I would like to generate a set of random real numbers within certain rang (-n to +n) and have predefined mean and standard deviation
Please Help
I would like to generate a set of random real numbers within certain rang (-n to +n) and have predefined mean and standard deviation
Please Help
You mean truncated normal random deviates?
Entia non sunt multiplicanda sine necessitate
No , the mean not equal to standard deviation
Pardon ?
Sorry;
Please ignore my last reply
It is truncated normal random deviates
To be clear in my question I need to generate 2300 random number between -1.65 to 2.2 with mean 0.5 and slandered deviation 1.2 is it possible by excel
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---A---- --B--- -----------------------------C------------------------------ 1 Mean 0.5 B1: Input 2 SD 1.2 B2: Input 3 Lower -1.65 B3: Input 4 Upper 2.2 B4: Input 5 6 Prob(L) 0.0366 B6: =NORMDIST(B3, B1, B2, TRUE) 7 Prob(U) 0.9217 B7: =NORMDIST(B4, B1, B2, TRUE) 8 9 Deviates 1.935 B8 and down: =NORMINV($B$5 + RAND()*($B$6-$B$5), $B$1, $B$2) 10 1.320 11 -0.681 12 1.100 13 2.193 14 0.108
Thank you very much![]()
I have a problem
The calculated Mean and SD is not as the input values. Please see attachment
Last edited by ahmedhattab; 02-21-2013 at 06:31 AM.
If you truncate a normal distribution, of course the SD of the surviving values changes -- you chopped off the tails.
If you truncate it asymmetrically, of course the mean of the surviving values changes -- you chopped more off one side than the other. Quantitatively, the mean changes to
= mean + SD^2 * (NORMDIST(lower, mean, SD, FALSE) - NORMDIST(upper, mean, SD, FALSE)) / Z
where Z = NORMDIST(upper, mean, SD, TRUE) - NORMDIST(lower, mean, SD, TRUE)
If you had a bunch of random values that had a certain mean and standard deviation, but a truncated distribution, those would not be normal deviates.
Last edited by shg; 02-21-2013 at 03:00 PM.
Excellent. Thank you for your help
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