Question

assume that adults have iq scores that are normally distributed with a mean of 100 and a standard deviation of 24. find the probability that a randomly selected adult has an iq greater than 143. (hint: draw a graph.)
the probability that a randomly selected adult from this group has an iq greater than 143 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 \(x = 143\), \(\mu=100\), and \(\sigma = 24\).

$$z=\frac{143 - 100}{24}=\frac{43}{24}\approx1.79$$

Step2: Find the probability using the standard normal distribution

We want \(P(X>143)\), which is equivalent to \(P(Z > 1.79)\) in the standard normal distribution (\(X\) is the original normal random variable and \(Z\) is the standard normal random variable).
Since \(P(Z>z)=1 - P(Z\leq z)\), and from the standard normal table (or using a calculator with a normal - distribution function), \(P(Z\leq1.79)=0.9633\)

$$P(Z > 1.79)=1-0.9633 = 0.0367$$

Answer:

\(0.0367\)