Module 3 — Discrete Distributions & Group Comparisons

Let \( X \) be the number of retakes a randomly chosen student needs after the first sitting, recorded as \( 0, 1, 2, 3, 4 \) (where 0 means the student passed on the first sitting and needs no retake). A probability table assigns one probability to each value:

That row is the PMF. Formally, the probability mass function is \( p(x) = P(X = x) \), and a table is a valid PMF exactly when \( 0 \le p(x) \le 1 \) for every value and \( \sum_x p(x) = 1 \). Here \( 0.08 + 0.20 + 0.34 + 0.26 + 0.12 = 1.00 \), so the table is valid. If the total were \( 0.97 \) or \( 1.05 \), nothing downstream — expected value, spread, or any chart — would be interpretable until the table was repaired.

If you worked Module 2's raw-tally example — the observed attempts \( 1, 1, 2, 3, 3, 3 \) became a table by tallying each value and dividing by \( n = 6 \) — you have already built a probability table the long way: count, then divide. The table above hands you the probabilities directly, and the two validity checks are the same either way. In Module 2 the entries came from data; here they describe a model of the process. The reactive probability table lab edits this same table live.

The cumulative distribution function answers “what is the probability of at most this many retakes?”: \( F(x) = P(X \le x) = \sum_{k \le x} p(k) \). You get it by running a cumulative sum down the PMF:

Three checks that the CDF is right: it never decreases, it ends at exactly 1, and each PMF value can be recovered as a difference of neighbours, \( p(x) = F(x) - F(x-1) \). For example \( P(X = 2) = F(2) - F(1) = 0.62 - 0.28 = 0.34 \). For a discrete variable the CDF is a step function — flat between the allowed values, jumping by \( p(x) \) at each one — which is what the right-hand panel of the lab draws.

Reading probabilities from the CDF instead of adding PMF cells:

• \( P(X \le 2) = F(2) = 0.62 \) — 62% of students need at most two retakes.
• \( P(X > 2) = 1 - F(2) = 0.38 \).
• \( P(1 \le X \le 3) = F(3) - F(0) = 0.88 - 0.08 = 0.80 \). Subtract \( F \) at the value just below the lower limit, because \( F(1) \) would remove \( P(X = 1) \) itself. Unpacked, the difference keeps exactly the interval's PMF cells — \( P(X=1) + P(X=2) + P(X=3) = 0.20 + 0.34 + 0.26 = 0.80 \) — so the subtraction rule and adding the cells are the same computation seen two ways.

The expected value is the probability-weighted average of the values, \( E(X) = \sum_x x\,p(x) \):

\[ E(X) = 0(0.08) + 1(0.20) + 2(0.34) + 3(0.26) + 4(0.12) = 0.20 + 0.68 + 0.78 + 0.48 = 2.14 \]

Over many students the mean number of retakes is 2.14. Note that \( E(X) = 2.14 \) is not a value \( X \) can take — an expected value is a long-run average, not a prediction for one student.

The variance uses the shortcut \( \operatorname{Var}(X) = E(X^2) - [E(X)]^2 \). First \( E(X^2) = \sum_x x^2 p(x) \):

\[ E(X^2) = 0(0.08) + 1(0.20) + 4(0.34) + 9(0.26) + 16(0.12) = 0.20 + 1.36 + 2.34 + 1.92 = 5.82 \]

\[ \operatorname{Var}(X) = 5.82 - 2.14^2 = 5.82 - 4.5796 = 1.2404, \qquad \mathrm{SD}(X) = \sqrt{1.2404} \approx 1.114 \]

Interpretation: over many students the mean number of retakes is 2.14, with a standard deviation of about 1.1 retakes. Open the reactive table lab without changing anything and confirm the rounded story: total probability is 1, the mean is about 2.14 retakes, the variance is about 1.24, and the standard deviation is about 1.11.

Module 3 asks you to defend or reject a model, which only makes sense once you separate two objects that both get drawn as bar charts:

Worked comparison. Suppose 40 students at one campus needed the following retakes: 0 retakes for 6 students, 1 for 11, 2 for 10, 3 for 8, 4 for 3, and exactly 5 for 2. The empirical table divides each count by 40; the sample mean is \( \bar{x} = 77/40 = 1.925 \), so a Poisson model with \( \hat{\lambda} = 1.925 \) (computed with dpois(0:5, 1.925)) gives:

Across the displayed exact counts, the two rows disagree by at most about 0.03. The fitted Poisson model also assigns about 0.014 probability to counts above 5; the observed sample has none. A single student changes an empirical cell by 0.025 when \( n = 40 \), so the displayed differences are consistent with ordinary sampling variation. Thus, “the Poisson model is consistent with these data” is defensible — not a proof that retake counts are Poisson. A model is challenged when the empirical row shows a feature the model cannot readily produce: a second peak, far more zeros than \( e^{-\hat{\lambda}} \) allows, or variance far larger than the mean. For a Poisson model, \( \operatorname{Var}(X) = \lambda \), so \( s^2 \gg \bar{x} \) is an overdispersion warning. Say which checks you used in your Project 1 defense.

# PMF -> CDF -> E(X) -> Var(X), from a probability table
x <- 0:4
p <- c(0.08, 0.20, 0.34, 0.26, 0.12)
sum(p)                        # validity check: must print 1
cdf <- cumsum(p); cdf         # F(x): 0.08 0.28 0.62 0.88 1.00
ex  <- sum(x * p); ex         # E(X) = 2.14
vx  <- sum(x^2 * p) - ex^2    # Var(X) = 1.2404
c(variance = vx, sd = sqrt(vx))

# Empirical table versus a fitted Poisson model
attempts <- rep(0:5, times = c(6, 11, 10, 8, 3, 2))   # the 40-student sample above
n <- length(attempts)
empirical <- as.numeric(table(factor(attempts, levels = 0:5))) / n
lambda_hat <- mean(attempts); lambda_hat            # 1.925
model <- dpois(0:5, lambda = lambda_hat)
round(rbind(x = 0:5, empirical = empirical, poisson = model), 3)
ppois(5, lambda = lambda_hat, lower.tail = FALSE) # model probability above 5: about 0.014
c(sample_mean = mean(attempts), sample_var = var(attempts))  # Poisson needs these close

Every line prints something you can quote in the “Discrete distribution evidence” field of the graded assignment: the validity total, the CDF, \( E(X) \), \( \operatorname{Var}(X) \), the empirical-versus-model table, and the mean-versus-variance check.