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Stats and Prob
| Question | Answer |
|---|---|
| When finding the P value for a hypothesis test | its 2xP(T or Z >whatever value of W) |
| X bar | sample mean |
| u | mean |
| What are the 3 Axioms of Probability? | 1) P(A) ≥ 0 for all events A. 2) P(S) = 1 (whole sample space). 3) If A1, A2,... are mutually exclusive, then P(A1 ∪ A2 ∪ ...) = P(A1) + P(A2) + ... |
| What is the Addition Rule P(A∪B)? | P(A∪B) = P(A) + P(B) − P(A∩B). Subtract the intersection to avoid double-counting the overlap between A and B. |
| What is De Morgan's Law? | (A∪B)^c = A^c ∩ B^c. The complement of a union equals the intersection of the complements. Flip the operation and complement each part. |
| What is Conditional Probability P(A|B)? | P(A|B) = P(A∩B) / P(B). Probability of A GIVEN B occurred. Numerator is the joint probability; denominator is the probability of the condition. |
| What is the Conditional Complement rule? | P(A^c | B) = 1 − P(A|B). The complement rule still holds inside a conditional — the probabilities of A and not-A given B must sum to 1. |
| What is the Law of Total Probability? | If {B1,...Bm} is a partition, then P(A) = P(A|B1)P(B1) + ... + P(A|Bm)P(Bm). Break a hard probability into weighted conditional probabilities over all cases. |
| What is Bayes' Theorem? | P(A|B) = P(B|A)·P(A) / P(B). Reverses conditional probabilities. P(A) is the prior; P(A|B) is the posterior after observing B. |
| When are events A and B independent? | A and B are independent iff P(A∩B) = P(A)·P(B). Equivalently P(A|B) = P(A) — knowing B happened gives no information about A. |
| What is the Permutations formula P(n,k)? | P(n,k) = n! / (n−k)!. Ordered arrangements of k objects from n distinct objects WITHOUT replacement. Order matters. |
| What is the Combinations formula C(n,k)? | C(n,k) = n! / (k!(n−k)!). Ways to choose k objects from n WITHOUT replacement where order does NOT matter. Also written "n choose k." |
| What is permutations WITH replacement? | n^k. The number of ordered arrangements of k objects chosen from n, where each object can be reused. Order matters and repetition is allowed. |
| What is the Equiprobable Probability formula? | P(Q) = |Q| / |S|. When all outcomes are equally likely, probability = number of favorable outcomes divided by total outcomes in sample space S. |
| What is the Binomial Probability formula? | P(X=k) = C(n,k)·p^k·(1−p)^(n−k). Probability of exactly k successes in n independent trials each with success probability p. |
| What is the Bernoulli Distribution — PMF, mean, variance? | PMF: P(X=1)=p, P(X=0)=1−p. Mean = p. Variance = p(1−p). Models a single trial with two outcomes: success (1) or failure (0). |
| What is the Geometric Distribution — PMF, mean, variance? | PMF: PX(k) = p(1−p)^(k−1) for k=1,2,... Number of trials until the FIRST success. Mean = 1/p. Variance = (1−p)/p². |
| What is the Binomial Distribution — PMF, mean, variance? | PMF: PX(k) = C(n,k)·p^k·(1−p)^(n−k) for k=0,...,n. Counts successes in n independent trials. Mean = np. Variance = np(1−p). |
| What is the Pascal Distribution — PMF, mean, variance? | PMF: PX(k) = C(k−1,m−1)·p^m·(1−p)^(k−m) for k=m,m+1,... Trials needed for m successes. Mean = m/p. Variance = m(1−p)/p². Geometric is Pascal with m=1. |
| What is the Poisson Distribution — PMF, mean, variance? | PMF: PX(k) = λ^k · e^(−λ) / k! for k=0,1,2,... Models rare events. Mean = λ. Variance = λ. The parameter λ is both mean AND variance. |
| What is the Discrete Uniform Distribution — PMF, mean, variance? | PMF: PX(k) = 1/(b−a+1) for k=a,a+1,...,b. All integers equally likely. Mean = (a+b)/2. Variance = ((b−a+1)²−1)/12. |
| What is the Continuous Uniform Distribution — PDF, mean, variance? | PDF: fX(x) = 1/(b−a) for a ≤ x ≤ b, else 0. Mean = (a+b)/2. Variance = (b−a)²/12. All values in [a,b] are equally likely. |
| What is the Exponential Distribution — PDF, mean, variance? | PDF: fX(x) = λe^(−λx) for x ≥ 0. Models time between events. Mean = 1/λ. Variance = 1/λ². λ is the rate parameter; higher λ means shorter average wait. |
| What is the Erlang Distribution — PDF, mean, variance? | PDF: fX(x) = λ^n · x^(n−1) · e^(−λx) / (n−1)! for x ≥ 0. Sum of n independent Exponential(λ) r.v.s. Mean = n/λ. Variance = n/λ². |
| What is the Gaussian (Normal) Distribution — PDF and CDF? | PDF: fX(x) = (1/σ√(2π))·e^(−(x−µ)²/2σ²). CDF: FX(x) = Φ((x−µ)/σ). Mean = µ. Variance = σ². Φ is the standard Normal CDF. |
| What does Φ(z) mean and what is its complement rule? | Φ(z) = P(Z ≤ z) for Z ~ Normal(0,1). Complement rule: Φ(−z) = 1 − Φ(z). Standardize any Normal: z = (x−µ)/σ, then look up Φ. |
| What is the relationship between PDF fX(x) and CDF FX(x)? | fX(x) = dFX(x)/dx. FX(x) = ∫[−∞ to x] fX(u)du. PDF is the derivative of CDF; CDF is the integral of PDF. |
| How do you compute P(a ≤ X ≤ b)? | P(a ≤ X ≤ b) = ∫[a to b] fX(x)dx = FX(b) − FX(a). Integrate the PDF over the interval, or subtract CDF values at the endpoints. |
