into the formula for
u,
and we get
Il
I
fL
= '"
x·
n.
pX(l
p)"X
~
x!(Ilx)!
x=O
Then, making use of the fact that
x
1
=
x!
(x 
I)!
•
•
96
Chapter 4 ProbabilityDistributions
EXAMPLE
Solution
Mean ofhypergeometric
distribution
EXAMPLE
and
n!
=
n(n 
I)!, and factoring out
nand p,
we obtain
/I
(nl)!
pxI(l_p)nx
fJ.
=
np
L
(x _ 1)1(n  x)!
x=l
th
ti
starts
with
x
=
1since the original summand iszerofor
x
:=
O.
where esumma ton·
.
If we now let
y
=
xI
and
m
=
n
1, we obtain
m
,
~
m.
pY(1
_
p)m
y
fJ.=npLJ
y!(my)!
y=o
and this last sum can easily be recognized
as tbat of all the terms of thebinomial
distribution with the parameters
m
and
p.
Hence, this sum equals 1 and Itfollows
that
u.
=
np.
Using
JL
=
lip
to findthe mean number of heads inthree tosses
Find the mean of the probability
distribution
of the number of heads obtainedin
3 flips of a balanced coin.
113
For a binomial distribution with
n
=
3 and
p
=
2'
we get
fJ.
=
3 .
2
=
2'
andthis
agrees with the result obtained on page 95.
•
The formula
fJ.
=
np
applies, of course, only to binomial distributions, butfor
other special distributions, we can, similarly, express the mean in termsoftheir
parameters. For instance, for the mean of the hypergeometric
distribution withthe
parameters
11,
a,
and
N,
we can write
a
fJ.=n·

N
In
Exercise 4.43, the reader will be asked
to
derive this formula by a method similar
to the one we used to derive the formula for the mean of a binomial distribution.
Usingthe formula for the mean of a hypergeometric distribution
With reference to the example on page 90 in which 5 of 20 car chargers were
defective, find the mean of the probability
distribution
of the number of defectives
in a sample of 10 randomly chosen for inspection.
Solution
Substituting
n
= 10,
a
= 5,
and
N
=
20 into the above formula for
u;
we get
5
"=10'=25
~
20'
In
other words, if we inspect
10
of the car chargers, we can expect
2.5
defectives,
where
expect
is to be interpreted in the sense, it represents
the longrun average
number of defectives if 10 chargers
are repeatedly
selected from 20 chargersof
which 5 are defective.
•
To study the second of the two properties of probability distributions mentioned
on page 94, their vanation, let us refer again to the two probability
distributionsof
Figure a.S. Forthe one where
n
=
4, there is a high probability of getting
valuesclcs
tothe mean, but for the one where
II
=
16, there is a high probability of getting valu~s
scattered over
considerable distances
away from the mean. Using this
property,
It
Varianceofprobability
distribution
Standarddeviation of
probabilitydistribution
EXAMPLE
Solution
Sec4.4 The Mean and the Variance of aProbability Distribution
97
may seem reasonable to measure thevariation ofaprobability distribution withthe
quantity
L
(x 
jh) .
f(x)
all
x
namely, theaverage amount bywhich thevalues oftherandom variable deviate from
the mean. Unfortunately,
.L
(x 
jh) .
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 Fall '09
 CHUNG
 Probability, Probability theory, Binomial distribution, Discrete probability distribution, Hypergeometric Distribution