probability distribution: the long-run average over many repetitions, not a value that must occur. It is usually taught in a statistics course or Precalculus.
(+) Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
Common Core State Standards for Mathematics · Domain: Using Probability to Make Decisions (MD) · Cluster: Calculate expected values and use them to solve problems Also written as HSS-MD.A.2 or S-MD.2 · Official standard
Students compute the expected value of a random variable, E(X) = Σ x · P(X = x), from its probability distribution. The lesson opens with a mean that students already know how to find, the mean price of ten café orders, and rewrites it as a sum of values times relative frequencies. Replacing relative frequencies with probabilities gives exactly the expected value formula, which is why the standard calls E(X) the mean of the probability distribution.
Students then interpret E(X) in three ways that support one another: as a weighted average of the values, as the balance point of the probability histogram, and as the long-run average of the variable over many repetitions. Along the way they meet the common misreadings: expected value is not the most likely value and does not have to be a value the variable can take.
Learning Objectives
By the end of this lesson, students will be able to:
Calculate the expected value of a random variable from its probability distribution, including variables with negative values
Explain why E(X) is the mean of the probability distribution by comparing it with the mean of a frequency table
Locate E(X) as the balance point of a probability histogram
Interpret E(X) in context as a long-run average, with units, and explain why it need not be a possible value
Prior Knowledge Required
Students should already be comfortable with:
Random variables and their probability distributions HSS.MD.A.1
Finding the mean of a data set, including from a frequency table 6.SP.B.5
Post the question and give students three minutes to answer it alone.
Warm-Up Prompt
"A café sells coffee in three sizes: small for $2, medium for $3 and large for $5. Of its last 10 orders, 5 were small, 3 were medium and 2 were large. What was the mean price of these 10 orders? Can you find it without writing out all 10 prices?"
Collect both methods. Listing all ten prices gives (2 + 2 + 2 + 2 + 2 + 3 + 3 + 3 + 5 + 5)/10 = 29/10 = $2.90. Grouping gives (5 · 2 + 3 · 3 + 2 · 5)/10 = $2.90. Then split the fraction: 2(5/10) + 3(3/10) + 5(2/10) = 2(0.5) + 3(0.3) + 5(0.2) = 2.90. Write this last form in a box and leave it on the board: each value is multiplied by the fraction of the time it occurs. That is the whole idea of expected value.
Direct Instruction20 minutes
Part 1: The formula. If a random variable X has values x1, x2, ..., xn with probabilities p1, p2, ..., pn, its expected value is E(X) = x1p1 + x2p2 + ... + xnpn. It is also written μ or μX, the Greek letter used for a mean. Connect it to the warm-up: there the fractions 0.5, 0.3 and 0.2 were relative frequencies of past orders; in a probability distribution they are probabilities. Give students the procedure:
Write the distribution as a table with one row per value and check that the probabilities add to 1.
Add a column for x · P(X = x) and fill it in, keeping negative signs.
Add that column: the total is E(X).
Check that the answer is reasonable: E(X) must lie between the smallest and largest values, and near where the histogram would balance.
Interpret it in context, with units: "Over many repetitions, the average value of X is about E(X)."
Equally likely values
Roll a fair number cube once; X is the number rolled. Each value from 1 to 6 has probability 1/6.
Equation: E(X) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5, a long-run average that no single roll can show
A table of probabilities
X is the number of library books a randomly chosen student checks out in a week (invented model; table below).
On a game-show wheel, a contestant gains 50 points with probability 1/4, gains 10 points with probability 1/2 and loses 40 points with probability 1/4.
X has values 1, 2, 3 with probabilities 0.2, 0.5, 0.3. Picture 10 repetitions that match the probabilities exactly: 1, 1, 2, 2, 2, 2, 2, 3, 3, 3.
Equation: Data mean = 21/10 = 2.1 and E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 2.1
Expected value of the number of library books, one row per value
x (books)
P(X = x)
x · P(X = x)
0
0.15
0
1
0.30
0.30
2
0.25
0.50
3
0.20
0.60
4
0.10
0.40
Total
1.00
E(X) = 1.80
Part 2: What E(X) means. Give three readings of the same number. (1) Weighted average: each value counts in proportion to its probability, exactly like the café mean. (2) Balance point: if the bars of the probability histogram were weights on a seesaw, the seesaw would balance at E(X). Show Diagram 1: the library histogram balances at 1.8, to the right of its tallest bar at 1, because the bars at 3 and 4 pull the balance point that way. (3) Long-run average: repeat the chance process many times and average the results. Diagram 2 shows a simulation of 200 rolls of a number cube: the average jumps around at first and then settles close to 3.5.
