Develop theoretical and empirical probability distributions and find their expected values

Learning goal: Develop a probability distribution for a random variable defined on a sample space in which theoretical probabilities can be calculated, and for one in which probabilities are assigned empirically from data; find the expected value in each case and use it to answer questions such as an expected grade under a grading scheme or an expected total across many households.

Scope note. S.MD.2, S.MD.3 and S.MD.4 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.

Two sources of probabilities. A probability distribution can be built in two ways. When the sample space has a known structure, such as fair coins, dice or random guessing among equally likely choices, the probabilities are theoretical: they follow from counting and the rules of probability, and they are exact within the model. When no such structure is available, probabilities are assigned empirically, as relative frequencies observed in data; they are estimates whose quality depends on how the data were collected and how much there is. The expected value is computed the same way in both cases, as Σ x · P(X = x), but its meaning inherits the source: exact for a theoretical model, estimated for an empirical one.

A theoretical distribution: guessing on a quiz. A quiz has five multiple-choice questions, each with four choices and exactly one correct answer. A student who has not studied guesses independently on every question, choosing each option with probability 1/4. Let X be the number of correct answers, from 0 to 5. Each question is correct with probability 1/4 and incorrect with probability 3/4, independently.

The probability of a particular pattern with k correct and 5 − k incorrect is (1/4)ᵏ (3/4)⁵⁻ᵏ by the multiplication rule for independent events. The number of patterns with exactly k correct is 5Ck, the number of ways to choose which k questions are the correct ones. So

P(X = k) = 5Ck · (1/4)ᵏ · (3/4)⁵⁻ᵏ, for k = 0, 1, …, 5.

Working over the common denominator 4⁵ = 1,024: P(X = 0) = 1 · 243/1,024; P(X = 1) = 5 · 81/1,024 = 405/1,024; P(X = 2) = 10 · 27/1,024 = 270/1,024; P(X = 3) = 10 · 9/1,024 = 90/1,024; P(X = 4) = 5 · 3/1,024 = 15/1,024; P(X = 5) = 1/1,024. The numerators 243 + 405 + 270 + 90 + 15 + 1 = 1,024, so the distribution is valid. As decimals: 0.237, 0.396, 0.264, 0.088, 0.015, 0.001. The most likely score is 1, and the chance of three or more correct is 106/1,024 ≈ 0.104.

Expected value of the theoretical distribution. E(X) = (0·243 + 1·405 + 2·270 + 3·90 + 4·15 + 5·1)/1,024 = (405 + 540 + 270 + 60 + 5)/1,024 = 1,280/1,024 = 1.25. A guesser expects 1.25 correct answers out of 5, which is 5 × 1/4, as one would hope: the expected number of successes in n independent trials with success probability p is n · p.

Expected grade under various grading schemes. A grading scheme is itself a random variable, a function of X, and its expected value follows from the distribution of X.

Scheme 1: 20 points per correct answer, out of 100. Grade G₁ = 20X, so E(G₁) = 20 · E(X) = 25 points. A pure guesser expects 25 out of 100.

Scheme 2: 1 point per correct answer, minus 1/3 point per wrong answer. G₂ = X − (5 − X)/3. E(G₂) = E(X) − (5 − E(X))/3 = 1.25 − 3.75/3 = 1.25 − 1.25 = 0. The penalty of 1/3 per wrong answer is chosen exactly so that guessing among four options has expected score 0; that is the purpose of choosing this particular penalty. Checking against the distribution directly: the scores for k = 0 to 5 are −5/3, −1/3, 1, 7/3, 11/3, 5, and the weighted sum is (−5·243 − 1·405 + 3·270 + 7·90 + 11·15 + 15·1)/(3 · 1,024) = (−1,215 − 405 + 810 + 630 + 165 + 15)/3,072 = 0/3,072 = 0.

Scheme 3: 4 points per correct, minus 1 per wrong. G₃ = 4X − (5 − X) = 5X − 5. E(G₃) = 5 · 1.25 − 5 = 1.25. The guesser expects a small positive score under this scheme, so it discourages guessing less strongly than Scheme 2. Comparing schemes by expected guess score is one way a teacher could decide which one to use.

