Create a vector z with 1000 elements generated by the rand-function. The generated values must have a mean value of 0.5 (mean(0.5)). And a standard deviation of sqrt(1/100*12), (std(x)).

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So my task is to determine mean of an random vector to be set to 0.5. And the set standard deviation to be sqrt(1/100*12)
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Adam
Adam 2018 年 9 月 7 日
And what have you done so far. I assume you have searched the Matlab help for functions such as 'rand' and followed the links from there to the other random number generation functions.
Rik
Rik 2023 年 6 月 28 日
What makes you think after almost 5 years that this question is unclear? Please provide an explanation in a comment.
I have removed your flag and will copy the question below.
Create a vector z with 1000 elements generated by the rand-function. The generated values must have a mean value of 0.5 (mean(0.5)). And a standard deviation of sqrt(1/100*12), (std(x)).
So my task is to determine mean of an random vector to be set to 0.5. And the set standard deviation to be sqrt(1/100*12)

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採用された回答

Cesar Antonio Lopez Segura
Cesar Antonio Lopez Segura 2018 年 9 月 7 日
Hi all
Here the code:
mu = 0.5;
sigma = 0.3464;
z = mu + randn(1,1000)*sigma
STDcalculated = std( z )
MuCalculated = mean( z )
  3 件のコメント
Vetle Normann
Vetle Normann 2018 年 9 月 7 日
Thank you! When I run loop on this script in Matlab STDcalculated and MuCalculated, the results are always close to 0.5 and 0.3463, but not the excact value. But is this as close as possible we get to 0.5 and 0.3464?
Cesar Antonio Lopez Segura
Cesar Antonio Lopez Segura 2018 年 9 月 7 日
Hi Vetle,
Using this values you can minimize your error:
mu = 0.50329141240781;
sigma = 0.347330944035507;

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その他の回答 (1 件)

Torsten
Torsten 2018 年 9 月 7 日
編集済み: Torsten 2018 年 9 月 7 日
0.5*(a+b) = 0.5
1/12*(b-a)^2 = 0.12
Solve for a and b.
This will give you the interval [a b] for which you must generate uniformly distributed random numbers.
Then you can use the rand-function as
z=a+(b-a)*rand(1000,1)
Best wishes
Torsten.

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