1. Problem 5: Find the probability that exactly 4 out of 6 telephones are busy, given the probability of a telephone being busy is $p=\frac{1}{10}=0.1$. 2. Use the binomial probability formula: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=6$, $k=4$, $p=0.1$. 3. Calculate the binomial coefficient: $$\binom{6}{4} = \frac{6!}{4!2!} = 15$$ 4. Calculate the probability: $$P(X=4) = 15 \times (0.1)^4 \times (0.9)^2 = 15 \times 0.0001 \times 0.81 = 0.001215$$ --- 5. Problem 6: Test if the average time spent on the website differs from 5 minutes using a one-sample t-test. 6. Given sample mean $\bar{x}=4.5$, population mean $\mu=5$, sample standard deviation $s=1.2$, and sample size $n=50$. 7. Calculate the t-statistic: $$t = \frac{\bar{x} - \mu}{s/\sqrt{n}} = \frac{4.5 - 5}{1.2/\sqrt{50}} = \frac{-0.5}{1.2/7.071} = \frac{-0.5}{0.1697} \approx -2.95$$ 8. Compare $t$ with critical t-value for $df=49$ at $\alpha=0.05$ (two-tailed). Since $|t|=2.95$ is greater than critical value $\approx 2.01$, reject null hypothesis. --- 9. Problem 13: Find combined standard deviation for two groups. 10. Given: Group 1: $n_1=50$, $\bar{x}_1=630$, $s_1=90$ Group 2: $n_2=40$, $\bar{x}_2=540$, $s_2=60$ 11. Calculate combined mean: $$\bar{x}_c = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2}{n_1 + n_2} = \frac{50 \times 630 + 40 \times 540}{90} = \frac{31500 + 21600}{90} = 590$$ 12. Calculate combined variance: $$s_c^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + n_1(\bar{x}_1 - \bar{x}_c)^2 + n_2(\bar{x}_2 - \bar{x}_c)^2}{n_1 + n_2 - 1}$$ 13. Compute terms: $$(50-1) \times 90^2 = 49 \times 8100 = 396900$$ $$(40-1) \times 60^2 = 39 \times 3600 = 140400$$ $$50 \times (630 - 590)^2 = 50 \times 1600 = 80000$$ $$40 \times (540 - 590)^2 = 40 \times 2500 = 100000$$ 14. Sum numerator: $$396900 + 140400 + 80000 + 100000 = 717300$$ 15. Divide by $n_1 + n_2 - 1 = 89$: $$s_c^2 = \frac{717300}{89} \approx 8056.18$$ 16. Combined standard deviation: $$s_c = \sqrt{8056.18} \approx 89.75$$ --- 17. Problem 14: Calculate ANOVA coefficient (F-ratio) for given data. 18. Data: Hibiscus: $n=5$, mean=12, $s=2$ Marigold: $n=5$, mean=16, $s=1$ Rose: $n=5$, mean=20, $s=4$ 19. Calculate overall mean: $$\bar{x} = \frac{5 \times 12 + 5 \times 16 + 5 \times 20}{15} = \frac{60 + 80 + 100}{15} = 16$$ 20. Calculate SST (total sum of squares): $$SST = \sum n_i (\bar{x}_i - \bar{x})^2 = 5(12-16)^2 + 5(16-16)^2 + 5(20-16)^2 = 5 \times 16 + 0 + 5 \times 16 = 80 + 0 + 80 = 160$$ 21. Calculate SSW (within groups sum of squares): $$SSW = \sum (n_i - 1) s_i^2 = 4 \times 2^2 + 4 \times 1^2 + 4 \times 4^2 = 4 \times 4 + 4 \times 1 + 4 \times 16 = 16 + 4 + 64 = 84$$ 22. Calculate degrees of freedom: Between groups: $df_b = 3 - 1 = 2$ Within groups: $df_w = 15 - 3 = 12$ 23. Calculate mean squares: $$MSB = \frac{SST}{df_b} = \frac{160}{2} = 80$$ $$MSW = \frac{SSW}{df_w} = \frac{84}{12} = 7$$ 24. Calculate ANOVA F-ratio: $$F = \frac{MSB}{MSW} = \frac{80}{7} \approx 11.43$$ --- 25. Problem 15: Test if observed frequencies fit expected probabilities using chi-square test. 26. Given: Observed: Red=32, Green=45, Blue=23, Total=100 Expected probabilities: Red=0.40, Green=0.35, Blue=0.25 27. Calculate expected frequencies: Red: $100 \times 0.40 = 40$ Green: $100 \times 0.35 = 35$ Blue: $100 \times 0.25 = 25$ 28. Calculate chi-square statistic: $$\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(32-40)^2}{40} + \frac{(45-35)^2}{35} + \frac{(23-25)^2}{25} = \frac{64}{40} + \frac{100}{35} + \frac{4}{25} = 1.6 + 2.857 + 0.16 = 4.617$$ 29. Degrees of freedom: $df = 3 - 1 = 2$ 30. Critical value at $\alpha=0.05$ and $df=2$ is 5.991. 31. Since $4.617 < 5.991$, we fail to reject the null hypothesis; the observed frequencies fit the expected probabilities. Final answers: - Problem 5: Probability = 0.001215 - Problem 6: t-statistic = -2.95, reject null hypothesis - Problem 13: Combined standard deviation $\approx 89.75$ - Problem 14: ANOVA F-ratio $\approx 11.43$ - Problem 15: Chi-square = 4.617, fail to reject null hypothesis