Lecture PDF Questions — Complete Worked Solutions

Course: IIM Lucknow IPMX 2025-26 — Probability & Distributions

Use: Each question is solved step-by-step, mapped to its source PDF, and linked to similar Ken Black book problems for further practice.

Note on Ken Black references:

KB Ch 4 = Probability; KB Ch 5 = Discrete RVs (Binomial, Poisson);

KB Ch 6 = Continuous RVs (Uniform, Exponential, Normal). KB problems are not solved here — attempt those after mastering the lecture problems below.

📘 PDF 1: 01_2_Probability.pdf

Q1.1 — Cable TV / Two TV sets (Exercise, slide 7)

Question: 68% of US households with TV have cable. 75% have two or more TV sets. 56% have both. A household is selected at random. (a) P(cable OR 2+ TVs)? (b) P(cable OR 2+ TVs but NOT both)? © P(neither)?

Setup: Let C = cable, B = 2+ TVs. P(C) = 0.68, P(B) = 0.75, P(C ∩ B) = 0.56.

Step 1 — Build contingency table:

(0.75 − 0.56 = 0.19; 0.68 − 0.56 = 0.12; rest by subtraction)

Step 2 — Apply formulas:

• (a) P(C ∪ B) = P(C) + P(B) − P(C ∩ B) = 0.68 + 0.75 − 0.56 = 0.87
• (b) "Or but not both" = P(C ∪ B) − P(C ∩ B) = 0.87 − 0.56 = 0.31
• (c) P(neither) = P(C' ∩ B') = 0.13 (read directly from table) = 0.13

Similar KB problems: Ch 4 §4.5 problems 4.13, 4.18 (cable/HDTV/DVR contingency).

Q1.2 — Stockholders / College Education (Exercise, slide 17)

Question: 43% of US adults are stockholders (Stk). 39% have some college education (Ed). 75% of stockholders have some college education. Find six probabilities (a–f).

Setup: P(Stk) = 0.43, P(Ed) = 0.39, P(Ed | Stk) = 0.75.

Step 1 — Find P(Stk ∩ Ed): $P(\text{Stk}\cap \text{Ed})=P(\text{Ed}\mid \text{Stk})·P(\text{Stk})=0.75\times 0.43=0.3225$

Step 2 — Build contingency table:

Step 3 — Read answers:

• (a) P(Stk') = 0.57
• (b) P(Stk ∩ Ed) = 0.3225
• (c) P(Stk ∪ Ed) = 0.43 + 0.39 − 0.3225 = 0.4975
• (d) P(Stk' ∩ Ed') = 0.5025
• (e) P(Stk' ∪ Ed') = 0.57 + 0.61 − 0.5025 = 0.6775
• (f) P(Stk' ∩ Ed) = 0.0675

Similar KB problems: Ch 4 §4.5 problem 4.19 (same stockholder setup).

Q1.3 — Brand A Soap (Slide 11)

Question: A market survey of soap preference across 4 cities yielded:

(a) P(prefers A)? (b) P(prefers A AND from Chennai)? (c) P(prefers A | Chennai)? (d) P(Mumbai | prefers A)?

Step-by-step:

• (a) P(A) = 210 / 390 = 0.5385
• (b) P(A ∩ Chennai) = 60 / 390 = 0.1538
• (c) P(A | Chennai) = 60 / 100 = 0.60 (out of 100 Chennai respondents, 60 said yes)
• (d) P(Mumbai | A) = 50 / 210 = 0.2381 (out of 210 yes-responders, 50 from Mumbai)

Similar KB problems: Ch 4 §4.4 (cross-tabulation) and §4.6 (conditional probability) problems 4.15, 4.16, 4.20.

Q1.4 — Warranty Repair / USA Cars (Exercise, slides 12–14)

Question: P(repair) = 0.04, P(USA-made) = 0.60, P(repair AND USA-made) = 0.025. (a) P(repair OR USA-made)? (b) P(repair AND NOT USA-made)?

Step 1 — Build contingency table:

Step 2 — Compute:

• (a) P(R ∪ S) = 0.04 + 0.60 − 0.025 = 0.615
• (b) P(R ∩ S') = 0.04 − 0.025 = 0.015

Similar KB problems: Ch 4 §4.5 problems 4.13 (same structure).

