1. Problem: Given the average TV viewing hours per household per week is 50.4 hours with a standard deviation of 11.8 hours and a sample size of 42 households. 1.a. Find the probability that the sample average is more than 52 hours. - Step 1: Calculate the standard error (SE) of the sample mean: $$SE = \frac{\sigma}{\sqrt{n}} = \frac{11.8}{\sqrt{42}} \approx 1.82$$ - Step 2: Calculate the z-score for 52 hours: $$z = \frac{52 - 50.4}{1.82} \approx 0.88$$ - Step 3: Find the probability that the sample mean is more than 52 hours: $$P(\bar{x} > 52) = 1 - P(Z \leq 0.88) \approx 1 - 0.8106 = 0.1894$$ 1.b. Find the probability that the sample average is between 47.5 and 52 hours. - Step 1: Calculate z-scores for 47.5 and 52: $$z_1 = \frac{47.5 - 50.4}{1.82} \approx -1.60$$ $$z_2 = 0.88 \text{ (from above)}$$ - Step 2: Find the probability between these z-scores: $$P(47.5 < \bar{x} < 52) = P(-1.60 < Z < 0.88) = P(Z < 0.88) - P(Z < -1.60) \approx 0.8106 - 0.0548 = 0.7558$$ 1.c. Given 71% of sample means are greater than 49 hours, population mean is 50.4, find population standard deviation. - Step 1: Let $\sigma$ be the population standard deviation. - Step 2: Find z-score corresponding to 71% greater than 49 means 29% less than 49, so $$P(\bar{x} < 49) = 0.29$$ - Step 3: Find z for 0.29: $$z = -0.55$$ - Step 4: Use formula: $$z = \frac{49 - 50.4}{\sigma / \sqrt{n}} = -0.55$$ $$-0.55 = \frac{-1.4}{\sigma / \sqrt{42}} \Rightarrow \sigma = \frac{1.4 \times \sqrt{42}}{0.55} \approx 15.5$$ 2. Problem: Dell controls 27% of PC market, sample size 130. 2.a. Probability more than 39 bought Dell. - Step 1: Calculate mean and standard deviation of binomial: $$\mu = np = 130 \times 0.27 = 35.1$$ $$\sigma = \sqrt{np(1-p)} = \sqrt{130 \times 0.27 \times 0.73} \approx 4.95$$ - Step 2: Use normal approximation with continuity correction: $$P(X > 39) = P(X \geq 40) \approx P\left(Z > \frac{39.5 - 35.1}{4.95}\right) = P(Z > 0.89) = 1 - 0.8133 = 0.1867$$ 2.b. Probability between 28 and 38 inclusive. - Step 1: Calculate z-scores with continuity correction: $$z_1 = \frac{27.5 - 35.1}{4.95} = -1.53$$ $$z_2 = \frac{38.5 - 35.1}{4.95} = 0.69$$ - Step 2: Probability: $$P(28 \leq X \leq 38) = P(-1.53 < Z < 0.69) = 0.7549 - 0.0630 = 0.6919$$ 2.c. Probability fewer than 23. - Step 1: Continuity correction: $$P(X < 23) = P(X \leq 22) \approx P\left(Z < \frac{22.5 - 35.1}{4.95}\right) = P(Z < -2.56) = 0.0052$$ 2.d. Probability exactly 33. - Step 1: Use normal approximation with continuity correction: $$P(X=33) \approx P(32.5 < X < 33.5)$$ - Step 2: Calculate z-scores: $$z_1 = \frac{32.5 - 35.1}{4.95} = -0.51$$ $$z_2 = \frac{33.5 - 35.1}{4.95} = -0.32$$ - Step 3: Probability: $$P(32.5 < X < 33.5) = P(-0.51 < Z < -0.32) = 0.3745 - 0.3050 = 0.0695$$ 3. Problem: 95% confidence interval for population variance given sample size 20, sample standard deviation 4.3. - Step 1: Degrees of freedom: $$df = n - 1 = 19$$ - Step 2: Chi-square critical values for 95% confidence: $$\chi^2_{0.025,19} = 32.852$$ $$\chi^2_{0.975,19} = 8.907$$ - Step 3: Calculate confidence interval for variance: $$\left( \frac{(n-1)s^2}{\chi^2_{0.025}}, \frac{(n-1)s^2}{\chi^2_{0.975}} \right) = \left( \frac{19 \times 4.3^2}{32.852}, \frac{19 \times 4.3^2}{8.907} \right) = (10.7, 39.4)$$ 4. Problem: Comparison of thermostat temperatures in Paris and Brussels households (data not provided, so no calculations). Final answers: 1.a. Probability sample mean > 52 hours: 0.1894 1.b. Probability sample mean between 47.5 and 52 hours: 0.7558 1.c. Population standard deviation estimate: 15.5 2.a. Probability more than 39 bought Dell: 0.1867 2.b. Probability between 28 and 38 bought Dell: 0.6919 2.c. Probability fewer than 23 bought Dell: 0.0052 2.d. Probability exactly 33 bought Dell: 0.0695 3. 95% confidence interval for population variance: (10.7, 39.4)