To find the probability that the mean IQ score of people in the sample is less than 98, first we need to convert the value into z-score by using the formula:
[tex]z=\frac{x-\mu}{\sigma}[/tex]Then, when x=98:
[tex]\begin{gathered} z=\frac{98-100.0}{15.0} \\ z=\frac{-2.0}{15.0} \\ z=-0.1333 \end{gathered}[/tex]By checking in a standard normal table z=-0.1333 has a p-value=0.4471.
Then the probability is 0.4471 that also can be expressed as 44.71%