Question

assume that adults have iq scores that are normally distributed with a mean of 96 and a standard deviation of 20. find the probability that a randomly selected adult has an iq greater than 135. (hint: draw a graph.) the probability that a randomly selected adult from this group has an iq greater than 135 is (round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 135\), \(\mu=96\), and \(\sigma = 20\).

$$z=\frac{135 - 96}{20}=\frac{39}{20}=1.95$$

Step2: Find the probability

We want to find \(P(X>135)\), which is equivalent to \(P(Z > 1.95)\) since \(X\) (IQ score) is normally distributed.
Using the property \(P(Z>z)=1 - P(Z\leq z)\).
From the standard normal table, \(P(Z\leq1.95)=0.9744\).

$$P(Z > 1.95)=1-0.9744 = 0.0256$$

Answer:

\(0.0256\)