Re: Standard deviation?

On May 16, 5:03 am, "saneman" <a...@xxxxxx> wrote:
"Chip Eastham" <hardm...@xxxxxxxxx> skrev i en meddelelsenews:bb0bbb0e-6f06-4fc8-9a1a-70d23c13144d@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx



On May 15, 8:42 pm, Virgil <Vir...@xxxxxxxxx> wrote:
In article <g0icma$mr...@xxxxxxxxxxxxxxxxx>, "saneman" <a...@xxxxxx>
wrote:

"Virgil" <Vir...@xxxxxxxxx> skrev i en meddelelse
news:Virgil-9EC47E.16001515052008@xxxxxxxxxxxxxxxxxxxxxxxxxxx
In article <g0i9oe$m3...@xxxxxxxxxxxxxxxxx>, "saneman" <a...@xxxxxx>
wrote:

I would like to calculate the standard deviation on this set of
data:

v = 0.5677 0.4792 0.4844 0.4870 0.5104 0.4870
0.4792
0.4974 0.4688 0.4870

But how do I accomplish this without any information about the
propabilities?

I know of no formula for the standard deviation of a set of values
that
involves an probabilities.

The standard deviation is the square root of the variance:

http://en.wikipedia.org/wiki/Variance

and the variance is formulated on the above link as the sum of the
products
of with probabilities and the deviation. Thats why I thought it would
be
necessary to now the probabilities too.

That is the variance of a probability distribution, not of a finite set
of values with no probabilities assigned to them.

It amounts to treating the finite set as having
equal probabilities summing to 1, i.e. uniform
distribution over the finite set.

regards, chip

Another thing. Assume that the standard deviation (std) is 0.5 for some
vector. If I want to conclude anything about the density of the data should
the median not be supplied too? If the median is 0.001 then the data is very
diverse but if its 0.45 then its very dense around the median.

It seems impossible to conclude anything using only that std = 0.5.

You seem to be assuming that all the values are
known to be nonnegative. Otherwise I don't see
how knowing the median affects the "density".

The standard deviation is often called "a measure
of central tendency", which I take to be the sort
of thing you mean by "density", i.e. how tightly
cluster around the "middle" are most observations.

Where that "middle" is is what the (arithmetic) mean
and (in a different way) the median are intended to
summarize. The standard deviation (or to use the
order statistics compatible with the median, say,
an upper and lower quartile) summarizes how tightly
packed around the middle most values are.

You are right that often the standard deviation
would be reported along with a mean, in order for
a rough picture of the data spread to be conveyed.
But one way to think about it is that all the
values can be translated by some constant, x_i + c,
which changes the mean (or median) by exactly +c
without changing the standard deviation at all.

So another point of view is that the utility
of the standard deviation as a measure of central
tendency stands on its own, at least in an abstact
sense of clustering around an unspecified middle.

regards, chip
.

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