20 value from a normal distribution..

joy
2015 5 7
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更新时间：2015 5 7
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Hi,
I have mu and sigma (stand deviation) i.e. mean=20 and sigma=4. I am generating data with normal distribution from these parameters.
I need to select 20 values ranges between 8 to 32 from this distribution..How to do so?
Thanks in advance
Mu = 20;
sigma = 4;
N = normrnd(Mu, sigma, 100000, 1);
N=round(N);
hist(N,0:1:60);

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Michael Haderlein
编辑：Michael Haderlein 2015-5-7

0 个投票

First, make sure that there are enough numbers available (with the values you use, this might not be a problem):
lo=8;hi=32;num=20;
while sum(N>=lo & N<=hi)<num
N(N<lo | N>hi)=round(normrnd(Mu,sigma,sum(N<lo | N>hi),1));
end
Then, take the first 20 values which are in the respective interval:
ind=find(N>=lo & N<hi,num,'first');
N(ind)

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joy
joy 2015-5-7
Hi, Thank you.
ind=find(N>=lo & N<hi,num,'first');
can you please, let me know the significance of 'first' word... if I need another set of 20 values from this distribution as well...where there shall be changes?
'first' means the first num indices are used. If you rather want the last num indices, use 'last' instead. If you want, let's say, 3x num values, you can set the respective parameter in the find function to 3*num (instead of only num). Then take N(1:num), N(num+1:2*num) and N(2*num+1:3*num).
joy
joy 2015-5-7
编辑：joy 2015-5-7
p1=20;
lo=8;hi=32;num=2*p1;
while sum(N>=lo & N<=hi)<num
N(N<lo | N>hi)=round(normrnd(Mu,sigma,sum(N<lo | N>hi),1));
end
ind=find(N>=lo & N<hi,num,'first');
q=round(N(ind));
q=q';
w1=q(1:num/2); %%%%%one set of 20 values
h1=q(num/2+1:end); %%%%%another set of 20 values
is it correct way? Did I get u correctly?
Looks good. You don't need to round again when you create q. Also, instead of num/2, you can use p1 again. When finding the indices, you have N<hi instead of N<=hi. The rest should be fine.
joy
joy 2015-5-7
Thank you. I shall make those corrections.

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