A Few Key Concepts to Take From One Complete Program and Apply to Another Very Similar Second Computer Program from an Exponential Distribution with Lambda Function Applied..

Hello, this may take a few minutes, be warned hehe.. but really shouldn't be too bad at all!

Here is first program:

nsamples=100;  % number of samples
n = 225;  % sample size 225
a = 1;
b = 2;
data = -3+9*rand(nsamples, n);  % random 100x225 data sample from a uniform (0,1) distribution
cMean = mean(data, 2);  % mean of each sample
answer_a = sum(cMean > a & cMean < b)/nsamples;  % how many data means are between a and b
answer_b = mean(cMean);
answer_c = std(cMean, 1);
hist(cMean);
title('Generating 100 Samples of Size 225 Simulated U(-3,6) Random Numbers Histogram Question 1 Part d)');  %title of histogram  
xlabel('Z');  %labeling x-axis
ylabel('Frequency');  %labeling y-axis

Here are the specifications for the first program:

1. Generate 100 samples of size 225 random numbers from U(-3,6). For each of these 100 samples calculate the mean.

a) Find the simulated probability that the mean is between 1 and 2.

b) Find the mean of the means.

c) Find the standard deviation of the means.

d) Draw the histogram of the means.

Here are the specifications for the second program with few differences from the first program to a second from an Exponential Distribution with Lambda function:

2. Generate 400 samples of size 625 random numbers from an Exponential distribution with lambda = 0.2. For each of these 400 samples calculate the mean.

a) Find the simulated probability that the mean is between 3 and 5.

b) Find the simulated mean of the means.

c) Find the simulated standard deviation of the means.

d) Draw the histogram of the means.

Thank you, Happy Holidays!

Answer 1

[James Tursa]

I think your question is how to generate exponential(lambda) random numbers. If so, you can get them from uniform(0,1) numbers with the following:

T = -log(U)/lambda

Where U are the uniform numbers and T are the exponential numbers.

[HrMmz]

Okay Great, so I change the data value from

-3+9*rand...

to

 T = -log(U)/lambda*rand...

I think all I have to do now is changing the uniform distribution between 3 and 5 for part (a) and the samples and size numbers to 400 and 625 from 100 and 225 for (b) and (c) should be okay and the histogram as well for (d)!!! awesome thanks

[HrMmz]

oh and define lambda :)

[James Tursa]

Almost. E.g., this for data given the other inputs:

data = -log(rand(nsamples, n)) / lambda;

[HrMmz]

Oh wow.. I feel dumb sometimes that felt so basic thanks, good to be humbled.

So just

data=-log(rand(nsamples, n))/lambda

and then changing the other specs.. I hope I'm not missing something now that I know how easy that one was haha I'll have to post again when I change it later tonight