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PR-4 Solutions: Discrete Random Variables

Solutions Reference · ← Back to Lesson PR-4

Section 5 — Guided Practice Solutions

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GP1 — Discrete or Continuous? (Variants 0–4)

Variant 0 — Number of phone calls a help-desk agent receives in a one-hour shift:

Answer: Discrete. You can count calls as whole numbers (0, 1, 2, 3, …). You cannot receive 2.5 calls. The arrival time of each call is continuous, but the count of calls is always a whole number.

Variant 1 — Height (in centimetres) of a randomly chosen student:

Answer: Continuous. Height can take any value in an interval (e.g., 170.3 cm, 170.31 cm, 170.314 cm, …). Rounding to the nearest centimetre is a measurement choice, not a property of the variable itself.

Variant 2 — Number of defective items in a quality-control sample of 50 products:

Answer: Discrete. The count of defectives is a whole number between 0 and 50. We are counting, not measuring on a continuous scale.

Variant 3 — Time (in minutes) a customer waits in line at a coffee shop:

Answer: Continuous. Waiting time can be any non-negative real number (e.g., 2.37 minutes). Time flows continuously — 2.37 minutes is a perfectly valid waiting time, as is 2.371 minutes.

Variant 4 — Number of siblings a randomly chosen person has:

Answer: Discrete. Sibling count is a whole number (0, 1, 2, 3, …). You cannot have 1.5 siblings. Regardless of how family size varies across the population, each individual's count is a whole number.

GP2 — Is This PMF Valid? (Variants 0–4)

Variant 0 — PMF: \( x \in \{1, 2, 3, 4\} \), \( P = \{0.25, 0.35, 0.30, 0.15\} \):

Answer: Not valid. Sum = \( 0.25 + 0.35 + 0.30 + 0.15 = 1.05 \neq 1 \). All non-negativity conditions hold, but the sum exceeds 1.

Fix: Reduce one probability so the total equals 1.00. For example, set \( P(X = 4) = 0.10 \): new sum = \( 0.25 + 0.35 + 0.30 + 0.10 = 1.00 \) ✓.

Variant 1 — PMF: \( x \in \{0, 1, 2, 3\} \), \( P = \{0.40, 0.30, 0.20, 0.10\} \):

Answer: Valid. All probabilities are non-negative ✓. Sum = \( 0.40 + 0.30 + 0.20 + 0.10 = 1.00 \) ✓. A PMF does not need to be symmetric or have equal probabilities — decreasing probabilities are perfectly acceptable.

Variant 2 — PMF: \( x \in \{1, 2, 3\} \), \( P = \{0.50, 0.60, -0.10\} \):

Answer: Not valid. \( P(X = 3) = -0.10 < 0 \), which violates the first PMF condition (all probabilities must be non-negative). Note: the sum \( 0.50 + 0.60 - 0.10 = 1.00 \) does equal 1, but that does not fix the negative probability.

Fix: A negative probability is impossible — the distribution must be completely restructured with non-negative values that sum to 1.

Variant 3 — PMF: \( x \in \{2, 4, 6, 8\} \), \( P = \{0.15, 0.35, 0.35, 0.15\} \):

Answer: Valid. All probabilities are non-negative ✓. Sum = \( 0.15 + 0.35 + 0.35 + 0.15 = 1.00 \) ✓. Outcomes do not need to be consecutive integers — skipping odd numbers is perfectly fine. There is also no minimum number of outcomes required.

Variant 4 — PMF: \( x \in \{-1, 0, 1, 2\} \), \( P = \{0.20, 0.30, 0.30, ?\} \) with claimed \( P(X = 2) = 0.25 \):

Answer: Not valid. Sum = \( 0.20 + 0.30 + 0.30 + 0.25 = 1.05 \neq 1 \). Note: negative outcome values are allowed — the error is the sum exceeding 1, not the value \( x = -1 \).

Fix: \( P(X = 2) = 1 - (0.20 + 0.30 + 0.30) = 0.20 \). With \( P(X = 2) = 0.20 \), the sum is exactly 1.00 ✓.

GP3 — Compute E(X) (Generator)

Solutions vary by generated PMF. The strategy is always the same:

\[ E(X) = \sum_{\text{all } x} x \cdot P(X = x) \]

Multiply each outcome \( x \) by its probability \( P(X = x) \) and sum all terms. Refer to Worked Example 2 in Section 4 for a complete worked example with narration.