Q1.5 — Gender and Pass Rate Independence (Slide 21)

Question: Check whether gender and passing are independent given:

Test: Independent if P(Pass) = P(Pass | Man) = P(Pass | Woman).

• P(Pass) = 16 / 40 = 0.40
• P(Pass | Man) = 6 / 15 = 0.40
• P(Pass | Woman) = 10 / 25 = 0.40

All equal → Gender and Passing are INDEPENDENT.

Similar KB problems: Ch 4 §4.6 — independence checks via cross-tabulation.

Q1.6 — Two Suppliers Bayes (Slides 26–29)

Question: Parts come from two suppliers. P(E₁) = 0.70, P(E₂) = 0.30. Defect rates: P(D | E₁) = 0.05, P(D | E₂) = 0.10. Machine breaks down due to a defective part. What's P(E₁ | D) and P(E₂ | D)?

Step 1 — Compute joint probabilities:

• P(D ∩ E₁) = P(D | E₁) × P(E₁) = 0.05 × 0.70 = 0.035
• P(D ∩ E₂) = P(D | E₂) × P(E₂) = 0.10 × 0.30 = 0.030

Step 2 — Total probability of defect: $P(D)=0.035+0.030=0.065$

Step 3 — Apply Bayes:

• P(E₁ | D) = 0.035 / 0.065 = 0.5385
• P(E₂ | D) = 0.030 / 0.065 = 0.4615

(Despite E₂ being a smaller supplier, its higher defect rate makes it almost equally likely to be the source of a defective part.)

Similar KB problems: Ch 4 §4.7 (Bayes' Rule) — all problems on revising prior probabilities.

Q1.7 — GMAT Prep Course Bayes (Exercise, slide 27)

Question: Among GMAT scorers ≥ 650, 52% took a prep course. Among scorers <650, 23% took a prep course. An applicant has 10% prior probability of scoring ≥ 650. He'll take the $500 prep course only if it doubles his probability of scoring ≥ 650. Should he take it?

Step 1 — Define events: H = score ≥ 650, PC = took prep course. Given: P(H) = 0.10, P(H') = 0.90, P(PC | H) = 0.52, P(PC | H') = 0.23.

Step 2 — Find joint probabilities:

• P(H ∩ PC) = 0.52 × 0.10 = 0.052
• P(H' ∩ PC) = 0.23 × 0.90 = 0.207

Step 3 — Marginal P(PC): $P(PC)=0.052+0.207=0.259$

Step 4 — Bayes for P(H | PC): $P(H\mid PC)=\frac{0.052}{0.259}\approx 0.2008$

Step 5 — Decision: New probability ≈ 0.20 = 2 × 0.10 = doubled. YES, he should take the course.

Similar KB problems: Ch 4 §4.7 (Bayes problems on diagnostic tests, marketing).

📘 PDF 2: 3_4_Random_Variables.pdf

Q2.1 — Coin Toss Game (Illustration, slides 8–10, 17)

Question: Toss a fair coin twice. Win Rs 100 if at least one head; lose Rs 200 otherwise. Long-term average gain?

Step 1 — Define X = gain. Sample space = {HH, HT, TH, TT}.

Step 2 — Probability of winning:

• P(at least one H) = 3/4 → X = +100
• P(no H = TT) = 1/4 → X = −200

Step 3 — Expected value: $E(X)=100\times \frac{3}{4}+(-200)\times \frac{1}{4}=75-50=\boxed{\text{Rs }25}$

So long-run, the player gains Rs 25 per game.

Similar KB problems: Ch 5 §5.1 — discrete RV expected value problems.

Q2.2 — Discrete RV with unknown k (Slide 11) — Find E(X)

Question: X has the following pmf. Find E(X).

Step 1 — Find k. Probabilities sum to 1: $0.4+k+0.2+0.3=1\Rightarrow k=0.1$

Step 2 — Compute E(X): $E(X)=(-2)(0.4)+(-1)(0.1)+(0)(0.2)+(1)(0.3)=-0.8-0.1+0+0.3=\boxed{-0.6}$

Similar KB problems: Ch 5 §5.2 problems on expected value of a discrete distribution.

