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.