| What is Expected Value E[X] for a discrete random variable? | E[X] = Σ x·PX(x) over all x in the range. Weighted average of all possible values, each weighted by its probability. |
| What is E[g(X)] for a continuous random variable? | E[g(X)] = ∫[−∞ to +∞] g(x)·fX(x)dx. Integrate g(x) weighted by the PDF. For discrete: E[g(X)] = Σ g(x)·PX(x). |
| What are the Linear Expectation and Variance rules? | E[aX+b] = a·E[X]+b. Var(aX+b) = a²·Var(X). Constants shift the mean but not variance; scaling multiplies variance by the square of the scale factor. |
| What is Variance Var(X) and its shortcut formula? | Var(X) = E[(X−µ)²] = E[X²] − µ². Average squared deviation from the mean. Shortcut: compute E[X²] then subtract the square of the mean. |
| What is the PMF Normalization condition? | Σ PX(x) = 1 over all x. All probabilities must sum to 1. For continuous: ∫ fX(x)dx = 1. A valid distribution must satisfy this. |
| What is the Joint CDF FX,Y(x,y)? | FX,Y(x,y) = P(X ≤ x, Y ≤ y). The probability that X is at most x AND Y is at most y simultaneously. |
| How do you find Marginal PDFs from a joint PDF? | fX(x) = ∫[−∞ to +∞] fX,Y(x,y)dy. Integrate out Y to get the marginal of X. Similarly fY(y) = ∫ fX,Y(x,y)dx. Removes dependence on the other variable. |
| What is the Conditional PDF fX|Y(x|y)? | fX|Y(x|y) = fX,Y(x,y) / fY(y). Density of X given Y=y. Divide joint density by the marginal of the conditioning variable Y. |
| When are X and Y independent (joint distributions)? | X and Y are independent iff fX,Y(x,y) = fX(x)·fY(y) for continuous, or PX,Y(x,y) = PX(x)·PY(y) for discrete. Joint equals the product of the marginals. |
| What is Linearity of Expectation for joint r.v.s? | E[g(X) + h(Y)] = E[g(X)] + E[h(Y)]. Always true regardless of whether X and Y are independent. Expectation distributes over sums. |
| What is Covariance Cov(X,Y)? | Cov(X,Y) = E[(X−µX)(Y−µY)] = E[XY] − µX·µY. Measures how X and Y vary together. Positive = move together; negative = move oppositely; zero = uncorrelated. |
| What is the Correlation Coefficient ρ(X,Y)? | ρ(X,Y) = Cov(X,Y) / (σX·σY). Normalized covariance; always in [−1,1]. ρ=1 perfect positive linear relationship; ρ=−1 perfect negative; ρ=0 uncorrelated. |
| What is Var(X+Y)? | Var(X+Y) = Var(X) + Var(Y) + 2·Cov(X,Y). If X and Y are independent, Cov=0, so Var(X+Y) = Var(X)+Var(Y). |
| What is E[Wn] and Var(Wn) for a sum of n i.i.d. r.v.s? | Wn = X1+...+Xn. E[Wn] = n·E[X]. Var(Wn) = n·Var(X). Std dev = σX·√n. The mean scales linearly with n but std dev only grows as √n. |
| What is the Sample Mean X̄ — its mean and variance? | X̄ = (1/n)·ΣXi. E[X̄] = µX (unbiased). Var(X̄) = Var(X)/n. Std dev = σX/√n. As n grows, X̄ becomes more concentrated around the true mean. |
| What is the Central Limit Theorem (CLT)? | For large n, Wn = ΣXi is approximately Normal(n·µX, σX·√n). Equivalently, X̄ ≈ Normal(µX, σX/√n). Written: FWn(w) ≈ Φ((w−nµX)/(σX·√n)). Works regardless of the original distribution. |
| What is the Continuity Correction for CLT? | P(a ≤ W ≤ b) ≈ P(a−0.5 ≤ WCLT ≤ b+0.5). Used when W is a discrete integer-valued r.v. approximated by a continuous Normal. Expand interval by 0.5 on each side. |
| What is the Weak Law of Large Numbers? | lim[n→∞] P(|X̄ − µX| < ε) = 1 for any ε > 0. The sample mean converges in probability to the true mean as n → ∞. Foundation of statistical estimation. |
| What is Bias of an estimator? | B(θ̂) = E[θ̂] − θ. Systematic error: how far the estimator's expected value is from the true parameter θ. Unbiased means B=0. Example: X̄ is unbiased for µ. |
| What is Mean Squared Error (MSE) of an estimator? | MSE(θ̂) = E[(θ̂−θ)²] = Var(θ̂) + [B(θ̂)]². Total error = variance + bias². When unbiased, MSE = Variance. Lower MSE means a better estimator. |
| What is the Sample Variance S² and why n−1? | S² = (1/(n−1))·Σ(Xi−X̄)². Dividing by n−1 (not n) makes S² an unbiased estimator of σ². The −1 corrects for estimating X̄ from the same data. |
| What is the Likelihood Function and MLE? | L(θ) = Π fX(xi;θ). The MLE is θ̂ = argmax L(θ) — the parameter making observed data most probable. Often maximize log L = Σ log fX(xi;θ) for easier computation. |
| What are the critical values zα and zα/2? | P(Z > zα) = α and P(Z > zα/2) = α/2 where Z ~ Normal(0,1). So zα = Φ⁻¹(1−α) and zα/2 = Φ⁻¹(1−α/2). For 95% CI: α=0.05 and zα/2 ≈ 1.96. |
| What is a Confidence Interval for the mean using z? | P(X̄ − δ ≤ µ ≤ X̄ + δ) ≥ 1−α, where δ = zα/2·S/√n. A 95% CI uses z=1.96. Interpretation: 95% of such intervals constructed this way will contain the true µ. |
| When do you use Student's t instead of z for a CI? | When n is small AND σ is unknown. Use δ = tα/2,n−1·S/√n where t has n−1 degrees of freedom. The t-distribution has heavier tails; it approaches Normal as n→∞. |
| What is the Confidence Interval for variance σ²? | xl = (n−1)S²/χ²(α/2,n−1) and xh = (n−1)S²/χ²(1−α/2,n−1). Uses the chi-squared distribution with n−1 degrees of freedom. The interval is asymmetric. |
