Question

assume that adults have iq scores that are normally distributed with a mean of \\( \mu = 105 \\) and a standard deviation \\( \sigma = 15 \\). find the probability that a randomly selected adult has an iq between 87 and 123. 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 87 and 123 is \\( \square \\). (type an integer or decimal rounded to four decimal places as needed.)

Explanation:

Step1: Calculate the z - scores

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 87\), \(z_1=\frac{87 - 105}{15}=\frac{- 18}{15}=-1.2\).
For \(x = 123\), \(z_2=\frac{123 - 105}{15}=\frac{18}{15}=1.2\).

Step2: Use the standard normal distribution table

We want to find \(P(87<X<123)\), which is equivalent to \(P(-1.2<Z<1.2)\).
Since the standard normal distribution is symmetric, \(P(-1.2<Z<1.2)=P(Z < 1.2)-P(Z<-1.2)\).
From the standard normal table, \(P(Z < 1.2)=0.8849\) and \(P(Z<-1.2)=0.1151\).

Step3: Calculate the probability

\(P(-1.2<Z<1.2)=0.8849-0.1151 = 0.7698\)

Answer:

\(0.7698\)