Mean: mean Standard Deviation: std

Hi SAKSHAM, I understand that you want to visualize what we call the empirical rule in the statistical theory. I reviewed the attached code and noticed that that the plot was not intersecting the lines specified by μ ± 2σ and μ ± 3σ. I perceive it as an issue with the range of x data. To be more clear, consider the following observations x = data(:,1); min(x); % 425 max(x); % 740 Further, observe the range for μ ± 3σ mu - 3*sd % 308.4033 mu + 3*sd % 856.5967 It can be easily observed that, min(x) >= mu - 3*sd and max(x) <= mu + 3*sd. This is not desired because, to illustrate the empirical rule effectively, the range of x should extend beyond the range of μ ± 3σ. To achieve this, the x data can be tweaked by incorporating the mean and standard deviation from the original data. Here is a code snippet that demonstrates this Corrected_UV_D4_spectra_1 = xlsread('corrected_uvd4_spectra_1','sheet1','A26:B341'); data = Corrected_UV_D4_spectra_1; x = data(:,1); y = data(:,2); mu = mean(x); xline(mu,'g--') sd = std(x); x0 = (mu-4*sd):0.1:(mu+4*sd); % tweaking x data to adjust the range % pdf of the normal distribution with mean mu and standard deviation sigma pdf_values = normpdf(x0, mu, sd); plot(x0, pdf_values); % plot normal distribution xline(mu + sd,'m--') xline(mu - sd,'m--') xline(mu + 2*sd,'b:') xline(mu - 2*sd,'b:') xline(mu + 3*sd,'k-.') xline(mu - 3*sd,'k-.') Have a look at the documentation page for better understanding https://in.mathworks.com/help/stats/normpdf.html?s_tid=doc_ta I hope this helps.

std(test,[],3) The second argument refers to wheter you want the std to be corrected for the degrees of freedom or not. For more info: doc std

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.

A = [1 2 3;4 5 6;7 8 9] A = 1 2 3 4 5 6 7 8 9 >> std(A) ans = 3 3 3 It works just fine. std([1 4 7]) ans = 3 sqrt(6) ans = 2.4495 I think what you misunderstand is there are indeed two ways to compute the standard deviation of a set of numbers. (This gets into basic probability and statistics. Hey, you are using the standard deviation. A stats class might be useful.) The difference is that betyween the standard deviation of a complete population, and the standard deviation of a random sample form a distribution. The two get different formulas, for a good reason. If you want to compute the standard deviation of the complete population, then used std, with the second argument as 1. std([1 4 7],1) ans = 2.4495 When does the difference matter? The difference is seen when we try to compute the standard deviation of a complete population. So if we have just the 6 numbers on the 6 faces of a fair die, the standard deviation of the population is x = 1:6; >> sqrt(sum((x - mean(x)).^2)/6) ans = 1.7078 >> std(1:6,1) ans = 1.7078 As you can see, std does that, if we use 1 as the second argument. That causes std to use the formula that apparently you know, dividing by n inside the sqrt. However, when you compute the standard deviation of a random sample of numbers, if you divide by n, thus the number of data points, this will give you a biased estimator. (Again, basic prob and stats.) X = randn(10,1e7); >> mean(std(X,1)) ans = 0.92264 >> mean(std(X)) ans = 0.97255 So above, I generated 1e7 sets of 10 numbers, sampled from a Gaussian with a unit standard deviation. If we use the formula where we divide by n (thus std(X,1)) we get a biased estimate. It tends to be a little low. However, the usual formula for a statistical sample, where the mean itself is ALSO an estimate, yields a better estimate of the standard deviation. Again, basic prob & stats. But this is why there are TWO formulas for the standard deviation, and why you need to know which one to use and when.

You simply use std() again on the whole group (vector or matrix) of standard deviations that you have. stdOfGroup = std(individualStDevs);

The standard deviation of the standard deviation of the rows will not be the same as the standard deviation of the whole matrix. So no, you cannot do |std(std(x))|. The simplest way to calculate the standard deviation (and the mean) is to reshape the input into a vector. The normal way this is done: %mean and standard deviation of n-dimensional x: mean(x(:)) %x(:) reshape into a column vector std(x(:)) However, you cannot chain the (:) indexing with your indexing, so you can either invoke reshape explicity: mean(reshape(data.emg1(j,1:129), 1, [])) std(reshape(data.emg1(j,1:129), 1, [])) Or use a helper function: flatten =@(x) x(:); mean(flatten(data.emg1(j,1:129))) std(flatten(data.emg1(j,1:129))) Also note that the |find| is totally unnecessary and only slows down your code. If you do: j = data.DotsDirection == -1 & data.CorrectTarget(:) == 1; %j is now a logical array mean(reshape(data.emg1(j, 1:129), 1, [])) std(reshape(data.emg1(j, 1:129), 1, [])) it will work just as well. See the .