Difficulty: Intermediate | Prerequisites: Binomial distribution, normal approximation, expected value and variance rules
These problems apply the core tools (expected value, variance, CLT, hypothesis testing) to concrete scenarios: scouting baseball players, insuring cars, evaluating salespeople, and sampling incomes. The recurring pattern is a group of individuals with hidden types (talented vs untalented, crash-prone vs safe), where you use sample data and hypothesis tests to make decisions or estimate aggregate outcomes.
Binomial distribution
The number of successes in n independent trials, each with success probability p. E[X] = np, Var(X) = np(1 − p). In simple terms, it counts how many times something happens in a fixed number of attempts.
Central Limit Theorem (CLT)
The sample mean of n independent observations approaches a normal distribution as n grows, regardless of the underlying distribution. This is what lets us use z-scores for hypothesis tests even when the data is not normal.
Sampling distribution
The probability distribution of a statistic (like X̄ or p̂) across all possible samples of a given size. Think of it as the distribution you would get if you repeated the sampling process infinitely many times.
Heterogeneous mixture (hidden types)
When a population contains subgroups with different parameters (e.g. 40% talented, 60% untalented), the overall expected value and variance must account for both groups. The expected value of each observation is a weighted average of the subgroup means.
Complement rule for "at least one"
P(at least 1) = 1 − P(none). This is nearly always easier than computing P(1) + P(2) + ... directly.
100 prospects: 40 talented (p = 0.30 per at-bat), 60 untalented (p = 0.25). Each gets n at-bats.
For a talented prospect: E[H_i] = 0.30n, Var(H_i) = n(0.30)(0.70) = 0.21n.
For an untalented prospect: E[H_i] = 0.25n, Var(H_i) = n(0.25)(0.75) = 0.1875n.
E[H̄] = [40(0.30n) + 60(0.25n)] / 100 = (12n + 15n)/100 = 0.27n.
Var(H̄) = [40 × 0.21n + 60 × 0.1875n] / 100² = (8.4n + 11.25n)/10,000 = 19.65n/10,000 = 0.001965n.
Problem 9 (salespeople): 60 good (p = 0.40), 40 bad (p = 0.30), each makes n pitches. E[T̄] = [60(0.40n) + 40(0.30n)]/100 = 0.36n. Var(T̄) = [60 × 0.24n + 40 × 0.21n]/10,000 = (14.4n + 8.4n)/10,000 = 0.00228n.
Problem 4(b): n = 49. E[H̄] = 0.27(49) = 13.23. Var(H̄) = 0.001965(49) = 0.09629. SE = √0.09629 ≈ 0.3103.
90% CI: 13.23 ± 1.645 × 0.3103 = 13.23 ± 0.5104 = (12.72, 13.74).
Problem 9(b): 97.8% CI uses z* = 2.29 (since P(|Z| < 2.29) ≈ 0.978). Range: E[T̄] ± 2.29 × SE(T̄).
Problem 4(c): For each prospect, H₀: p = 0.25, H₁: p > 0.25, α = 0.05.
Under H₀ with n = 49 at-bats: E[p̂] = 0.25, SE = √(0.25 × 0.75/49) = √(0.003827) ≈ 0.06186.
Cutoff: 0.25 + 1.645 × 0.06186 ≈ 0.3518. In terms of hits: 0.3518 × 49 ≈ 17.24, so need at least 18 hits to reject.
P(at least 1 of 60 untalented gets a contract) = 1 − P(none of 60 gets a contract). Each untalented prospect has P(reject H₀ | p = 0.25) = α = 0.05. P(none) = (1 − 0.05)^60 = 0.95^60 ≈ 0.0461. So P(at least 1) ≈ 1 − 0.0461 = 0.9539.
Problem 4(e): A talented prospect has p = 0.30. The cutoff is p̂ > 0.3518. Under p = 0.30: SE = √(0.30 × 0.70/49) ≈ 0.06547. z = (0.3518 − 0.30)/0.06547 ≈ 0.791. P(fail to reject) = P(Z < 0.791) ≈ 0.786. So a talented prospect has about a 78.6% chance of not receiving an offer.
900 cars, each has 5% crash probability, payout = $10,000 per crash.
Let X = number of crashes. X ~ Binomial(900, 0.05).
E[payout] = 900 × 0.05 × $10,000 = $450,000.
Var(X) = 900 × 0.05 × 0.95 = 42.75. SD(X) = 6.538. SD(payout) = 6.538 × $10,000 = $65,383.
P(payout > $750,000) = P(X > 75) = P(Z > (75 − 45)/6.538) = P(Z > 4.59) ≈ essentially 0.
Problem 6(d): Average payout over 4 years ≥ $750,000. The average of 4 i.i.d. years has E = $450,000 and SD = $65,383/√4 = $32,691. P(avg > $750,000) = P(Z > (750,000 − 450,000)/32,691) = P(Z > 9.18) ≈ 0.
Population: μ = $50,000, σ = $10,000, n = 400.
SE(X̄) = 10,000/√400 = $500.
P(|X̄ − 50,000| > 1,000) = P(|Z| > 2) = 2 × 0.0228 = 0.0456.
For P(|X̄ − μ| > 50) ≤ 0.05: need z × σ/√n ≤ 50, so 1.96 × 10,000/√n ≤ 50, √n ≥ 392, n ≥ 153,664.
