Question

assume that adults have iq scores that are normally distributed with a mean of \\( \mu = 100 \\) and a standard deviation \\( \sigma = 20 \\). find the probability that a randomly selected adult has an iq less than 132.
click to view page 1 of the table. click to view page 2 of the table.
the probability that a randomly selected adult has an iq less than 132 is
(type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 132\), \(\mu=100\), and \(\sigma = 20\).

$$z=\frac{132 - 100}{20}=\frac{32}{20}=1.6$$

Step2: Find the probability using the standard normal table

We want to find \(P(X\lt132)\), which is equivalent to \(P(Z\lt1.6)\) in the standard normal distribution (\(X\) is the IQ score and \(Z\) is the standard normal variable).
Looking up the value of \(z = 1.6\) in the standard - normal table (the cumulative - distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\lt1.6)=0.9452\)

Answer:

\(0.9452\)