Question

assume that adults have iq scores that are normally distributed with a mean of 102 and a standard deviation of 24. find the probability that a randomly selected adult has an iq greater than 144 (hint: draw a graph.) the probability that a randomly selected adult from this group has an iq greater than 144 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 $x = 144$, $\mu=102$, and $\sigma = 24$.
$z=\frac{144 - 102}{24}=\frac{42}{24}=1.75$

Step2: Find the probability using the standard normal distribution table

The standard normal distribution table gives $P(Z\leq z)$. We want $P(Z > 1.75)$. Since the total area under the normal - curve is 1, we use the property $P(Z>z)=1 - P(Z\leq z)$.
From the standard normal distribution table, $P(Z\leq1.75)=0.9599$.
So, $P(Z > 1.75)=1 - 0.9599 = 0.0401$

Answer:

$0.0401$