A normal distribution has a mean of µ = 80 with σ = 20. What score separates the highest 40% of the distribution from the rest of the scores?

To find the score that separates the highest 40% of the distribution from the rest, you can use the z-score formula and then convert it back to the original scale of the distribution. 1. Find the z-score corresponding to the top 40% of the distribution using the standard normal distribution table or calculator. 2. Once you have the z-score, use it to find the corresponding score in the original distribution. The z-score formula is: \[ z = \frac{x - \mu}{\sigma} \] Where: - \( x \) is the score in the distribution - \( \mu \) is the mean of the distribution - \( \sigma \) is the standard deviation of the distribution - \( z \) is the z-score Given: - \( \mu = 80 \) - \( \sigma = 20 \) We need to find the z-score corresponding to the top 40% of the distribution, which is \( z_{0.40} \). You can find \( z_{0.40} \) using a standard normal distribution table or calculator. Once you have \( z_{0.40} \), you can find the corresponding score \( x \) using the formula: \[ x = \mu + z_{0.40} \times \sigma \] Let's do the calculations: 1. Find \( z_{0.40} \): Using a standard normal distribution table or calculator, \( z_{0.40} \approx 0.25 \). 2. Find the corresponding score \( x \): \[ x = 80 + 0.25 \times 20 = 80 + 5 = 85 \] So, the score that separates the highest 40% of the distribution from the rest of the scores is approximately 85.