Q2.3 — Pizzas Delivered (Slide 11) — Find E(X)

Question: X = number of pizzas delivered per month. Find E(X).

Solution: $E(X)=0(0.1)+1(0.3)+2(0.4)+3(0.2)=0+0.3+0.8+0.6=\boxed{1.7}$

Similar KB problems: Ch 5 §5.2.

Q2.4 — Painting vs $1000 Cash (Exercise, slide 12)

Question: Choose between $1000 cash OR a painting worth: $2000 (P=0.20), $1000 (P=0.50), $500 (P=0.30). What's the rational choice?

Step 1 — Compute E(painting): $E(\text{painting})=2000(0.20)+1000(0.50)+500(0.30)=400+500+150=\$1050$

Step 2 — Compare: $1050 (painting) > $1000 (cash).

Decision: Take the painting (in expected-value terms). Note: a risk-averse person who needs the money may still prefer the certain $1000.

Similar KB problems: Ch 5 §5.2 — decision-making with expected values.

Q2.5 — Discrete RV (Slide 11 again) — Find Var(X)

Question: Same pmf as Q2.2. Find Var(X).

Step 1 — Recall E(X) = −0.6 (from Q2.2).

Step 2 — Compute E(X²): $E(X^2)=(-2{)}^2(0.4)+(-1{)}^2(0.1)+(0{)}^2(0.2)+(1{)}^2(0.3)=1.6+0.1+0+0.3=2.0$

Step 3 — Apply variance formula: $V(X)=E(X^2)-[E(X){]}^2=2.0-(-0.6{)}^2=2.0-0.36=\boxed{1.64}$

Similar KB problems: Ch 5 §5.2 problems on variance.

Q2.6 — Pizzas Variance (Slide 11)

Question: Same pmf as Q2.3. Find V(X).

Step 1 — E(X) = 1.7 (from Q2.3).

Step 2 — E(X²): $E(X^2)=0^2(0.1)+1^2(0.3)+2^2(0.4)+3^2(0.2)=0+0.3+1.6+1.8=3.7$

Step 3: $V(X)=3.7-{1.7}^2=3.7-2.89=\boxed{0.81}$

Similar KB problems: Ch 5 §5.2.

Q2.7 — Stock Portfolio (Slide 13)

Question: Investor puts 40% in stock A and 60% in stock B. E(A) = 9.5%, E(B) = 7.6%, SD(A) = 12.93%, SD(B) = 8.20%, Cov(A, B) = 18.60. Find expected return and variance of the portfolio.

Setup: Portfolio return R = 0.4 A + 0.6 B.

Step 1 — Expected return (linear in expectations): $E(R)=0.4\times 9.5+0.6\times 7.6=3.8+4.56=\boxed{8.36\%}$

Step 2 — Variance (using Var(aX + bY) = a²V(X) + b²V(Y) + 2ab·Cov(X,Y)):

• V(A) = 12.93² = 167.18
• V(B) = 8.20² = 67.24 $V(R)=(0.4{)}^2(167.18)+(0.6{)}^2(67.24)+2(0.4)(0.6)(18.60)$ $=0.16\times 167.18+0.36\times 67.24+0.48\times 18.60$ $=26.75+24.21+8.93=\boxed{59.89}$

Step 3: SD® = √59.89 ≈ 7.74%.

Similar KB problems: Ch 5 §5.2 — laws of variance and covariance.

Q2.8 — Bivariate Distribution (Slide 24)

Question: Joint pmf:

Compute E(X), E(Y), V(X), V(Y), Cov(X,Y), E(X+Y), V(X+Y).

⚠️ Watch the table orientation! The lecture has the columns labeled as Y values and rows as X values. Marginal of X = row totals (0.6, 0.3, 0.1); Marginal of Y = column totals (0.4, 0.5, 0.1).

Step 1 — E(X) using row marginals: $E(X)=0(0.6)+1(0.3)+2(0.1)=0+0.3+0.2=0.5$

Hmm wait — let me re-read the lecture answer key: a) 0.7. So the table layout in the lecture has X across columns. Let me re-orient:

Re-orienting: The first row is X = 0,1,2 with totals 0.4, 0.5, 0.1 → so X marginal = (0.4, 0.5, 0.1). Then Y marginal = (0.6, 0.3, 0.1) (rows).

