math-2017:S.MD.1math-2017:S.MD.2
math-2017:S.MD.1 Define a random variable for a quantity of interest by assigning a numerical value to each event in a sample space; graph the corresponding probability distribution using the same graphical displays as for data distributions.
math-2017:S.MD.2 Calculate the expected value of a random variable; interpret it as the mean of the probability distribution.
Learning goal: Define a random variable for a quantity of interest by assigning a numerical value to each outcome or event in a sample space, graph the corresponding probability distribution with the same displays used for data distributions, calculate the expected value of the random variable, and interpret it as the mean of the probability distribution.
Scope note. S.MD.1 and S.MD.2 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.
From outcomes to numbers. Many questions about a chance process concern a number rather than an outcome: how many heads, what total, how much money. A random variable is a rule that assigns a numerical value to each outcome of the sample space. It is usually written with a capital letter such as X, and its possible values with lowercase letters or plain numbers. Defining a random variable is a modeling decision: you decide which quantity is of interest and write down its value for every outcome.
Toss three fair coins. The sample space has 8 equally likely outcomes: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. Let X be the number of heads. Then X assigns 3 to HHH; 2 to HHT, HTH and THH; 1 to HTT, THT and TTH; and 0 to TTT. The event "X = 2" is the subset {HHT, HTH, THH}, so P(X = 2) = 3/8. In this way every value of X names an event, and the probability of the value is the probability of that event.
The probability distribution. The probability distribution of a discrete random variable lists each possible value together with its probability. For X, the number of heads in three tosses:
P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8.
Two properties must hold for any probability distribution: every probability is between 0 and 1, and the probabilities add to exactly 1, since the values of X partition the sample space. Here 1/8 + 3/8 + 3/8 + 1/8 = 8/8. If your list does not add to 1, an outcome has been missed or double counted.
Graphing the distribution. A probability distribution is displayed exactly as a data distribution is: a bar graph (probability histogram) with the values of X on the horizontal axis and probability on the vertical axis, or a dot plot with heights proportional to probability. The shape can be read as for data: the three-coin distribution is symmetric about 1.5, with its two tallest bars at 1 and 2. The only difference from a data histogram is what the heights mean. In a data histogram a bar height is a frequency or relative frequency in a particular sample; in a probability histogram it is the model's probability, what the relative frequency would settle toward over a very long run of trials.
Expected value. The expected value of X, written E(X), is the weighted average of its values, each weighted by its probability:
E(X) = Σ x · P(X = x), summed over every possible value x.
For three coins: E(X) = 0 · 1/8 + 1 · 3/8 + 2 · 3/8 + 3 · 1/8 = (0 + 3 + 6 + 3)/8 = 12/8 = 1.5 heads.
Interpreting E(X) as the mean of the distribution. The formula is the same computation as the mean of a data set organized in a frequency table, where each value is multiplied by its relative frequency and the products are added. The expected value is therefore the mean of the probability distribution: the balance point of the probability histogram, the point at which the bars would balance if probability were weight. It is also the long-run average: if three coins were tossed a great many times and the number of heads recorded each time, the average of the recorded numbers would settle near 1.5. Note that 1.5 is not a possible value of X. Expected value is a mean, and means need not be attainable values; no toss of three coins can show 1.5 heads, just as the mean number of children per family can be 2.4 although no family has 2.4 children.
A fully worked model: the sum of two dice. Let T be the sum when a red and a blue die are rolled. The sample space has 36 equally likely ordered pairs; T takes the values 2 through 12. Counting pairs for each sum: T = 2 has 1 pair, 3 has 2, 4 has 3, 5 has 4, 6 has 5, 7 has 6, 8 has 5, 9 has 4, 10 has 3, 11 has 2, 12 has 1. Total 36, so the distribution is P(T = t) = (number of pairs)/36. The histogram is a symmetric triangle peaking at 7. The expected value:
E(T) = (2·1 + 3·2 + 4·3 + 5·4 + 6·5 + 7·6 + 8·5 + 9·4 + 10·3 + 11·2 + 12·1)/36 = (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)/36 = 252/36 = 7.
E(T) = 7 is also visible from symmetry: the distribution is symmetric about 7, so its balance point is 7. Alternatively, each die has expected value (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5 and the expected sum is 3.5 + 3.5 = 7; expected values add even when the variables are not independent, a fact that often shortens the arithmetic.
Random variables with money values. A random variable can assign any numbers, not just counts. Suppose a hypothetical game pays 10 dollars if a fair die shows 6, 2 dollars if it shows 4 or 5, and nothing otherwise. Let W be the payout. P(W = 10) = 1/6, P(W = 2) = 2/6, P(W = 0) = 3/6. E(W) = 10 · 1/6 + 2 · 2/6 + 0 · 3/6 = 14/6 ≈ 2.33 dollars. The mean payout per play, over a long run, is about 2.33 dollars, although no single play ever pays 2.33.
Misconceptions. The first is treating the expected value as the most likely value. For T the two agree (7 is both), but for W the most likely payout is 0 while E(W) ≈ 2.33. The expected value is a mean, not a mode. The second is computing E(X) as the plain average of the possible values, (0 + 1 + 2 + 3)/4 = 1.5 for the coins, which happens to give the right answer here because this distribution is symmetric about 1.5 (other distributions can also agree by coincidence, but the method is not valid in general); for W the plain average (10 + 2 + 0)/3 = 4 is wrong. Always weight by probability.
