Let's assume you intend to sum a series of observations, and then discuss the distributions associated with each individual observation, as well as the distribution of the sum.
THEN: The mean of the sum will equal the sum of the means: mu(total) = sum of(mu(i)), from i = 1 to n AND the variance of the sum will equal the sum of the variances of the observations. sigma squared (total) = sum of(sigma sq(i)) from i = 1 to n Note that the stdev dev(i) = square root(sigma squared(i)) Assuming that the individual observations are Normally distributed, you can complete the description of each distribution by giving the mean and stdev of each obs. does this help any? Jay abc wrote: > Hi All, > > First of All, I am not a Statistics Expert by myself, however, I would like > to raise another angle of this issue. > > If instead of divide (or partition) into two separate normal distribution, I > would > do this into n separate normal distribution where n is the number of points > constituting the original distribution N(50,5). > > What would happen then to the individual means and std ??? > > "gilgames" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > << > > Folks, > > > > I have one question. Please let me know if any of you know the answer to > > my question. > > > > I have a normal distribution with mean = 50 and standard deviation = 5 > > I want to divide this distribution into two separate normal > > distributions each with mean = 25 so that when I add them I will get my > > original distribution. > > > > Do any one know how to find the two separate distributions? > > > > Is the answer distribution 1 : N(25,3) and distribution 2: N(25,4) correct > ? > > >> > > > > If you have two independent normal random set N(25,3) and N(25,4) and > > add them together piecevise, you will get N(50, 5). In general for two > > independent set the mean is the sum of the components and also the > > variances are the sum of the variances of the components. > > > > If you reverse it and from a normal random set N(50,5) substract an > > independent N(25,3) normal random set piecevise, then you will get N(25, > > 5.831) being the new std = sqrt(std1**2 + std2**2) > > > > http://localhost/cgi-bin/igperl/igp.pl?dir=test&name=addeddistr > > > > > > Now the insteresting point: If you have a normal random set N(50,5) and > > make two distributions on that manner, that divide each number on the > > ration of the expected new means ($a = $orig * $mean1 / $meanorig and > > $b = $orig - $a) then the resulted standard deviation also will be > > divided by the ration of the expected new means, on that manner that the > > sum of the two standard deviations will be the original (if the given > > new mean is larger than the orginal, then that std will be larger also), > > so N(50,5) divided onto two equal half each element will result > > N(50,2.5), N(50,2.5) sets > > > > http://localhost/cgi-bin/igperl/igp.pl?dir=test&name=dividedistr > > > > . > . > ================================================================= > Instructions for joining and leaving this list, remarks about the > problem of INAPPROPRIATE MESSAGES, and archives are available at: > . http://jse.stat.ncsu.edu/ . > ================================================================= -- Jay Warner Principal Scientist Warner Consulting, Inc. 4444 North Green Bay Road Racine, WI 53404-1216 USA Ph: (262) 634-9100 FAX: (262) 681-1133 email: [EMAIL PROTECTED] web: http://www.a2q.com The A2Q Method (tm) -- What do you want to improve today? . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