Common mistake: Adding the outcome values without weighting by probability. Every term in the sum must include both \( x \) and \( P(X = x) \).

GP4 — Compute Var(X) and σX (Generator)

Solutions vary by generated PMF. Use the four-step workflow:

  1. Step 1: \( E(X) = \sum x \cdot P(X = x) \)
  2. Step 2: \( E(X^2) = \sum x^2 \cdot P(X = x) \) — square each \( x \) value first, then weight
  3. Step 3: \( \text{Var}(X) = E(X^2) - [E(X)]^2 \)
  4. Step 4: \( \sigma_X = \sqrt{\text{Var}(X)} \)

Refer to Worked Example 3 in Section 4 for a complete worked example.

Critical pitfall: \( E(X^2) \neq [E(X)]^2 \). You must square each \( x \) value individually before multiplying by its probability. Computing \( [E(X)]^2 \) instead of \( E(X^2) \) gives \( \text{Var}(X) = 0 \), which is almost always wrong.

Section 6 — Independent Practice Solutions

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IP1 — Full E(X) + Var(X) Workflow (Generator)

Solutions vary by generated PMF and context. Apply the complete six-step workflow:

  1. (a) Verify the PMF: Check all \( P(X = x) \geq 0 \) and \( \sum P(X = x) = 1 \).
  2. (b) Compute E(X): \( E(X) = \sum x \cdot P(X = x) \) — write out every term.
  3. (c) Compute E(X²): \( E(X^2) = \sum x^2 \cdot P(X = x) \) — square each outcome before weighting.
  4. (d) Compute Var(X): \( \text{Var}(X) = E(X^2) - [E(X)]^2 \) — substitute numbers from steps (b) and (c).
  5. (e) Compute σX: \( \sigma_X = \sqrt{\text{Var}(X)} \).
  6. (f) Interpret σX: "The [context variable] typically deviates from the mean of [E(X)] by about [σX] [units]."

IP2 — CDF Practice (Variants 0–4)

Variant 0 — PMF: \( x \in \{0, 1, 2, 3, 4\} \), \( P = \{0.05, 0.20, 0.45, 0.25, 0.05\} \):

(a) CDF table:

\( x \)\( F(x) = P(X \leq x) \)
00.05
10.25
20.70
30.95
41.00

(b) \( P(X \leq 2) = F(2) = 0.70 \)

(c) \( P(X > 1) = 1 - F(1) = 1 - 0.25 = 0.75 \)

(d) \( P(1 < X \leq 3) = F(3) - F(1) = 0.95 - 0.25 = 0.70 \)

Variant 1 — PMF: \( x \in \{1, 2, 3, 4, 5\} \), \( P = \{0.10, 0.25, 0.30, 0.25, 0.10\} \):

(a) CDF table:

\( x \)\( F(x) \)
10.10
20.35
30.65
40.90
51.00

(b) \( P(X \leq 3) = F(3) = 0.65 \)

(c) \( P(X \geq 3) = 1 - F(2) = 1 - 0.35 = 0.65 \) (equals \( P(X \leq 3) \) because the distribution is symmetric about \( x = 3 \))

(d) \( P(2 < X \leq 4) = F(4) - F(2) = 0.90 - 0.35 = 0.55 \)

Variant 2 — PMF: \( x \in \{0, 2, 4, 6\} \), \( P = \{0.30, 0.40, 0.20, 0.10\} \):

(a) CDF table:

\( x \)\( F(x) \)
00.30
20.70
40.90
61.00

(b) \( P(X \leq 4) = F(4) = 0.90 \)

(c) \( P(X > 2) = 1 - F(2) = 1 - 0.70 = 0.30 \)

(d) \( P(0 < X \leq 4) = F(4) - F(0) = 0.90 - 0.30 = 0.60 \) (includes \( x = 2 \) and \( x = 4 \); \( x = 0 \) is excluded by the strict inequality)

Variant 3 — PMF: \( x \in \{-2, -1, 0, 1, 2\} \), \( P = \{0.10, 0.20, 0.40, 0.20, 0.10\} \):

(a) CDF table:

\( x \)\( F(x) \)
−20.10
−10.30
00.70
10.90
21.00

(b) \( P(X \leq 0) = F(0) = 0.70 \)

