Question

IQ test scores are standardized to produce a normal distribution with a mean of μ= 100 and a standarddeviation of o=15. Find the proportion of the popula-tion in each of the following IQ categories.a. Genius or near genius: IQ over 140b. Very superior intelligence: IQ from 120 to 140c. Average or normal intelligence: IQ from 90 to 109

Answer 1

$a)0.0038\:b)0.088\:c)\:0.4743$

1) Let's begin with finding the Z-score for this.

a) X> 140

$\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ \\ Z=\frac{140-100}{15}=2.67 \\ \end{gathered}$

Now, that we have found this Z-score we need to resort to the Z-table:

$P(Z>2.67)=0.0038$

This is equivalent to saying that 0.38% of this population are taken to be genius.

b) Very superior intelligence 120

Since we know the Z-score for 140 let's find the Z-score for 120

[tex]\begin{gathered} Z=\frac{120-100}{15}=1.33 \\ \\ P(120So we can tell that approximately 8.8% are taken to possess very superior intelligence.

c) Normal 90

[tex]\begin{gathered} Z=\frac{90-100}{15}=-0.67 \\ \\ Z=\frac{109-100}{15}=0.6 \\ \\ P(-0.67