1. The problem involves calculating the z-score for a list of values relative to a population and a sample. 2. First, let's define the dataset: $108, 200, 310, 150, 180, 119, 160, 180, 201, 190, 280, 202$. 3. We will calculate the mean ($\mu$ for population, $\bar{x}$ for sample) and the standard deviation ($\sigma$ for population, $s$ for sample). 4. Calculate the population mean: $$\mu = \frac{108 + 200 + 310 + 150 + 180 + 119 + 160 + 180 + 201 + 190 + 280 + 202}{12} = \frac{2290}{12} = 190.83$$ 5. Calculate the population variance: $$\sigma^2 = \frac{\sum (x_i - \mu)^2}{N}$$ Compute each term: $$(108-190.83)^2 = 6890.49, (200-190.83)^2 = 83.61, (310-190.83)^2 = 14001.49,$$ $$(150-190.83)^2 = 1670.49, (180-190.83)^2 = 117.36, (119-190.83)^2 = 5218.19,$$ $$(160-190.83)^2 = 951.19, (180-190.83)^2 = 117.36, (201-190.83)^2 = 104.89,$$ $$(190-190.83)^2 = 0.69, (280-190.83)^2 = 7913.89, (202-190.83)^2 = 124.69$$ Sum of squared differences = $42984.31$ $$\sigma^2 = \frac{42984.31}{12} = 3582.03$$ 6. Calculate the population standard deviation: $$\sigma = \sqrt{3582.03} = 59.85$$ 7. For the sample calculations, mean $\bar{x} = 190.83$ (same as population mean as data is the full set) and sample variance: $$s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1} = \frac{42984.31}{11} = 3907.66$$ 8. Sample standard deviation: $$s = \sqrt{3907.66} = 62.53$$ 9. Z-score formula for a value $x$ given population mean and standard deviation: $$z = \frac{x - \mu}{\sigma}$$ 10. Let's calculate the z-score for $x=200$ (for example): $$z = \frac{200 - 190.83}{59.85} = \frac{9.17}{59.85} = 0.15$$ Final Answer: Population mean $\mu = 190.83$ Population standard deviation $\sigma = 59.85$ Sample standard deviation $s = 62.53$ Example z-score for $x=200$ is approximately $0.15$.