Answer: Poisson (lowest) and Binomial (highest)

## Explanation Let's calculate the variances for each distribution: 1. **Standard Normal**: Variance = 1 2. **Student's t with 10 df**: Variance = df/(df-2) = 10/(10-2) = 10/8 = 1.25 3. **Bernoulli with p = 0.04**: Variance = p(1-p) = 0.04 × 0.96 = 0.0384 4. **Poisson with λ = 5**: Variance = λ = 5 5. **Binomial with n = 50, p = 0.02**: Variance = np(1-p) = 50 × 0.02 × 0.98 = 0.98 Now let's order them from lowest to highest variance: - Bernoulli: 0.0384 (lowest) - Binomial: 0.98 - Standard Normal: 1 - Student's t: 1.25 - Poisson: 5 (highest) But the question asks for "the lowest value and highest value as its variance among the set", and looking at the options: A: Standard normal (lowest) and Bernoulli (highest) - **Incorrect** B: Binomial (lowest) and Student's t (highest) - **Incorrect** C: Bernoulli (lowest) and Poisson (highest) - **Correct based on calculations** D: Poisson (lowest) and Binomial (highest) - **Incorrect** Wait, my calculations show Bernoulli has the lowest variance (0.0384) and Poisson has the highest variance (5), which would be option C. But the answer key shows D as correct. Let me double-check the binomial variance: Binomial: n = 50, p = 0.02 Variance = np(1-p) = 50 × 0.02 × 0.98 = 0.98 Poisson: λ = 5, Variance = 5 So indeed, Bernoulli has the lowest variance (0.0384) and Poisson has the highest variance (5). This corresponds to option C, not D. There might be an error in the source material or my interpretation.

Author: LeetQuiz Editorial Team