Assume that adults have IQ scores that are normally distributed with a mean of 100 and a standard deviation of 15. Find the IQ score separating the top 37% from the others. Answer in one decimal place.
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X = 100 + 15(.33) = 105.0
To find the IQ score that separates the top 37% from the others, we need to calculate the z-score associated with the top 37% and then convert it back to the original IQ scale using the mean and standard deviation.
Step 1: Convert the percentile (37%) to a z-score.
The percentile is given as 37%, which means 37% of the scores are below the desired IQ score. To find the z-score corresponding to this percentile, we subtract the percentile from 1 (to find the percentage above the desired IQ score) and then look up the z-score from a standard normal distribution table.
1 - 0.37 = 0.63
Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a cumulative percentage of 0.63 is approximately 0.325.
Step 2: Convert the z-score back to the IQ scale.
Now that we have the z-score, we can convert it back to the IQ scale using the mean and standard deviation.
IQ = (z-score * standard deviation) + mean
IQ = (0.325 * 15) + 100
IQ = 5.85 + 100
IQ ≈ 105.9
Therefore, the IQ score that separates the top 37% from the others is approximately 105.9, rounded to one decimal place.