Question

. 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 5,000 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 average cost of the promotion?

b. What is the probability that Grear will refund more than $25 for a tire?

Answer 1

Answer:

a)

Given,

Mean = 36500

Standard deviation = 5000

P(X < 30000) = P((x-mu)/s < (30000 - 36500)/5000)

= P(z < - 1.30)

= 0.0968005 [since from z table]

P(X < 30000) = 0.0968

Average cost that it may not reach = 30000*0.0968

= 2904

Now for $1 per 100 , so Average cost = 2904/100

= 29.04

Average cost = 29.04*0.0968

= 2.81

b)

Refund more than 25 = 25*100

= 2500

Now consider,

= P(z < (27500 - 36500)/5000)

= P(z < -1.8)

= 0.0359303 [since from z table]

= 0.0359