Answer: a. 0.1056 b. 0.1056 c. 0.8882
Step-by-step 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. [tex]\mu=65[/tex] and [tex]\sigma=4[/tex]
Let x represents the distance traveled on 1 gallon of fuel .
a. The probability that the car travels more than 70 miles per gallon :
[tex]P(x>70)=P(\dfrac{x-\mu}{\sigma}>\dfrac{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[/tex]
b. The probability that the car travels less than 60 miles per gallon :
[tex]P(x<60)=P(\dfrac{x-\mu}{\sigma}<\dfrac{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[/tex]
c. The probability that the car travels between 55 and 70 miles per gallon:
[tex]P(55<x<70)=P(\dfrac{55-65}{4}<\dfrac{x-\mu}{\sigma}<\dfrac{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[/tex]