(c) \( P(X > 0) = 1 - F(0) = 1 - 0.70 = 0.30 \)

(d) \( P(-1 < X \leq 1) = F(1) - F(-1) = 0.90 - 0.30 = 0.60 \) (includes \( x = 0 \) and \( x = 1 \))

Variant 4 — PMF: \( x \in \{5, 10, 15, 20\} \), \( P = \{0.40, 0.30, 0.20, 0.10\} \):

(a) CDF table:

\( x \)\( F(x) \)
50.40
100.70
150.90
201.00

(b) \( P(X \leq 10) = F(10) = 0.70 \)

(c) \( P(X \geq 10) = 1 - F(5) = 1 - 0.40 = 0.60 \)

(d) \( P(5 < X \leq 15) = F(15) - F(5) = 0.90 - 0.40 = 0.50 \) (includes \( x = 10 \) and \( x = 15 \))

CDF strategy tip: Always build the full CDF table first. Then any probability question is just a lookup or a subtraction. For \( P(X > k) \), use \( 1 - F(k) \). For \( P(X \geq k) \), use \( 1 - F(k-1) \) — or equivalently \( 1 - F \) evaluated at the outcome just below \( k \).

IP3 — Is the Game Fair? (Variants 0–4)

Variant 0 — Raffle: 100 tickets, prizes \$200 and \$50, ticket costs \$5.

Net winnings: \( x \in \{-5, 45, 195\} \), \( P = \{98/100, 1/100, 1/100\} \).

\[ E(X) = (-5)\!\left(\tfrac{98}{100}\right) + (45)\!\left(\tfrac{1}{100}\right) + (195)\!\left(\tfrac{1}{100}\right) = -4.90 + 0.45 + 1.95 = -2.50 \]

Conclusion: \( E(X) = -\$2.50 \). The game is unfavorable — on average, you lose \$2.50 per ticket.

Variant 1 — Card game: pay \$3, win \$15 for ace (4/52), \$5 for face card (12/52), \$0 otherwise (36/52).

Net winnings: \( x \in \{-3, +2, +12\} \), \( P = \{36/52, 12/52, 4/52\} \).

\[ E(X) = (-3)\!\left(\tfrac{36}{52}\right) + (2)\!\left(\tfrac{12}{52}\right) + (12)\!\left(\tfrac{4}{52}\right) = \frac{-108 + 24 + 48}{52} = \frac{-36}{52} \approx -0.69 \]

Conclusion: \( E(X) \approx -\$0.69 \). The game is unfavorable — you lose about 69 cents per play on average.

Variant 2 — Find the Error — Insurance scenario: pay \$10 to enter, prizes \$0, \$20, \$50 with probabilities 0.60, 0.30, 0.10.

Net winnings: \( x \in \{-10, +10, +40\} \), same probabilities.

Errors in the student's work:

Error 1 (net winnings and E(X)): The student's net winnings calculation and \( E(X) = 1.0 \) are actually correct: \[ E(X) = (-10)(0.60) + (10)(0.30) + (40)(0.10) = -6.0 + 3.0 + 4.0 = 1.0 \] The conclusion that \( E(X) = 1 > 0 \) makes the game favorable is also logically correct.

Error 2 (Critical — variance formula): The student wrote \( \text{Var}(X) = [E(X)]^2 = 1^2 = 1 \). This is completely wrong — the student used \( [E(X)]^2 \) instead of the correct formula \( \text{Var}(X) = E(X^2) - [E(X)]^2 \).

Correct variance computation:

\[ E(X^2) = (-10)^2(0.60) + (10)^2(0.30) + (40)^2(0.10) = 60 + 30 + 160 = 250 \] \[ \text{Var}(X) = 250 - (1.0)^2 = 249, \qquad \sigma_X = \sqrt{249} \approx 15.78 \]

The true variance is enormous — the student's \( \text{Var}(X) = 1 \) underestimates the spread by a factor of 249. Even though the game is favorable on average, a single play can result in a \$10 loss or a \$40 gain.

Variant 3 — 4-sided die game: pay \$2, win \$6 on 4 (prob 1/4), \$2 on 3 (prob 1/4), \$0 on 1–2 (prob 1/2).

Net winnings: \( x \in \{-2, 0, +4\} \), \( P = \{0.50, 0.25, 0.25\} \).

\[ E(X) = (-2)(0.50) + (0)(0.25) + (4)(0.25) = -1 + 0 + 1 = 0 \]

Conclusion: \( E(X) = 0 \). The game is perfectly fair — neither player gains an advantage on average.

