How to plot a pdf and cdf for my code

The cdf for this can be computed using the pdf.

norm_cdf = cumsum(norm)./sum(norm); %make sure your variable "sum" is cleared before doing this operation...
figure;
plot(x,norm_cdf,'-k')

If you run the code that I typed above, you will get the same results as shown in your images (with the only difference being in the resolution along the x-axis). Except the image has the pdf and cdf reversed.

Yes, if you compute area under the pdf over a specified interval, then this is equal to the probability of the event occurring in that interval (i.e., P(f|t1<t<t2) = int_{t1}^{t2} pdf(t)dt). The total area under the pdf, however, will always equal 1.

The idea that this is the probability of failure does not make sense when you use a normal distribution because a normal distribution always has a greater than zero probability at negative values. This implies that a component or system could fail in negative time, which is nonsensical. Most of the time in reliability engineering/analysis, the Weibull distribution is used (also the log-normal is used occasionally).

The plot that you showed in Capture2.JPG looks exactly like the Figure 9.6 shown in Capture1.JPG, except that the resolution along the x-axis is higher.

You typed this code in your OP:

sum = 0; for i = 1:trials y1(i) = x1(i)*sigma1 + mu1; % transform the normally distributed randon numbers for input 1 y2(i) = x2(i)*sigma2 + mu2; % transform the normally distributed randon numbers for input 2 Y(i) = (3-((4*1000000)/(30000000*2*4))*(sqrt((y2(i)./16)^2+(y1(i)./4)^2))); % The output, where "failure" is defined as y2 is less than y1 if Y(i) < 0 sum = sum + 1; end end

In this code, you have a variable called "sum". I suggested that you use the following code to form your cdf:

norm_cdf = cumsum(norm)./sum(norm); %make sure your variable "sum" is cleared before doing this operation...

You must clear your variable "sum" before you use the Matlab command sum().