Continuous Probability Distributions
1 of
5
Objectives: By the end of this
lesson, you should be able to do the following:
1.
State the properties of a
probability density function (continuous pmf).
2.
State the properties of a
continuous cdf.
3.
Calculate probabilities of events in continuous situations.
4.
Apply this general thinking to the special case of the
Normal Distribution Function
.
Review
We looked at
discrete probability mass functions
. Aside from a complicated notation, they really boil down to
addition and subtraction. Underlying all the calculations was the idea that probabilities of the discrete sort can
be related to an area calculation. The critical next step in going to a continuous situation is comparable to
moving from a Riemann Sum to a definite integral.
Example:
The function
for
creates the geometric sequence
.
1
( )
2
x
f
x
=
{
}
1,2,3,
x
∈
{
}
1 1 1
,
,
,
2 4 8
S
=
1.
This is a discrete, infinite sequence.
2.
You might recall that the sum of this infinite series has a limit of 1.
3.
Further
all
values of
x
are bounded:
.
0
( )
1
f
x
<
<
Let’s use it by saying that for the random variable
, the probability for any whole number
x
is given
X
S
∈
by
. Now we have a
pmf
!
1
(
)
2
x
f
X
x
=
=
We can easily answer questions like this set:
1.
Calculate the probability that
.
.
5
x
=
5
1
1
(
5)
32
2
f
X
=
=
=
2.
Calculate the probability that
x
is 2, 4, or 6.
2
4
6
1
1
1
21
(
{2,4,6})
64
2
2
2
f
X
∈
=
+
+
=
3.
Write the associated cumulative distribution function.
1
1
(
)
2
x
k
k
F X
x
=
=
=
4.
Calculate the probability that
. This equate to
. This might seem like it’ll be
10
X
≤
(
10)
F X
=
tough. However, the sum for the first
n
terms of finite geometric sequences is calculated by
.
(
)
(
)
0
1
1
n
n
a
r
S
r
−
=
−
In our case
.
(
)
(
)
(
)
(
)
1
1
2
2
1
2
1
2
1
(
)
1
1
x
x
x
F X
x
S
−
=
=
=
=
−
−
From there it’s short jump to
.
(
)
10
1
2
1023
(
10)
1
1024
F X
=
=
−
=
© 2010, Arizona State University, School of Mathematical and Statistical Sciences

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