Difficulty: Intermediate to Advanced | Prerequisites: Normal distribution, z-scores, basic probability rules
A confidence interval gives a range of plausible values for a parameter based on sample data. Mixture distributions arise when a random variable is drawn from one of several different distributions (e.g. a professor randomly picks between an easy and a hard exam). Bayes' theorem lets you reverse the direction of conditioning: given an observed outcome, work backwards to figure out which scenario produced it.
Confidence interval (CI)
A range of values, computed from sample data, that is expected to contain the true population parameter a specified percentage of the time across repeated samples. A 99% CI does not mean there is a 99% probability the parameter is inside; it means the procedure captures the truth 99% of the time.
Mixture distribution
A probability distribution formed by combining two or more component distributions, each selected with a given probability. In simple terms, nature first flips a (possibly unfair) coin to pick which distribution to use, then draws a value from that distribution.
Law of total probability
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + ... for a partition {B₁, B₂, ...}. Think of it as breaking a probability into cases and adding them up.
Bayes' theorem
P(B|A) = P(A|B) × P(B) / P(A). This reverses the conditioning: given you observed A, what is the probability that scenario B produced it?
Conditional probability
The probability of an event given that another event has occurred. P(A|B) = P(A ∩ B) / P(B).
A CI for μ with known σ: X̄ ± z* × (σ/√n), where z* corresponds to the confidence level.
For 99% confidence, z* = 2.576. For 95%, z* = 1.96.
Problem 3(a): σ = 10, n = 100. The 99% CI extends ± 2.576 × (10/√100) = ± 2.576 above and below the class average.
Problem 3(b): You built a CI assuming σ = 10, but the exam is wacky with σ = 20. The actual standard error is 20/√100 = 2, not 1. Your CI of ± 2.576 is too narrow relative to the true variability.
The interval ± 2.576 corresponds to z = 2.576/2 = 1.288 standard errors under the true distribution. P(|Z| < 1.288) ≈ 0.8023. So you should have only about 80% confidence, not 99%.
A professor gives easy tests 25% of the time and hard tests 75% of the time (Problem 5). Or normal exams 95% and wacky 5% (Problem 3). Or hard 40% and insanely hard 60% (Problem 10).
The overall distribution of scores is not normal, even when each component is normal. It is a weighted combination.
Use the law of total probability: P(score > x) = P(score > x | easy) × P(easy) + P(score > x | hard) × P(hard).
Problem 5(a): Average score = 0.25 × 85 + 0.75 × 70 = 73.75.
Problem 5(b): P(score > 90) = 0.25 × P(Z > (90−85)/5) + 0.75 × P(Z > (90−70)/10) = 0.25 × P(Z > 1) + 0.75 × P(Z > 2) = 0.25(0.1587) + 0.75(0.0228) = 0.0397 + 0.0171 = 0.0568.
Problem 5(c): P(score < 80) = 0.25 × P(Z < (80−85)/5) + 0.75 × P(Z < (80−70)/10) = 0.25 × P(Z < −1) + 0.75 × P(Z < 1) = 0.25(0.1587) + 0.75(0.8413) = 0.0397 + 0.6310 = 0.6707.
Given an observed outcome, figure out which component distribution likely produced it.
Problem 5(d): A friend scored above 90. Was it easy or hard?
P(easy | score > 90) = P(score > 90 | easy) × P(easy) / P(score > 90) = (0.1587 × 0.25) / 0.0568 = 0.0397/0.0568 ≈ 0.699.
P(hard | score > 90) = 1 − 0.699 ≈ 0.301.
It is more likely the test was easy, which makes intuitive sense: scoring above 90 is much more plausible on an easy test.
You assume the exam is normal (σ = 10) and set your rejection region at α = 0.01 for H₀: μ = 50. But there is a 5% chance the exam is wacky (σ = 20).
Under the normal exam: P(reject | normal) = 0.01 (by design).
Under the wacky exam: the rejection boundary is 50 ± 2.576 × 1 = (47.424, 52.576). But the true SE is 2 when σ = 20. P(|X̄ − 50| > 2.576 | wacky) = P(|Z| > 2.576/2) = P(|Z| > 1.288) ≈ 0.1977.
Total P(Type I error) = 0.95 × 0.01 + 0.05 × 0.1977 = 0.0095 + 0.00989 ≈ 0.0194.
Average score = 0.40 × 70 + 0.60 × 50 = 58.
P(score > 70) = 0.40 × P(Z > 0) + 0.60 × P(Z > (70−50)/20) = 0.40(0.5) + 0.60 × P(Z > 1) = 0.20 + 0.60(0.1587) = 0.20 + 0.0952 = 0.2952.
