Question

A manufacturer of laptop computers offers a one-year warranty. If the laptop computer fails for any reason during this period, it is replaced. The length of satisfactory service (years) is a random variable having the probability density function f(x)=1/3.3 e^(-x⁄3.3) x>0 What percentage of the laptops will fail within the warranty period? Find the probability that one of these laptops will provide satisfactory service for 3 to 4 years. Find the probability that one of these laptops will provide satisfactory service for at least 4 years.   The manufacturing cost of a laptop computer is $100, and the profit per sale is $300. What is the effect of warranty replacement on profit? Calculate the mean life-time of the laptops.  

Answer 1

a) What percentage of the laptops will fail within the warranty period?

P(X < x) = 1 - e^(-x/mean)

P(X < 1) = 1 - e^(-1/3.3) = 0.2614

=26.14 %

b) Find the probability that one of these laptops will provide satisfactory service for 3 to 4 years.

P( 3 < X < 4)

= e^(-3/3.3) - e^(-4/3.3)

= 0.1053

c) Find the probability that one of these laptops will provide satisfactory service for at least 4 years.

P( X >4)

=e^(-4/3.3)

= 0.29756

d)

when warranty is given then

expected profit will decrease from 300-100 = 200 to

300 - 100 - 100 * Probability that it will fail during warranty period

= 200 - 100* 0.2614

= 173.86

e)

mean life time of laptops is 3.3 years