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)

2 Kommentare

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.
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)

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Hi all
Here the code:
mu = 0.5;
sigma = 0.3464;
z = mu + randn(1,1000)*sigma
STDcalculated = std( z )
MuCalculated = mean( z )

3 Kommentare

"rand"-function, not "randn"-function should be used.
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?
Hi Vetle,
Using this values you can minimize your error:
mu = 0.50329141240781;
sigma = 0.347330944035507;

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (1)

Torsten
Torsten am 7 Sep. 2018
Bearbeitet: Torsten am 7 Sep. 2018
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.

Kategorien

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by