Biomedical Engineering Reference
In-Depth Information
This definition implies that
N
n
2
P
n
-2
2
P
n
2
σ
=
µnP
n
+
µ
(E.6)
n
=
0
N
N
N
n
2
P
n
-2
µ
2
=
nP
n
+
µ
P
n
.
(E.7)
n
=
0
n
=
0
n
=
0
The first summation gives the expected value of
n
2
, the square of the number of
disintegrations. From Eq. (E.4) it follows that the second term is
-2
µ
2
.Thesumin
the last term is unity [Eq. (11.14)]. Thus, we can write in place of Eq. (E.7)
N
N
2
n
2
P
n
-2
2
+
2
n
2
P
n
-
2
.
σ
=
µ
µ
=
µ
(E.8)
n
=
0
n
=
0
We have previously evaluated
µ
[Eq. (E.4)]; it remains to find the sum involving
n
2
.
To this end, we differentiate both sides of Eq. (E.3) with respect to
x
:
N
N
(
N
-1)
p
2
(
px
+
q
)
N
-2
n
(
n
-1)
x
n
-2
P
n
.
(E.9)
=
n
=
0
Letting
x
= 1
with
p
+
q
= 1
, as before, implies that
N
N
(
N
-1)
p
2
=
n
(
n
-1)
P
n
(E.10)
n
=
0
N
N
N
n
2
P
n
-
n
2
P
n
-
µ
.
=
nP
n
=
(E.11)
n
=
0
n
=
0
n
=
0
Thus,
N
n
2
P
n
=
N
(
N
-1)
p
2
+
µ
.
(E.12)
n
=
0
Substituting this result into Eq. (E.8) and remembering that
µ
=
Np
,wefindthat
2
N
(
N
-1)
p
2
+
Np
-
N
2
p
2
σ
=
=
Np
(1 -
p
)
=
Npq
.
(E.13)
The standard deviation of the binomial distribution is therefore
σ
=
Npq
.
(E.14)
Poisson Distribution
As stated at the beginning of Section 11.5, we consider the binomial distribution
when
N
1
,
N
n
,and
p
1
. Under these conditions, the binomial coefficient
