Question

assume that adults have iq scores that are normally distributed with a mean of μ = 100 and a standard deviation σ = 15. find the probability that a randomly selected adult has an iq less than 118. the probability that a randomly selected adult has an iq less than 118 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 \(x = 118\), \(\mu=100\), and \(\sigma = 15\).

$$z=\frac{118 - 100}{15}=\frac{18}{15}=1.2$$

Step2: Find the probability using the standard normal distribution table

We want to find \(P(X\lt118)\), which is equivalent to \(P(Z\lt1.2)\) in the standard normal distribution.
Looking up the value of \(z = 1.2\) in the standard normal distribution table (or using a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(1.2, TRUE)), we get \(P(Z\lt1.2)=0.8849\)

Answer:

\(0.8849\)