how do i plot (histogram and normal plot) Uniform distribution for uniformly distributed height of a building between 10 to 200?

height of building uniformly distributed

Answer 1

[Image Analyst]

Sounds like homework so we can't just give you our code to hand in as your own. So look at this example from the help

% Generate a 10-by-1 column vector of uniformly distributed numbers in the interval (-5,5).
r = -5 + (5+5)*rand(10,1)

Adapt the interval from (-5, 5) to (10, 200) in the obvious way (make -5 10 and make the +5 to be 200). Then just pass r into the histogram() function.

[Sudeep Gyawali]

I am actually trying to plot the Rayleigh distribution in suburban environment, but first for that I am trying to plot a uniform distribution. I tried in a similar way as you have mentioned but without using histogram().

x = a+(b-a).*rand(1,N);%unifrnd(a,b) %uniformly distributed random variable
c = linspace(a,b,10); %linearly spaced 
count(size(c))=0; %count initialize
for i = 1:length(c)-1
    for j = 1:length(x) %random variable input
        if x(j)>=c(i) && x(j)<c(i+1)
            count(i) = count(i) + 1;
        end
    end
    y(i) = (c(i) + c(i+1))/2 ;
end
Z = count(1:end-1)/N;
subplot(211);
bar(y,Z);
subplot(212);
plot(y,Z);

[Image Analyst]

OK. Though that's a funny way to get the histogram (using a double for loop) even if you are doing it manually. You know that 'Rayleigh' is an option for the random() -- not rand() -- function, right?