Question

Some IQ tests are standardized to a Normal model N(100,19). What is the cutoff value for the top 5% of all IQs

Answer 1

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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:

$\mu = 100, \sigma = 19$

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.

$Z = \frac{X - \mu}{\sigma}$

$1.645 = \frac{X - 100}{19}$

$X - 100 = 19*1.645$

$X = 131.255$

The cutoff value for the top 5% of all IQs is 131.255.