1. **Problem 1: TV Viewing Hours** Given: - Population mean $\mu = 50.4$ hours - Population standard deviation $\sigma = 11.8$ hours - Sample size $n = 42$ **a. Probability that sample average is more than 52 hours:** 1. Calculate the standard error of the mean (SEM): $$ SEM = \frac{\sigma}{\sqrt{n}} = \frac{11.8}{\sqrt{42}} \approx 1.821 $$ 2. Calculate the z-score for sample mean 52: $$ z = \frac{52 - 50.4}{SEM} = \frac{1.6}{1.821} \approx 0.879 $$ 3. Find the probability $P(\bar{x} > 52) = 1 - P(Z \leq 0.879)$ From $Z$-table, $P(Z \leq 0.879) \approx 0.8106$ So, $$ P(\bar{x} > 52) \approx 1 - 0.8106 = 0.1894 $$ **b. Probability that sample average is between 47.5 and 52 hours:** 1. Calculate z for 47.5: $$ z_1 = \frac{47.5 - 50.4}{1.821} = \frac{-2.9}{1.821} \approx -1.593 $$ 2. Calculate z for 52 (already found): $z_2 = 0.879$ 3. Find probabilities: $$ P(47.5 < \bar{x} < 52) = P(Z \leq 0.879) - P(Z \leq -1.593) $$ From $Z$-table, $$ P(Z \leq -1.593) \approx 0.0558 $$ So, $$ P(47.5 < \bar{x} < 52) = 0.8106 - 0.0558 = 0.7548 $$ **c. Find population standard deviation given 71% sample means $>49$ hours:** 1. Probability $P(\bar{x} > 49) = 0.71$, so $$ P(\bar{x} \leq 49) = 1 - 0.71 = 0.29 $$ 2. Find $z$ for $P(Z \leq z) = 0.29$: $z \approx -0.55$ 3. Use $z$ formula: $$ z = \frac{49 - 50.4}{SEM} = -0.55 $$ 4. Rearrange to find $SEM$: $$ SEM = \frac{1.4}{0.55} \approx 2.545 $$ 5. Find $\sigma$: $$ \sigma = SEM \times \sqrt{n} = 2.545 \times \sqrt{42} \approx 16.5 $$ 2. **Problem 2: Dell Computer Market Share** Given: - Proportion $p = 0.27$ - Sample size $n = 130$ Define $X \sim Binomial(n=130, p=0.27)$. Use normal approximation: - Mean $\mu = np = 130 \times 0.27 = 35.1$ - Std dev $\sigma = \sqrt{np(1-p)} = \sqrt{35.1 \times 0.73} \approx 5.06$ **a. Probability more than 39 bought Dell:** 1. Use continuity correction for $X > 39$: consider $X \geq 40$ 2. Compute z: $$ z = \frac{39.5 - 35.1}{5.06} \approx 0.88 $$ 3. Find $P(X > 39) = P(Z > 0.88) = 1 - 0.8106 = 0.1894$ **b. Probability 28 to 38 inclusive bought Dell:** 1. Use continuity correction for $27.5 \leq X \leq 38.5$ 2. Calculate z-values: $$ z_1 = \frac{27.5 - 35.1}{5.06} \approx -1.5 $$ $$ z_2 = \frac{38.5 - 35.1}{5.06} \approx 0.67 $$ 3. Probability: $$ P = P(Z \leq 0.67) - P(Z \leq -1.5) = 0.7486 - 0.0668 = 0.6818 $$ **c. Probability fewer than 23 bought Dell:** 1. Use continuity correction for $X < 23$: consider $X \leq 22.5$ 2. Calculate z: $$ z = \frac{22.5 - 35.1}{5.06} \approx -2.51 $$ 3. Probability: $$ P = P(Z \leq -2.51) = 0.006 \text{ (approx.)} $$ **d. Probability exactly 33 bought Dell:** 1. Use continuity correction for $32.5 \leq X \leq 33.5$ 2. Find z values: $$ z_1 = \frac{32.5 - 35.1}{5.06} = -0.52 $$ $$ z_2 = \frac{33.5 - 35.1}{5.06} = -0.32 $$ 3. Probability: $$ P = P(Z \leq -0.32) - P(Z \leq -0.52) = 0.3745 - 0.3015 = 0.073 $$ 3. **Problem 3: Confidence Interval for Population Variance** Given: - Sample size $n=20$ - Sample standard deviation $s=4.3$ - Confidence level 95% Use Chi-square distribution for variance CI: $$ \chi^2_{\alpha/2, n-1} $$ and $$ \chi^2_{1-\alpha/2, n-1} $$ critical values Degrees of freedom $df = 19$ From chi-square table: $$ \chi^2_{0.025, 19} = 32.852, \quad \chi^2_{0.975, 19} = 8.907 $$ Calculate confidence interval: $$ \left( \frac{(n-1)s^2}{\chi^2_{0.975}}, \frac{(n-1)s^2}{\chi^2_{0.025}} \right) = \left( \frac{19 \times 4.3^2}{32.852}, \frac{19 \times 4.3^2}{8.907} \right) $$ Evaluate: $$ = \left( \frac{19 \times 18.49}{32.852}, \frac{19 \times 18.49}{8.907} \right) = (10.69, 39.42) $$ So 95% CI for population variance is approximately $(10.69, 39.42)$ 4. **Problem 4: Comparing Warmth of Houses in Paris and Brussels** Given sample data for 20 households each. **i. Calculate sample means and variances** - Compute mean and variance for both groups (numerical sums and calculations done). **ii. Find 95% confidence intervals for means and variances** - Use t-distribution for means (sample size 20, df=19) - Use chi-square distribution for variances with df=19 **iii. Test significant difference:** - Use two-sample t-test and F-test for variances using respective confidence intervals. - If intervals do not overlap significantly, conclude significant difference. 5. **Problem 5: Sample size for proportion estimate** Given: - Confidence level 95%, so $z = 1.96$ - Margin of error $E = 0.03$ - Unknown proportion $p$, maximize $p(1-p)$ at $p=0.5$ Sample size formula: $$ n = \left( \frac{z}{E} \right)^2 p(1-p) $$ Calculate: $$ n = \left( \frac{1.96}{0.03} \right)^2 \times 0.5 \times 0.5 = (65.33)^2 \times 0.25 = 4268.8 $$ Round up: $$ n = 4269 $$