Let me redo:

Step 1 (corrected) — Marginals:

• P(X = 0) = 0.4, P(X = 1) = 0.5, P(X = 2) = 0.1
• P(Y = 0) = 0.6, P(Y = 1) = 0.3, P(Y = 2) = 0.1

Step 2 — E(X) and E(Y): $E(X)=0(0.4)+1(0.5)+2(0.1)=0.7$ $E(Y)=0(0.6)+1(0.3)+2(0.1)=0.5$

Step 3 — E(X²) and E(Y²): $E(X^2)=0+1(0.5)+4(0.1)=0.9$ $E(Y^2)=0+1(0.3)+4(0.1)=0.7$

Step 4 — Variances: $V(X)=0.9-{0.7}^2=0.9-0.49=0.41$ $V(Y)=0.7-{0.5}^2=0.7-0.25=0.45$

Step 5 — E(XY) (sum of x·y·pᵢⱼ over the table):

Step 6 — Covariance: $\text{Cov}(X,Y)=E(XY)-E(X)E(Y)=0.20-(0.7)(0.5)=0.20-0.35=-0.15$

Step 7 — Sum: $E(X+Y)=0.7+0.5=1.2$ $V(X+Y)=V(X)+V(Y)+2\text{Cov}(X,Y)=0.41+0.45+2(-0.15)=0.86-0.30=0.56$

Final answers: E(X) = 0.7, E(Y) = 0.5, V(X) = 0.41, V(Y) = 0.45, Cov = −0.15, E(X+Y) = 1.2, V(X+Y) = 0.56.

Similar KB problems: Ch 5 §5.2 — bivariate / joint distribution problems.

📘 PDF 3: 05_Binomial_Distribution.pdf

Q3.1 — 10% Defective Items, n = 20 (Slide 13)

Question: Production process has 10% defectives. Sample of 20 items. (a) P(0 defectives)? (b) P(exactly 1)? © P(< 3)?

Setup: X ~ Binomial(n = 20, p = 0.10), q = 0.90.

Step 1 — P(X = 0): $P(X=0)=\left(\genfrac{}{}{0ex}{}{20}{0}\right)(0.10{)}^0(0.90{)}^{20}=(0.90{)}^{20}=0.1216$

Step 2 — P(X = 1): $P(X=1)=\left(\genfrac{}{}{0ex}{}{20}{1}\right)(0.10{)}^1(0.90{)}^{19}=20\times 0.10\times 0.1351=0.2702$

Step 3 — P(X = 2): $P(X=2)=\left(\genfrac{}{}{0ex}{}{20}{2}\right)(0.10{)}^2(0.90{)}^{18}=190\times 0.01\times 0.1501=0.2852$

Step 4 — P(X < 3) = P(0) + P(1) + P(2): $0.1216+0.2702+0.2852=\boxed{0.6770}$

Excel: =BINOM.DIST(2, 20, 0.1, TRUE) gives 0.6769.

Similar KB problems: Ch 5 §5.3 problems on binomial defectives.

Q3.2 — Tomato Seeds 90% Germinate, n = 25 (Slide 17)

Question: Seeds germinate 90%. Plant 25. (a) P(exactly 20 germinate)? (b) P(20 or more)? © P(24 or fewer)? (d) Expected number?

Setup: X ~ Binomial(n = 25, p = 0.90).

Step 1 — (a) P(X = 20): $P(X=20)=\left(\genfrac{}{}{0ex}{}{25}{20}\right)(0.90{)}^{20}(0.10{)}^5=53,130\times 0.1216\times 0.00001\approx \boxed{0.0646}$

Step 2 — (b) P(X ≥ 20): Use Excel: =1 - BINOM.DIST(19, 25, 0.9, TRUE) ≈ 0.9666

Step 3 — © P(X ≤ 24) = 1 − P(X = 25): $P(X=25)=(0.90{)}^{25}=0.0718$ $P(X\le 24)=1-0.0718=\boxed{0.9282}$

Step 4 — (d) E(X) = np = 25 × 0.90 = 22.5

Similar KB problems: Ch 5 §5.3 — binomial seeds/germination/quality problems.

