1. Problem: Carla rolls a fair six-sided die. She wins 12 if she rolls a 6, but loses 3 for any other number. Find the expected value. Formula: Expected value $E(X) = \sum [x \cdot P(x)]$ Calculation: - Probability of rolling 6 is $\frac{1}{6}$, winning 12. - Probability of rolling other numbers is $\frac{5}{6}$, losing 3. $$E(X) = 12 \times \frac{1}{6} + (-3) \times \frac{5}{6} = 2 - 2.5 = -0.5$$ Answer: B. -0.50 2. Problem: Find the mean (expected value) of $X$ with distribution $P(X)$ given. Formula: $E(X) = \sum x P(x)$ Calculation: $$E(X) = 1 \times 0.2 + 2 \times 0.5 + 3 \times 0.3 = 0.2 + 1 + 0.9 = 2.1$$ Answer: B. 2.1 3. Problem: Charity sells 100 tickets at 20 each; one ticket wins 1000. Find expected net gain for one ticket. Calculation: - Probability of winning: $\frac{1}{100}$, gain = 1000 - 20 = 980 - Probability of losing: $\frac{99}{100}$, loss = -20 $$E = 980 \times \frac{1}{100} + (-20) \times \frac{99}{100} = 9.8 - 19.8 = -10$$ Answer: A. -10 4. Problem: Find variance of $X$ with values 2,4,6 and probabilities 0.3,0.4,0.3. Steps: - Mean: $\mu = 2(0.3) + 4(0.4) + 6(0.3) = 0.6 + 1.6 + 1.8 = 4.0$ - Variance: $\sigma^2 = \sum P(x)(x - \mu)^2$ Calculation: $$= 0.3(2-4)^2 + 0.4(4-4)^2 + 0.3(6-4)^2 = 0.3(4) + 0 + 0.3(4) = 1.2 + 0 + 1.2 = 2.4$$ Answer: B. 2.40 5. Problem: Insurance costs 500, pays 10000 if accident occurs with probability 0.02. Find expected value. Calculation: $$E = 0.02(10000 - 500) + 0.98(-500) = 0.02(9500) - 490 = 190 - 490 = -300$$ Answer: A. -300 6. Problem: Spinner points 0 (0.2), 5 (0.5), 10 (0.3). Find expected value. Calculation: $$E = 0(0.2) + 5(0.5) + 10(0.3) = 0 + 2.5 + 3 = 5.5$$ Answer: B. 5.5 7. Problem: $X$ takes 0 (0.5), 1 (0.3), 2 (0.2). Find variance. Steps: - Mean: $\mu = 0(0.5) + 1(0.3) + 2(0.2) = 0 + 0.3 + 0.4 = 0.7$ - Variance: $$= 0.5(0-0.7)^2 + 0.3(1-0.7)^2 + 0.2(2-0.7)^2 = 0.5(0.49) + 0.3(0.09) + 0.2(1.69) = 0.245 + 0.027 + 0.338 = 0.61$$ Answer: B. 0.61 8. Problem: Raffle ticket costs 5; prizes 50 (0.05), 10 (0.15), else 0. Find expected net gain. Calculation: $$E = 0.05(50-5) + 0.15(10-5) + 0.8(0-5) = 0.05(45) + 0.15(5) - 4 = 2.25 + 0.75 - 4 = -1.0$$ Answer: B. -1.00 9. Problem: Mean of sampling distribution of sample mean for masks with mean 40, sample size 64. Answer: B. 40 (mean of sampling distribution equals population mean) 10. Problem: Formula for standard error of mean. Answer: A. Standard deviation ÷ square root of sample size 11. Problem: Find standard error with population std dev 5, sample size 64. Calculation: $$SE = \frac{5}{\sqrt{64}} = \frac{5}{8} = 0.625$$ Answer: C. 0.625 12. Problem: True statement about sampling distribution of mean. Answer: A. It always has a smaller spread than the population. 