Assume that adults have IQ scores that are normally distributed with a mean of \( \mu=100 \) and a standard deviation \( \sigma=15 \). Find the probability that a randomly selected adult has an IQ less than 118 . 1 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 118 is (Type an integer or decimal rounded to four decimal places as needed.)

To find the probability that a randomly selected adult has an IQ less than 118, we first need to calculate the z-score using the formula: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the IQ score (118), \( \mu \) is the mean (100), and \( \sigma \) is the standard deviation (15). Calculating the z-score: \[ z = \frac{118 - 100}{15} = \frac{18}{15} = 1.2 \] Next, we look up the z-score in the standard normal distribution table or use a calculator to find the probability. A z-score of 1.2 corresponds to a probability of approximately 0.8849. Therefore, the probability that a randomly selected adult has an IQ less than 118 is approximately **0.8849**.