1. **Problem Statement:** We are given data about systolic and diastolic blood pressure for non-smoking females aged 18-35 and asked to use the empirical rule to fill in values and estimate counts. Then, we analyze a sample about high blood pressure in Irish people over 50. --- ### Part (a)(i): Fill in missing numbers on the horizontal axis for systolic pressure 2. The empirical rule states that for a normal distribution: - About 68% of data lies within $\pm 1\sigma$ of the mean - About 95% lies within $\pm 2\sigma$ - About 99.7% lies within $\pm 3\sigma$ 3. Given mean $\mu = 120$ and standard deviation $\sigma = 5$ for systolic pressure. 4. Calculate the values at $-2\sigma$, $-\sigma$, $\sigma$, and $2\sigma$: $$-2\sigma = 120 - 2 \times 5 = 120 - 10 = 110$$ $$-\sigma = 120 - 5 = 115$$ $$\sigma = 120 + 5 = 125$$ $$2\sigma = 120 + 10 = 130$$ 5. The missing numbers are: - At $-2\sigma$: 110 - At $-\sigma$: 115 (given) - At $\sigma$: 125 - At $2\sigma$: 130 --- ### Part (a)(iii): Estimate number of females with systolic pressure $\geq 125$ mmHg 6. $125$ mmHg corresponds to $\mu + \sigma$. 7. By the empirical rule, about 68% lie between $115$ and $125$ (within $\pm 1\sigma$), so 32% lie outside this range. 8. Since the normal distribution is symmetric, half of 32% = 16% lie above $125$ mmHg. 9. Number of females with systolic pressure $\geq 125$: $$1600 \times 0.16 = 256$$ --- ### Part (a)(iv): Estimate number of females with diastolic pressure $\leq 56$ mmHg 10. Given mean $\mu = 68$ and standard deviation $\sigma = 6$ for diastolic pressure. 11. Calculate how many standard deviations $56$ is below the mean: $$z = \frac{56 - 68}{6} = \frac{-12}{6} = -2$$ 12. By the empirical rule, about 2.5% of data lies below $\mu - 2\sigma$. 13. Number of females with diastolic pressure $\leq 56$: $$1600 \times 0.025 = 40$$ --- ### Part (a)(v): Estimate number of females with diastolic pressure between 62 and 80 mmHg 14. Calculate $z$-scores for 62 and 80: $$z_{62} = \frac{62 - 68}{6} = -1$$ $$z_{80} = \frac{80 - 68}{6} = 2$$ 15. By the empirical rule: - About 68% lie within $\pm 1\sigma$ (62 to 74) - About 95% lie within $\pm 2\sigma$ (56 to 80) 16. We want between 62 and 80, which is from $-1\sigma$ to $+2\sigma$. 17. The area between $-1\sigma$ and $+2\sigma$ is: $$P(-1 < Z < 2) = P(Z < 2) - P(Z < -1)$$ 18. Using empirical rule approximations: - $P(Z < 2) \approx 0.975$ - $P(Z < -1) \approx 0.16$ 19. So, $$0.975 - 0.16 = 0.815$$ 20. Number of females: $$1600 \times 0.815 = 1304$$ --- ### Part (b)(i): Margin of error for population proportion 21. Given sample size $n=625$, confidence level 95%, and population proportion $p=0.64$. 22. Margin of error formula: $$ME = z^* \times \sqrt{\frac{p(1-p)}{n}}$$ 23. For 95% confidence, $z^* = 1.96$. 24. Calculate: $$ME = 1.96 \times \sqrt{\frac{0.64 \times 0.36}{625}} = 1.96 \times \sqrt{\frac{0.2304}{625}} = 1.96 \times \sqrt{0.00036864} = 1.96 \times 0.0192 = 0.0376$$ 25. Margin of error is approximately 3.76%, rounded to 4%. --- ### Part (b)(ii): Percentage of sample with high blood pressure 26. Number with high blood pressure = 385 out of 625. 27. Percentage: $$\frac{385}{625} \times 100 = 61.6\%$$ --- ### Part (b)(iii): 95% confidence interval for true proportion 28. Sample proportion: $$\hat{p} = \frac{385}{625} = 0.616$$ 29. Margin of error from (b)(i) is 0.04. 30. Confidence interval: $$0.616 \pm 0.04 = (0.576, 0.656)$$ 31. As percentages: $$57.6\% \text{ to } 65.6\%$$ --- ### Part (b)(iv): Hypothesis test at 5% significance level 32. Null hypothesis $H_0$: $p = 0.64$ (the cardiologist's claim) 33. Alternative hypothesis $H_1$: $p \neq 0.64$ (claim is not true) 34. Test statistic: $$z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} = \frac{0.616 - 0.64}{\sqrt{\frac{0.64 \times 0.36}{625}}} = \frac{-0.024}{0.0192} = -1.25$$ 35. Critical z-values for 5% significance (two-tailed) are $\pm 1.96$. 36. Since $-1.96 < -1.25 < 1.96$, we fail to reject $H_0$. 37. **Conclusion:** There is not enough evidence to reject the cardiologist's claim at the 5% significance level. 38. **Reason:** The sample proportion is within the margin of error expected if the true proportion is 64%.