PR-4: Solutions — Discrete Random Variables

Problem 1 — Discrete or Continuous? (Variants 0–4)

• Variant 0 (help-desk calls per shift): discrete — counts are whole numbers (0, 1, 2, …); you can’t receive 2.5 calls.
• Variant 1 (student height in cm): continuous — any value in an interval (170.314 cm…); rounding is a measurement choice, not a property.
• Variant 2 (defectives in a sample of 50): discrete — a whole number from 0 to 50.
• Variant 3 (waiting time in minutes): continuous — any non-negative real (2.371 minutes).
• Variant 4 (number of siblings): discrete — a whole number; you can’t have 1.5 siblings.

Problem 2 — Is This PMF Valid? (Variants 0–4)

• Variant 0 (): not valid — sum . Fix: e.g. set .
• Variant 1 (): valid — non-negative, sum . A PMF need not be symmetric.
• Variant 2 (): not valid — . The sum is 1, but a negative probability is impossible; restructure entirely.
• Variant 3 (): valid — non-negative, sum . Outcomes need not be consecutive.
• Variant 4 (): not valid — sum . Negative outcome values () are fine; the error is the sum. Fix: .

Problem 3 — Compute E(X) (Generator)

Multiply each outcome by its probability and sum.

Problem 4 — Compute Var(X) and σ (Generator)

1. .
2. — square each first, then weight.
3. .
4. .

Problem 1 — Full E(X) + Var(X) Workflow (Generator)

(a) Verify the PMF (, ). (b) . (c) . (d) . (e) . (f) Interpret: “the variable typically deviates from the mean of by about units.”

Problem 2 — CDF Practice (Variants 0–4)

Variant 0 — on :

0	0.05
1	0.25
2	0.70
3	0.95
4	1.00

(b) . (c) . (d) .

Variant 1 — on :

1	0.10
2	0.35
3	0.65
4	0.90
5	1.00

(b) . (c) (symmetric about ). (d) .

Variant 2 — on :

0	0.30
2	0.70
4	0.90
6	1.00

(b) . (c) . (d) .

Variant 3 — on :

−2	0.10
−1	0.30
0	0.70
1	0.90
2	1.00

(b) . (c) . (d) .

Variant 4 — on :

5	0.40
10	0.70
15	0.90
20	1.00

(b) . (c) . (d) .

Problem 3 — Is the Game Fair? (Variants 0–4)

• Variant 0 (raffle, 100 tickets, prizes $200 and $50, ticket $5): net , . → unfavorable (lose $2.50/ticket).
• Variant 1 (card game, pay $3, win $15 for ace, $5 for face): net , . → unfavorable (lose ~69¢/play).
• Variant 2 (Find the Error) (pay $10, prizes $0/$20/$50 at 0.60/0.30/0.10): net . The student’s and “favorable” conclusion are correct. The error is the variance: they wrote . Correct: , so , — the spread is huge despite the favorable mean.
• Variant 3 (4-sided die, pay $2, win $6 on 4, $2 on 3): net , . → perfectly fair.
• Variant 4 (coin flip, win $5 heads / lose $5 tails, +$1 bonus): net , . → favorable (the $1 bonus tips it).

Problem 4 — Find the Missing Probability, Then Compute E(X) (Generator)

1. Use : .
2. Substitute back, then compute .

Problem 5 — Compute E(X) and σ, Then Interpret (Generator)

After computing and : interpret as the long-run average per period, and as the typical deviation from that average.

Problem 6 — Multi-Step Synthesis

Let = number of defective items in a box of 5.

	0	1	2	3	4	5
	0.60	0.20	0.10	0.06	0.03	0.01

(a) Verify: ✓; all non-negative ✓.

(b) E(X): — about 0.75 defectives per box (≈75 per 100 boxes).

(c) Var(X) and σ: ; ; .

(d) : — a 20% chance a box is flagged.

Problem 1 — Feynman Test: Why E(X) ≠ Mode

is a probability-weighted average of all outcomes; the mode is just the most probable single outcome. Because accounts for both the likelihood and the magnitude of every outcome, the two measure different things.

Example: a lottery winning $1,000 with probability 0.001 and losing $1 with probability 0.999 has mode −$1 but . Classic example: a fair die has , which isn’t even an outcome (and has no mode). The expected value is the distribution’s balance point — no obligation to coincide with the tallest bar.

Problem 2 — Apply: Medical Testing Revenue

PMF for = tests per order: on ; revenue $80 per test.

MCQ 1: compute first, then multiply by 80 (linearity ).

MCQ 2: tests, so $160.

Variance: ; ; , so $80. The hospital averages $160 per order with SD $80.

Problem 3 — Error Analysis

PMF: , each probability . The student found ✓ but wrote ✗ — they never computed .

Correct: ; ; . The correct variance is ~2.67, not 16 (overestimated 6×).

Path A — The Analyst: Carnival Game

Costs $5 to play. Prizes: $20 (0.05), $10 (0.15), $3 (0.30), $0 (0.50).

Task 1 — PMF for net winnings:

(net)
+15 ($20 − $5)	0.05
+5 ($10 − $5)	0.15
−2 ($3 − $5)	0.30
−5 ($0 − $5)	0.50

Sum ✓.

Task 2 — E(X): → unfavorable for the player (lose $1.60/play; the house has the edge).

Task 3 — Var(X) and σ: ; ; .

Task 4 — Model response: missing is the most common single outcome (50%), but that’s not the whole story. The expected net is −$1.60/play, so the game favors the house in the long run. The SD of $5.11 is large relative to the $5 fee, so a single play could yield +$15 or −$5 — short sessions can profit, but over many plays losses dominate.

Path B — The Architect: Design a PMF

Design a 4-outcome PMF with and . One valid example: , .

Verify: non-negative ✓; sum ✓; ✓.

Var(X): ; ✓; .

Model explanation: an expected daily payout of $2.00, with SD ≈ $1.26 meaning day-to-day results usually land between roughly $0.74 and $3.26 — a relatively predictable product.

Challenge 1 — Derive the Shortcut Formula

From with : Substituting :

Why always: since for every value of , its expectation is , forcing . A negative computed variance is always an arithmetic error.

Bonus: iff with probability 1, i.e. with probability 1 — a random variable with zero variance is a constant.

Challenge 2 — Regenerable Stretch Problem (Generator)

Apply the same six-step workflow as Problem 1 of Section 6. Compute with the shortcut, then verify against the definition — both must agree.