1. Problem 1: Find the probability that the sample mean score is less than 70 given $X \sim N(65, 23^2)$ and sample size $n=45$. 2. Use the Central Limit Theorem: The sampling distribution of the sample mean $\bar{X}$ is normal with mean $\mu=65$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{23}{\sqrt{45}}$. 3. Calculate $\sigma_{\bar{X}}$: $$\sigma_{\bar{X}}=\frac{23}{\sqrt{45}}=\frac{23}{6.708}=3.43$$ 4. Find the $z$-score for $\bar{X}=70$: $$z=\frac{70-65}{3.43}=\frac{5}{3.43}=1.46$$ 5. Use standard normal tables or calculator to find $P(Z<1.46)=0.9279$. 6. So, the probability that the sample mean is less than 70 is approximately 0.9279. --- 7. Problem 2: Construct a 99% confidence interval for the mean waiting time given $n=40$, sample mean $\bar{x}=2$, and population standard deviation $\sigma=1.2$. 8. The formula for confidence interval when $\sigma$ is known: $$\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ 9. For 99% confidence, $z_{\alpha/2}=2.576$. 10. Calculate margin of error: $$ME=2.576 \times \frac{1.2}{\sqrt{40}}=2.576 \times \frac{1.2}{6.324}=2.576 \times 0.1897=0.4889$$ 11. Confidence interval: $$2 \pm 0.4889 = (1.5111, 2.4889)$$ --- 12. Problem 3: Construct a 95% confidence interval for the proportion of students suffering from stress given $n=150$, $x=81$. 13. Sample proportion: $$\hat{p}=\frac{81}{150}=0.54$$ 14. For 95% confidence, $z_{\alpha/2}=1.96$. 15. Standard error: $$SE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.54 \times 0.46}{150}}=\sqrt{0.001656}=0.0407$$ 16. Margin of error: $$ME=1.96 \times 0.0407=0.0798$$ 17. Confidence interval: $$(0.54 - 0.0798, 0.54 + 0.0798) = (0.4602, 0.6198)$$ 18. Interpretation: We are 95% confident that the true proportion of final year students suffering from stress is between 46.02% and 61.98%. --- 19. Problem 4: Test if the mean age of teen vaping has become younger at 5% significance level. 20. Given: $n=50$, sample mean $\bar{x}=13$, sample standard deviation $s=4$, population mean $\mu_0=15$. 21. Hypotheses: $$H_0: \mu = 15$$ $$H_a: \mu < 15$$ 22. Use t-test since population standard deviation unknown: $$t=\frac{\bar{x} - \mu_0}{s/\sqrt{n}}=\frac{13-15}{4/\sqrt{50}}=\frac{-2}{4/7.071}=\frac{-2}{0.566}= -3.53$$ 23. Degrees of freedom: $df=49$. 24. Critical t-value for one-tailed test at 5% significance and 49 df is approximately $-1.68$. 25. Since $t=-3.53 < -1.68$, reject $H_0$. 26. Conclusion: There is sufficient evidence at 5% significance level to conclude that the mean age of teen vaping has become younger.