Hi, I have S=[2 4 4 4 5 5 7 9]; can any one tell me how can I count and plot standard deviation in this values?

2 Kommentare

Image Analyst
Image Analyst am 28 Sep. 2011
Sorry, no. Not until you supply more info on what you mean, like why you expect more than one standard deviation from that set of observations.
Firoz Alam
Firoz Alam am 29 Sep. 2011
or if the data set is s = [1:5; 6:10; 11:15], what will b the Std and std plot plz ? each row represent freq and column measurement valus... hope u undstd now

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Walter Roberson
Walter Roberson am 29 Sep. 2011

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numdev = floor((mean(S)-min(S))./std(S));
Determining the position of each standard deviation and plotting is left as a homework exercise.
Though possibly you want
numdev = floor((mean(S,2)-min(S,[],2))./std(S,[],2));

4 Kommentare

Firoz Alam
Firoz Alam am 29 Sep. 2011
@Walter - its generate one number only. hoe can I plot that? can u plz come more details ?
Walter Roberson
Walter Roberson am 29 Sep. 2011
The one number it generates is the number of standard deviations that "fit" in to the actual range of the data. There are always an equal number of standard deviations around the mean value. If S is one standard deviation,
mean - 3 * S, mean - 2 * S, mean - S, mean, mean + S, mean + 2 * S, mean + 3 * S
or as appropriate: the above would be for numdev of 3.
Firoz Alam
Firoz Alam am 30 Sep. 2011
thnx :)
Shivakumar
Shivakumar am 25 Sep. 2013
Thank you. I was helped by the answer.

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the cyclist
the cyclist am 29 Sep. 2011

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The standard deviation of the values in S can be calculated with the std() command:
>> std(S)
I'm not sure what you mean by plotting that standard deviation.

3 Kommentare

Firoz Alam
Firoz Alam am 29 Sep. 2011
Sir, I know this one but I need to plot ! :(
the cyclist
the cyclist am 29 Sep. 2011
Sorry, but I am very confused. The standard deviation is one number. What exactly do you want to plot? What are you trying to do, see, or understand about that set of numbers by plotting? Type "help plot" for help on one simple plotting command.
Firoz Alam
Firoz Alam am 30 Sep. 2011
its done ! thnx :)

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