To find the 95% confidence interval for the population mean when the population standard deviation is known, we use the Z-distribution. The formula for the confidence interval is: $CI = \bar{x} \pm Z \times \frac{\sigma}{\sqrt{n}}$ Where:

• Sample mean ($\bar{x}$) = 72,000
• Population standard deviation ($\sigma$) = 7,800
• Sample size ($n$) = 120
• Z-value for 95% confidence = 1.96

Standard error $= \frac{7,800}{\sqrt{120}} = \frac{7,800}{10.95445} \approx 712.04$ Margin of error $= 1.96 \times 712.04 \approx 1,395.60$

Lower bound $= 72,000 - 1,395.60 = 70,604.40$ Upper bound $= 72,000 + 1,395.60 = 73,395.60$

The 95% confidence interval is (USD 70,604.40, USD 73,395.60).