What score in a distribution that has a mean of 500 and a standard deviation of 100, would identify the top 60% of scores?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion and its Z score. Use value in equation above to find the score.
To determine the score that would identify the top 60% of scores, we need to find the corresponding z-score using the standard normal distribution.
Step 1: Convert the desired percentile to a z-score.
Since we want the top 60% (the upper tail), we subtract the desired percentile from 1:
Percentile = 1 - 0.60 = 0.40
Step 2: Find the z-score corresponding to the desired percentile.
Consulting a standard normal distribution table or using a calculator, we can find that the z-score for a percentile of 0.40 is approximately 0.253.
Step 3: Use the formula z = (x - μ) / σ to solve for x.
Rearranging the formula, we have x = z * σ + μ.
Given that the mean (μ) is 500 and the standard deviation (σ) is 100, we substitute these values into the formula:
x = 0.253 * 100 + 500 = 25.3 + 500 = 525.3
Therefore, a score of approximately 525.3 would identify the top 60% of scores in a distribution with a mean of 500 and a standard deviation of 100.