The distribution of scores on a standardized aptitude test is approximately normal with a mean of 510 and a standard deviation of 95 What is the minimum score needed to be in the top \( 20 \% \) on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer.

To determine the minimum score needed to be in the top 20% on the standardized aptitude test, we'll follow these steps: 1. **Understand the Problem:** - **Mean (μ):** 510 - **Standard Deviation (σ):** 95 - **Desired Percentile:** Top 20%, which corresponds to the 80th percentile. 2. **Find the Z-Score Corresponding to the 80th Percentile:** - The z-score represents the number of standard deviations a data point is from the mean. - Using standard normal distribution tables or a calculator, the z-score for the 80th percentile is approximately **0.8416**. 3. **Calculate the Required Score (X):** \[ X = μ + z \times σ \] \[ X = 510 + 0.8416 \times 95 \] \[ X = 510 + 79.952 = 589.952 \] 4. **Round to the Nearest Integer:** - Rounding **589.952** to the nearest integer gives **590**. **Answer:** 590