Name two misreadings out loud. E(X) is not the most likely value (the library example: 1 book is most likely, 1.8 is expected), and it does not have to be a possible value (nobody checks out 1.8 books or rolls a 3.5).
Guided Practice15-20 minutes
Work two problems with the class, with pairs doing each step before you show it.
(1) The number of goals G a school soccer team scores in a game follows this invented model: P(G = 0) = 0.20, P(G = 1) = 0.35, P(G = 2) = 0.25, P(G = 3) = 0.15, P(G = 4) = 0.05. Pairs build the x · P column: 0, 0.35, 0.50, 0.45, 0.20, so E(G) = 1.5 goals per game. Ask: "What does 1.5 mean for a 20-game season?" (About 30 goals in total, on average.)
(2) An ice cream shop finds that a customer orders 1 scoop with probability 0.5, 2 scoops with probability 0.4, and 3 scoops otherwise. Pairs first find the missing probability (0.1), then E(S) = 0.5 + 0.8 + 0.3 = 1.6 scoops. Ask pairs to sketch the histogram and mark 1.6 with a triangle under the axis, checking that it looks like the balance point.
Watch for students who divide the x · P total by the number of values, and for students who average the values without the probabilities.
Independent Practice10-15 minutes
Students work alone on four tasks: (1) a spinner with four equal sections labeled 2, 4, 4 and 11 (E = 21/4 = 5.25); (2) the number of eggs in a bird's nest, invented model P(2) = 0.25, P(3) = 0.40, P(4) = 0.35 (E = 0.5 + 1.2 + 1.4 = 3.1 eggs); (3) a card game where a player gains 3 points with probability 0.2, scores 0 with probability 0.5 and loses 1 point with probability 0.3 (E = 0.6 - 0.3 = 0.3 points); (4) a written answer: for a distribution whose histogram is skewed right, is E(X) usually to the left or to the right of the tallest bar? Why? For each of (1) to (3), students write one sentence interpreting E in context.
Closure5 minutes
Exit ticket: (1) X takes the value 10 with probability 0.7 and the value 20 with probability 0.3. Find E(X). (Answer: 7 + 6 = 13.) (2) Finish the sentence: "E(X) = 13 means that ..." A strong answer mentions many repetitions and an average, and says that a single value of X will be 10 or 20, never 13.
Differentiation Strategies
For Struggling Students
Give a three-column template (x, P(X = x), x · P) with a total row, and have students check the probability column adds to 1 before multiplying
Start every problem by imagining 10 or 100 repetitions: "In 100 games, how many times does each value happen?" Then find the mean of those 100 imagined results
Use money contexts first, where students already have a feel for averages
For Advanced Students
Show that E(aX + b) = aE(X) + b for the number cube by computing both sides for a = 2, b = 1, then explain why it works in general
Build a distribution with three values whose expected value is 4 but whose most likely value is 1
Estimate how many rolls of a number cube it takes for the running average to stay within 0.1 of 3.5, using a spreadsheet simulation
Assessment Guidance
What to Look For
Check the x · P column first: many errors come from dropping a negative sign or from dividing the total by the number of values. Ask students to say where E(X) sits on the histogram before they calculate; an answer outside the range of the values, or far from the balance point, signals an arithmetic slip. In interpretations, look for the idea of many repetitions and an average, with units. Push back on answers such as "you will get 1.8 books" or "1.8 is the most likely number."
02
Classroom Activities
3 Activities
1
Class Mean, Then Expected Value
20 minWhole class, then pairs
The class collects one number from each student, and pairs compute its mean in two ways. Then the class treats "choose one student at random" as a chance process. The probability distribution of that student's number is exactly the class's relative frequency table, so its expected value must equal the class mean. This makes the interpretation in the standard concrete.