P(insanely hard | score > 70) = P(score > 70 | insanely hard) × P(insanely hard) / P(score > 70) = (0.0952 × ... wait, let me be precise) = [P(Z > 1) × 0.60] / 0.2952 = 0.0952/0.2952 ≈ 0.3225.
\text{CI: } \bar{X} \pm z^* \times \frac{\sigma}{\sqrt{n}}P(A) = \sum_i P(A \mid B_i) \cdot P(B_i) \quad \text{(law of total probability)}P(B_j \mid A) = \frac{P(A \mid B_j) \cdot P(B_j)}{P(A)} \quad \text{(Bayes' theorem)}E[X] = \sum_i E[X \mid B_i] \cdot P(B_i) \quad \text{(expected value of a mixture)}Common z* values: 90% CI: 1.645. 95% CI: 1.96. 99% CI: 2.576.
Students often interpret a 95% CI as "there is a 95% probability the true mean is in this interval." The parameter is fixed; it is the interval that is random. The correct interpretation: 95% of intervals constructed this way will capture the true parameter.
Forgetting to use the law of total probability when computing probabilities from a mixture. You cannot just use one component; you must weight each component by its probability.
Applying Bayes' theorem but putting the wrong quantity in the denominator. The denominator P(A) must be the total probability, computed via the law of total probability.
Assuming a mixture of normals is itself normal. It is not. The mixture can be bimodal or have heavier tails than any individual component.
⚠️ Mixture distribution problems appear in at least three different questions on this practice exam. The pattern is always the same: state the components and their weights, then apply the law of total probability.
⚠️ The Bayes' theorem reversal (given an outcome, which component?) is a high-frequency exam question. Set up the fraction carefully; the denominator is the total probability you already computed.
⚠️ The "wrong σ" problem (Problem 3b,c) tests whether you understand that a CI's coverage probability depends on the true σ, not the assumed one. If your assumed σ is too small, your CI is too narrow and covers less than stated.
⚠️ Know the difference between the CI formula and the hypothesis test rejection region. They use the same z* values but answer different questions.
Fill in the blank: A 95% confidence interval uses z* = ______. (1.96)
True or false: A mixture of two normal distributions is always normal. (False)
Fill in the blank: The law of total probability says P(A) = P(A|B)P(B) + P(A|B^c) × ______. (P(B^c))
True or false: If you build a 99% CI using σ = 10 but the true σ is 20, the actual coverage is less than 99%. (True)
Fill in the blank: Bayes' theorem gives P(B|A) = P(A|B) × P(B) / ______. (P(A))
Q: A professor gives easy tests (25%, μ = 85, σ = 5) and hard tests (75%, μ = 70, σ = 10). What is the average score across all tests?
A: E[score] = 0.25(85) + 0.75(70) = 21.25 + 52.50 = 73.75.
Q: What is P(score > 90) for the professor above?
A: P(score > 90) = 0.25 × P(Z > (90 − 85)/5) + 0.75 × P(Z > (90 − 70)/10) = 0.25 × P(Z > 1) + 0.75 × P(Z > 2) = 0.25(0.1587) + 0.75(0.0228) ≈ 0.0568.
Q: A friend scored above 90 on the test above. What is the probability it was an easy test?
A: P(easy | > 90) = P(> 90 | easy) × P(easy) / P(> 90) = (0.1587 × 0.25) / 0.0568 = 0.0397/0.0568 ≈ 0.699. About 70% likely it was the easy test.
Q: Assuming a normal exam (σ = 10, n = 100), how wide is a 99% CI around the class average?
A: Half-width = 2.576 × 10/√100 = 2.576. The CI extends ± 2.576 points from the sample mean.
Q: The exam turns out to be wacky (σ = 20). What is the true coverage of the interval from the question above?
A: The interval is ± 2.576 around X̄. True SE = 20/10 = 2. The interval covers ± 2.576/2 = ± 1.288 true standard errors. P(|Z| < 1.288) ≈ 0.802. True coverage is about 80%.
Q: Hard tests (40%, μ = 70, σ = 10) and insanely hard tests (60%, μ = 50, σ = 20). What is the overall average score?
A: 0.40(70) + 0.60(50) = 28 + 30 = 58.
Q: A student scores above 70 on the exam above. What is P(insanely hard test)?
A: P(> 70) = 0.40 × P(Z > 0) + 0.60 × P(Z > 1) = 0.40(0.5) + 0.60(0.1587) = 0.2 + 0.0952 = 0.2952. P(insanely hard | > 70) = 0.0952/0.2952 ≈ 0.3225.
Confidence intervals are the mirror image of hypothesis tests: a value outside the CI would be rejected at the corresponding significance level. Mixture distributions appear in any setting where there is hidden heterogeneity (different exam types, different hospital quality). Bayes' theorem connects forward to Bayesian statistics and posterior probability calculations. The law of total probability is a building block for nearly every probability result in the course.
Related Terms / Search Tags
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