You are the manufacturer of chain saws and are prepared to replace free all saws that fail while under guarantee. If you are willing to replace only 3% of the saws that fail, how long a guarantee should you offer? Assume that you know that the lives of the saws follow a normal distribution with an average of 12 years and a standard deviation of 2 years.

DOES This look right?
0.301 corresponds to a z-value of -1.88

= -1.88 = (xi-12)/2
xi = 2(-1.88)+12
= 8.74 years

Question

You are the manufacturer of chain saws and are prepared to replace free all saws that fail while under guarantee. If you are willing to replace only 3% of the saws that fail, how long a guarantee should you offer? Assume that you know that the lives of the saws follow a normal distribution with an average of 12 years and a standard deviation of 2 years.

DOES This look right?
0.301 corresponds to a z-value of -1.88

 = -1.88 = (xi-12)/2
xi = 2(-1.88)+12
 = 8.74 years