1. **State the problem:** We have a sample of 20 HDL cholesterol levels: 35, 49, 52, 54, 65, 51, 51, 47, 86, 36, 46, 33, 39, 45, 39, 63, 95, 35, 30, 48. We want to find better estimates for the population mean and standard deviation based on this sample. 2. **Formula for sample mean:** The sample mean $\bar{x}$ is given by $$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$$ where $n=20$ is the sample size and $x_i$ are the observations. 3. **Calculate the sample mean:** $$\bar{x} = \frac{35 + 49 + 52 + 54 + 65 + 51 + 51 + 47 + 86 + 36 + 46 + 33 + 39 + 45 + 39 + 63 + 95 + 35 + 30 + 48}{20}$$ Calculate the sum: $$\sum x_i = 35 + 49 + 52 + 54 + 65 + 51 + 51 + 47 + 86 + 36 + 46 + 33 + 39 + 45 + 39 + 63 + 95 + 35 + 30 + 48 = 1004$$ So, $$\bar{x} = \frac{1004}{20} = 50.2$$ 4. **Formula for sample standard deviation:** The sample standard deviation $s$ is $$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}$$ 5. **Calculate each squared deviation:** Calculate $(x_i - 50.2)^2$ for each observation and sum them: - $(35 - 50.2)^2 = 231.04$ - $(49 - 50.2)^2 = 1.44$ - $(52 - 50.2)^2 = 3.24$ - $(54 - 50.2)^2 = 14.44$ - $(65 - 50.2)^2 = 219.04$ - $(51 - 50.2)^2 = 0.64$ - $(51 - 50.2)^2 = 0.64$ - $(47 - 50.2)^2 = 10.24$ - $(86 - 50.2)^2 = 1297.44$ - $(36 - 50.2)^2 = 201.64$ - $(46 - 50.2)^2 = 17.64$ - $(33 - 50.2)^2 = 295.84$ - $(39 - 50.2)^2 = 125.44$ - $(45 - 50.2)^2 = 27.04$ - $(39 - 50.2)^2 = 125.44$ - $(63 - 50.2)^2 = 163.84$ - $(95 - 50.2)^2 = 2016.04$ - $(35 - 50.2)^2 = 231.04$ - $(30 - 50.2)^2 = 408.04$ - $(48 - 50.2)^2 = 4.84$ Sum of squared deviations: $$\sum (x_i - \bar{x})^2 = 1001.6 + 1297.44 + 2016.04 + ... = 5531.6$$ 6. **Calculate sample standard deviation:** $$s = \sqrt{\frac{5531.6}{20 - 1}} = \sqrt{\frac{5531.6}{19}} = \sqrt{291.14} \approx 17.06$$ 7. **Interpretation:** The sample mean $50.2$ and sample standard deviation $17.06$ are better estimates for the population mean and standard deviation because they use all data points and correct for sample size with $n-1$ in the denominator for standard deviation. **Final answers:** - Estimated population mean $\approx 50.2$ - Estimated population standard deviation $\approx 17.06$