Answer:
The cutoff value for the top 5% of all IQs is 131.255.
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 = 19[/tex]
What is the cutoff value for the top 5% of all IQs
This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 100}{19}[/tex]
[tex]X - 100 = 19*1.645[/tex]
[tex]X = 131.255[/tex]
The cutoff value for the top 5% of all IQs is 131.255.
