Download Q7.R.10 Normal Distribution

Probability and X value of Normal Distribution
We selected Q7.R.10 (p.362) as an example of finding the probability of a normally distributed random
variable and finding the x value of a given normal probability.
Q7.R.10
Tire Wear Supposed that Dunlop Tire manufactures a tire with a lifetime that approximately follows a normal
distribution with mean 70,000 miles and standard deviation 4400 miles.
(a) What proportion of the tires will last at least 75,000 miles?
(b) Suppose that Dunlop warrants the tires for 60,000 miles. What proportion of the tires will last 60,000 miles
or less?
(c) What is the probability that a randomly selected Dunlop tire lasts between 65,000 and 80,000 miles?
(d) Suppose that Dunlop wants to warrant no more than 2% of its tires. What mileage should the company
advertise as its warranty mileage?
(a) What proportion of the tires will last at least 75,000 miles?
---> Find P( X  75000) .
Normal Distribution,  =70000,  =4400
Step 1: 1) Log onto StatCrunch and get a blank data sheet.
2) Click Stat → Calculators → Normal.
Step 2: 1) When the normal distribution dialogue box pops up. Click the Standard tab.
2) For this X variable, input 7000 for Mean: and input 4400 for Std. Dev. : .
3) Use  to select  → Input 75000.
4) Click Compute.
The shaded region represents P( X  75000) with  =70000 and  =4400.
P( X  75000)  0.1279022  0.1279 .
(b) Suppose that Dunlop warrants the tires for 60,000 miles. What proportion of the tires will last 60,000 miles
or less?
---> Find P( X  60000) .
Normal Distribution,  =70000,  =4400
Use the same normal distribution dialogue box with Mean: 70000 and Std. Dev.: 4400.
Step 1: 1) Use  to select  → Input 60000.
2) Click Compute.
The shaded region represents P( X  60000) with  =70000 and  =4400.
P( X  60000)  0.01152131  0.0115 .
(c) What is the probability that a randomly selected Dunlop tire lasts between 65,000 and 80,000 miles?
---> Find P(65000  X  80000) .
Normal Distribution,  =70000,  =4400
Use the same normal distribution dialogue box with Mean: 70000 and Std. Dev.: 4400.
Step 1: 1) Click the Between tab.
2) Input 65000 and 80000 in the boxes for the boundaries of x .
3) Click Compute.
The shaded region represents P(60000  X  85000) with  =70000 and  =4400.
P(60000  X  85000)  0.86057649  0.8606
(d) Suppose that Dunlop wants to warrant no more than 2% of its tires. What mileage should the company
advertise as its warranty mileage?
Dunlop wants to warrant no more than 2% ---> P( X  what mileage? )  0.02 .
Use the same normal distribution dialogue box with Mean: 70000 and Std. Dev.: 4400.
Step 1: 1) Click the Standard tab.
2) Select  → Move the cursor to the last box of the same line and input 0.02.
3) Click Compute.
The X value is 60963.505  60964 miles.
The company should advertise the warranty mileage for Dunlop is 60964 miles.