Finding the standard deviation of a matrix

I'm trying to find the standard deviation of a matrix, but I'm getting a different answer when I do it myself than what matlab gives me.

I know that to find the standard deviation you:

1.Take the mean of all the #s

2.Then for each of those #s you subtract the mean and then square it

3.Then find the mean for all those new #s and then take the square root of that #

Heres what I have for the code:

 a=[3 -2 1;4 0 5;1 2.2 -3]
 m=numel(a)
 sum=0;
 for i=1:m;
    sum=sum+a(i);
    avg=sum/m;
end
avg

 o=length(a)
 sd=0;
for i=1:o;
    sd=sd+(a(i)-avg)^2;
    sd=(sd/(o-1))^1/2;
end
sd

I got 2.49 for the standard deviation but matlab is giving me 0.5377. I have to use loops to do this I can't just use matlab commands for standard deviation. Any insight as to what I might be doing wrong is appreciated. Thank you.

Answer 1

[John D'Errico]

You are getting a different answer because you are not even doing what you said you wanted to do. Should you be surprised? Your code does some nasty stuff inside that loop that does not do what you apparently think you wanted to do.

By the way, a REALLY bad idea is to use the letter o for a variable name. One day you will end up with bugs in your code because you typed the number 0 instead of o someplace.

Anyway, there are two possible standard deviations to consider. A population standard deviation, or a sample standard deviation. The difference depends on whether you divide by n or n-1. What you described, taking a mean, implies a population standard deviation. But then you divided by o-1.

std(a(:))

does what you want, unless you really did want a population standard deviation. In that case, use

std(a(:),1)

Do you really feel the need to verify that MATLAB knows how to compute a standard deviation, so that you need to do it by hand to check?

std(a(:))
ans =
        2.64344051905425

sqrt(sum((a(:) - mean(a(:))).^2)/(numel(a)-1))
ans =
        2.64344051905425

Yes. You could do it using loops. But if you intend to learn MATLAB, then learn to use it as it should be used.

Answer 2

[Azzi Abdelmalek]

you can use std function

std(a(:))

If you want to use your for loop, you have to correct some errors

a=[3 -2 1;4 0 5;1 2.2 -3]
m=numel(a)
som=0;
for i=1:m;
  som=som+a(i);
end
  avg=som/m;
o=numel(a)
sd=0;
for i=1:o;
  sd=sd+(a(i)-avg)^2;
end
sd=(sd/(o-1))^(1/2)