Answer :
Answer:
The IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent is 107.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
[tex]\mu = 100, \sigma = 15[/tex]
Find IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent.
This is X when Z has a pvalue of 0.68. So X when Z = 0.47.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.47 = \frac{X - 100}{15}[/tex]
[tex]X - 100 = 0.47*15[/tex]
[tex]X = 107.05[/tex]
So, rounding to the nearest integer:
The IQ score that separates the top 32 percent of adult IQ scores from the bottom 68 percent is 107.