M= 500 SD= 100 What X value, separates the top 15% of the distribution from the rest?
Use the same table indicated in your later post to find the Z score, then insert into equation indicated.
rly think its 1?
To find the X value that separates the top 15% of the distribution from the rest, you can use the concept of z-scores.
A z-score represents the number of standard deviations a data point is from the mean. It can be calculated using the formula:
z = (X - μ) / σ
Where:
X is the data point
μ is the mean
σ is the standard deviation
In this case, you have M = 500 and SD = 100. Since the distribution is assumed to be normal, you can use the standard normal distribution table (also known as the z-table) to find the corresponding z-score for the top 15% of the distribution.
Since you want to find the X value that separates the top 15% of the distribution, the remaining 85% will lie below this X value. This corresponds to a z-score of 1.036 on the standard normal distribution table, as 0.85 is the closest value to 0.8508 (which represents the cumulative probability up to 1.036).
Now that you have the z-score, you can rearrange the z-score formula to solve for X:
X = z * σ + μ
Plugging in the values:
X = 1.036 * 100 + 500
Calculating this expression:
X ≈ 603.6
Therefore, the X value that separates the top 15% of the distribution from the rest is approximately 603.6.