Question

assume that adults have iq scores that are normally distributed with a mean of $mu = 105$ and a standard deviation $sigma = 20$. find the probability that a randomly selected adult has an iq between 89 and 121. 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 between 89 and 121 is (type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$.
For $x = 89$, $z_1=\frac{89 - 105}{20}=\frac{-16}{20}=- 0.8$.
For $x = 121$, $z_2=\frac{121 - 105}{20}=\frac{16}{20}=0.8$.

Step2: Use the standard normal distribution table

We want to find $P(-0.8From the standard - normal table, $P(Z < 0.8)=0.7881$ and $P(Z<-0.8) = 0.2119$.
Then $P(-0.8 < Z < 0.8)=P(Z < 0.8)-P(Z<-0.8)$.
$P(-0.8 < Z < 0.8)=0.7881 - 0.2119=0.5762$.

Answer:

$0.5762$