Procedure
Each student writes on a sticky note how many pets live in their home (or use the invented table below if the question is not suitable for your class)
Build a frequency table on the board, and pairs compute the class mean as total pets divided by the number of students
Pairs convert each frequency to a relative frequency and compute the sum of value × relative frequency
Now define X = the number of pets of one student chosen at random from this class. Pairs write the distribution of X and find E(X)
Invented Class Data Table
A class of 25 students: 8 students have 0 pets, 9 have 1, 5 have 2, 2 have 3 and 1 has 4. Total pets = 0 + 9 + 10 + 6 + 4 = 29, so the class mean is 29/25 = 1.16 pets. The distribution of X is P(0) = 8/25, P(1) = 9/25, P(2) = 5/25, P(3) = 2/25, P(4) = 1/25, and E(X) = 29/25 = 1.16 as well.
Discussion Questions
Why did the two calculations give exactly the same number?
Would the mean of another class be the same? What would change in the distribution?
Is 1.16 a number of pets any student could have? What does it describe?
Modification for Distance Learning
Collect the numbers with an anonymous online form, paste the results into a shared spreadsheet, and have pairs build both calculations in adjacent columns.
2
Chasing the Long Run
20 minPairs
Pairs spin a spinner many times, keep a running average, and watch it approach the expected value. Pooling the class's spins shows the long-run meaning of E(X) more clearly than any one pair can.
Setup
Each pair draws a spinner on paper: half of the circle is labeled 1, one quarter is labeled 2 and one quarter is labeled 9. A paper clip spun around a pencil point is the pointer
After 10, 20, 30 and 40 spins, they compute the running average and plot it against the number of spins, as in Diagram 2
The class adds all spins together on the board and computes the pooled average
Discussion Questions
After 10 spins, how far was your average from 3.25? After 40?
Did any pair's average move away from 3.25 for a while? Does that mean the expected value was wrong?
Why is 3.25 not a number the spinner can ever land on, and why is that fine?
Challenge Variation
Simulate 1,000 spins with a spreadsheet: use a random number between 0 and 1, map values below 0.5 to 1, from 0.5 to 0.75 to 2, and above 0.75 to 9, then chart the running average.
3
Expected Value Card Sort
15-20 minGroups of 3-4
Groups compute the expected value of eight distributions, order the cards from smallest to largest expected value, and find pairs of cards that share an expected value but look different. The sort pushes students to see E(X) as one summary of a distribution, not the whole story.
Each student computes two cards, and the group checks each other's work
The group orders the cards on a number line by expected value: E 0.5, A 1, B 1, C 2, H 2, D 3, G 3.5, F 5
The group sketches the histograms of A and B, and of C and H, and writes how each pair differs even though the expected values match
Challenge Variation
Each group writes a ninth card whose expected value is 1 but whose largest value is 20, then trades with another group to check it.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Expected Value as the Balance Point
The probability histogram of the library example, drawn to scale. A seesaw under the axis would balance at E(X) = 1.8, to the right of the tallest bar, because the bars at 3 and 4 are farther from the center than the bar at 0.
Diagram 2: Expected Value as a Long-Run Average
One computer simulation of 200 rolls of a fair number cube, made for this page. After 10 rolls the average was 4.0; after 200 rolls it was 3.46. Over more and more rolls the average tends to settle near E(X) = 3.5, though any single run wanders.
04
Homework Assignment
~30 min
HSS.MD.A.2 Homework: Calculating and Interpreting Expected Value
Directions: Show the distribution table and an x · P column for every calculation. Give units, and write every interpretation as a complete sentence that mentions many repetitions. All data are invented for practice.
Part 1: Calculating Expected Value (Problems 1-3)
A spinner has five equal sections labeled 1, 1, 2, 3 and 9. Let X be the number spun. Write the probability distribution of X and find E(X).
At a food truck, the number of tacos T in a randomly chosen order has this distribution: P(T = 1) = 0.25, P(T = 2) = 0.40, P(T = 3) = 0.25, P(T = 4) = 0.10. Find E(T) and interpret it. If the truck takes 150 orders on a Saturday, about how many tacos should it plan to make?
A random variable Y has the values -5, 0, 5 and 20 with probabilities 0.4, 0.3, 0.2 and k. Find k, then find E(Y).
Part 2: Interpreting Expected Value (Problems 4-6)
Find E(X) for one roll of a fair four-sided die numbered 1 to 4. A classmate says your answer must be wrong because the die can never land on it. Write a response that uses the words mean and long run.
