Answer: 0.1974
Step-by-step explanation:
Let X be a random variable that denotes the gas mileage (in miles per gallon) of the individual sedan.
Given: Population mean: [tex]\mu=20[/tex]
Standard deviation: [tex]\sigma=2[/tex]
The probability that the gas mileage (in miles per gallon) of your single, individual sedan is between 19.5 and 20.5:
[tex]P(19.5<x<20.5)=P(\dfrac{19.5-20}{2}<\dfrac{x-\mu}{\sigma}<\dfrac{20.5-20}{2})\\\\=P(-0.25<z<0.25)=2P(0.25)-1 [P(-z<Z<z)=2P(Z<z)-1]\\\\=2(0.5987)-1=0.1974[/tex]
The required probability=0.1974
