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.

a) Calculate the probability of an engine operating for less than 12 years.
Calculate the probability of an engine operating for more than 25 years.
Calculate the probability that the life of an engine is between 10 and 20 years.

In order to promote the sale of their engines, the company wants to issue a guarantee on the engines which means 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 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 NOK 1200 on a motor that does not fail before the warranty period, while it has a loss of NOK 4500 (ie a profit of -4500 kroner) 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

This is a normal distribution question with



a) P(x < 12.0)=?
The z-score at x = 12.0 is,

z = -1.5417
This implies that



P(x > 25.0)=?
The z-score at x = 25.0 is,

z = 1.1667
This implies that
P(x > 25.0) = P(z > 1.1667) = 1 - 0.8783342285842126




P(10.0 < x < 20.0)=?

This implies that
P(10.0 < x < 20.0) = P(-1.9583 < z < 0.125) = P(Z < 0.125) - P(Z < -1.9583)
P(10.0 < x < 20.0) = 0.5497382248301129 - 0.02509741040604356



b) Given in the question
P(X < x) = 0.975
This implies that
P(Z < 1.96) = 0.975
With the help of formula for z, we can say that

x = 28.8078
PS: you have to refer z score table to find the final probabilities.


Since we know that,

Please hit thumps up if the answer helped you