Question

I need help with my homework PLEASE CHECK WORK WHEN DONE NUMBER 7

Answer 1

Given:

`\begin{gathered} mean(\mu)=28mpg \\ Standard\text{ }deviation(\sigma)=2mpg \end{gathered}`

To Determine: The percentage with greater than 24mpg

Solution

`Z=(x-\mu)/(\sigma)`
`\begin{gathered} P(x>Z)=P(x>(24-28)/(2)) \\ P(x>Z)=P(x>-2) \end{gathered}`