Question

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?

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Miles

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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.

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