A normal distribution has a mean of µ = 28 with σ = 5. What is the minimum raw score needed to be in the top 25% of the distribution?
Use z-scores.
Determine the z-value using a z-table for the top 25%. Substitute the z into the formula, along with the mean and standard deviation. Solve for x.
Formula:
z = (x - mean)/sd
I hope this will help get you started.
To find the minimum raw score needed to be in the top 25% of a normal distribution, you first need to convert the percentage into a Z-score. Then, you can use the Z-score formula to find the corresponding raw score.
Here's how you can do it step by step:
Step 1: Determine the Z-score.
The top 25% of the distribution corresponds to the area under the curve to the right of that point. Since the normal distribution is symmetric, the top 25% is the same as the lower 75%. In other words, we need to find the Z-score that corresponds to the 75th percentile.
To find the Z-score for the 75th percentile, you can use a standard normal distribution table or a statistical calculator. The Z-score for the 75th percentile is approximately 0.674.
Step 2: Use the Z-score formula to find the raw score.
The Z-score formula is: Z = (X - µ) / σ
Rearranging the formula to solve for X, the raw score, we get: X = Z * σ + µ
Substituting the given values, we have: X = 0.674 * 5 + 28
X ≈ 31.37
Therefore, the minimum raw score needed to be in the top 25% of the distribution is approximately 31.37.