An empirical distribution: television sets per household. Now suppose the quantity of interest is the number of television sets in a randomly chosen household. There is no theoretical model for this; the probabilities must come from data. The following survey is hypothetical, invented for this lesson; it is not a real national statistic, and a student working the real problem should locate a current published data distribution and cite its source. Suppose 400 randomly selected households reported: 0 sets, 20 households; 1 set, 120; 2 sets, 160; 3 sets, 72; 4 sets, 28. Let Y be the number of sets in a randomly chosen household. The empirical distribution assigns each value its relative frequency: P(Y = 0) = 20/400 = 0.05, P(Y = 1) = 0.30, P(Y = 2) = 0.40, P(Y = 3) = 0.18, P(Y = 4) = 0.07. Sum: 1.00.

Expected value of the empirical distribution. E(Y) = 0(0.05) + 1(0.30) + 2(0.40) + 3(0.18) + 4(0.07) = 0 + 0.30 + 0.80 + 0.54 + 0.28 = 1.92 sets per household. This is the sample mean of the survey, and it is an estimate of the population mean. In 100 randomly selected households from the same population you would expect about 100 × 1.92 = 192 sets in total, not exactly 192, and the estimate is only as good as the survey behind it.

What differs between the two cases. In the quiz model, 1.25 is exact given the assumptions (independent guesses, four equally likely choices). If the assumptions fail, for example because a student can eliminate one option, the model changes and so does the answer. In the household model, 1.92 is an estimate; a second survey of 400 would give a slightly different distribution and mean, and a survey that was not random, say of households near an electronics store, could be badly wrong for the population. State which kind of distribution you have built, and state its assumptions or its data source.

Misconception. Students sometimes treat an empirical distribution as exact because the arithmetic is exact. The arithmetic is exact for the sample; the claim about the population is not. Conversely, a theoretical distribution is only as good as its assumptions. Both deserve a stated condition: "assuming independent guesses with four equally likely options" or "based on this survey of 400 households".

Worked theoretical distribution with a different n: Three questions, each with four choices, guessed independently. P(X = 0) = (3/4)³ = 27/64; P(X = 1) = 3(1/4)(3/4)² = 27/64; P(X = 2) = 3(1/4)²(3/4) = 9/64; P(X = 3) = 1/64. Sum 64/64. E(X) = (0·27 + 1·27 + 2·9 + 3·1)/64 = 48/64 = 0.75 = 3 × 1/4, as the n·p rule predicts. Under the 1 point per correct, −1/3 per wrong scheme, E = 0.75 − (3 − 0.75)/3 = 0.75 − 0.75 = 0, again zero.

Worked empirical distribution from a frequency table: A hypothetical count of cars per household in 50 homes: 0 cars in 5, 1 car in 20, 2 cars in 18, 3 cars in 7. Relative frequencies: 0.10, 0.40, 0.36, 0.14, sum 1.00. E(C) = 0(0.10) + 1(0.40) + 2(0.36) + 3(0.14) = 0.40 + 0.72 + 0.42 = 1.54 cars per household. Expected total in 200 similar households: about 308 cars. The estimate rests on 50 homes and on how they were chosen.

Worked expected total from a theoretical model: If 40 students each guess on all five quiz questions, the expected total number of correct answers across the class is 40 × 1.25 = 50, because expected values add across students. Under Scheme 2 the expected class total is 40 × 0 = 0 points, though individual students will score above and below zero.

Represent and record: Build the six-row table for X, the number of correct guesses on the five-question quiz: k, 5Ck, (1/4)ᵏ(3/4)⁵⁻ᵏ as a fraction over 1,024, and P(X = k). Draw the probability histogram. Compute E(X) in a fourth column of products and confirm 1.25. Then add three columns for the three grading schemes, listing the score at each k and the expected grade for each scheme. Separately, tabulate the hypothetical television-set survey with relative frequencies and compute E(Y) = 1.92.

Independent performance: Part 1: for a quiz of four true-false questions guessed independently, develop the theoretical distribution of the number correct, find its expected value, and find the expected grade under a scheme of your own design that makes guessing worth exactly 0. Part 2: find a current, published data distribution for a count per household or per person (cite the source and date) or collect one from at least 30 classmates or families; build the empirical distribution, find the expected value, and state the expected total for 100 randomly chosen units. Success criteria: theoretical probabilities are computed from combinations and powers and sum to 1; the guessing-worth-zero scheme is verified by computation; the empirical source is cited or the collection method described; each expected value is labeled exact-within-model or estimate.