Question

A race car company  developed a new car battery that has a longer life span than that of a traditional battery. From the date of purchase of a race car, the distribution of the life span of the new battery is approximately normal with mean 30 months and standard deviation 8 months. For the price of $50, the company offers a two-year warranty on the new battery for customers who purchase a race car. The warranty guarantees that the car will be replaced at no cost to the customer if the battery no longer works within 24 months from the date of purchase.

(a) In how many months from the date of purchase is it expected that 25 percent of the batteries will no longer work? Show work.

(b) Suppose one customer who purchases the warranty is selected at random. What is the probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works? Show your work and label your answer with appropriate probability notation.

(C). The company has a gain of $50 for each customer who purchases a warranty but does not require a replacement. The company has a loss (negative gain) of $150 for each customer who purchases a warranty and does require a replacement. What is the expected value of the gain for the company for each warranty purchased? Show work

Answer 1

Let life span of a new battery in months be denoted as T.

It is expected that 25% of the batteries will no longer work in 't' months from the date of purchase.

From the Standard Normal distribution table it is obtained that :-

Thus it is expected that 25% of the batteries will no longer work in 27 months from the date of purchase.

Probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works

[ The above probability is obtained from the Standard Normal distribution table. ]

Probability of gaining $50 = Probability that a customer who purchases a warranty does not require a replacement

= P(T24) = 1 - P(T24) = 1 - 0.226627 = 0.773373

Probability of losing $150 = Probability that a customer who purchases a warranty requires a replacement

= P(T24) = 0.773373

Thus the expected value of the gain for the company for each warranty purchased

= $500.773373 - $1500.226627 $4.67