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Find the score that separates the top 59 percent from the bottom 41 percent, given:

- Mean: 37.6
- Standard deviation: 7.6

Answer :

Final answer:

The score that separates the top 59% from the bottom 41% in a normally distributed set with a mean score of 37.6, and a standard deviation of 7.6 is approximately 35.88. This is calculated using Z-scores and the formula for Z-score transformation.

Explanation:

The student is asking for the score that separates the top 59% from the bottom 41% in a normally distributed set. This question is dealing with Z-scores in statistics. Z-score is a measure of how many standard deviations an element is from the mean. It helps us identify where a score lies in a distribution.

In this case, the mean score is 37.6, and the standard deviation is 7.6. We need the Z-score corresponding to the 41st percentile (as it separates the lower 41% from the top 59%). The Z-score for the 41st percentile can be found using a standard normal distribution table or a calculator with inverse Normal Distribution functions, and it's generally about -0.2.

Once we know that, we can apply the formula for Z-score transformation, which is Z = (X - μ) / σ. Rearranging for X gives us X = Z*σ + μ. Substituting the known values we get X = -0.2*7.6 + 37.6, which is about 35.88. So, the score that separates the top 59% from the bottom 41% is approximately 35.88.

Learn more about Z-score here:

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