Question

Problem 16-03

Grear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes that the standard deviation is 5000 miles and that tire mileage is normally distributed. To promote the new tire, Grear has offered to refund some money if the tire fails to reach 30,000 miles before the tire needs to be replaced. Specifically, for tires with a lifetime below 30,000 miles, Grear will refund a customer $1 per 100 miles short of 30,000.

1. For each tire sold, what is the expected cost of the promotion? If required, round your answer to two decimal places.
   
   
   
2. What is the probability that Grear will refund more than $50 for a tire? If required, round your answer to three decimal places.
   
   
   
3. What mileage should Grear set the promotion claim if it wants the expected cost to be $2.00? If required, round your answer to the hundreds place.
   
   miles

Answer 1

Answer:-

Given That:-

Grear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes that the standard deviation is 5000 miles and that tire mileage is normally distributed. To promote the new tire, Grear has offered to refund some money if the tire fails to reach 30,000 miles before the tire needs to be replaced. Specifically, for tires with a lifetime below 30,000 miles, Grear will refund a customer $1 per 100 miles short of 30,000.

(a). For each tire sold, what is the expected cost of the promotion? If required, round your answer to two decimal places.

Given

= 36500

= 5000

Let X be the no. of miles reached.
The probability that the tire fails to reach 30,000 miles is

  

  

= 0.0968 (From Normal distribution table)

The Probability that it will not reach 30000 miles is = 0.0968

Now E(X) = 30000 0.0968

= 2904

As $ 1 per 100 m, the expected cost of the promotion per tire = 2904/100

= 29.04

(b) What is the probability that Grear will refund more than $50 for a tire? If required, round your answer to three decimal places.

Given

Grear will refund more than $ 50 for a tire.

As $ 1 per 100 miles

50 100 = 5000

  

= P(Z < -2.3)

= 0.0107

= 0.011

(c). What mileage should Grear set the promotion claim if it wants the expected cost to be $2.00? If required, round your answer to the hundreds place.
miles

As per $ 1 per 100 miles, it should limit as 25000 for cost of promotion to $ 2.00

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