X has the values 1, 2 and 3, each with probability 1/3. Y has the values -4, 2 and 8 with probabilities 0.25, 0.5 and 0.25. Find E(X) and E(Y), draw both probability histograms, and mark each balance point. What is the same about the two distributions, and what is different?
In a group of 20 households, 2 have no car, 6 have 1 car, 9 have 2 cars and 3 have 3 cars. (a) Find the mean number of cars per household. (b) One of these households is chosen at random, and X is its number of cars. Write the distribution of X and find E(X). (c) Explain why your answers to (a) and (b) agree.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Distribution Table
Complete, probabilities add to 1
One value or probability missing
No table
Expected Value
Correct x · P column and total
Method right, one arithmetic error
Values averaged without probabilities
Interpretation
Long-run average in context, with units
Correct idea but no context or units
Treated as a guaranteed or most likely value
Mean Connection
Explains E(X) as the mean or balance point
Connection stated without reason
No connection
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
A random variable X takes the values 1, 2 and 3 with probabilities 0.5, 0.3 and 0.2. What is E(X)?
Answer: B
E(X) = 1(0.5) + 2(0.3) + 3(0.2) = 0.5 + 0.6 + 0.6 = 1.7. Choice A is the plain mean of the three values, which ignores the probabilities. Choice C is the most likely value. Choice D divides 1.7 by 3, a step the formula does not have.
Question 2 of 20 · Multiple Choice
A spinner has four equal sections labeled 3, 3, 6 and 12. What is the expected value of one spin?
Answer: D
Each section has probability 1/4, so E = (3 + 3 + 6 + 12)/4 = 24/4 = 6. Equivalently 3(1/2) + 6(1/4) + 12(1/4) = 6. Choice A averages only the distinct values 3, 6 and 12, which ignores that 3 covers half of the spinner. Choice C forgets to multiply by the probabilities.
Question 3 of 20 · Multiple Choice
A random variable takes the values -10, 0 and 30 with probabilities 0.5, 0.3 and 0.2. What is E(X)?
Answer: C
E(X) = -10(0.5) + 0(0.3) + 30(0.2) = -5 + 0 + 6 = 1. Choice A drops the negative sign on -10, giving 5 + 6. Choice B uses only the first product. Choice D is the plain mean of the three values, 20/3.
Question 4 of 20 · Multiple Choice
X is the number of customers who arrive at a food cart in a 5-minute period, and E(X) = 2.4. Which statement is the best interpretation?
Answer: A
Expected value is the mean of the distribution, so it describes the average over many periods. Choice B treats it as a guaranteed value, and no period can have 2.4 customers. Choice C confuses the mean with the most likely value. Choice D reads it as a percent.
Question 5 of 20 · Multiple Choice
On a probability histogram, where is the expected value located?
Answer: C
E(X) is the mean of the distribution, and the mean is the balance point: each bar acts as a weight equal to its probability. Choice A describes the most likely value. Choice B is the midrange, which equals E(X) only for special shapes, such as symmetric distributions.
Question 6 of 20 · Multiple Choice
X has the values 0, 1, 2 and 3 with probabilities 0.2, 0.1, 0.3 and 0.4. What is E(X)?
Answer: A
E(X) = 0(0.2) + 1(0.1) + 2(0.3) + 3(0.4) = 0 + 0.1 + 0.6 + 1.2 = 1.9. Choice B is the plain mean of 0, 1, 2 and 3. Choice C is the most likely value. Choice D divides the correct total by 4.
Question 7 of 20 · Multiple Choice
A random variable takes only the values 2 and 8, and E(X) = 3.5. What is P(X = 8)?
Answer: D
Let p = P(X = 8). Then 2(1 - p) + 8p = 3.5, so 2 + 6p = 3.5 and p = 0.25. Check: 2(0.75) + 8(0.25) = 1.5 + 2 = 3.5. Choice A is P(X = 2), the two probabilities swapped. Choice B would give E(X) = 5, the midpoint. Choice C is 3.5/8, which assumes X = 8 is the only value.
Question 8 of 20 · Multiple Choice
X has the values 0, 1, 2, 3 and 4 with probabilities 0.40, 0.30, 0.15, 0.10 and 0.05. Which statement is correct?
Answer: D
E(X) = 0 + 0.30 + 0.30 + 0.30 + 0.20 = 1.1. The histogram is skewed right, and the long right tail pulls the balance point to the right of the tallest bar at 0. Choice A confuses the mean with the mode, and choice B with the middle of the list of values.