Q3.3 — Discount Broker, n = 9 (Slide 17)

Question: 30% of investors use a discount broker. Sample of 9. (a) P(X = 2)? (b) P(X ≤ 3)? © P(X ≥ 3)?

Setup: X ~ Binomial(n = 9, p = 0.30).

Step 1 — (a) P(X = 2): $P(X=2)=\left(\genfrac{}{}{0ex}{}{9}{2}\right)(0.30{)}^2(0.70{)}^7=36\times 0.09\times 0.0824=\boxed{0.2668}$

Step 2 — (b) P(X ≤ 3) — Excel: =BINOM.DIST(3, 9, 0.3, TRUE): By summing P(0) + P(1) + P(2) + P(3):

• P(0) = (0.70)⁹ = 0.0404
• P(1) = 9(0.30)(0.70)⁸ = 0.1556
• P(2) = 0.2668
• P(3) = C(9,3)(0.30)³(0.70)⁶ = 84 × 0.027 × 0.1176 = 0.2668
• Sum ≈ 0.7297

Step 3 — © P(X ≥ 3) = 1 − P(X ≤ 2): $1-(0.0404+0.1556+0.2668)=1-0.4628=\boxed{0.5372}$

Similar KB problems: Ch 5 §5.3.

📘 PDF 4: 6_Poisson_Distribution.pdf

Q4.1 — 1% Defectives in Package of 200 (Slide 9)

Question: Factory output 1% defective. Package of 200 units. (a) P(at most 2 defectives)? (b) P(at least 2)?

Setup: Use Poisson approximation (n large, p small): λ = np = 200 × 0.01 = 2.

Step 1 — Compute individual Poisson probabilities using P(X = x) = e⁻ᵠ λˣ / x!. With e⁻² ≈ 0.1353:

• P(0) = e⁻² = 0.1353
• P(1) = 2 × e⁻² = 0.2707
• P(2) = (4/2) × e⁻² = 0.2707

Step 2 — (a) P(X ≤ 2): $0.1353+0.2707+0.2707=\boxed{0.6767}$

Step 3 — (b) P(X ≥ 2) = 1 − P(X ≤ 1): $1-(0.1353+0.2707)=1-0.4060=\boxed{0.5940}$

Similar KB problems: Ch 5 §5.4 problems 5.17, 5.20 (Poisson with given λ).

Q4.2 — City Deaths Sum of Two Poissons (Slide 11)

Question: Daily deaths from road accidents ~ Poisson(2), from other causes ~ Poisson(6), independent. P(total deaths ≤ 2)?

Step 1 — Sum of independent Poissons is Poisson: $Z=X+Y~\text{Poisson}(2+6)=\text{Poisson}(8)$

Step 2 — Compute P(Z ≤ 2) using e⁻⁸ ≈ 0.000335:

• P(0) = e⁻⁸ = 0.000335
• P(1) = 8 × e⁻⁸ = 0.002684
• P(2) = (64/2) × e⁻⁸ = 0.010735

$P(Z\le 2)=0.000335+0.002684+0.010735=\boxed{0.0138}$

Excel: =POISSON.DIST(2, 8, TRUE) ≈ 0.0138.

Similar KB problems: Ch 5 §5.4 — additive property of Poisson.

Q4.3 — Defective Bottles in Boxes (Slide 17)

Question: 0.1% bottles defective. 500 bottles per box. Drug manufacturer buys 100 boxes. Expected number of boxes with (a) 0 defectives, (b) at most 2 defectives?

Setup: λ per box = 500 × 0.001 = 0.5.

Step 1 — P(0) per box with e⁻⁰·⁵ ≈ 0.6065: $P(0)=e^{-0.5}=0.6065$

Step 2 — (a) Expected boxes with 0 defectives: $100\times 0.6065\approx \boxed{60.65\text{ (i.e., about 61 boxes)}}$

Step 3 — P(X ≤ 2) per box:

• P(0) = 0.6065
• P(1) = 0.5 × e⁻⁰·⁵ = 0.3033
• P(2) = (0.25 / 2) × e⁻⁰·⁵ = 0.0758 $P(X\le 2)=0.6065+0.3033+0.0758=0.9856$

Step 4 — (b) Expected boxes with at most 2 defectives: $100\times 0.9856\approx \boxed{98.56\text{ (about 99 boxes)}}$

Similar KB problems: Ch 5 §5.4.

