Question

assume that adults have iq scores that are normally distributed with a mean of 96 and a standard deviation of 18. find the probability that a randomly selected adult has an iq greater than 121. (hint: draw a graph.)
the probability that a randomly selected adult from this group has an iq greater than 121 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 $x = 121$, $\mu=96$, and $\sigma = 18$.
$z=\frac{121 - 96}{18}=\frac{25}{18}\approx1.39$

Step2: Find the probability using the standard normal distribution

We want $P(X>121)$, which is equivalent to $P(Z > 1.39)$ in the standard normal distribution. Since the total area under the standard - normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$.
From the standard normal table, $P(Z\leq1.39) = 0.9177$.
So $P(Z>1.39)=1 - 0.9177=0.0823$

Answer:

$0.0823$