Answer: [$193,247, $206,753]

The correct answer is A) [$193,247, $206,753]. **Explanation:** To construct a 95% confidence interval for the population mean when the sample size is large (n=100), we use the formula: $$\bar{X} \pm z_{\alpha/2} \times \frac{s}{\sqrt{n}}$$ Where: - $\bar{X}$ = sample mean = $200,000$ - $s$ = sample standard deviation = $34,456$ - $n$ = sample size = 100 - $z_{\alpha/2}$ = critical value for 95% confidence = 1.96 **Calculation:** 1. Standard error = $\frac{s}{\sqrt{n}} = \frac{34,456}{\sqrt{100}} = \frac{34,456}{10} = 3,445.6$ 2. Margin of error = $1.96 \times 3,445.6 = 6,753.376$ 3. Lower bound = $200,000 - 6,753.376 = 193,246.624 \approx 193,247$ 4. Upper bound = $200,000 + 6,753.376 = 206,753.376 \approx 206,753$ Therefore, the 95% confidence interval is approximately [$193,247, $206,753]. **Why other options are incorrect:** - **B:** Incorrect because it uses an asymmetric interval and doesn't properly calculate the margin of error. - **C:** Incorrect because the values are too small and don't match the scale of the data. - **D:** Incorrect because it's too wide and doesn't use the proper statistical formula for confidence interval construction.

Author: Nikitesh Somanthe