1. **Problem statement:** Calculate a 96% confidence interval for the population mean height of a species of deer given a sample mean of 1.42 m, sample size 150, and population standard deviation 0.35 m. 2. **Formula and explanation:** The confidence interval for the population mean when the population standard deviation is known is given by: $$\bar{x} \pm z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}}$$ where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the z-score for the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. 3. **Find the z-score for 96% confidence:** The confidence level is 96%, so $\alpha = 1 - 0.96 = 0.04$. The critical value $z_{\alpha/2}$ corresponds to $z_{0.02}$ (since $\alpha/2=0.02$). From standard normal tables, $z_{0.02} \approx 2.054$. 4. **Calculate the margin of error:** $$\text{Margin of error} = z_{\alpha/2} \times \frac{\sigma}{\sqrt{n}} = 2.054 \times \frac{0.35}{\sqrt{150}}$$ Calculate $\sqrt{150} \approx 12.247$. So, $$\text{Margin of error} = 2.054 \times \frac{0.35}{12.247} = 2.054 \times 0.02857 \approx 0.0587$$ 5. **Calculate the confidence interval:** $$1.42 \pm 0.0587$$ So the interval is: $$\left(1.42 - 0.0587, 1.42 + 0.0587\right) = (1.3613, 1.4787)$$ 6. **Interpretation:** We are 96% confident that the true population mean height lies between 1.3613 m and 1.4787 m. **Final answer:** The 96% confidence interval for the population mean height is $$\boxed{(1.361, 1.479)}$$ (rounded to three decimal places). **Slug:** confidence interval **Subject:** statistics **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 7