a government agency checked the gasoline mileage of a partic

a government agency checked the gasoline mileage of a particular make of automobile and found the mileages to be normally distributed, with a mean of 28.6 mpg and a standard deviation of 2.3 mpg. For one of these automobiles, what is the probability that the mileage will be

a) at least 30 mpg?

b) between 28 and 32 mpg?

c) at most 32 mpg?

Solution

Mean ( u ) =28.6
Standard Deviation ( sd )=2.3
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 30) = (30-28.6)/2.3
= 1.4/2.3= 0.6087
= P ( Z <0.6087) From Standard Normal Table
= 0.7286                  
P(X > = 30) = (1 - P(X < 30)
= 1 - 0.7286 = 0.2714                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 28) = (28-28.6)/2.3
= -0.6/2.3 = -0.2609
= P ( Z <-0.2609) From Standard Normal Table
= 0.3971
P(X < 32) = (32-28.6)/2.3
= 3.4/2.3 = 1.4783
= P ( Z <1.4783) From Standard Normal Table
= 0.93033
P(28 < X < 32) = 0.93033-0.3971 = 0.5332                  

c)
P(X > 32) = (32-28.6)/2.3
= 3.4/2.3 = 1.4783
= P ( Z >1.478) From Standard Normal Table
= 0.0697                  
P(X < = 32) = (1 - P(X > 32)
= 1 - 0.0697 = 0.9303