Suppose that Dunlop Tire manufactures a tire with a lifetime

Suppose that Dunlop Tire manufactures a tire with a lifetime that approximately follows a normal disttribution with mean 70,000 miles and standard deviation 4,400 miles.

7. What proportion of tires will last at least 75,000 miles?

8. What is the probability that a randomly selected Dunlop tire lasts between 65,000 and 80,000 miles?

Solution

Normal Distribution
Mean ( u ) =70000
Standard Deviation ( sd )=4400
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 75000) = (75000-70000)/4400
= 5000/4400= 1.1364
= P ( Z <1.1364) From Standard Normal Table
= 0.8721                  
P(X > = 75000) = (1 - P(X < 75000)
= 1 - 0.8721 = 0.1279                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 65000) = (65000-70000)/4400
= -5000/4400 = -1.1364
= P ( Z <-1.1364) From Standard Normal Table
= 0.1279
P(X < 80000) = (80000-70000)/4400
= 10000/4400 = 2.2727
= P ( Z <2.2727) From Standard Normal Table
= 0.98848
P(65000 < X < 80000) = 0.98848-0.1279 = 0.8606