Verizon offers a simple extended warranty plan. If your phone is damaged, they will repair it for up to $300.

Question

Verizon offers a simple extended warranty plan. If your phone is damaged, they will repair it for up to $300. If you lose or destroy your phone, they will give you a $800 voucher towards a new phone. The company believes that 5% of customers will need the replacement voucher and 10% will request a repair each year.

1. If the company charges $20 per month for this extended warranty (i.e. $240 per year), what is the expected value of the profit they will earn each year?

2. What is the standard deviation of their profit?

3. Suppose the company collects 10 warranty plans on one day. What is the mean of the company's total profit?

4. What is the standard deviation of the 10 total warranty plans? What assumption does this calculation require? Do you think this assumption is reasonable?

5. What are the mean and standard deviation for the profit on a 1000 plans?

6. What do your answers to the previous question tell you about the company's likelihood of making a profit?

7. Is the warranty a wise purchase for you? Given that you will probably buy dozens of devices over the next decade, are these types of warranties a wise purchase for you?

Answer 1

The assumption here is that “customers” are the customers who purchased the extended warranty plan.

1.   If the company charges $20 per month for this extended warranty (i.e. $240 per year), what is the expected value of the profit they will earn each year?

If x represents the number of covered customers within a year, then we would have this equation for the expected value of the profit they will earn each year:

 240x – [300(.1x) + 800(.05x)]

= 240x -30x – 40x

= 170x

2.   What is the standard deviation of their profit?

The standard deviation is given by SQRT(p*q/n), where n is equal to the total number in the plan => n = x as used in question 1.

Now, notice that the expected profit is 170x, and the sale of the warranty is 240x. So,

(170x)/(240x)

= 170/240 = .7083 to four decimals.

So, p, the proportion of profit gained is equal to the above => p = .7083

Therefore, q = 1 – p = .2917 to four decimals.

Plugging all the values into the formula for the standard deviation (SD) we get:

 SD = SQRT([.7083*.2917]/x) = .4545/sqrt(x)

3.   Suppose the company collects 10 warranty plans on one day. What is the mean of the company's total profit?

Using 365 days per year, we have x = 365*10 = 3650

So, the mean of the company’s profit is the same as the expected value of the profit earned each year. So, just plug in the value for x = 3650 into the general formula for the expected value found in the answer for question 1:

    = 170*3650 = 620500  

This is $620,500 mean profit (expected value of the profit) from this warranty program.

1.   What is the standard deviation of the 10 total warranty plans? What assumption does this calculation require? Do you think this assumption is reasonable?

Assuming the mean warranty collections per day for the entire year is 10 per day (thus 3,650 for the entire year), we have the SD = .4545/sqrt(x), where x = 3650:

 SD = .0075 to four decimal places for the proportion of the mean profits made for the year,

So, the actual SD in terms of profit is .0075*620500 = 4653.75.

This SD equates to plus or minus $4,653.75 per year.

SD = $4,653.75

Its best for the company to collect data for at least 30 scattered days throughout the year on warranty plans sold or perhaps based on a previous year's collection to determine a more accurate mean of warranty plans sold per day (to get a more accurate annual total of warranties per year).

2.   What are the mean and standard deviation for the profit on a 1000 plans?

Mean profit for 1000 plans = $ 170,000

SD = .4545/sqrt(1000) of the mean = .0144*170000 (rounded to 4 decimals) = SD = $2,448

3.   What do your answers to the previous question tell you about the company's likelihood of making a profit?

Yes, they are very highly likely to make a profit.

4.   Is the warranty a wise purchase for you? Given that you will probably buy dozens of devices over the next decade, are these types of warranties a wise purchase for you?

No, a customer, like myself, is not wise to purchase the warranty. The odds are in favor of the company that sells the warranty.