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 a portion of the purchase price 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?
b. What is the probability that Grear will refund more than $50 for a tire?
c. What mileage should Grear set the promotion claim if it wants the expected cost to be $2

$0.013

0.010724

Explanation:

Given that :

Mean, m = 36500

Standard deviation, s = 5000

Refund of $1 per 100 mile short of 30,000 miles

A.) Expected cost of the promotion :

P(X < 30,000)

Using the Zscore relation :

Zscore = (x - m) / s

Zscore = (30000 - 36500) / 5000

= - 6500 / 5000

= - 1.3

100 miles = $1

1.3 / 100 = $0.013

b. What is the probability that Grear will refund more than $50 for a tire?

100 miles = $1

$50 = (100 * 50) = 5000 miles

Hence, more than $50 means x < (30000 - 5000) = x < 25000 miles

P(x < 25000) :

(25000 - 36500) / 5000

-11500 / 5000

= - 2.3

P(z < - 2.3) = 0.010724 (Z probability calculator)

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