13. Problem: Effect on standard error if sample size increases. Answer: B. It decreases 14. Problem: Mean (expected value) of sampling distribution for residents' weight. Answer: C. 150 15. Problem: Standard deviation of sample mean with population std dev 27, sample size 16. Calculation: $$SE = \frac{27}{\sqrt{16}} = \frac{27}{4} = 6.75$$ Answer: C. 6.75 16. Problem: Average weight per person to reach elevator limit 2500 with 16 persons. Calculation: $$\text{Average} = \frac{2500}{16} = 156.25$$ Answer: A. 156.25 17. Problem: Relationship between margin of error and sample size. Answer: C. A lower margin of error requires a larger sample size. 18. Problem: Effect of population std dev on required sample size. Answer: B. Lower σ reduces the required sample size. 19. Problem: Suggestion for researcher wanting minimum margin of error. Answer: C. Increase the sample size. 20. Problem: Sample size for 98% confidence, σ=5, margin error=2. Formula: $$n = \left(\frac{Z \sigma}{E}\right)^2$$ For 98%, $Z \approx 2.33$ Calculation: $$n = \left(\frac{2.33 \times 5}{2}\right)^2 = (5.825)^2 = 33.93 \approx 34$$ Answer: B. 34 21. Problem: Sample size for 95% confidence, σ=5, margin error=2. For 95%, $Z \approx 1.96$ Calculation: $$n = \left(\frac{1.96 \times 5}{2}\right)^2 = (4.9)^2 = 24.01 \approx 25$$ Closest answer: D. 34 (given options) 22. Problem: Sample size for 99% confidence, σ=8, margin error=3. For 99%, $Z \approx 2.576$ Calculation: $$n = \left(\frac{2.576 \times 8}{3}\right)^2 = (6.87)^2 = 47.2 \approx 47$$ Answer: B. 47 23. Problem: Effect on sample size if margin of error is halved. Answer: C. Quadruples 24. Problem: Sample size for population proportion with 95% confidence, margin error 0.05, p=0.5. Formula: $$n = \frac{Z^2 p(1-p)}{E^2}$$ For 95%, $Z=1.96$ Calculation: $$n = \frac{(1.96)^2 \times 0.5 \times 0.5}{0.05^2} = \frac{3.8416 \times 0.25}{0.0025} = \frac{0.9604}{0.0025} = 384.16$$ Answer: A. 384 25. Problem: Purpose of null hypothesis. Answer: B. To assume no effect or no difference 26. Problem: Action if p-value < significance level. Answer: B. Reject the null hypothesis 27. Problem: Location of critical region in two-tailed test. Answer: C. In both tails 28. Problem: What does Type I error represent? Answer: A. Rejecting a true null hypothesis 29. Problem: Null and alternative hypotheses if mean is greater than 75. Answer: B. $H_0: \mu \leq 75$, $H_1: \mu > 75$ 30. Problem: Last step in hypothesis testing. Answer: A. Draw a conclusion 31. Problem: p-value 0.03 with $\alpha=0.05$, what to do? Answer: A. Reject the null hypothesis 32. Problem: What separates critical from non-critical region? Answer: C. Critical value 33. Problem: What does correlation coefficient $r$ indicate? Answer: B. The relationship strength and direction between two variables 34. Problem: Who introduced Pearson's correlation coefficient? Answer: C. Karl Pearson 35. Problem: Range of correlation coefficient $r$. Answer: C. -1 to 1 36. Problem: Interpretation of $r=0.75$. Answer: A. Strong positive correlation 37. Problem: Value representing strong negative correlation. Answer: B. -0.88 38. Problem: Conclusion if $r=0$. Answer: C. No correlation 39. Problem: Scatter plot points clustering from lower-left to upper-right indicate? Answer: C. Positive correlation 40. Problem: If $r \approx 0.96$ between hours studied and exam scores, correct statement? Answer: C. Scores tend to increase as hours studied increase