Sounds as though you are confusing a couple of things, as some of the
responders to your message have suggested (though none has said it
explicitly). The idea of "area under a curve" applies to a continuous
curve, and thus to continuous distributions. It doesn't make sense for
discrete distributions, because a discrete distribution does not have a
curve for there to be an area under.
(I know, one isn't supposed to end sentences with prepositions,
but sometimes I agree with Winston Churchill, who is reported to have
said, "That is the sort of nonsense up with which I will not put.")
Anyway: for discrete distributions, the probability is not
continuously spread out over the range of the distribution, but is
concentrated at a finite number of points, each of which is said to have
a "point mass" probability P(x). As one respondent pointed out, these
probabilities must SUM to 1: in your case,
SUM(i = 1 to 6) of P(X_i) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1.
For a continuous distribution whose range is from a to b, the
corresponding total probability is represented by the integral
INTEGRAL(x = a to b) of f(x)dx = 1
where f(x) is the probability density function in question; graphically,
an integral corresponds to the area under the f(x) curve.
To graph a discrete distribution, choose a suitable scale for the
ordinate, in probability units, and plot each mass point AS a point
with coordinates (x, P(x)). If you like, put a vertical line under the
point, extending to the abscissa; but this line (analogous to a bar in a
histogram) has zero width. Something like this:
0.2_|
P(x) |
| o o o o o o
| | | | | | |
| | | | | | |
0.1_| | | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
0.0_|________|____|____|_________|____|____|_______
^ ^ ^ ^ ^
1 2 3 4 5
On Wed, 9 Aug 2000, Sheila King wrote (edited):
> I'm teaching a GE stat course, my first time teaching stat, and am
> having some points of confusion. Here is one of my questions:
> Suppose I have a probability distribution as follows:
>
> Sample space: 1.5, 2.0, 2.5, 3.5, 4.0, 4.5
>
> and each of these outcomes is equally likely. So, if my random variable
> is x, then P(x=1.5) = 1/6, P(x=2) = 1/6, and so on...
>
> To draw a probability distribution histogram, I wanted to make the bar
> for each outcome have a height of 1/6, but I became confused over this
> point:
> for x=2.0, the bar can only be one half unit wide, because of the
> neighboring outcomes 1.5 and 2.5 ...
Well, no. The "bar" has zero width, because P(x = 1.9) = 0,
P(x = 1.99) = 0, P(x = 1.999) = 0, ...; in fact, for c = 1.5 to 2.0,
exclusive, P(x = c) = 0. Only at the particular discrete values you've
specified in the sample space is P(x) NOT equal to zero.
> (until I had encountered this particular problem, I had always made the
> bars for each outcome a width of one unit wide, with a height equal to
> P(x=that outcome) and with the outcome value centered horizontally on
> the bar).
> But it seems to me, each of the bars should have an equal width.
>
> But if there are six bars (because of six outcomes) and each is only
> half a unit wide, and 1/6 of a unit tall, then the total area under the
> distribution is only 1/2 and not 1. This bothered me.
As remarked above, the concept of "area" for discrete
distributions is meaningless or nonsensical.
> But the solution manual shows the probability distribution histograms
> for this problem exactly as I have described above.
Yes, well, solutions manuals aren't infallible either.
> Shouldn't the total area under the distribution equal one?
Yes: for continuous distributions, for which one can properly
display an area.
-- DFB.
------------------------------------------------------------------------
Donald F. Burrill [EMAIL PROTECTED]
348 Hyde Hall, Plymouth State College, [EMAIL PROTECTED]
MSC #29, Plymouth, NH 03264 603-535-2597
184 Nashua Road, Bedford, NH 03110 603-471-7128
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