Question

assume that adults have iq scores that are normally distributed with a mean of \\( \mu = 105 \\) and a standard deviation \\( \sigma = 15 \\). find the probability that a randomly selected adult has an iq less than 120. 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 120 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 \(z=\frac{x-\mu}{\sigma}\), where \(x = 120\), \(\mu=105\), and \(\sigma = 15\).

$$z=\frac{120 - 105}{15}=\frac{15}{15}=1$$

Step2: Find the probability using the standard normal table

We want to find \(P(X\lt120)\), which is equivalent to \(P(Z\lt1)\) (since \(z = 1\) for \(x = 120\)).
Looking up the value of \(z = 1\) in the standard - normal table (the cumulative distribution function of the standard normal distribution \(\varPhi(z)\)), we find that \(P(Z\lt1)=0.8413\)

Answer:

\(0.8413\)