| How do you test H0: µ=µ0 vs H1: µ≠µ0 (two-sided)? | Test statistic W = (X̄−µ0)/(σ/√n). Reject H0 if |W| > zα/2. If σ unknown use S; if n small use tα/2,n−1. Deviations in either direction count as evidence. |
| How do you test H0: µ≤µ0 vs H1: µ>µ0 (one-sided upper)? | Test statistic W = (X̄−µ0)/(σ/√n). Reject H0 if W > zα. One-sided test — only large positive values of W provide evidence against H0. |
| How do you test H0: µ≥µ0 vs H1: µ<µ0 (one-sided lower)? | Test statistic W = (X̄−µ0)/(σ/√n). Reject H0 if W < −zα. One-sided test — only large negative values of W provide evidence against H0. |
| How do you compare two means H0: µA=µB vs H1: µA≠µB? | Test statistic W = (X̄A−X̄B) / √((σA²+σB²)/n). Reject H0 if |W| > zα/2. Assumes equal n; use S if σ unknown; use tα/2,2n−2 if n is small. |
| What is the Chi-squared goodness-of-fit test? | X² = Σ(Oi−Ei)²/Ei where Oi=observed count, Ei=expected count in bin i. Reject H0 (data fits the distribution) if X² > χ²(α,k−1) where k=number of bins. |
| What is the p-value in hypothesis testing? | The p-value is the lowest significance level α at which H0 would be rejected. If p-value < α, reject H0. A smaller p-value means stronger evidence against H0. |
| What is the Simple Linear Regression model? | Y = β0 + β1·X + ε. β0 = intercept (Y when X=0). β1 = slope (change in Y per unit increase in X). ε = random error with mean 0. |
| What are the OLS formulas for β̂0 and β̂1? | β̂1 = sxy/sxx where sxy=Σ(xi−x̄)(yi−ȳ) and sxx=Σ(xi−x̄)². β̂0 = ȳ − β̂1·x̄. The regression line always passes through (x̄, ȳ). |
| What are sxx, syy, and sxy in linear regression? | sxx = Σ(xi−x̄)² (spread of X). syy = Σ(yi−ȳ)² (spread of Y). sxy = Σ(xi−x̄)(yi−ȳ) (co-spread of X and Y). Building blocks of OLS regression. |
| What are predicted values ŷi and residuals ei? | ŷi = β̂0 + β̂1·xi (fitted value on the regression line). ei = yi − ŷi (residual = actual minus predicted). Residuals represent the unexplained part of Y. |
| What is r² in linear regression and what does it mean? | r² = sxy² / (sxx·syy). Proportion of variance in Y explained by X. Ranges 0 to 1. r²=1 means perfect linear fit; r²=0 means X explains none of the variation in Y. |
| What is the Standard Error of the slope Sβ1? | Sβ1 = √( (1/(n−2))·Σei² / Σ(xi−x̄)² ). Measures uncertainty in the slope estimate. Used in test: W = β̂1/Sβ1 to test if β1=0. |
| How do you test whether X and Y are correlated in regression? | H0: β1=0 vs H1: β1≠0. Test statistic W = β̂1/Sβ1. Reject H0 if |W| > zα/2. If n is small, use tα/2,n−2. Rejecting H0 means slope is significantly nonzero. |
| Mid1 A Q1 SB-a — A={X is even}={2,4,6}, B={X>4}={5,6}. What is (A∩B)^c? | First find A∩B = {6}, then take the complement. (A∩B)^c = {1,2,3,4,5}. Key: De Morgan's is NOT needed here — just complement the intersection directly. |
| Mid1 A Q1 SB-b — A={2,4,6}, B={5,6}, die roll. What theory do you need to find P(A|B)? | Conditional Probability: P(A|B) = P(A∩B)/P(B). A∩B={6}, so P(A∩B)=1/6. P(B)=2/6. Answer = (1/6)/(2/6) = 1/2. |
| Mid1 A Q1 SB-c — A={2,4,6}, B={5,6}, die roll. What theory do you need to find P(B|A)? | Conditional Probability: P(B|A) = P(A∩B)/P(A). A∩B={6}, P(A∩B)=1/6. P(A)=3/6. Answer = (1/6)/(3/6) = 1/3. |
| Mid1 A Q1 SB-d — Events A={2,4,6}, B={5,6}, C={1,2,3}. Which pairs are disjoint? | Two events are disjoint if their intersection is empty. B∩C = {} (no overlap between {5,6} and {1,2,3}). So B and C are disjoint. Check all pairs: A∩B={6}≠∅, A∩C={2}≠∅, B∩C=∅. |
| Mid1 A Q1 SB-e — A={2,4,6}, B={5,6} on a fair die. How do you prove A and B are independent? | Independence check: P(A)·P(B) must equal P(A∩B). P(A)=1/2, P(B)=1/3, P(A∩B)=1/6. Since (1/2)·(1/3) = 1/6 = P(A∩B), A and B ARE independent. |
| CONCEPT — How do you identify a Conditional Probability problem? | Look for the word "given" or the | symbol. Setup: you already know one event occurred, and want the probability of another. Formula: P(A|B) = P(A∩B)/P(B). Always find the joint and the conditioning event's probability. |
| CONCEPT — How do you identify an Independence vs Disjoint problem? | Disjoint: events CANNOT happen together — P(A∩B)=0. Independent: events don't AFFECT each other — P(A∩B)=P(A)·P(B). WARNING: disjoint events with nonzero probability are NEVER independent. |
| Mid1 A Q2 SB-a — X~Geometric(0.4). Calculate P(X<2 | X<3). | Conditional probability. P(X<2|X<3) = P(X<2 AND X<3)/P(X<3) = P(X=1)/P(X=1 or X=2). PX(1)=0.4, PX(2)=0.4(0.6)=0.24. Answer = 0.4/(0.4+0.24) = 0.625. |
| Mid1 A Q2 SB-b — Y~Binomial(8, 0.2). Calculate P(Y≥2). | Use complement: P(Y≥2) = 1 - P(Y<2) = 1 - [P(Y=0)+P(Y=1)]. P(Y=0)=C(8,0)(0.2)^0(0.8)^8=0.1677. P(Y=1)=C(8,1)(0.2)^1(0.8)^7=0.3355. Answer = 1 - 0.1677 - 0.3355 = 0.4968. |
| Mid1 A Q2 SB-c — Y~Binomial(8,0.2). Make 10 measurements of Y, find mean number of times Y=1. | This is a Binomial WITHIN a Binomial. Each measurement has P(Y=1)=0.3355. N = number of times Y=1 in 10 trials ~ Binomial(10, 0.3355). Mean = n·p = 10 × 0.3355 = 3.355. |
