Question

A company produces refrigerator motors. These engines have a life expectancy of 19.4 years with a standard deviation of 4.8 years. Assume that the service life of the motors is normally distributed.

In order to promote the sale of its engines, the company wants to go out with a guarantee on the engines which assumes that the customer can replace the engine free of charge if it breaks before a certain time.

b) How many years of warranty can the company expire if they do not want to have to replace more than 2.5% of the engines? (that is, the warranty period should be such that the probability that an engine's service life is less than the warranty period is 0.025)
The company has a profit of $ 1200 on a motor that fails before the warranty period, while it has a loss of $ 4,500 (that is, a profit of $ 4500) on a motor that fails before the warranty period. If the company uses the warranty period calculated, what is the expected profit from the sale of an engine?
Briefly explain what this expected profit in practice tells us.

A company produces refrigerator motors. These engines have a life expectancy of 19.4 years with a standard deviation of 4.8 years. Assume that the service life of the motors is normally distributed. In order to promote the sale of its engines, the company wants to go out with a guarantee on the engines which assumes that the customer can replace the engine free of charge if it breaks before a certain time. b) How many years of warranty can the company expire if they do not want to have to replace more than 2.5% of the engines? (that is, the warranty period should be such that the probability that an engine's service life is less than the warranty period is 0.025) The company has a profit of $ 1200 on a motor that fails before the warranty period, while it has a loss of $ 4,500 (that is, a profit of $ 4500) on a motor that fails before the warranty period. If the company uses the warranty period calculated, what is the expected profit from the sale of an engine? Briefly explain what this expected profit in practice tells us.

Answer 1

a) The mean life and standard deviation of the refrigerator motor is 19.4 years and 4.8 years respectively.

We need to find the number of years n such that:

P(N<n) = 0.025

Using the normal tables we get:

n = 9.9921 or 10 years

Thus, the number of years of warranty the company can give such that the probability of replacement is 2.5% is 10 years.

b) The expected profit form the failure of the motor is:

1200*(1-0.025) -4500*0.025 = $1057.5

There is a positive expected profit which is really close to the value of the profit which the company can make when the motor fails after the warranty period.

This is the amount of money the company expects to earn by putting a warranty year strategy in place.