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Oops. In rechecking my numbers I noted a typo. The average or mean height
is 6' 7", not 6' 8". Thus going one SD either side yields 77.5% of the
players, two SDs yields 98.3% and three SDs still yields 99.5%.

At 07:01 AM 04/30/2000 -0400, you wrote:
>Here's an update on the NBA stats...
>
>My original data entry was based on a tallying of NBA player heights. I
>didn't enter each height as an individual data point but instead entered
>the frequency counts. I think that's why Excel gave me a different answer
>for median the first time. In any event, I went back and entered all 410
>player heights as discrete data points and the median changed
>slightly. Here are the new stats, plus the standard deviations.
>
>Average or Mean 6' 8" (Actually, 79.1878 inches)
>
>Median 6' 8"
>
>Range 5' 3" to 7' 7"
>
>Mode 6' 9"
>
>SD(S) 3.805 (SD for the 410 as a sample)
>
>SD(P) 3.801 (SD for the 410 as the entire population)
>
>Going one SD either side picks up 77% of the players; going two SDs either
>side picks up 97% and three SDs either side picks up 99%.
>
>>On 29 Apr 00, at 13:07, Fred Nickols wrote:
>>
>> > Just for the heck of it, I went to the NBA player site and obtained the
>> > height of all 410 NBA players. I put them in a spreadsheet and took at
>> > look at some basic stats:
>> >
>> > Mode 6' 9"
>> > Average Height 6' 8"
>> > Median Height 6' 7"
>> >
>> > Range 5' 3" to 7' 7"
>> >
>> > The spreadsheet can be found at http://home.att.net/~nickols/nba.xls if
>> > anyone wants to do any additional crunching. A bar chart is included. It
>> > is skewed toward the 6' 9" side of things but you can also see the good
>> > old bell curve in there.
>>
>>Ok, actually you don't determine whether a distribution is normal
>>by looking at it you do it with numbers. So, let's assume that we
>>haven't ruled out the possibility it isn't normal.
>>
>>(here's a bonus question. Can you have a distribution with the
>>same median and mean and have it not be a normal distribution?
>>
>>Ok, if you answer that right you get to go to the next round of data
>>testing.
>>
>>Calculate the standard deviation. Then determine what percentage
>>of people in the group fall + or - one standard deviation to the
>>mean. Compare that with the characteristics of a normal
>>distribution.
>>
>>Do the same for two SD's and three SD's from the mean.
>>
>>For those of you who can't figure out what this is about, we are
>>actually going through what can be an excellent teaching tool if you
>>ever have to teach about data and normal distributions.
>>
>>If it conforms to those characteristics, we probably have a normal
>>distribution. If it doesn't, it ain't. We could also do other tests for
>>skewness.
>>
>>
>>Visit the work911.com supersite at http://www.work911.com
>>for work related articles, or to find almost anything including
>>book reviews and suggestions, discussion lists and more.
>
>Fred Nickols
>The Distance Consulting Company
>"Assistance at A Distance"
>http://home.att.net/~nickols/distance.htm
>nickols@worldnet.att.net
>(609) 490-0095

Fred Nickols
The Distance Consulting Company
"Assistance at A Distance"
http://home.att.net/~nickols/distance.htm
nickols@worldnet.att.net
(609) 490-0095

On 30 Apr 00, at 10:09, Fred Nickols wrote:

> Oops. In rechecking my numbers I noted a typo. The average or mean
> height is 6' 7", not 6' 8". Thus going one SD either side yields 77.5% of
> the players, two SDs yields 98.3% and three SDs still yields 99.5%.

Ok, so assuming all the numbers are correct, you have a
distribution that is typical of a reduced variation distribution as a
result of selection. A normal distribution would come in with 68% of
people within one SD. This particular distribution is quite different
from a normal one.

There are tests to measure skewness, but I don't recall how to
calculate them, and there are also tests for statistical signficance
of "skew".


Visit the work911.com supersite at http://www.work911.com
for work related articles, or to find almost anything including
book reviews and suggestions, discussion lists and more.

Well, I hate dragging this out but we're not quite done yet. I'm going to
take the raw data in to one of the statisticians at ETS and have my work
checked by a pro. I've been using the stat functions in Excel and I don't
know if I've used them properly or not. I'm particularly concerned about a
standard deviation of 3.80 (inches), which I simply rounded to four and
used that to go four intervals either side of the mean (said intervals
being in inches). Anyway, once a real statistician takes a look at the
data I'll get back to y'all with a qualified read on the data.

At 11:52 AM 04/30/2000 -0500, you wrote:
>On 30 Apr 00, at 10:09, Fred Nickols wrote:
>
> > Oops. In rechecking my numbers I noted a typo. The average or mean
> > height is 6' 7", not 6' 8". Thus going one SD either side yields 77.5% of
> > the players, two SDs yields 98.3% and three SDs still yields 99.5%.
>
>Ok, so assuming all the numbers are correct, you have a
>distribution that is typical of a reduced variation distribution as a
>result of selection. A normal distribution would come in with 68% of
>people within one SD. This particular distribution is quite different
>from a normal one.
>
>There are tests to measure skewness, but I don't recall how to
>calculate them, and there are also tests for statistical signficance
>of "skew".
>
>
>Visit the work911.com supersite at http://www.work911.com
>for work related articles, or to find almost anything including
>book reviews and suggestions, discussion lists and more.

Fred Nickols
The Distance Consulting Company
"Assistance at A Distance"
http://home.att.net/~nickols/distance.htm
nickols@worldnet.att.net
(609) 490-0095