| CONCEPT — How do you identify a Geometric distribution problem? | Key phrases: "first success," "keep trying until," "number of trials until." PMF: PX(k)=p(1-p)^(k-1). Mean=1/p. The experiment stops as soon as you succeed. |
| CONCEPT — How do you identify a Binomial distribution problem? | Key phrases: "exactly k successes," "out of n trials," fixed number of trials, each trial is independent with same probability p. PMF: C(n,k)·p^k·(1-p)^(n-k). Mean=np. |
| CONCEPT — When should you use the complement to find a probability? | Use complement when computing "at least one" or "at least k" directly is hard. P(at least one) = 1 - P(none). P(Y≥2) = 1 - P(Y=0) - P(Y=1). The complement is often far fewer terms to calculate. |
| Mid1 A Q3 SB-a — 10 M&Ms chosen, P(blue)=0.3. What is P(exactly 4 blue)? | Binomial distribution with n=10, p=0.3. P(N=4) = C(10,4)·(0.3)^4·(0.7)^6 = 210·0.0081·0.1176 ≈ 0.2001. |
| Mid1 A Q3 SB-b — Choose 2 M&Ms, P(R)=0.5, P(W)=0.2, P(B)=0.3. What is P(same color)? | P(same color) = P(RR)+P(WW)+P(BB) = (0.5)²+(0.2)²+(0.3)² = 0.25+0.04+0.09 = 0.38. Key: treats each candy as independent draws. |
| Mid1 A Q3 SB-c — Choose 3 M&Ms, P(R)=0.5, P(W)=0.2, P(B)=0.3. What is P(one of each color)? | P(one specific order like RWB) = 0.5×0.2×0.3 = 0.03. There are 3! = 6 permutations of 3 colors. So P = 6×0.03 = 0.18. Key: multiply by number of orderings! |
| CONCEPT — How do you recognize a problem needing permutations of arrangements? | When outcome has multiple distinguishable items and ORDER doesn't matter for the event, count the number of orderings (k! for k distinct items) and multiply by the probability of one specific order. |
| Mid1 A Q4 SB-a — 75% parts from supplier A (1% defective), 25% from B (2.5% defective). What is P(defective)? | Law of Total Probability: P(D) = P(D|A)·P(A) + P(D|B)·P(B) = 0.01·0.75 + 0.025·0.25 = 0.0075 + 0.00625 = 0.01375. |
| Mid1 A Q4 SB-b — Same setup. Given a part is defective, what is P(it came from supplier B)? | Bayes' Theorem: P(B|D) = P(D|B)·P(B)/P(D) = (0.025·0.25)/0.01375 = 0.00625/0.01375 ≈ 0.454. |
| Mid1 A Q4 SB-c — Same setup. What is P(part from supplier A | part is NOT defective)? | Bayes' with complement: P(A|D^c) = P(D^c|A)·P(A)/P(D^c) = (0.99·0.75)/(1-0.01375) = 0.7425/0.98625 ≈ 0.7529. |
| CONCEPT — How do you identify a Law of Total Probability problem? | Look for a partition (mutually exclusive, exhaustive groups like supplier A vs B, or disease vs no disease) and you want the overall probability of an event. Formula: P(event) = sum of P(event|group_i)·P(group_i). |
| CONCEPT — How do you identify a Bayes' Theorem problem? | You're asked to REVERSE the conditioning: you know P(B|A) but want P(A|B). Key phrase: "given that it IS defective/positive/outcome, what is the probability it CAME FROM group X?" Formula: P(A|B) = P(B|A)·P(A)/P(B). |
| Mid1 A Q5 SB-a — Baseball throw, p=0.1, max 3 throws, stop if you win. Find the PMF of X ( | throws made). |
| Mid1 A Q5 SB-b — Same baseball game. What is P(you win the prize)? | P(win) = 1 - P(NNN) = 1 - (0.9)^3 = 1 - 0.729 = 0.271. Always check: can you use the complement to avoid summing multiple cases? |
| Mid1 A Q5 SB-c — Same baseball game. What is P(X=2 | you win)? | Conditional probability: P(X=2|win) = P(X=2 AND win)/P(win). If X=2 you must have won on throw 2 (NY), so P(X=2 AND win) = P(X=2) = 0.09. P(X=2|win) = 0.09/0.271 ≈ 0.332. |
| CONCEPT — How do you set up a non-standard sequential probability problem? | Draw a tree diagram. At each branch: success (probability p, stops) or failure (probability 1-p, continues). Multiply along branches for each outcome. Check probabilities sum to 1. Only use Geometric when there is NO cap on trials. |
| Mid1 A Q6 SB-a — Pull 3 tiles from 26-letter bag. What is P(at least one of {x,y,z,p,d,q})? | Complement method: P(at least one) = 1 - P(none of them). P(none) = (20/26)·(19/25)·(18/24) ≈ 0.438. Answer = 1 - 0.438 = 0.562. Key: sampling WITHOUT replacement so probabilities change each draw. |
| Mid1 A Q6 SB-b — Best-of-7 series, C wins each game with p=0.6. What is P(series ends in exactly 5 games)? | Pascal distribution: series ends in 5 games means one team wins game 5 (their 4th win) with 3 wins in first 4 games. P(C wins in 5) = C(4,3)·(0.6)^4·(0.4)^1. P(D wins in 5) = C(4,3)·(0.4)^4·(0.6)^1. Total = 0.2074+0.06144 = 0.2688. |
| CONCEPT — How do you identify a Pascal (Negative Binomial) distribution problem? | Look for: "series ends when one side reaches k wins," "mth success occurs on the nth trial." The last trial MUST be the mth success. Formula: P(X=n) = C(n-1,m-1)·p^m·(1-p)^(n-m). The key: fix the last event as the success, permute the earlier ones. |
| CONCEPT — How do you identify sampling WITHOUT replacement vs WITH replacement? | Without replacement: each draw changes the pool (like drawing tiles, cards, people). Probabilities change each draw. Use products of fractions: (n/N)·((n-1)/(N-1))... With replacement: pool stays the same, use p^k. "Randomly select from a bag/deck/group" |
