Given:
[tex]\begin{gathered} mean(\mu)=28mpg \\ Standard\text{ }deviation(\sigma)=2mpg \end{gathered}[/tex]
To Determine: The percentage with greater than 24mpg
Solution
[tex]Z=\frac{x-\mu}{\sigma}[/tex][tex]\begin{gathered} P(x>Z)=P(x>\frac{24-28}{2}) \\ P(x>Z)=P(x>-2) \end{gathered}[/tex]