Worked distribution from a sample space: Let D be the larger of the two numbers when two dice are rolled (the value shown if they tie). D = 1 only for (1, 1): 1 pair. D = 2 for pairs with max 2: (1, 2), (2, 1), (2, 2): 3 pairs. In general D = k has 2k − 1 pairs: 1, 3, 5, 7, 9, 11 for k = 1 to 6, total 36. E(D) = (1·1 + 2·3 + 3·5 + 4·7 + 5·9 + 6·11)/36 = (1 + 6 + 15 + 28 + 45 + 66)/36 = 161/36 ≈ 4.47. The bars rise toward 6, so the distribution is skewed left (its tail is toward the small values), and its mean, 4.47, lies below the most likely value, 6.
Worked expected value as a long-run average: A fair die is rolled; X is the number shown. E(X) = (1 + 2 + 3 + 4 + 5 + 6)/6 = 3.5. If the die is rolled 600 times, the total of all rolls is expected to be about 600 × 3.5 = 2,100, and the average roll about 3.5, although any particular 600 rolls will give an average somewhat above or below.
Worked money variable: A hypothetical raffle sells 500 tickets; one wins 300 dollars, two win 50 dollars each, five win 10 dollars each, and the rest win nothing. Let V be the prize on one random ticket. P(V = 300) = 1/500, P(V = 50) = 2/500, P(V = 10) = 5/500, P(V = 0) = 492/500. E(V) = (300 + 100 + 50)/500 = 450/500 = 0.90 dollars. The mean prize per ticket is 90 cents; the most likely prize is 0.
Represent and record: Write out the eight outcomes of three coin tosses and the value of X, the number of heads, beside each. Build the distribution table, draw the probability histogram with the vertical axis from 0 to 1/2, and mark the balance point at 1.5. Then do the same for T, the sum of two dice, using the 36-cell grid to count each sum, and mark the balance point at 7.
Independent performance: Choose a chance process with at least 12 equally likely outcomes and define two different random variables on it (for example, with two dice: the product, and the absolute difference). For each, list the distribution, verify the probabilities sum to 1, draw the probability histogram, compute the expected value and interpret it as the mean of the distribution and as a long-run average. Success criteria: every value of the random variable is matched to its subset of outcomes; the histogram uses the same conventions as a data histogram and its bars sum to 1; the expected value is computed as a probability-weighted sum, shown in full; one sentence states whether the expected value is an attainable value and why that does not matter.
An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.
What is a random variable, and how does the event "X = 2" arise from it when X counts heads in three coin tosses?
A random variable assigns a number to every outcome of the sample space. X = 2 is the set of outcomes assigned the value 2: {HHT, HTH, THH}. Its probability, 3/8, is the probability of that event.
Write the probability distribution of X, the number of heads in three fair coin tosses, and verify it is a valid distribution.
P(X = 0) = 1/8, P(X = 1) = 3/8, P(X = 2) = 3/8, P(X = 3) = 1/8. Each probability lies between 0 and 1, and the sum is 8/8 = 1, because the four values partition the eight equally likely outcomes.
Compute E(X) for the three-coin variable and explain why 1.5 is a sensible answer even though no toss produces 1.5 heads.
E(X) = 0(1/8) + 1(3/8) + 2(3/8) + 3(1/8) = 12/8 = 1.5. Expected value is the mean of the distribution, the balance point of its histogram and the long-run average number of heads per three tosses; a mean need not be an attainable value.
Let T be the sum of two dice. Find E(T) from the distribution, and give a second argument for the same value.
E(T) = (2·1 + 3·2 + … + 12·1)/36 = 252/36 = 7. Alternatively, the distribution is symmetric about 7, so its balance point is 7; or each die has mean 3.5 and expected values add, giving 3.5 + 3.5 = 7.
How does a probability histogram differ from a data histogram, and how is it the same?
It uses the same display: values on the horizontal axis, bars whose heights are compared and whose shape (symmetric, skewed) is read the same way. The difference is meaning: a data histogram shows frequencies from one sample; a probability histogram shows the model probabilities, which the relative frequencies approach over a long run.
A hypothetical game pays 10 dollars on a 6, 2 dollars on a 4 or 5, and 0 otherwise on one fair die. Find the expected payout and compare it with the most likely payout.
E(W) = 10(1/6) + 2(2/6) + 0(3/6) = 14/6 ≈ 2.33 dollars. The most likely payout is 0 dollars, with probability 1/2. Expected value is a mean, not the most probable value.
A student finds E(X) for the three-coin variable as (0 + 1 + 2 + 3)/4 = 1.5 and claims the method is correct. Explain the flaw using the dice-payout variable.
The method ignores the probabilities; it gave the right answer for the coins because that distribution happens to be symmetric about 1.5: the values 0, 1, 2 and 3 are evenly spaced about 1.5 and their probabilities 1/8, 3/8, 3/8 and 1/8 mirror about it too. It is not a valid method in general. For the payout variable it gives (10 + 2 + 0)/3 = 4 dollars, but the correct expected value is 14/6 ≈ 2.33 dollars. Expected value must weight each value by its probability.