| Mid1 B Q1 SB-a — A={X is odd}={1,3,5}, B={X<2}={1}, die roll. Find (A∩B)^c. | A∩B = {1}. Complement = everything not in {1} = {2,3,4,5,6}. |
| Mid1 B Q1 SB-e — A={1,3,5}, B={1,2}, fair die. Verify A and B are independent. | P(A)=3/6=1/2, P(B)=2/6=1/3, A∩B={1}, P(A∩B)=1/6. Check: P(A)·P(B) = 1/2·1/3 = 1/6 = P(A∩B). ✓ Independent. |
| Mid1 B Q2 SB-a — 80% parts from A (1% defective), 20% from B (2.5% defective). What is P(defective)? | Law of Total Probability: P(D) = P(D|A)·P(A) + P(D|B)·P(B) = 0.01·0.8 + 0.025·0.2 = 0.008 + 0.005 = 0.0130. |
| Mid1 B Q2 SB-b — Same setup (80/20 split). Given defective, what is P(from supplier B)? | Bayes': P(B|D) = P(D|B)·P(B)/P(D) = (0.025·0.2)/0.0130 = 0.005/0.0130 ≈ 0.3846. |
| Mid1 B Q2 SB-c — Same setup (80/20 split). Given NOT defective, what is P(from supplier A)? | Bayes' with complement: P(A|D^c) = P(D^c|A)·P(A)/P(D^c) = (0.99·0.8)/(1-0.0130) = 0.792/0.9870 ≈ 0.8024. |
| Mid1 B Q3 SB-b — Best-of-7 series, C wins with p=0.70. What is P(series ends in exactly 5 games)? | Pascal: P(C wins in 5) = C(4,3)·(0.7)^4·(0.3)^1 = 4·0.2401·0.3 = 0.2881. P(D wins in 5) = C(4,3)·(0.3)^4·(0.7)^1 = 4·0.0081·0.7 = 0.0227. Total ≈ 0.3108. |
| CONCEPT — What is the key difference between Exam A Q4 and Exam B Q2 (supply chain)? | Same problem structure (Bayes' + Total Probability) but different split percentages: Version A uses 75%/25% while Version B uses 80%/20%. The METHOD is identical — always set up P(D) first with Total Probability, then use Bayes' for the reverse conditiona |
| CONCEPT — General strategy: given a probability word problem, how do you pick the right tool? | Step 1: Is there a condition ("given")? → Conditional probability or Bayes'. Step 2: Fixed n trials, count successes? → Binomial. Step 3: Trials until first success, no cap? → Geometric. Step 4: Trials until mth success / series ends? → Pascal. Step 5: Mu |
| Mid2 A Q1 SB-a — Bears: µ=450 kg, σ=90 kg, Normal. What fraction weigh MORE than 550 kg? | Standardize: z = (550−450)/90 = 1.11. P(X>550) = 1 − Φ(1.11) = 1 − 0.8665 = 0.1335. Key step: always standardize first with z=(x−µ)/σ, then use the complement 1−Φ(z). |
| Mid2 A Q1 SB-b — Same bear distribution. Given a bear is below-average weight (X<450), what is P(X>300)? | Conditional on X<450: P(X>300|X<450) = P(300<X<450)/P(X<450). P(X<450)=0.5. Standardize 300: z=(300−450)/90=−1.67. Φ(−1.67)=1−0.9525=0.0475. Numerator = 0.5−0.0475 = 0.4525. Answer = 0.4525/0.5 = 0.905. |
| Mid2 A Q1 SB-c — Same bear distribution. What minimum weight puts a bear in the heaviest 5%? | Set P(X>x)=0.05, so P(X≤x)=0.95. Inverse normal: (x−µ)/σ = Φ⁻¹(0.95) = 1.65. Solve: x = 450 + 1.65×90 = 598.5 kg. Key: "heaviest 5%" means top 5%, so set CDF = 0.95 and solve for x. |
| Mid2 B Q1 SB-b — Bears µ=450, σ=90. Given below-average weight (X<450), what is P(X>400)? | P(X>400|X<450) = P(400<X<450)/P(X<450). P(X<450)=0.5. z=(400−450)/90=−0.53. Φ(−0.53)=1−0.7019=0.2981. Numerator=0.5−0.2981=0.2019. Answer=0.2019/0.5=0.409 (≈0.409). |
| Mid2 B Q1 SB-c — Same bear distribution. What minimum weight is in the heaviest 10%? | P(X>x)=0.10, so FX(x)=0.90. Φ⁻¹(0.90)=1.28. x = 450 + 1.28×90 = 565 kg. Compare to Version A (5% → 1.65 → 598.5 kg): higher percentile cutoff means higher weight. |
| CONCEPT — How do you identify a Normal Distribution / standardization problem? | Key phrases: "normally distributed," "mean µ and std dev σ," "what fraction," "what probability," "what weight/score/value." Steps: 1) Standardize z=(x−µ)/σ. 2) Look up Φ(z). 3) Use complement (1−Φ) for upper tail. 4) For inverse ("what value"), set Φ(z)= |
| CONCEPT — How do you find a percentile / inverse Normal value? | "Top k%" → P(X>x)=k/100 → FX(x)=1−k/100 → z=Φ⁻¹(1−k/100) → x=µ+z·σ. Key z-values: top 5%: z=1.645. Top 10%: z=1.28. Top 1%: z=2.33. Top 2.5%: z=1.96. |
| CONCEPT — How do you handle a conditional Normal probability? | P(a<X<b | X<c) = P(a<X<b AND X<c) / P(X<c). Simplify the numerator by finding the overlap of the two conditions. Then standardize both boundaries and subtract CDF values. P(X<µ)=0.5 always for symmetric Normal. |
| Mid2 A Q3 — fY(y)=4y(1−y²) for 0≤y≤1. What is the mean E[Y]? | E[Y] = ∫[0 to 1] y·4y(1−y²)dy = ∫[0 to 1] (4y²−4y⁴)dy = [4y³/3 − 4y⁵/5] from 0 to 1 = 4/3 − 4/5 = 8/15 ≈ 0.533. |
| Mid2 A Q3 — fY(y)=4y(1−y²). What is Var(Y)? | First find E[Y²] = ∫[0 to 1] y²·4y(1−y²)dy = ∫(4y³−4y⁵)dy = [y⁴−4y⁶/6] from 0 to 1 = 1−2/3 = 1/3. Var(Y) = E[Y²]−(E[Y])² = 1/3−(8/15)² = 1/3−64/225 ≈ 0.0489. σY = 0.221. |
| Mid2 B Q6 — fX(x)=4x(1−x²) for 0≤x≤1. Calculate mean, variance, and std dev of X. | E[X]=8/15 (same as Version A Q3 by symmetry of the PDF form). E[X²]=1/3. Var(X)=1/3−(8/15)²=0.0489. σX=0.221. Both versions have the SAME PDF, just different variable names Y vs X. |
