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. Required:
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?

1. The expected cost of production for each tire sold is $0.013 per tire.

2. Probability that Grear will refund more than $50 for a tire is 0.0107

Explanation;

1. Mileage is 36,500 miles

Standard deviation is 5,000 miles

Observed miles is 30,000 miles

100 miles failed at $1

Therefore;

(36,500 - 30,000) /5,000 = 1.3

To get the cost of production,

Since 100 miles equals $1 if fail

1.3 × 1 / 100

= $0.013 per tire.

2. P(Z<25,000 - 36,500/5,000)

= P(Z<-11,500/5,000)

=Z<2.3

Therefore,

1-0.9893

=0.0107

The probability that Grear will refund more than $50 for a tire is 0.0107

you will have $2593.74 in 10 years if you set aside $1,000 at 10%

you will have $9321.91 in 20 years if you set aside $2000 at 8%

you will have $1469.32 in 5 years if you set aside $1000 at 8%

you will have $1296.87 in 10 years if you set aside $500 at 10%

you will have $239.66 in 15 years if you set aside $100 at 6%

you will have $4660.96 in 20 years if you set aside $1000 at 8%