In article <[EMAIL PROTECTED]> in
sci.stat.edu, Tony <[EMAIL PROTECTED]> wrote:
>One of these
>statistics collected was how many weeks did it take to fill a vacancy.
>... The reply from
>her supervisor was "if the results are normally distributed then use
>the mean otherwise use the median". I am sure this is sage advice,
>but why? I would have thought that if the distribution was normally
>distributed then the mean and the median would be roughly similar
Not just roughly similar, but identical. The same is true for any
symmetric distribution, whether normal or not.
>since the normal curve has the frequency distributed around
>the mean.
I do not understand what this means.
> My further thinking about this was that for any
>distribution the mean will always be the upper bound of the median.
>Is this correct?
No.
When a distribution is skewed (lots of data bunched up at one end,
and a thin tail extending in the other direction), the mean and
median are different. The mean is "pulled" toward the tail. So for
instance with salaries at a given corporation, there are only a few
individuals with high salaries, and the great mass of employees have
low salaries. The distribution is skewed right, and the mean is
greater than the median.
When the distribution is skewed left, it works the opposite way: the
mean will be less than the median. I'm having trouble thinking of a
straightforward example of a distribution that is skewed left, but
an easy (if somewhat contrived) example would be "number of hours
work to earn $10,000". For most people the number would be high; for
a few it would be low. Here the mean is less than the median.
--
Stan Brown, Oak Road Systems, Cortland County, New York, USA
http://OakRoadSystems.com/
It's not necessary to send me a copy of anything you post
publicly, but if you do please identify it explicitly to avoid
confusion.
.
.
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