How to calculate mean and variance?
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SHRIVATHSA S.
on 19 Jan 2017
Answered: Nikos Lambadakis
on 1 Apr 2022
Greetings. I have generated a vector of 10,000 complex samples. I want to calculate the Mean and Variance of the samples. How can I do it without using the built-in functions _ mean()_ and * var()* ? and how do I do that in a loop? Please help.
Thank you
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Ana Soph
on 6 May 2020
How can convert to 1 minute to 10 minutes, a wind directio data ? please, i have a matrix fo 1440x31
Accepted Answer
Jorge Mario Guerra González
on 19 Jan 2017
Edited: Jorge Mario Guerra González
on 19 Jan 2017
Since you want to do it without using the functions, just do:
A=rand(1000,1); %your array
sum1=0;
for i=1:length(A)
sum1=sum1+A(i);
end
M=sum1/length(A); %the mean
sum2=0;
for i=1:length(A)
sum2=sum2+ (A(i)-M)^2;
end
V=sum2/length(A); %Varaince
4 Comments
Jorge Mario Guerra González
on 19 Jan 2017
Edited: Jorge Mario Guerra González
on 19 Jan 2017
@Stephen Cobeldick absolutely agree, that was just a fast script I wrote to show the proccedure. I edited the answer
More Answers (3)
Walter Roberson
on 19 Jan 2017
Mean is the average -- the sum divided by the number of entries.
Variance is the sum of the squares of (the values minus the mean), then take the square root and divided by the number of samples
You can vectorize the calculation using sum().
To use a for loop to calculate sums, initialize a running total to 0, and then each iteration of the loop, add the current value to the running total.
3 Comments
Jurgen
on 19 Dec 2017
This answer contradicts the variance formula on the MATLAB website (no square root). Oh and wikipedia disagrees. The MATLAB variance assumes a uniform probability distribution. A very important assumption.
John D'Errico
on 24 Jul 2020
@ Jurgen - you make no sense in some of what you said, although you are correct about the square root being incorrect in this answer. Variance has no square root in it.
A variance is something you can compute from the data, or for a population, but there is no assumption about the underlying distribution, nor is there any need to make such an assumption.
Olayemi Akinsanya
on 16 Sep 2017
Edited: Olayemi Akinsanya
on 16 Sep 2017
P=[10:2:70]
i wrote a script for variance, it gives me a value of 320. when i check with var(P), it gives me a value of 330.6667. can someone advice what i did wrong.
1 Comment
Walter Roberson
on 16 Sep 2017
You probably divided by N (number of items) instead of dividing by (N-1) (number of degrees of freedom). See https://stats.stackexchange.com/questions/100041/how-exactly-did-statisticians-agree-to-using-n-1-as-the-unbiased-estimator-for
Nikos Lambadakis
on 1 Apr 2022
A=rand(1000,1); %your array
sum((A(1:end)-sum(A(1:end))./length(A)).^2)/(length(A)-1) % the same result as
% var(A)
var(A) == sum((A(1:end)-sum(A(1:end))./length(A)).^2)/(length(A)-1)
ans =
logical
1
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