After reading the example above where one can see how probability
Unit 4
Practical Exercise
The following example illustrates how probability values using areas under the standard normal distribution can be used to help make business decisions.
Example:
Let’s suppose that the Goodyear Tire Company has just developed a tire to be sold. Since the tire is new and Goodyear wants to introduce it into the consumer market, company executives believe that a mileage guarantee offered with the tire will be an important factor in buying the product. Goodyear managers want probability information about the number of miles the tire will last.
From actual test data, Goodyear engineers estimate the mean () tire mileage to be:
μ = 36,500 miles
σ = 5,000 miles
A. What is the probability that tire mileage will exceed 40,000 miles?
Step1. At x = 40,000 .70
Step 2. The area to the left of .70 (listed in Table V on page A11 in the appendix in your text) is equal to .7580
Step 3. Subtract .7580 from 1.0 (the total area under the curve is equal to 1.0)
1.0 – .7580 = .2420
or
24.20% of the tires will exceed 40,000 in mileage
B. Let’s now assume that Goodyear is considering a guarantee that will provide a discount on replacement tires if the original tires do not exceed the mileage in the guarantee. What should the guaranteed mileage be if Goodyear wants no more than 10% of the tires to be eligible for a discount?
The information we have available right now to help in this decision is that we have a population mean of 36,500 with a standard deviation of 5,000. Goodyear management wants no more than 10% of the tires to be eligible for a discount guarantee. If we draw a normal distribution curve, 10% of the tires will fall in the lower left tail of the curve. Thus, we are looking for the zscore value, such that the area to the left of this zscore under the standard normal distribution curve is 10%, or 0.1000. Since the value 0.1000 does not appear exactly in Table V, we use the closest value that is in the table, which is 0.1003. It appears at the intersection of row 1.2 and column .08, which corresponds to the zscore of 1.28.
To find the mileage of an unknown value of x that corresponds to a z = 1.28 we solve algebraically:
Follow each step:
Method 1 or Method 2


1. 
1. 
2. 
2. 
3. 30,100 = 36,500 – 1.28 (5,000) 
3. x = 30,100 
Either method to solve for x works. Use the method you’re most comfortable with.
A guarantee of 30,000 will meet the requirement that 10% of the tires will be eligible for the guarantee.
From this example, we can see how probability can help the decision maker in reaching a good decision.
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