```Date: Sun, 6 Jan 2008 22:19:59 -0500 Reply-To: Richard Ristow Sender: "SPSSX(r) Discussion" From: Richard Ristow Subject: Re: can SD be greater than AM? Comments: To: Pushpender Nath In-Reply-To: <4b72890a0801060241p5d18d7bem5b31dd2dfcd715ad@mail.gmail.co m> Content-Type: text/plain; charset="us-ascii"; format=flowed At 05:41 AM 1/6/2008, Pushpender Nath wrote: >I have a question in my mind. I want to know "Whether Standard >Deviation can be greater than Arithmetic Mean?" >I asked this to many fellows. Some say yes , some say no. The >fellows who say yes, could never give me any example. >I suppose, SD can never be greater than AM. Am I right? If not then >please give me some data example. It's very easy indeed. There's no required relationship whatever. Take any sample or population, subtract any constant a from all values, and if the old arithmetic mean (AM) was A, the new one is A-a, without changing the SD at all. You can make the AM as much smaller than the SD, or as much larger, as you like. Now, this can easily result in negative data values. Your advisors maybe are thinking of data with only positive values, for which it isn't quite as easy. But you can still get a SD much larger than the mean. Here's a log-normal distribution: SET RNG = MT /* 'Mersenne twister' random number generator */ . SET MTINDEX = 2069 /* Providence, RI telephone book */ . INPUT PROGRAM. . STRING City (A10). . NUMERIC Person (N3). . NUMERIC Income (DOLLAR10). . LEAVE City Person . . COMPUTE City = 'Barrington'. . LOOP Person = 1 TO 500. . COMPUTE Income = 25E3 * 2**RV.NORMAL(0,1.4825*SQRT(2.0)). . END CASE. . END LOOP. END FILE. END INPUT PROGRAM. DESCRIPTIVES VARIABLES=Income /STATISTICS=MEAN STDDEV MIN MAX . Descriptives |---------------------------|---------------------| |Output Created |06-JAN-2008 21:02:49 | |---------------|---|-------|----------|----------|--------------| | |N |Minimum|Maximum |Mean |Std. Deviation| |---------------|---|-------|----------|----------|--------------| |Income |500|\$322 |\$1,868,068|\$76,739.48|\$168,062.673 | |---------------|---|-------|----------|----------|--------------| But you needn't be that fancy; you can do it with a simple two-point distribution: If Z is x with probability p, y with probability q, where (1) p+q=1 then, skipping most of the algebra, (2) AM=px+qy (3) SD**2=VAR=pq(x-y)**2 If y=1, AM=2, (4) x-1=1/p and (5) VAR=(1-p)/p (see below). So, keeping AM=2, the variance and the SD can be made arbitrarily large, by decreasing p (the probability of choosing x) and increasing x correspondingly. ................................. Here's the algebra for (5), using y=1 plus (1) through (4): (3) Var=pq(x-y)**2 By (1), and because y=1, this gives (6) Var=p(1-p)(x-1)**2 Since (4) gives x-1=1/p, this becomes (5) Var=p(1-p)(1/p)**2=(1-p)/p QED ===================== To manage your subscription to SPSSX-L, send a message to LISTSERV@LISTSERV.UGA.EDU (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD ```

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