1. **Problem 5.a:** Test if the survival rate is greater than 85% at 5% significance level. Given: Sample size $n=20$, survivors $x=18$, hypothesized proportion $p_0=0.85$. 2. Calculate sample proportion: $\hat{p} = \frac{18}{20} = 0.9$. 3. Compute test statistic for one-proportion z-test: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.9 - 0.85}{\sqrt{\frac{0.85 \times 0.15}{20}}} = \frac{0.05}{\sqrt{0.006375}} = \frac{0.05}{0.0798} \approx 0.626$$ 4. Critical value for one-tailed test at 5% level: $z_{0.05} = 1.645$. 5. Since $0.626 < 1.645$, we **fail to reject** the null hypothesis. There is insufficient evidence to say survival rate is more than 85%. 6. **Problem 5.b:** Test if difference between means of two independent samples is significant. Sample 1: $n_1=8$, values: 9,11,13,11,15,9,12,14 Sample 2: $n_2=7$, values: 10,12,10,14,9,8,10 7. Calculate sample means: $$\bar{x}_1 = \frac{9+11+13+11+15+9+12+14}{8} = \frac{94}{8} = 11.75$$ $$\bar{x}_2 = \frac{10+12+10+14+9+8+10}{7} = \frac{73}{7} \approx 10.43$$ 8. Calculate sample variances: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1} = \frac{(9-11.75)^2 + ... + (14-11.75)^2}{7} = 5.36$$ $$s_2^2 = \frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1} = \frac{(10-10.43)^2 + ... + (10-10.43)^2}{6} = 4.10$$ 9. Calculate pooled variance: $$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2} = \frac{7 \times 5.36 + 6 \times 4.10}{13} = \frac{37.52 + 24.6}{13} = 4.77$$ 10. Calculate t-statistic: $$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{s_p^2(\frac{1}{n_1} + \frac{1}{n_2})}} = \frac{11.75 - 10.43}{\sqrt{4.77(\frac{1}{8} + \frac{1}{7})}} = \frac{1.32}{\sqrt{4.77 \times 0.2679}} = \frac{1.32}{1.13} = 1.17$$ 11. Degrees of freedom $df = n_1 + n_2 - 2 = 13$. 12. Critical t-value for two-tailed test at 5% level and 13 df is approximately 2.16. 13. Since $1.17 < 2.16$, we **fail to reject** the null hypothesis. The difference between means is not significant. 14. **Problem 6:** Calculate regression equation of $Y$ on $X$ and correlation coefficient. Data: $X = [48,33,40,9,16,16,65,24,16,57]$ $Y = [13,13,24,6,15,4,20,9,16,19]$ 15. Calculate means: $$\bar{X} = \frac{48+33+40+9+16+16+65+24+16+57}{10} = 32.4$$ $$\bar{Y} = \frac{13+13+24+6+15+4+20+9+16+19}{10} = 13.9$$ 16. Calculate sums: $$S_{XY} = \sum (X_i - \bar{X})(Y_i - \bar{Y}) = 1026.4$$ $$S_{XX} = \sum (X_i - \bar{X})^2 = 4624.4$$ 17. Calculate regression slope and intercept: $$b = \frac{S_{XY}}{S_{XX}} = \frac{1026.4}{4624.4} = 0.222$$ $$a = \bar{Y} - b\bar{X} = 13.9 - 0.222 \times 32.4 = 6.69$$ 18. Regression equation: $$Y = 6.69 + 0.222X$$ 19. Calculate correlation coefficient: $$r = \frac{S_{XY}}{\sqrt{S_{XX} \times S_{YY}}}$$ Where $$S_{YY} = \sum (Y_i - \bar{Y})^2 = 454.9$$ $$r = \frac{1026.4}{\sqrt{4624.4 \times 454.9}} = \frac{1026.4}{\sqrt{2103627}} = \frac{1026.4}{1450.7} = 0.707$$ 20. **Problem 7:** M/M/1 queue with exponential inter-arrival and service times. Given: Mean inter-arrival time $= \frac{1}{2}$ min $\Rightarrow \lambda = 2$ cars/min Mean service time $= \frac{1}{5}$ min $\Rightarrow \mu = 5$ cars/min 21. Calculate traffic intensity: $$\rho = \frac{\lambda}{\mu} = \frac{2}{5} = 0.4$$ 22. Average number of cars waiting (queue length): $$L_q = \frac{\rho^2}{1-\rho} = \frac{0.4^2}{1-0.4} = \frac{0.16}{0.6} = 0.267$$ 23. Average number of cars in system: $$L = \frac{\rho}{1-\rho} = \frac{0.4}{0.6} = 0.667$$ 24. Proportion of time pump is idle: $$P_0 = 1 - \rho = 1 - 0.4 = 0.6$$ 25. **Problem 8.a:** Define Markov chain. A Markov chain is a stochastic process where the probability of moving to the next state depends only on the current state and not on the sequence of events that preceded it. 26. **Problem 8.b:** Ball passing among boys A, B, C. - A always throws to B. - B always throws to C. - C throws to B or A with equal probability 0.5. 27. States: A, B, C. Transition matrix $P$: $$P = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0.5 & 0.5 & 0 \end{bmatrix}$$ 28. This process is Markovian because the next state depends only on the current state. 29. Classification: - All states communicate. - The chain is irreducible. - Since the chain can return to each state, all states are recurrent.