| CONCEPT — How do you calculate E[X] and Var(X) from a custom PDF? | Step 1: E[X] = ∫ x·fX(x)dx. Multiply x by the PDF and integrate over the support. Step 2: E[X²] = ∫ x²·fX(x)dx. Step 3: Var(X) = E[X²] − (E[X])². Step 4: σ = √Var(X). Always check normalization: ∫fX(x)dx = 1 first. |
| Mid2 A Q4 SB-a — Customers arrive at 20/hr. What is P(next customer arrives within 5 minutes)? | Exponential distribution. Convert rate: λ = 20/60 = 1/3 per minute. P(X≤5) = 1 − e^(−λ·5) = 1 − e^(−5/3) ≈ 0.811. Alternatively via Poisson: λ for 5 min = 5/3, P(at least 1) = 1 − e^(−5/3) = 0.811. |
| Mid2 A Q4 SB-b — Same arrival rate. What is P(at least 2 customers arrive in 10 minutes)? | Use Poisson with λ = 20/60 × 10 = 10/3 per 10-min window. P(X≥2) = 1−P(X=0)−P(X=1) = 1−e^(−λ)−λe^(−λ) = 1−[0.0357+0.1189] = 0.845. |
| Mid2 A Q4 SB-c — Same arrival rate, 10 minutes since last customer. What are mean and std dev of wait for next arrival? | Exponential is memoryless — past waiting time doesn't matter! λ = 1/3 min⁻¹. Mean = 1/λ = 3 min. Std dev = 1/λ = 3 min. Key: for Exponential, mean = std dev = 1/λ. |
| CONCEPT — How do you identify an Exponential vs Poisson problem? | Both involve random arrivals/events. Exponential: models the TIME between events (continuous). Poisson: models the COUNT of events in a fixed time window (discrete). Same λ, different questions. "Time until next" → Exponential. "How many in t minutes" → P |
| CONCEPT — What is the memoryless property of the Exponential distribution? | P(X > s+t | X > s) = P(X > t). If you've already waited s minutes with no arrival, the distribution of remaining wait time is the same as if you just started. "It's been 10 minutes" is irrelevant — mean and std dev are still 1/λ. |
| CONCEPT — How do you convert arrival rates between time units? | If rate is given per hour (e.g., 20/hr), convert to per-minute by dividing by 60: λ=20/60=1/3 min⁻¹. For a Poisson count over t minutes: λ_window = λ·t. For Exponential (time to next): use λ in consistent units. Always match units before plugging into for |
| Mid2 A Q2 SB-a — Boilers A and B: P(A=1)=0.4, P(B=1)=0.7, P(A=1,B=1)=0.2. Fill in the joint PMF table. | P(A=0,B=0)=0.1, P(A=1,B=0)=0.2 (column sum for B=0 is 0.3, P(A)=0.4 total). P(A=0,B=1)=0.5, P(A=1,B=1)=0.2. Row sums: P(B=0)=0.3, P(B=1)=0.7. Column sums: P(A=0)=0.6, P(A=1)=0.4. Use constraints: each row and column must sum to marginals. |
| Mid2 A Q2 SB-b — Same boiler joint PMF. What is P(exactly one boiler running)? | P(exactly one) = P(A=0,B=1) + P(A=1,B=0) = 0.5 + 0.2 = 0.7. Read directly from off-diagonal cells of the joint PMF table. |
| Mid2 A Q2 SB-c — Same boilers. Find Cov(A,B) and ρ(A,B). | E[AB] = ΣΣ a·b·P(A=a,B=b) = 1·1·0.2 = 0.2. µA=0.4, µB=0.7. Cov(A,B) = E[AB]−µA·µB = 0.2−0.28 = −0.08. σA=√(0.4·0.6)=0.490, σB=√(0.7·0.3)=0.458. ρ = −0.08/(0.490·0.458) = −0.356. |
| CONCEPT — How do you build a joint PMF table from marginals and one joint value? | You're given P(A=1), P(B=1), and P(A=1,B=1). Fill table using: joint cell = given. Other cells = marginal minus joint. Example: P(A=1,B=0) = P(A=1)−P(A=1,B=1). P(A=0,B=1) = P(B=1)−P(A=1,B=1). P(A=0,B=0) = 1−P(A=1)−P(B=1)+P(A=1,B=1). |
| CONCEPT — How do you calculate Covariance from a joint PMF? | Cov(X,Y) = E[XY]−µX·µY. Find E[XY] = ΣΣ x·y·PX,Y(x,y). Only nonzero when both x≠0 AND y≠0. Get µX and µY from marginals. Negative covariance means when one goes up the other tends to go down. |
| Mid2 A Q5 SB-a — Z~Geometric(0.4), B={Z<3}={1,2}. Find the conditional PMF PZ|B(z). | P(B)=P(Z=1)+P(Z=2)=0.4+0.24=0.64. PZ|B(1) = 0.4/0.64 = 0.625. PZ|B(2) = 0.24/0.64 = 0.375. PZ|B(z) = 0 for z≥3. Key: divide each in-event PMF value by P(B). |
| Mid2 B Q5 SB-a — Z~Geometric(0.5), B={Z<3}={1,2}. Find the conditional PMF PZ|B(z). | P(Z=1)=0.5, P(Z=2)=0.5×0.5=0.25. P(B)=0.75. PZ|B(1)=0.5/0.75=2/3≈0.667. PZ|B(2)=0.25/0.75=1/3≈0.333. Same method as Version A, different p. |
| Mid2 A Q5 SB-b — Y~Binomial(2, 0.5). Calculate E[Y³]. | Range: RY={0,1,2}. P(Y=0)=0.25, P(Y=1)=0.50, P(Y=2)=0.25. E[Y³] = 0³·0.25 + 1³·0.50 + 2³·0.25 = 0 + 0.5 + 2 = 2.5. Key: for E[g(Y)], compute g(y) for each value, multiply by probability, sum up. |
| Mid2 A Q5 SB-c — X~Poisson(λ=2.0). Calculate E[4X+2]. | Linearity of expectation: E[4X+2] = 4·E[X]+2. For Poisson, E[X]=λ=2.0. Answer = 4·2+2 = 10. Never expand the Poisson PMF for linear functions — always use E[aX+b]=aµ+b. |
| Mid2 B Q5 SB-c — X~Poisson(λ=2.0). Calculate E[3X−1]. | E[3X−1] = 3·E[X]−1 = 3·λ−1 = 3·2−1 = 5. Same linearity rule as Version A. |
| CONCEPT — How do you calculate E[g(X)] for a non-linear function like g(X)=X³? | You CANNOT use E[X³] = (E[X])³. Instead use E[g(X)] = Σ g(x)·PX(x) for discrete, or ∫g(x)fX(x)dx for continuous. List all values in the range, compute g at each, multiply by probability, sum. Linearity only works for g(X)=aX+b. |
| Mid2 A Q6 SB-a — X~Uniform(0,2). Write the PDF and CDF of X. | PDF: fX(x) = 1/2 for 0≤x≤2, 0 otherwise. CDF: FX(x) = 0 for x<0; x/2 for 0≤x≤2; 1 for x>2. To get CDF, integrate PDF from −∞ to x: ∫[0 to x] (1/2)du = x/2. |