Question 9 of 20 · Multiple Choice
How are the expected value of a random variable and the mean of its probability distribution related?
Answer: B
They are two names for one number, E(X) = μ. The formula Σ x · P(X = x) is the mean of a frequency table with probabilities in place of relative frequencies. Choice C describes when E(X) equals the midpoint of the values, and choice D describes when E(X) equals the plain average of the listed values. Neither condition is needed: E(X) and the mean of the distribution are always the same number.
Question 10 of 20 · Multiple Choice
In a game, a player gains 4 points with probability 0.25 and loses 1 point otherwise. What is the expected number of points per play?
Answer: C
E = 4(0.25) + (-1)(0.75) = 1 - 0.75 = 0.25 points per play. Choice B ignores the loss. Choice A averages 4 and -1 as if they were equally likely. Choice D adds the two values.
Question 11 of 20 · Multiple Choice
The number of absences X per student in a week at a school has E(X) = 0.6. The school has 500 students. About how many absences in total should the school expect in a typical week?
Answer: B
E(X) is the mean number of absences per student, so the total is about 500 × 0.6 = 300. Choice A is the per-student mean, not the total. Choice D divides 500 by 0.6 instead of multiplying.
Question 12 of 20 · Multiple Choice
A fair eight-sided die numbered 1 to 8 is rolled once. What is the expected value of the roll?
Answer: C
E = (1 + 2 + ... + 8)/8 = 36/8 = 4.5. Choice D is the sum of the values without multiplying by 1/8. Choices A and B are the two middle values, but the balance point lies halfway between them.
Question 13 of 20 · Multiple Choice
Two random variables have the same expected value. Which statement must be true?
Answer: A
The expected value is the balance point of the histogram, so equal expected values mean equal balance points. Nothing else has to match: for example, the value 3 with probability 1 and the values 0 and 6 with probability 1/2 each both have E = 3, but different values, modes and shapes.
Question 14 of 20 · Multiple Choice
X has the values 10, 20 and 30 with probabilities 0.2, 0.5 and 0.3. A student writes E(X) = (10 + 20 + 30)/3 = 20. What is the error?
Answer: D
E(X) = 10(0.2) + 20(0.5) + 30(0.3) = 2 + 10 + 9 = 21. The plain average works only when all values are equally likely, and here 30 is more likely than 10, which pulls the mean above 20. Choice C adds the probabilities, which always gives 1 for any distribution.
Question 15 of 20 · Short Answer
A random variable X has the values 1, 4 and 9 with probabilities 0.5, 0.25 and 0.25. Find E(X).
E(X) = 1(0.5) + 4(0.25) + 9(0.25) = 0.5 + 1 + 2.25 = 3.75. It lies between 1 and 9, closer to 1 because half of the probability sits there.
Question 16 of 20 · Short Answer
At a bakery, the number of cupcakes C bought by a customer has the distribution P(C = 1) = 0.45, P(C = 2) = 0.25, P(C = 6) = 0.20 and P(C = 12) = 0.10. Find E(C) and interpret it. Can a single customer buy E(C) cupcakes?
E(C) = 0.45 + 0.50 + 1.20 + 1.20 = 3.35 cupcakes. Over many customers, the bakery sells an average of about 3.35 cupcakes per customer. No single customer buys 3.35 cupcakes; it is a mean, and it is larger than the most likely value (1) because a few customers buy 6 or 12.
Question 17 of 20 · Short Answer
Explain why the expected value of a random variable must be between its smallest and largest values.
E(X) is a weighted average: the probabilities are nonnegative and add to 1, so E(X) = Σ x · p is at least (smallest value)(Σ p) = smallest value and at most (largest value)(Σ p) = largest value. On the histogram, a balance point cannot lie outside all of the weights.
Question 18 of 20 · Short Answer
A random variable X takes the values 0, 1 and 2. You know that P(X = 0) = 0.3 and E(X) = 1.2. Find P(X = 1) and P(X = 2).
Let a = P(X = 1) and b = P(X = 2). Then a + b = 0.7 and E(X) = a + 2b = 1.2. Subtracting gives b = 0.5, so a = 0.2. P(X = 1) = 0.2 and P(X = 2) = 0.5. Check: 0(0.3) + 1(0.2) + 2(0.5) = 1.2.
