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B–7 Examples of Important Distributions 697
The combinations AkB
n which are also the binomial coefficients, can be evaluated by using
Pascal’s triangle:
For a particular value of n, the combinations Ak n B for k = 0, 1,..., n, are the elements in
the nth row. For example, for n = 3,
or
The mean value of the binomial distribution is evaluated by the use of Eq. (B–23):
m = xq =
n
a
n
x k P(x k ) = a kP(k)
k = 0
k=0
Using the identity
or
we have
n = 0
n = 1
n = 2
n = 3
n = 4
n = 5
o
ka n k b =
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
o
a 3 0 b = 1, a3 1 b = 3, a3 2 b = 3, and a3 3 b = 1
= nc
m =
n
a ka n
k = 1 k bpk q n-k
kn!
(n - k)!k!
(n - 1)!
((n - 1)(k - 1))!(k - 1)! d
ka n k b = nan - 1
k - 1 b
n
m = a na n - 1
k = 1
k - 1 b pk q n-k
Making a change in the index, let j = k - 1. Thus,
n - 1
m = a na n - 1 bp j+1 q n-(j+1)
j = 0
j
n - 1
= npB a a n - 1 bp j q (n-1)-j
j = 0
j
R = np C(p + q) n-1 D
=
n(n - 1)!
(n - k)!(k - 1)!
(B–36)
(B–37)
(B–38)
(B–39)