Answer: a. 0.1056 b. 0.1056 c. 0.8882
Explanation:
Given : The distance traveled on 1 gallon of fuel is normally distributed with a mean of 65 miles and a standard deviation of 4 miles.
i.e.
and

Let x represents the distance traveled on 1 gallon of fuel .
a. The probability that the car travels more than 70 miles per gallon :
![P(x>70)=P((x-\mu)/(\sigma)>(70-65)/(4))\\\\=P(z>1.25)\\\\=1-P(z\leq1.25)\ \ [\because\ P(Z>z)=1-P(Z\leq z)]\\\\=1-0.8944=0.1056](https://img.qammunity.org/2020/formulas/mathematics/college/3m2rabb8h13j7tr3cql5b7jy635dxcnuua.png)
b. The probability that the car travels less than 60 miles per gallon :
![P(x<60)=P((x-\mu)/(\sigma)<(60-65)/(4))\\\\=P(z<-1.25)\\\\=1-P(z\leq1.25)\ \ [\because\ P(Z<-z)=1-P(Z\leq z)]\\\\=1-0.8944=0.1056](https://img.qammunity.org/2020/formulas/mathematics/college/dnnfhzmzm4ofu0qd6mwzjgmdgpavp8tiy9.png)
c. The probability that the car travels between 55 and 70 miles per gallon:
![P(55<x<70)=P((55-65)/(4)<(x-\mu)/(\sigma)<(70-65)/(4))\\\\=P(-2.5<z<1.25)\\\\=P(z<1.25)-P(z<-2.5)\\\\=P(z<1.25)-(1-P(z\leq2.5)\ \ [\because\ P(Z<-z)=1-P(Z\leq z)]\\\\=0.8944-(1-0.9938)=0.8882](https://img.qammunity.org/2020/formulas/mathematics/college/1ags3let1wbvibj48xzuwhnw7ow3tbl5qd.png)