Q4.4 — Toll-Free Calls 0.4/min (Slide 13)

Question: Calls average 0.4 per minute. (a) P(0 calls in 2 minutes)? (b) P(≥ 3 calls in 3 minutes)?

Step 1 — Scale λ for 2-min: λ = 0.4 × 2 = 0.8. $P(X=0)=e^{-0.8}=\boxed{0.4493}$

Step 2 — Scale λ for 3-min: λ = 0.4 × 3 = 1.2.

• P(0) = e⁻¹·² = 0.3012
• P(1) = 1.2 × e⁻¹·² = 0.3614
• P(2) = (1.44 / 2) × e⁻¹·² = 0.2169
• Sum = 0.8795

Step 3: $P(X\ge 3)=1-0.8795=\boxed{0.1205}$

Similar KB problems: Ch 5 §5.4 problem 5.20 (telephone calls Poisson).

Q4.5 — Car Hire Firm with 2 Cars (Slide 17)

Question: Demand for cars per day ~ Poisson(1.5). Firm has 2 cars. (a) P(no car used today)? (b) P(some demand refused)?

Setup: X ~ Poisson(λ = 1.5). e⁻¹·⁵ ≈ 0.2231.

Step 1 — (a) Neither car used = P(X = 0): $P(X=0)=e^{-1.5}=\boxed{0.2231}$

Step 2 — (b) Demand refused = X exceeds capacity 2 = P(X > 2) = P(X ≥ 3):

• P(0) = 0.2231
• P(1) = 1.5 × 0.2231 = 0.3347
• P(2) = (2.25 / 2) × 0.2231 = 0.2510
• P(X ≤ 2) = 0.8088 $P(X>2)=1-0.8088=\boxed{0.1912}$

Similar KB problems: Ch 5 §5.4 problem 5.20 (bank teller, capacity questions).

Q4.6 — Book Typographical Errors (Slide 17)

Question: 520 pages, 390 typos. (a) P(no error in 4-page sample)? (b) P(no error in 2-page sample)?

Step 1 — Average errors per page: λ₁ = 390 / 520 = 0.75.

Step 2 — (a) For 4 pages: λ₄ = 4 × 0.75 = 3. $P(X=0)=e^{-3}=\boxed{0.0498}$

Step 3 — (b) For 2 pages: λ₂ = 2 × 0.75 = 1.5. $P(X=0)=e^{-1.5}=\boxed{0.2231}$

Similar KB problems: Ch 5 §5.4.

📘 PDF 5: 07_UNIEXP_Distribution.pdf

Q5.1 — Plastic Module Assembly U(27, 39) (Slide 6)

Question: Assembly time uniformly distributed between 27 and 39 seconds. (a) P(30 ≤ X ≤ 35)? (b) P(X < 30)?

Setup: X ~ U(27, 39). PDF height = 1 / (39 − 27) = 1/12.

Step 1 — (a): $P(30\le X\le 35)=\frac{35-30}{39-27}=\frac{5}{12}=\boxed{0.4167}$

Step 2 — (b): $P(X<30)=\frac{30-27}{39-27}=\frac{3}{12}=\boxed{0.25}$

Similar KB problems: Ch 6 §6.2 (Uniform Distribution) problems.

Q5.2 — Firing Pin Specs U(0.85, 1.05) (Slide 8)

Question: Firing pin length uniform on [0.85, 1.05] inches. Discard if length < 0.90 OR > 1.00. What fraction is discarded?

Step 1 — Compute each tail:

• P(X < 0.90) = (0.90 − 0.85) / (1.05 − 0.85) = 0.05 / 0.20 = 0.25
• P(X > 1.00) = (1.05 − 1.00) / 0.20 = 0.05 / 0.20 = 0.25

Step 2 — Total discarded: $P(\text{discard})=0.25+0.25=\boxed{0.50=50\%}$

Similar KB problems: Ch 6 §6.2.

Q5.3 — DOT Bid Distribution U(2d/5, 2d) (Slide 9)

Question: Bid X uniform on (2d/5, 2d). What fraction of bids are less than DOT estimate d?

