Solution :
Given the distribution :
we compute all the probabilities theoretically as shown below :
=========================================================================
Note we will use the integral :
=========================================================================
==================================================================================
simulation based computations :
%*********************
clear all;
clc;
close all;
len = 5000;
X1 = rand(1,len);
X2 = rand(1,len);
X = log(X1./X2);
hist(X,100);
p1 = sum( X>=(-0.5) & X<=(0.5))/len
p2 = sum( X>=(-1) & X<=(1))/len
p3 = sum( X>=(-1.5) & X<=(1.5))/len
p4 = sum( X>=(-2) & X<=(2))/len
p5 = sum( X>=(-2.5) & X<=(2.5))/len
p6 = sum( X>=(-3) & X<=(3))/len
p7 = sum( X>=(-3.5) & X<=(3.5))/len
p8 = sum( X>=(-4) & X<=(4))/len
%*******************
this gives output :
p1 = 0.38640
p2 = 0.62520
p3 = 0.77720
p4 = 0.86720
p5 = 0.91680
p6 = 0.95060
p7 = 0.97040
p8 = 0.98180
so we have :
==================================================================
similarly for gaussian distribution :
we compute all the probabilities theoretically as shown below :
=========================
Matlab based simulation :
clear all;
clc;
close all;
len = 5000;
X = randn(1,len);
hist(X,100);
p1 = sum( X>=(-0.5) & X<=(0.5))/len
p2 = sum( X>=(-1) & X<=(1))/len
p3 = sum( X>=(-1.5) & X<=(1.5))/len
p4 = sum( X>=(-2) & X<=(2))/len
p5 = sum( X>=(-2.5) & X<=(2.5))/len
p6 = sum( X>=(-3) & X<=(3))/len
p7 = sum( X>=(-3.5) & X<=(3.5))/len
p8 = sum( X>=(-4) & X<=(4))/len
%**************
p1 = 0.39180
p2 = 0.68400
p3 = 0.87000
p4 = 0.95380
p5 = 0.98820
p6 = 0.99760
p7 = 0.99980
p8 = 1
So we have :
info here is given to help solve #3 below this photo: You may use any computer...
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