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Is it possible to have a dataset having a normal distribution with mean close to 0 and standard deviation less than 1?

for eg, my dataset have mean 0.0004 and Standard deviation of 0.0089. The reason for asking this, is that normally the standardized normal dist have mean 0 and standard dev of 1.

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    $\begingroup$ Just change the units of measurement. For instance, if these data are in kilometers, convert them to millimeters: the mean becomes $400$ and the standard deviation becomes $8900$. Although the mean is no longer "close to $0$" and the SD is definitely greater than $1$, your data haven't changed: we have only changed how they are described. Therefore there couldn't possibly be anything invalid about the original situation. $\endgroup$ – whuber Feb 16 '17 at 18:21
  • $\begingroup$ FYI to help you visualize standardization... changing the mean "shifts" the graft to the origin (addition or subtraction, applied first), changing the s.d. by scaling it squeezes or broadens it (multiplication or division, applied 2nd). $\endgroup$ – Brad Thomas Feb 17 '17 at 5:00
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Yes, it is perfectly possible. It is true that if you standardise it you get a mean of 0 and an sd of 1 but you have not standardised it so that does not apply to you. You can have any mean and any non-negative standard deviation.

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Simple example: mean 0 and std. dev. 0 corresponds to a variable whose value is 0 with probability 1. (This is referred to as the degenerate case.)

So, yes, this is perfectly possible and sensible.

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    $\begingroup$ Is it a normal distribution then? $\endgroup$ – Richard Hardy Feb 17 '17 at 8:40
  • $\begingroup$ @RichardHardy: Of course. Why wouldn't it be? $\endgroup$ – Mehrdad Feb 17 '17 at 8:43
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    $\begingroup$ I don't know if this can still be considered as normal, but it is a matter of definition. Probably you are right, I just wondered. $\endgroup$ – Richard Hardy Feb 17 '17 at 8:54
  • $\begingroup$ @RichardHardy: Zero is definitely nonnegative, so there's no problem there. I can see why it might trip someone up who's just learning about it, but I'm not actually aware of an accepted definition that doesn't allow it. $\endgroup$ – Mehrdad Feb 17 '17 at 8:56
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    $\begingroup$ A compromise might be to stress that this is a limiting or degenerate case. I wouldn't say that it was an easy example compared with those already given. $\endgroup$ – Nick Cox Feb 17 '17 at 9:04

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