Please step by step. Including formulas that were used! Thank You! 5.6 A certain
ID: 3316093 • Letter: P
Question
Please step by step.
Including formulas that were used!
Thank You!
5.6 A certain electrical appliance is sold with a 5-year guarantee. This provides that the full purchase price is refunded if the product fails within 2 years, and half of the purchase price is refunded if the product fails in the following 3 years. A study shows that out of a typical batch of 100 items, there will be 2 failures in the first year, 3 failures in the second year, and 4 failures per year after that. Assuming that interest is a constant 5% and that reimbursement is made at the end of the year of failure, what is the cost of this guarantee to the manufacturer, as a per- centage of the purchase price?Explanation / Answer
Out of a typical batch of 100 items, there will be 2 failures in the first year, 3 in the second year, so a total of 5 failures in first two year and 4 failures per year after that So, total of 12 failures in the next three years.
Therefore, probability that the product fails in first year = 2/100 = 0.05 where full price is refunded
probability that the product fails in second year = 3/100 = 0.05 where full price is refunded
probability that the product fails in the third year = 4/100 = 0.04 where half of price is refunded.
probability that the product fails in the fourth year = 4/100 = 0.04 where half of price is refunded.
probability that the product fails in the fifth year = 4/100 = 0.04 where half of price is refunded.
Let the purchase price be 'p'.
expected cost to the manufacturer = expected cost if the product fails in first two years + expected cost if the product fails in following three years + expected cost if product fails after that(which is zero)
Assuming compound interest
expected cost(accounting for interest) if the product fails in 1st and 2nd year = 0.02*p*1.05 + 0.03*p*(1.05)^2 = 0.054p
expected cost(accounting for interest) if the prodcut fails in 3rd,4th, or 5th year
= 0.04*(p/2)*((1.05)^3+(1.05)^4+(1.05)^5) = 0.073p
therefore, total expected cost = 0.054p+0.073p = 0.127p = 12.7 % of p