Step 1 — Range = 2d − 2d/5 = 8d/5.

Step 2 — Favorable interval = d − 2d/5 = 3d/5.

Step 3: $P(X<d)=\frac{3d/5}{8d/5}=\frac{3}{8}=\boxed{0.375=37.5\%}$

Similar KB problems: Ch 6 §6.2.

Q5.4 — Wiley Breakdowns (Slides 7–8)

Question: Machine breaks down once every 4 weeks on average; time between breakdowns ~ Exponential. P(no breakdown for at least 6 weeks)?

Step 1 — Find λ: Mean = 1/λ = 4 weeks → λ = 1/4 = 0.25 per week.

Step 2 — Apply tail formula: $P(X>6)=e^{-\lambda ·6}=e^{-0.25\times 6}=e^{-1.5}=\boxed{0.2231}$

Excel: =1 - EXPON.DIST(6, 0.25, TRUE) = 0.2231.

Similar KB problems: Ch 6 §6.4 (Exponential) problem 6.37.

Q5.5 — Phone Call Length (Slide 14)

Question: Phone call length X ~ Exponential(λ = 1/10 per minute). Someone is ahead of you. (a) P(wait > 10 min)? (b) P(10 < wait < 20)?

Step 1 — (a) Tail probability: $P(X>10)=e^{-(1/10)\times 10}=e^{-1}=\boxed{0.3679}$

Step 2 — (b) Use F(20) − F(10): $P(10<X<20)=(1-e^{-2})-(1-e^{-1})=e^{-1}-e^{-2}=0.3679-0.1353=\boxed{0.2326}$

Similar KB problems: Ch 6 §6.4.

Q5.6 — Battery Life Exp(λ = 0.05) (Slide 16)

Question: Battery life ~ Exp(0.05) hours. (a) Mean & SD? (b) P(10 < X < 15)? © P(X > 20)?

Step 1 — (a): Mean = 1/λ = 1/0.05 = 20 hrs. SD = 1/λ = 20 hrs (Exp has SD = mean).

Step 2 — (b) P(10 < X < 15): $=e^{-0.5}-e^{-0.75}=0.6065-0.4724=\boxed{0.1341}$

Step 3 — © P(X > 20): $=e^{-0.05\times 20}=e^{-1}=\boxed{0.3679}$

Similar KB problems: Ch 6 §6.4 problems 6.37 a–d.

📘 PDF 6: 08_09_Normal_Distribution.pdf

Q6.1 — Achievement Scores N(76, 4) (Slide 11)

Question: Scores ~ N(76, σ = 4). P(at least 80)?

Step 1 — Standardize: z = (80 − 76)/4 = 1.0.

Step 2 — P(Z ≥ 1.0) = 1 − Φ(1.0): $P(X\ge 80)=1-0.8413=\boxed{0.1587}$

Excel: =1 - NORM.DIST(80, 76, 4, TRUE) ≈ 0.1587.

Similar KB problems: Ch 6 §6.3 problem 6.6 (forward Normal probabilities).

Q6.2 — Cement Production N(50, 3.5) — Days/Year (Slide 10)

Question: Daily cement production ~ N(50 tons, σ = 3.5). On how many days per year does it exceed 55 tons?

Step 1 — z-score: z = (55 − 50)/3.5 = 1.43.

Step 2 — P(X ≥ 55) = 1 − Φ(1.43): $1-0.9236=0.0764$

Step 3 — Multiply by days in a year: $365\times 0.0764\approx \boxed{28\text{ days}}$

Similar KB problems: Ch 6 §6.3 problem 6.7 (Tompkins warehouse) and 6.8 (cell-phone bill).

Q6.3 — Bank Teller Wait N(5, 0.8) (Slides 14–15)

Question: Wait time ~ N(5 min, σ = 0.8). Find: (a) P(< 6 min)? (b) P(> 3.5 min)? © P(3.4 ≤ X ≤ 6.2)?

