Question

The life of a certain small motor has been determined to be a random variable that follows a normal distribution with an expected value of 10 years and a standard deviation of two years. The manufacturing company offers a 7 year warranty, during which time it replaces for free any motor that stops working.

a. What proportion of motors will have to be replaced under warranty?

b. What is the probability that it will be damaged the same day the warranty expires?

c. When is the life time exceeded by 5% of the motors?

part b

The company has made its economic analysis and has determined that it has to make adjustments so that it can only offer replacement to 1.5% of customers.

What is the maximum guarantee time that you can offer?

If you sell 25 of these engines this month, what is the probability that at least 5 of them will return in less than 5 years to claim a warranty?

Answer 1

We are given the distribution here as:

a) The proportion that has to be replaced is computed here as:
P(X < 7) since 7 year is the warranty period

Converting it to a standard normal variable, we get here:
P(Z < (7 - 10)/2)

P(Z < -1.5)

Getting it from the standard normal tables, we get here:
P(Z < -1.5) = 0.0668

Therefore 0.0668 is the required proportion here.

b) Probability that it would be damaged the same day the warranty expires is computed here as:

Putting all the values, we have here:

therefore 0.0648 is the required probability here.

c) From standard normal tables, we have:
P(Z > 1.645) = 0.05

Therefore the lifetime exceeded by only 5% motors is computed here as:
= Mean + 1.645* Std Dev

= 10 + 1.645*2

= 13.29 years

Therefore 13.29 years is the required lifetime here.

d) From standard normal tables, we have:
P( Z < -2.17 ) =0.015

Therefore the maximum guarantee time that can be offerred here is computed as:
= Mean - 2.17* Std Dev

= 10 - 2.17*2

= 5.66

Therefore 5.66 years is the required maximum guarantee time here.

e) First we compute the probability here:
P(X < 5)

Converting it to a standard normal variable, we have here:
P(Z < (5 - 10)/2)

= P(Z < -2.5)

Getting it from the standard normal tables, we have here :
P(Z < -2.5) = 0.0062

Therefore the number of motors out of 25 that would be returned before 5 years is modelled here as:

The required probability that at least 5 of them would be replaced is computed here as:
P(Y >= 5) = 1 - P(Y <= 4)

This is computed in EXCEL as:
=1-binom.dist(4,25,0.0062,TRUE)

therefore 0.000000439 is the required probability here.