Answer: The percentage of the vehicles produced by Mazda have a gas mileage greater than 24 = 84.13%
Explanation:
We assume that gas mileage is normal distribution.
Let x reprsents the gas mileage.
Given: Mean
= 26 mpg , standard deviation
= mpg
The probability that the vehicles produced by Mazda have a gas mileage greater than 24:-
![P(x>24)=P((x-\mu)/(\sigma)>(24-26)/(2))\\\\=P(z>-1)\ \ \ [z=(x-\mu)/(\sigma)]\\\\=P(z<1)=0.8413](https://img.qammunity.org/2022/formulas/mathematics/high-school/r709rogzqi1458w3vk211adspraidih8po.png)
Hence, the percentage of the vehicles produced by Mazda have a gas mileage greater than 24 = 84.13%