1. **State the problem:** We have a confidence interval for the mean salary of business administration graduates: $7800$ to $9400$. The sample size is $n=40$, and the population standard deviation is $\sigma=1800$. We need to find the confidence level of this interval. 2. **Recall the formula for a confidence interval for the mean when population standard deviation is known:** $$\bar{x} \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$$ where $\bar{x}$ is the sample mean, $z_{\alpha/2}$ is the critical z-value for the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. 3. **Find the sample mean $\bar{x}$:** $$\bar{x} = \frac{7800 + 9400}{2} = \frac{17200}{2} = 8600$$ 4. **Calculate the margin of error (ME):** $$ME = 9400 - 8600 = 800$$ 5. **Use the margin of error formula to find $z_{\alpha/2}$:** $$ME = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} \Rightarrow z_{\alpha/2} = \frac{ME \times \sqrt{n}}{\sigma} = \frac{800 \times \sqrt{40}}{1800}$$ Calculate $\sqrt{40} \approx 6.3246$: $$z_{\alpha/2} = \frac{800 \times 6.3246}{1800} \approx \frac{5059.68}{1800} \approx 2.81$$ 6. **Find the confidence level corresponding to $z_{\alpha/2} = 2.81$:** The confidence level is $1 - \alpha$, where $\alpha$ is the total area in the two tails. From standard normal tables or using a calculator: $$P(Z < 2.81) \approx 0.9975$$ So, $$\alpha/2 = 1 - 0.9975 = 0.0025$$ $$\alpha = 2 \times 0.0025 = 0.005$$ 7. **Calculate the confidence level:** $$1 - \alpha = 1 - 0.005 = 0.995 = 99.5\%$$ **Final answer:** The confidence level of the interval is approximately **99.5%**.