Variant 4 — Coin flip bet: win \$5 on heads, lose \$5 on tails, plus \$1 bonus for playing.

Net winnings: \( x \in \{+6, -4\} \), \( P = \{0.50, 0.50\} \).

\[ E(X) = (6)(0.50) + (-4)(0.50) = 3 - 2 = 1 \]

Conclusion: \( E(X) = \$1 \). The arrangement is favorable — the \$1 bonus tips the average to \$1 per flip in your favor.

IP4 — Find the Missing Probability, Then Compute E(X) (Generator)

Solutions vary by generated PMF. Use the two-step strategy:

  1. Step 1: Use \( \sum P(X = x) = 1 \). Sum the known probabilities, then solve \( p = 1 - (\text{sum of known probabilities}) \).
  2. Step 2: Substitute the solved \( p \) back into the PMF and compute \( E(X) = \sum x \cdot P(X = x) \), writing out every term.

Common mistake: Forgetting to substitute the solved value of \( p \) back before computing \( E(X) \). The \( p \) in the PMF table must be replaced with its numerical value for the weighted sum.

IP5 — Compute E(X) and σX, Then Interpret (Generator)

Solutions vary by generated PMF and context. After computing \( E(X) \) and \( \sigma_X \):

IP6 — Multi-Step Synthesis (Fixed Problem)

Let \( X \) = number of defective items in a box of 5 products.

\( x \)012345
\( P(X = x) \)0.600.200.100.060.030.01

(a) Verify the PMF:

\[ 0.60 + 0.20 + 0.10 + 0.06 + 0.03 + 0.01 = 1.00 \checkmark \]

All values non-negative ✓. Valid PMF.

(b) Compute E(X) and interpret:

\[ E(X) = 0(0.60) + 1(0.20) + 2(0.10) + 3(0.06) + 4(0.03) + 5(0.01) \] \[ = 0 + 0.20 + 0.20 + 0.18 + 0.12 + 0.05 = 0.75 \]

Interpretation: On average, there are 0.75 defective items per box. In the long run, the engineer expects about 75 defectives per 100 boxes.

(c) Compute Var(X) and σX:

\[ E(X^2) = 0^2(0.60) + 1^2(0.20) + 2^2(0.10) + 3^2(0.06) + 4^2(0.03) + 5^2(0.01) \] \[ = 0 + 0.20 + 0.40 + 0.54 + 0.48 + 0.25 = 1.87 \] \[ \text{Var}(X) = 1.87 - (0.75)^2 = 1.87 - 0.5625 = 1.3075 \] \[ \sigma_X = \sqrt{1.3075} \approx 1.143 \]

The number of defectives per box typically varies from the mean of 0.75 by about 1.14 defects.

(d) Probability a box is flagged (\( X \geq 2 \)):

\[ P(X \geq 2) = 1 - P(X \leq 1) = 1 - [P(X=0) + P(X=1)] = 1 - (0.60 + 0.20) = 0.20 \]

There is a 20% chance a given box is flagged for reinspection.

Section 7 — Mastery Check Solutions

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

Model answer (3–5 sentences):

The expected value \( E(X) \) is a weighted average of all possible outcomes, where each outcome is weighted by its probability. The mode is simply the outcome with the highest probability — it tells you which value occurs most often, but says nothing about the magnitude of other outcomes. Because \( E(X) \) accounts for both how likely each outcome is and how large it is, the two concepts measure different things.

For example, a lottery where you win \$1,000 with probability 0.001 and lose \$1 with probability 0.999: the mode is −\$1 (it happens 99.9% of the time), but \( E(X) = 1000(0.001) + (-1)(0.999) = 0 \). The tiny probability of winning \$1,000 is enough to bring the expected value to zero, even though the most likely outcome is a loss.

A classic example: a fair six-sided die has \( E(X) = 3.5 \), which is not even one of the outcomes. There is no mode (all six faces are equally likely). The expected value is the balance point of the distribution — the point where it would balance if each bar had weight equal to its probability — and has no obligation to coincide with the tallest bar.

Question 2 — Apply: Medical Testing Revenue

PMF for \( X \) = number of individual tests per order: \( x \in \{1, 2, 3, 4\} \), \( P = \{0.40, 0.30, 0.20, 0.10\} \).

