any logic to do this programming on random number
古いコメントを表示
Hi,
I have generated 1000 random numbers from a normal distribution with std deviation=0.10 and mean=0.36
like
r=0.36+0.10*randn(1,1000)
now, I want to select 10 numbers from this random numbers, the summation of which will be in between 3.55 to 3.65..
I need 30 sets of such 10 numbers which sum will lie between this..3.55 to 3.65
any logic to do so
thanks
2 件のコメント
Geoff Hayes
2014 年 6 月 11 日
編集済み: Geoff Hayes
2014 年 6 月 11 日
Note that the above equation is not quite correct (or your standard deviation and mean are reversed). If std is 0.1 and the mean is 0.36, then
r = 0.36+0.1*randn(1,1000);
Try std(r) and mean(r) to verify this.
joy
2014 年 6 月 11 日
採用された回答
その他の回答 (1 件)
Sean de Wolski
2014 年 6 月 11 日
編集済み: Sean de Wolski
2014 年 6 月 11 日
I would take a different approach. First generate the 30 numbers who you must sum to:
xsum = rand(1,30)+3.55;
Example:
%%Desired sums
xsum = rand(1,30)+3.55;
%%build random numbers
rn = zeros(10,30);
for ii = 1:30
rn(:,ii) = randfixedsum(10,1,xsum(ii),0,xsum(ii));
end
%%verify
assert(norm(sum(rn)-xsum)<10^-14)
3 件のコメント
John D'Errico
2014 年 6 月 11 日
編集済み: John D'Errico
2014 年 6 月 11 日
A flaw here is the numbers won't be at all normally distributed. Given the bound constraint on the mean, they won't be normally distributed anyway, but this gives up on any semblance of normality.
Sean de Wolski
2014 年 6 月 11 日
Ahh! Good catch.
Sean de Wolski
2014 年 6 月 11 日
I'll have to think about it a little more.
カテゴリ
ヘルプ センター および File Exchange で Resampling Techniques についてさらに検索
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!