To find the probability you can use the z-score formula, to convert 123.7 into a z-score and then read it on standard normal tables:
[tex]z=\frac{x-\mu}{\sigma}[/tex]Where:
[tex]\begin{gathered} z\text{ is the z-score} \\ x\text{ is the }value\text{ to evalute 123.7} \\ \mu\text{ is the mean} \\ \sigma\text{ is the standard deviation} \end{gathered}[/tex]By replacing the values you have:
[tex]z=\frac{123.7-97.1}{17.1}=1.56[/tex]Now, you need to find the probability that an adult has an IQ greater than 123.7, it means.
[tex]P(z>1.56)=1-P(z\leq1.56)[/tex]By using the standard normal tables, you will find that P(z<=1.56) is 0.9406:
Replace this value:
[tex]\begin{gathered} P(z>1.56)=1-P(z\leq1.56)=1-0.9406 \\ P(z>1.56)=0.0594 \end{gathered}[/tex]