MCQ 1: Most efficient approach = compute \( E(X) \) first, then multiply by 80 (linearity property \( E(aX) = a \cdot E(X) \)).

MCQ 2 — Full computation:

\[ E(X) = 1(0.40) + 2(0.30) + 3(0.20) + 4(0.10) = 0.40 + 0.60 + 0.60 + 0.40 = 2.00 \text{ tests} \] \[ E(R) = 80 \times E(X) = 80 \times 2.00 = \$160 \]

The correct answer is E(X) = 2.00 tests, E(R) = \$160.

Full solution including variance:

\[ E(X^2) = 1^2(0.40) + 2^2(0.30) + 3^2(0.20) + 4^2(0.10) = 0.40 + 1.20 + 1.80 + 1.60 = 5.00 \] \[ \text{Var}(X) = 5.00 - (2.00)^2 = 5.00 - 4.00 = 1.00 \] \[ \text{Var}(R) = 80^2 \cdot \text{Var}(X) = 6400 \times 1.00 = 6400, \qquad \sigma_R = \sqrt{6400} = \$80 \]

The hospital earns an average of \$160 per test order, with a standard deviation of \$80.

Question 3 — Error Analysis

PMF: \( x \in \{2, 4, 6\} \), each with probability \( 1/3 \).

Student's work: \( E(X) = 4 \) ✓ (correct). Then the student wrote \( \text{Var}(X) = [E(X)]^2 = 16 \) ✗.

Error identified: The student confused \( \text{Var}(X) = E(X^2) - [E(X)]^2 \) with \( \text{Var}(X) = [E(X)]^2 \). They computed \( [E(X)]^2 \) and used that as the variance directly — never computing \( E(X^2) \) at all.

Correct computation:

\[ E(X^2) = 2^2\!\left(\tfrac{1}{3}\right) + 4^2\!\left(\tfrac{1}{3}\right) + 6^2\!\left(\tfrac{1}{3}\right) = \frac{4 + 16 + 36}{3} = \frac{56}{3} \approx 18.67 \] \[ \text{Var}(X) = \frac{56}{3} - 4^2 = \frac{56}{3} - 16 = \frac{56 - 48}{3} = \frac{8}{3} \approx 2.67 \] \[ \sigma_X = \sqrt{8/3} \approx 1.63 \]

The correct variance is approximately 2.67, not 16. The student overestimated the spread by a factor of 6. Always compute \( E(X^2) \) separately by squaring each \( x \) value before multiplying by its probability.

Section 8 — Boss Fight Solutions

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Path A — The Analyst: Carnival Game

Game costs \$5 to play. Prizes and probabilities: \$20 red bottle (prob 0.05), \$10 blue bottle (prob 0.15), \$3 yellow bottle (prob 0.30), \$0 miss (prob 0.50).

Task 1 — PMF for net winnings X:

\( x \) (net)\( P(X = x) \)
+15 (red bottle: \$20 − \$5)0.05
+5 (blue bottle: \$10 − \$5)0.15
−2 (yellow bottle: \$3 − \$5)0.30
−5 (miss: \$0 − \$5)0.50

Verify: \( 0.05 + 0.15 + 0.30 + 0.50 = 1.00 \) ✓. All non-negative ✓.

Task 2 — Compute E(X):

\[ E(X) = (15)(0.05) + (5)(0.15) + (-2)(0.30) + (-5)(0.50) \] \[ = 0.75 + 0.75 - 0.60 - 2.50 = -1.60 \]

Conclusion: \( E(X) = -\$1.60 \). The game is unfavorable for the player — you expect to lose \$1.60 per play on average. The house has the edge.

Task 3 — Compute Var(X) and σX:

\[ E(X^2) = (15)^2(0.05) + (5)^2(0.15) + (-2)^2(0.30) + (-5)^2(0.50) \] \[ = 225(0.05) + 25(0.15) + 4(0.30) + 25(0.50) \] \[ = 11.25 + 3.75 + 1.20 + 12.50 = 28.70 \] \[ \text{Var}(X) = 28.70 - (-1.60)^2 = 28.70 - 2.56 = 26.14 \] \[ \sigma_X = \sqrt{26.14} \approx 5.11 \]

Task 4 — Model response using both E(X) and σX:

Your friend is right that missing is the most common single outcome (50%), but that alone does not tell the full story. The expected net winnings are −\$1.60 per play, meaning the game systematically favors the house in the long run — this is the most important number for deciding whether to play repeatedly. However, the standard deviation of \$5.11 is large relative to the \$5 entry fee, so outcomes are highly variable: any single play could yield +\$15 or −\$5. The high variance means a single play or short session can easily produce a profit, but the negative expected value ensures that over many plays, losses will dominate.