E[X] = np, \quad \text{Var}(X) = np(1-p) \quad \text{(binomial)}E[\bar{H}] = \frac{1}{N} \sum_{\text{groups}} n_g \cdot E[H_i \mid \text{group } g]\text{Var}(\bar{H}) = \frac{1}{N^2} \sum_{\text{groups}} n_g \cdot \text{Var}(H_i \mid \text{group } g)P(\text{at least 1}) = 1 - P(\text{none}) = 1 - (1-p)^n\text{SE}(\bar{X}) = \frac{\sigma}{\sqrt{n}} \quad \text{(standard error of the sample mean)}\text{SD(total payout)} = \text{payout per event} \times \sqrt{np(1-p)}Students often compute E[H̄] using only one subgroup's probability, forgetting to weight by the proportion of each type. When there are 40 talented and 60 untalented, the overall expected value is a weighted average, not a simple average of 0.30 and 0.25.
Confusing the variance of the total with the variance of the average. If Var(total) = V, then Var(average) = V/N², not V/N.
For "at least 1" problems, students sometimes try to compute P(1) + P(2) + ... instead of using the complement rule 1 − P(none). The complement approach is nearly always what the exam expects.
In the insurance problem, students sometimes compute SD of the total payout as √(Var(X)) × payout instead of SD(X) × payout. These are equivalent, but the intermediate steps look different and can cause confusion.
⚠️ The "hidden types" pattern (talented vs untalented, good vs bad) appears in at least two problems. Know how to compute E[H̄] and Var(H̄) by summing across groups.
⚠️ P(at least 1 untalented prospect gets a contract) = 1 − (0.95)^60 is a classic exam calculation. Memorise the setup.
⚠️ Insurance problems test whether you can move between the number of events and the dollar amount. SD(payout) = per-event cost × SD(number of events).
⚠️ The sample size problem (Problem 11b) requires solving for n. The algebra is z × σ/√n = tolerance, then squaring both sides. Expect large n values.
Fill in the blank: For a binomial with n trials and success probability p, Var(X) = ______. (np(1 - p))
True or false: P(at least 1 success in 60 trials) = 1 - (1 - p)^60. (True)
Fill in the blank: If 900 cars each have a 5% crash rate with 10,000 payout, the expected total payout is ______. (450,000)
True or false: Doubling the sample size halves the standard error. (False. It divides it by the square root of 2, roughly 1.414.)
Fill in the blank: The overall expected hit rate for 40% talented (p = 0.30) and 60% untalented (p = 0.25) is ______. (0.27)
Q: 100 baseball prospects, 40 talented (p = 0.30), 60 untalented (p = 0.25), each gets n at-bats. What is E[H-bar] and Var(H-bar)?
A: E[H-bar] = 0.27n. Var(H-bar) = [40(0.21n) + 60(0.1875n)] / 10,000 = 19.65n / 10,000 = 0.001965n.
Q: With n = 49 at-bats, what is the 90% confidence interval for H-bar?
A: E[H-bar] = 13.23, SE = sqrt(0.001965 x 49) = sqrt(0.09629) = 0.310. 90% CI: 13.23 +/- 1.645 x 0.310 = (12.72, 13.74).
Q: Testing H0: p = 0.25, H1: p > 0.25, alpha = 0.05, n = 49. What is the cutoff for rejecting?
A: Cutoff p-hat = 0.25 + 1.645 x sqrt(0.25 x 0.75 / 49) = 0.25 + 1.645 x 0.06186 = 0.3518. Need at least 18 hits out of 49.
Q: What is P(at least 1 of 60 untalented prospects gets a contract offer)?
A: P(at least 1) = 1 - 0.95^60 = 1 - 0.0461 = 0.954.
Q: 900 cars insured at 10,000 each, 5% crash rate. What is the expected payout and its SD?
A: E[payout] = 900 x 0.05 x 10,000 = 450,000. SD = 10,000 x sqrt(900 x 0.05 x 0.95) = 10,000 x 6.538 = 65,383.
Q: What is P(payout > 750,000)?
A: z = (75 - 45) / 6.538 = 4.59. P(Z > 4.59) is essentially 0.
Q: NYC income: mu = 50,000, sigma = 10,000, n = 400. What is P(sample mean is off by more than 1,000)?
A: SE = 10,000/20 = 500. P(|Z| > 1,000/500) = P(|Z| > 2) = 0.0456.
Q: How large a sample to have less than 5% chance of being off by more than 50 from the true income mean?
A: 1.96 x 10,000 / sqrt(n) = 50. sqrt(n) = 392. n = 153,664.
The binomial proportion tests here are direct applications of the hypothesis testing framework from the second set of notes. The "hidden types" setup connects to mixture distributions from the third set. The CLT justifies using normal approximations throughout. Insurance problems are a practical application of expected value and variance that connects to risk management and actuarial science.
Related Terms / Search Tags
Binomial distribution, binomial proportion test, CLT, central limit theorem, sampling distribution, expected value, variance of average, insurance payout, complement rule, at least one, hidden types, heterogeneous population, baseball scouting, sales evaluation, sample size determination, ASTATS NYU, analytical statistics midterm 2