Gumbel Distribution Homework: Proportion of Switches Failing Spec

[mikeliebelt]

Homework Statement

I am not very familiar with this distribution.It asking me the proportion of switches that fails to meet specification.The questoin is:

The specification for the output rise time of a bipolar Hall switch is between 0.23 and 0.27 microseconds. Fly Electronics can only achieve a process capability ( ) of 0.8.
(i) If Fly electronics achieve a process mean of 0.25, what proportion of switches would fail to meet specification if rise times are normally distributed?
(ii) Suppose Fly Electronics manufactured switches with a mean that is off centre and only achieve a process performance ( ) of 0.7. What proportion of switches would fail to meet specification if rise times are normally distributed?

(iii) If Fly electronics achieve a process mean of 0.25, what proportion of switches would fail to meet specification if rise times have a Gumbel distribution?

I found out that standard deviation is 8.33*10^-3.I did first 2 parts,not sure how to do 3.

Homework Equations

Gumbel distr. has cdf
F(x)=exp(-exp(-(x-e)/d))
mean=e+0.57722*d
sd=1.28255*d

The Attempt at a Solution

 

[kurt.physics]

I have done (i) and (ii),not sure how to do (iii).For (i) I did z1=(0.23-0.25)/(8.33*10^-3)= -0.24 and z2=(0.27-0.25)/(8.33*10^-3)= 0.24. Then I used the excel NORMSDIST function to get the probability of each z score,which gave me 0.41 and 0.41 respectively.Then I subtracted one from another, which gave me 0.09.For (ii) I did z1=(0.23-0.25)/(12.5*10^-3)= -0.16 and z2=(0.27-0.25)/(12.5*10^-3)= 0.16. Then I used the excel NORMSDIST function to get the probability of each z score,which gave me 0.54 and 0.54 respectively.Then I subtracted one from another, which gave me 0.08.