| Mid2 B Q2 SB-a — X~Uniform(0,3). Write the PDF and CDF of X. | PDF: fX(x) = 1/3 for 0≤x≤3, 0 otherwise. CDF: FX(x) = 0 for x<0; x/3 for 0≤x≤3; 1 for x≥3. General pattern for Uniform(0,b): PDF=1/b, CDF=x/b. |
| Mid2 A Q6 SB-c — X~Uniform(0,2), Y=e^X. Derive the PDF of Y. | CDF method: FY(y) = P(Y≤y) = P(e^X≤y) = P(X≤ln y) = FX(ln y) = (ln y)/2. Differentiate: fY(y) = d/dy[(ln y)/2] = fX(ln y)·(1/y) = (1/2)·(1/y) = 1/(2y) for 1≤y≤e². Key: Y=e^X ranges from e^0=1 to e^2 as X ranges from 0 to 2. |
| Mid2 B Q2 SB-c — X~Uniform(0,3), Y=e^X. Derive the PDF of Y. | Same CDF method: FY(y) = P(X≤ln y) = FX(ln y) = (ln y)/3. fY(y) = 1/(3y) for 1≤y≤e³. The only difference from Version A: divide by 3 instead of 2, and range extends to e³. |
| CONCEPT — How do you derive the PDF of a transformed random variable Y=g(X)? | CDF method (always works): 1) Write FY(y)=P(Y≤y)=P(g(X)≤y). 2) Solve for X: P(X≤g⁻¹(y))=FX(g⁻¹(y)). 3) Differentiate: fY(y) = fX(g⁻¹(y))·|d/dy[g⁻¹(y)]|. For Y=e^X: g⁻¹(y)=ln y, derivative=1/y. New support: y ranges from g(a) to g(b). |
| CONCEPT — How do you find the CDF from a PDF by integration? | FX(x) = ∫[−∞ to x] fX(u)du. For piecewise PDFs: in each region, integrate the PDF from the lower boundary to x and add any accumulated probability from previous regions. Always check: FX(−∞)=0 and FX(+∞)=1. |
| CONCEPT — Midterm 2 master topic map: what concept does each question test? | Q1 (both versions): Normal distribution, standardizing, inverse normal (percentiles). Q2/Q3: Joint PMF table OR continuous PDF mean/variance/std dev. Q4: Exponential/Poisson (arrivals). Q5: Conditional PMF, E[g(X)], linearity of expectation. Q6 (A) / Q2 ( |
| Mid3 A Q1 SB-a — Containers: µ=21 tons, σ=2.3 tons. What are the mean and std dev of the TOTAL weight of 100 containers? | CLT for sums: E[W100] = n·µ = 100·21 = 2100 tons. σ(W100) = σ·√n = 2.3·√100 = 23.0 tons. Key: mean scales by n, std dev scales by √n — NOT by n. |
| Mid3 A Q1 SB-b — Same containers. What is P(total weight of 100 > 2130 tons)? | Standardize the sum: z = (2130−2100)/23 = 30/23 = 1.304. P(W>2130) = 1−Φ(1.304) = 1−0.9032 = 0.0968. Key: use the CLT sum's mean=2100 and σ=23 as the Normal parameters. |
| Mid3 A Q1 SB-c — 50 existing containers weigh 1070 tons total. 50 new ones are added. What is P(grand total > 2130)? | The NEW 50 containers have mean=50·21=1050, σ=√50·2.3=16.26. Need the new 50 to weigh more than 2130−1070=1060 tons. z=(1060−1050)/16.26=0.615. P(W_new>1060)=1−Φ(0.615)=1−0.7291=0.27. |
| CONCEPT — What is the CLT formula for a SUM of n i.i.d. r.v.s? | Wn = X1+...+Xn. By CLT: Wn ≈ Normal(n·µ, σ·√n). So P(Wn > w) = 1−Φ((w−nµ)/(σ√n)). Mean grows as n, std dev grows as √n. Use this when asked about a total/sum of many independent items. |
| CONCEPT — How do you handle a "fixed amount already accumulated + new arrivals" CLT problem? | Subtract what's already there from the target. Only apply the CLT to the NEW random portion. New total target = overall target − fixed amount. Then use the CLT on just the new n items with their own mean and σ√n. |
| Mid3 A Q2 — Joint PDF fX,Y(x,y)=x+y for 0≤x≤1, 0≤y≤1. Calculate P(X+Y>1). | Sketch: region is upper-right triangle above the line y=1−x. Set up iterated integral: ∫[0 to 1]dx ∫[1−x to 1] (x+y)dy. Inner integral = [xy+y²/2] from 1−x to 1 = (x+1/2)−(x(1−x)+(1−x)²/2). Simplify to ∫[0 to 1](x²+x−x²/2)dx = 2/3. |
| CONCEPT — How do you set up a double integral over a non-rectangular region? | Step 1: Sketch the region. Find the boundary line (e.g., y=1−x means X+Y=1). Step 2: Determine limits — for region above y=1−x: x from 0 to 1, y from (1−x) to 1. Step 3: Integrate inner integral first (over y), then outer (over x). Always sketch first. |
| Mid3 A/B Q3 SB-a — n=9, X̄=12.05, S²=1.87. Calculate 95% CI using CLT (z). | S=√1.87=1.367. For 95%, α=0.05, zα/2=z0.025=1.96. δ = 1.96·1.367/√9 = 1.96·1.367/3 = 0.893. CI: µ ∈ [12.05−0.89, 12.05+0.89] = [11.16, 12.94]. |
| Mid3 A/B Q3 SB-b — Same data. Calculate 95% CI using the t-distribution. | n=9 → df = n−1 = 8 → t(0.025, 8) = 2.306. δ = 2.306·1.367/√9 = 2.306·1.367/3 = 1.051. CI: µ ∈ [12.05±1.05] = [11.00, 13.10]. Wider than z-interval because t has heavier tails for small n. |
| Mid3 A/B Q3 SB-c — Same data. Calculate 95% CI for the variance σ². | Use χ² with df=n−1=8. χ²(0.025,8)=17.53 (upper), χ²(0.975,8)=2.18 (lower). xl=(n−1)S²/χ²upper = 8·1.87/17.53 = 0.85. xh=(n−1)S²/χ²lower = 8·1.87/2.18 = 6.86. σ² ∈ [0.85, 6.86]; σ ∈ [0.92, 2.62]. |
| CONCEPT — When do you use z vs t for a confidence interval? | Use z (1.96 for 95%): when n is large (≥30) OR σ is known. Use t (tα/2,n−1): when n is small AND σ is unknown. Same formula: δ=critical value · S/√n. The t-value is always larger than z for the same α, giving a wider interval — reflects extra uncertainty |
| CONCEPT — How do you calculate a CI for variance vs a CI for the mean? | CI for mean uses z or t. CI for variance uses χ² distribution with df=n−1. xl=(n−1)S²/χ²(α/2,n−1) and xh=(n−1)S²/χ²(1−α/2,n−1). Then CI for σ is [√xl, √xh]. The variance CI is NOT symmetric around S². |