Step 1 — (a) z = (6 − 5)/0.8 = 1.25: $P(X<6)=\Phi (1.25)=\boxed{0.8944}$

Step 2 — (b) z = (3.5 − 5)/0.8 = −1.875: $P(X>3.5)=1-\Phi (-1.875)=\Phi (1.875)=\boxed{0.9696}$

Step 3 — © z₁ = −2.0, z₂ = 1.5: $P(3.4\le X\le 6.2)=\Phi (1.5)-\Phi (-2.0)=0.9332-0.0228=\boxed{0.9104}$

Similar KB problems: Ch 6 §6.3 problem 6.6 a–f.

Q6.4 — Food Expenditure N(125, 25) + Binomial Combination (Slide 17)

Question: Weekly food expenditure ~ N(₹125, σ = ₹25). (a) P(family ≤ ₹157)? (b) Out of 8 families, P(exactly 5 spend ≤ ₹157)?

Step 1 — (a) z = (157 − 125)/25 = 1.28: $P(X\le 157)=\Phi (1.28)=\boxed{0.8997}$

Step 2 — (b) Now Y = number of families (out of 8) with X ≤ 157. Y ~ Binomial(n = 8, p = 0.8997).

$P(Y=5)=\left(\genfrac{}{}{0ex}{}{8}{5}\right)(0.8997{)}^5(0.1003{)}^3$ $=56\times 0.5896\times 0.001009\approx \boxed{0.0333}$

💡 This is a hybrid Normal + Binomial problem. First find p from the Normal, then plug into Binomial.

Similar KB problems: Ch 5 §5.3 + Ch 6 §6.3 — combined problems on hybrid distributions.

Q6.5 — Bakery Cupcakes — Inverse Normal (Slide 18)

Question: Daily demand ~ N(850, σ = 90). Bakery wants P(running short) ≤ 20%. How many cupcakes to make?

Step 1 — Translate "running short": Run short means demand exceeds supply x. Want P(X > x) ≤ 0.20, i.e., P(X ≤ x) ≥ 0.80.

Step 2 — Find z for cumulative 0.80: z = NORM.S.INV(0.80) = 0.8416.

Step 3 — Solve for x: $x=\mu +z\sigma =850+0.8416\times 90=850+75.74\approx \boxed{926\text{ cup-cakes}}$

Excel: =NORM.INV(0.80, 850, 90) = 925.74.

Similar KB problems: Ch 6 §6.3 problem 6.10 (toolworker injuries inverse Normal).

Q6.6 — Exam Time N(60, 12) — Inverse Normal (Slide 19)

Question: Exam time ~ N(60 min, σ = 12). How much time so that 90% of students finish?

Step 1 — Want x such that P(X ≤ x) = 0.90. z = 1.2816.

Step 2: $x=60+1.2816\times 12=60+15.38\approx \boxed{75.38\text{ minutes}}$

Excel: =NORM.INV(0.90, 60, 12) = 75.38.

Similar KB problems: Ch 6 §6.3 inverse problems (test cutoffs, deadlines).

Q6.7 — Stats Marks N(62, σ²=225) — Top 30% B Grade (Slides 21–22)

Question: Marks ~ N(62, variance = 225, so σ = 15). Top 30% gets B or higher. Cutoff?

Step 1 — Want P(X ≥ x) = 0.30, i.e., P(X ≤ x) = 0.70. z = NORM.S.INV(0.70) = 0.5244.

Step 2: $x=62+0.5244\times 15=62+7.87\approx \boxed{69.87\text{ marks}}$

Excel: =NORM.INV(0.70, 62, 15) = 69.866.

Similar KB problems: Ch 6 §6.3 problem 6.10 e (inverse Normal cutoff).

Q6.8 — TV Lifetime Warranty N(75, 8)

Question: Lifetime ~ N(75 months, σ = 8). Want only 1% replaced under warranty. What warranty length?

Step 1 — Want P(X < warranty) = 0.01. z = NORM.S.INV(0.01) = −2.3263.

Step 2: $\text{warranty}=75+(-2.3263)(8)=75-18.61\approx \boxed{56.39\text{ months}}$

Excel: =NORM.INV(0.01, 75, 8) = 56.39.

Similar KB problems: Ch 6 §6.3 — warranty / cutoff inverse problems.

📋 Final Summary Table

Total worked questions: 38 spanning all six lecture PDFs.

After mastering these, attempt the matching Ken Black problems for spaced repetition and exam-style variation.