Question 19 of 20 · Short Answer
The expected value of one roll of a fair twelve-sided die numbered 1 to 12 is 6.5. A student rolls it four times and gets 2, 3, 1 and 5, an average of 2.75. Does this show that the expected value is wrong? Explain.
No. First confirm E = (1 + 2 + ... + 12)/12 = 78/12 = 6.5. Expected value describes the average over many rolls. Four rolls are far too few: averages of a few rolls vary a lot, and an average of 2.75 is not surprising. Over hundreds of rolls, the average should settle near 6.5.
Question 20 of 20 · Short Answer
Spinner P lands on 0 or 10, each with probability 1/2. Spinner Q lands on 4, 5 or 6, each with probability 1/3. Find the expected value of each spinner. On a typical spin, which spinner gives results closer to its expected value?
E(P) = 0(1/2) + 10(1/2) = 5 and E(Q) = (4 + 5 + 6)/3 = 5, so both expected values are 5. Spinner Q always lands within 1 of 5, while spinner P always lands exactly 5 away from it. Spinner Q gives results closer to its expected value: equal means do not say how spread out the values are.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSS.MD.A.2 mean?
HSS.MD.A.2 means students can calculate the expected value of a random variable and explain what it tells you. They multiply each value by its probability and add, and they interpret the result as the mean of the probability distribution: the average value over many repetitions of the chance process.
How do you calculate expected value?
Multiply each value of the random variable by its probability, then add the products. For a variable with values 0, 3 and 5 and probabilities 0.6, 0.3 and 0.1, the expected value is 0(0.6) + 3(0.3) + 5(0.1) = 0 + 0.9 + 0.5 = 1.4. A table with columns x, P(X = x) and x · P(X = x) keeps the work organized.
Why is expected value called the mean of the distribution?
Because it is computed the same way as the mean of a frequency table. The mean of data can be written as the sum of each value times the fraction of the data with that value. In a probability distribution, those fractions are replaced by probabilities, which are the long-run fractions. So E(X) is the mean you would get from an ideal, very long run of the chance process.
Can the expected value be a number that cannot actually happen?
Yes, and it often is. One roll of a number cube has expected value 3.5, but no face shows 3.5. The expected value is an average over many trials, just as a data set of car trips can have a mean of 1.3 riders per car, so it does not need to be a possible value of a single trial.
Is HSS.MD.A.2 taught in Algebra 2 or Statistics?
It varies by school. HSS.MD.A.2 is marked (+), the Common Core label for additional mathematics that students need for advanced courses. It is usually taught in a statistics course or Precalculus, and some Algebra II courses include expected value in a probability unit.
Is the expected value the same as the most likely value?
No. The most likely value is the one with the largest probability (the tallest bar), while the expected value is the balance point of all the bars. They match for some symmetric distributions with a single peak, but in a skewed distribution the expected value is pulled toward the long tail.
What does the notation E(X) or μ mean?
Both stand for the expected value of the random variable X. E(X) is read "the expected value of X," and μ (mu), sometimes written μX, is the Greek letter statisticians use for the mean of a population or a probability distribution. The sample mean of data is written x̄ instead.
What are common mistakes when finding expected value?
Averaging the values without using the probabilities, dividing the final sum by the number of values, dropping negative signs on losses, and using a distribution whose probabilities do not add to 1. In interpretation, a common error is to say the variable "will be" the expected value on the next trial.
Where is expected value used outside the classroom?
Anywhere an average outcome over many cases matters. Insurance companies set prices using expected claim costs, stores estimate how much stock they need from the expected number of items per order, and game designers check the expected points per turn. HSS.MD.B.5 and HSS.MD.B.7 use expected value to compare decisions.
How does HSS.MD.A.2 connect to other standards?
It builds on HSS.MD.A.1, where students define random variables and graph their distributions, and on the mean of data from HSS.ID.A.2. The next standards in the cluster, HSS.MD.A.3 and HSS.MD.A.4, ask students to build distributions from theoretical and from empirical probabilities and find their expected values. The HSS.MD.B cluster then uses expected values to weigh decisions.
07
Related Standards
6 standards
These standards connect to HSS.MD.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSS.MD.A.1Prerequisite
Define a random variable on a sample space and graph its probability distribution
