Question

assume that adults have iq scores that are normally distributed with a mean of 105 and a standard deviation of 19. find the probability that a randomly selected adult has an iq greater than 130. (hint: draw a graph.)

the probability that a randomly selected adult from this group has an iq greater than 130 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 \(x = 130\), \(\mu=105\), and \(\sigma = 19\).

$$z=\frac{130 - 105}{19}=\frac{25}{19}\approx1.32$$

Step2: Find the probability using the standard normal table

We want \(P(X>130)\). Since \(P(X > x)=1 - P(X\leq x)\), and for a normal distribution \(P(X\leq x)\) corresponds to the cumulative distribution function of the standard normal distribution at the z - score.
From the standard normal table (or using a calculator with a normalcdf function: normalcdf\((z,\infty,\mu = 0,\sigma = 1)\)), \(P(Z\leq1.32)=0.9066\) (using a standard normal table or technology).

$$P(X>130)=1 - P(Z\leq1.32)=1 - 0.9066 = 0.0934$$

Answer:

\(0.0934\)