Let's continue solving questions (c) and (d) **in the same style as the examples provided in your screenshot. --- ### (c) Minimum and Maximum Starting Salaries for the Middle 95% of LU Graduates The middle 95% of a normal distribution lies between the 2.5th percentile and the 97.5th percentile. These correspond to Z-scores of approximately -1.96 and +1.96. \[ \text{Minimum Salary} = \mu + (Z_{min} \cdot \sigma) \] \[ \text{Maximum Salary} = \mu + (Z_{max} \cdot \sigma) \] Given: - \(\mu = 20,000\) - \(\sigma = 8,000\) - \(Z_{min} = -1.96\) - \(Z_{max} = +1.96\) \[ \text{Minimum} = 20,000 + (-1.96 \times 8,000) = 20,000 - 15,680 = \$4,320 \] \[ \text{Maximum} = 20,000 + (1.96 \times 8,000) = 20,000 + 15,680 = \$35,680 \] **Thus, the minimum and maximum starting salaries for the middle 95% of graduates are \$4,320 and \$35,680.** --- ### (d) Number of Graduates if 189 Have Salaries ≥ $32,240 First, find the probability that a graduate has a salary of at least $32,240. \[ P(X \geq 32,240) = P\left(Z \geq \frac{32,240 - 20,000}{8,000}\right) = P(Z \geq 1.53) \] Using the standard normal table or Excel: \[ P(Z \geq 1.53) = 1 - P(Z < 1.53) = 1 - .93699 = .063 \] Let \(N\) be the total number of graduates: \[ .063 \times N = 189 \] \[ N = \frac{189}{.063} \approx 3,000 \] **So, approximately 3,000 students graduated this year from this university.** --- ## **Summary Table** | Part | Calculation | Answer | |------|-------------|--------| | c | Min: \(20,000 - 1.96 \times 8,000\) <br> Max: \(20,000 + 1.96 \times 8,000\) | Min: \$4,320 <br> Max: \$35,680 | | d | \(N = 189 / .063\) | \(N \approx 3,000\) |