Answer: The percentage of the vehicles produced by Mazda have a gas mileage greater than 24 = 84.13%
Step-by-step explanation:
We assume that gas mileage is normal distribution.
Let x reprsents the gas mileage.
Given: Mean [tex]\mu[/tex] = 26 mpg , standard deviation [tex]\sigma[/tex] = mpg
The probability that the vehicles produced by Mazda have a gas mileage greater than 24:-
[tex]P(x>24)=P(\dfrac{x-\mu}{\sigma}>\dfrac{24-26}{2})\\\\=P(z>-1)\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\=P(z<1)=0.8413[/tex]
Hence, the percentage of the vehicles produced by Mazda have a gas mileage greater than 24 = 84.13%