Path B — The Architect: Design a PMF

Design a 4-outcome PMF with positive integer payouts, \( E(X) = 2 \), and \( \text{Var}(X) \leq 3 \).

Many valid PMFs satisfy these constraints. Here is one verified example:

Outcomes: \( x \in \{0, 1, 3, 4\} \), probabilities \( P = \{0.10, 0.40, 0.40, 0.10\} \).

Task 2 — Verify:

Task 3 — Compute Var(X):

\[ E(X^2) = 0^2(0.10) + 1^2(0.40) + 3^2(0.40) + 4^2(0.10) = 0 + 0.40 + 3.60 + 1.60 = 5.60 \] \[ \text{Var}(X) = 5.60 - (2.00)^2 = 5.60 - 4.00 = 1.60 \]

\( \text{Var}(X) = 1.60 \leq 3 \) ✓. \( \sigma_X = \sqrt{1.60} \approx 1.26 \).

Task 4 — Model investor explanation:

This financial product has an expected daily payout of \$2.00, meaning that if you invested in it day after day over a long period, you would earn about \$2.00 per day on average. The standard deviation of approximately \$1.26 tells you how much day-to-day results typically vary from that average — on a given day, the payout will usually be somewhere between roughly \$0.74 and \$3.26. This is a relatively predictable product: the variance is small enough that outcomes cluster near the expected value rather than swinging wildly.

Note for Path B: Many different PMFs satisfy \( E(X) = 2 \) and \( \text{Var}(X) \leq 3 \) — your answer is valid as long as you show the verification steps. The key insight is that you have two constraints (sum of probabilities = 1 and \( E(X) = 2 \)) but four unknowns (the four probabilities), so there are infinitely many solutions. The variance constraint then filters out PMFs with too much spread.

Section 9 — Challenge Problem Solutions

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Challenge 1 — Derive the Shortcut Formula

Starting from the definition \( \text{Var}(X) = E\!\left[(X - \mu)^2\right] \) where \( \mu = E(X) \):

Step 1: Expand the square inside the expectation.

\[ \text{Var}(X) = E\!\left[X^2 - 2\mu X + \mu^2\right] \]

Step 2: Apply linearity of expectation — \( E[A + B + C] = E[A] + E[B] + E[C] \) and \( E[cA] = c \cdot E[A] \) for any constant \( c \).

\[ = E(X^2) - 2\mu \cdot E(X) + \mu^2 \]

Step 3: Substitute \( E(X) = \mu \).

\[ = E(X^2) - 2\mu \cdot \mu + \mu^2 \] \[ = E(X^2) - 2\mu^2 + \mu^2 \] \[ = E(X^2) - \mu^2 \]

Step 4: Replace \( \mu \) with \( E(X) \).

\[ \boxed{\text{Var}(X) = E(X^2) - [E(X)]^2} \]

Why Var(X) ≥ 0 always: Since \( (X - \mu)^2 \geq 0 \) for every possible value of \( X \), its expected value must also be \( \geq 0 \). This forces \( E(X^2) \geq [E(X)]^2 \) — the shortcut formula can never produce a negative variance. A negative computed variance is always an arithmetic error.

Bonus insight: \( \text{Var}(X) = 0 \) if and only if \( E[(X - \mu)^2] = 0 \). Since \( (X - \mu)^2 \geq 0 \), this expectation equals zero only when \( (X - \mu)^2 = 0 \) with probability 1, meaning \( X = \mu \) with probability 1. A random variable with zero variance is a constant.

Challenge 2 — Regenerable Stretch Problem (Generator)

Solutions vary by generated PMF. Apply the same six-step workflow as IP1 — no scaffolding is provided in the lesson. Use the shortcut formula to compute \( \text{Var}(X) \), then verify your answer using the definition formula \( \text{Var}(X) = E[(X-\mu)^2] = \sum (x - \mu)^2 \cdot P(X = x) \). Both must give the same result.