Question

assume that adults have iq scores that are normally distributed with a mean of μ = 105 and a standard deviation σ = 20. find the probability that a randomly selected adult has an iq between 88 and 122. 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 88 and 122 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 $x = 88$, $z_1=\frac{88 - 105}{20}=\frac{- 17}{20}=-0.85$.
For $x = 122$, $z_2=\frac{122 - 105}{20}=\frac{17}{20}=0.85$.

Step2: Find probabilities from z - table

We know that $P(-0.85From the standard normal distribution table, $P(Z < 0.85)=0.8023$ and $P(Z<-0.85)=0.1977$.

Step3: Calculate the final probability

$P(-0.85 < Z < 0.85)=0.8023-0.1977 = 0.6046$.

Answer:

$0.6046$