| Mid3 A Q4 SB-a — Y~Poisson, estimated mean=6.3, so σ≈√6.3≈2.51. How many measurements n needed for CI width ±1% of mean (α=0.05)? | Want δ=1% of 6.3=0.063. Formula: δ=zα/2·σ/√n → 0.063=(1.96·2.51)/√n → √n=1.96·2.51/0.063=78.0 → n=(78.0)²=6098≈6100. |
| Mid3 B Q3 SB-a — Y~Poisson, estimated mean=6.1, σ≈√6.1≈2.47. How many measurements for CI width ±1% (α=0.05)? | Want δ=1% of 6.1=0.061. 0.061=(1.96·2.47)/√n → √n=1.96·2.47/0.061=79.3 → n≈6299≈6300. Same method as Version A — for Poisson σ=√λ. |
| CONCEPT — How do you find the sample size n needed for a target CI width? | Set δ = zα/2·σ/√n and solve for n: n = (zα/2·σ/δ)². For ±δ CI (half-width), use this directly. For Poisson with unknown λ, approximate σ≈√λ̂. Round UP to the next integer. Larger target precision (smaller δ) → much larger n needed. |
| Mid3 A Q4 SB-b — Existing CI: 5.05±0.23 at α=0.05, n=1000. What is the new δ at α=0.01 using same data? | Back out S: 0.23=z0.025·S/√1000=1.96·S/31.62 → S=3.71. New δ at α=0.01: z0.005=2.576. δ=2.576·3.71/√1000=0.302. Shortcut: new δ = old δ·(z_new/z_old) = 0.23·(2.576/1.96) = 0.302. |
| Mid3 B Q3 SB-b — Existing CI: 5.05±0.28 at α=0.05, n=1000. Find new δ at α=0.01. | Back out S: 0.28=1.96·S/√1000 → S=4.52. New δ=2.576·4.52/√1000=0.368. Or shortcut: 0.28·(2.576/1.96)=0.368. Wider CI because α=0.01 requires more confidence than α=0.05. |
| CONCEPT — How does changing α affect the confidence interval width? | Smaller α (more confidence) → larger critical value z → WIDER interval. δ scales proportionally with z. Shortcut: new δ = old δ × (z_new / z_old). Example: going from 95% to 99% CI multiplies width by 2.576/1.96 ≈ 1.315. |
| Mid3 A Q5 SB-a — fX(x)=2x⁻³ for x≥1. Calculate P(X>2). | P(X>2) = ∫[2 to ∞] 2x⁻³ dx = [2·x⁻²/(−2)] from 2 to ∞ = [−x⁻²] from 2 to ∞ = 0−(−1/4) = 1/4. |
| Mid3 A Q5 SB-b — Same fX(x)=2x⁻³, B={X>2}. Find conditional PDF fX|B(x) and calculate E[X|B]. | fX|B(x) = fX(x)/P(B) = (2x⁻³)/(1/4) = 8x⁻³ for x≥2. E[X|B] = ∫[2 to ∞] x·8x⁻³dx = ∫[2 to ∞] 8x⁻²dx = [8·x⁻¹/(−1)] from 2 to ∞ = [−8/x] from 2 to ∞ = 0−(−4) = 4. |
| CONCEPT — How do you find a conditional PDF given an event B? | fX|B(x) = fX(x)/P(B) for x in B, and 0 otherwise. Step 1: Find P(B) by integrating fX over B. Step 2: Divide fX by P(B) — this re-normalizes the PDF to the restricted region. Step 3: Use this new PDF to compute conditional expectations. |
| Mid3 A Q6 SB-a — MLE for Uniform(a,b): â=min{xi}. Is this estimator positively biased, negatively biased, or unbiased? | Positively biased. Every observation xi > a (since P(X=a)=0 for continuous), so min{xi} > a always. Therefore E[min{xi}] > a, meaning E[â]−a > 0. Positive bias: estimator overshoots the true value on average. |
| Mid3 B Q6 SB-a — MLE for Uniform(a,b): b̂=max{xi}. Is this estimator positively biased, negatively biased, or unbiased? | Negatively biased. Every observation xi < b (since P(X=b)=0), so max{xi} < b always. Therefore E[max{xi}] < b, meaning E[b̂]−b < 0. Negative bias: estimator undershoots the true value on average. |
| CONCEPT — What is bias and how do you determine its sign? | Bias B(θ̂) = E[θ̂]−θ. Positive bias: estimator tends to OVERSHOOT (be larger than) the true value. Negative bias: estimator tends to UNDERSHOOT. Unbiased: E[θ̂]=θ exactly. For min/max estimators of boundaries, ask: is it physically possible for the estima |
| Mid3 A Q6 SB-b — Z~Exponential(λ), observations Z=2 and Z=3. Find the MLE of λ. | L(λ) = fZ(2)·fZ(3) = λe^(−2λ)·λe^(−3λ) = λ²e^(−5λ). Set dL/dλ=0: 2λe^(−5λ)−5λ²e^(−5λ)=0 → divide by λe^(−5λ): 2−5λ=0 → λ̂=2/5=0.4. Note: MLE of λ for Exponential = 1/x̄ = 2/(2+3) = 2/5. ✓ |
| Mid3 B Q6 SB-b — Z~Exponential(λ), observations Z=2 and Z=4. Find the MLE of λ. | L(λ) = λe^(−2λ)·λe^(−4λ) = λ²e^(−6λ). dL/dλ = 2λe^(−6λ)−6λ²e^(−6λ) = 0 → 2−6λ=0 → λ̂=1/3. Check: 1/x̄ = 2/(2+4) = 2/6 = 1/3. ✓ Same method as Version A, different observations. |
| CONCEPT — How do you find the MLE for the Exponential rate parameter λ? | For n observations {x1,...,xn}: L(λ) = Π λe^(−λxi) = λⁿ·e^(−λΣxi). Take dL/dλ=0 (or d/dλ of log L = 0): nλⁿ⁻¹e^(−λΣxi) − Σxi·λⁿe^(−λΣxi)=0 → n = λ·Σxi → λ̂ = n/Σxi = 1/x̄. MLE for Exponential rate = reciprocal of the sample mean. |
| CONCEPT — What is the general MLE procedure? | Step 1: Write the likelihood L(θ) = Π fX(xi;θ) (product of PDFs/PMFs at each observation). Step 2: Often easier to maximize log-likelihood: ℓ(θ) = Σ log fX(xi;θ). Step 3: Differentiate with respect to θ, set equal to 0. Step 4: Solve for θ̂. Step 5: Verif |
| CONCEPT — Midterm 3 master topic map: what concept does each question test? | A: Q1=CLT for sums/totals. Q2=double integral over joint PDF. Q3=CIs (z, t, chi-squared). Q4=sample size calculation + adjusting CI width. Q5=conditional PDF and E[X|B]. Q6=MLE bias and deriving MLE. B: same topics in different order (Q1↔Q4, Q2↔Q5, Q3↔Q3, |
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