Download 22.7 Binomial, Poisson, and Hypergeometric Distributions

Binomial, Poisson,and HypergeometricDistributions
12.7
1079
Show that for a symmetricdistribution (whosethird central momentexists) the skewness
is zero.
(e) Find the skewnessof the distribution with density f(x) = xe-z when x > 0 and f(x) = 0
otherwise.Sketchf(x).
(f) Calculatethe skewnessof a few simple discretedistributionsof your own choice.
(g) Find a nonsymmetricdiscretedistribution with 3 possiblevalues,mean0, and skewnessO.
22.7 Binomial, Poisson, and
Hypergeometric Distributions
These are the three most important discrete distributions, with numerous applications.
Binomial Distribution
The binomial distribution occurs in games of chance (rolling a die, see below, etc.),
quality inspection (e.g., count of the number of defectives), opinion polls (number of
employees favoring certain schedule changes, etc.), medicine (e.g., number of patients
recovered by a new medication), and so on. The conditions of its occurrence are as follows.
We are interested in the number of times an event A occurs in n independent trials. In
each trial the event A has the same probability P(A) = p. Then in a trial, A will not occur
with probability q
=I
- p. In n trialstherandomvariablethatinterestsus is
x = Numberof timesA occurs in n trials.
X can assumethe values 0, I,..',
n, and we want to determine the corresponding
probabilities.Now X = x meansthat A occursin x trials and in n - x it doesnot occur.
This may look as follows.
(1)
Here B
~. .
BB...B.
AA"'A
- -~
-~~
xtimes
n -xtimes
= AC is the complementof A, meaningthat A doesnot occur. We now use the
assumption that the trials are independent. that is. they do not influence each other. Hence
(I) has the probability (see Sec. 22.3 on independent events)
--i-;". ~..- .
(i-)
pp"
-~ 'p
.\: times
qq
-
. . . q = pZq"-Z.
~~
n-xtimes
Now (1) is just one order of arrangingx A's and n - x B's. We now use Theorem 1(b)
in Sec.22.4, which gives the numberof permutationsof n things (the n outcomesof the
n trials) consistingof 2 classes,class 1 containingthe nl = x A's and class2 containing
."-then - nl = ~ - x B's. This numberis
()
;1("n!- x)i = nx .
1080
Chap. 22
Data Analysis.Probability Theory
Accordingly, (I *) multiplied by this number gives the probability P(X = x) of X = x,
that is, of obtaining preciselyx A's in n trials. HenceX hasthe probability function
(x.- 0, 1. . . . .11)
(1.)
and f(x)
= 0 otherwise. The distribution of X with probability function (2) is called the
binomial distribution or Bernoulli distribution. The occurrence of A is called success
(regardless of what it actually is; it may mean that you miss your plane or lose your watch)
and the nonoccurrence of A is called failure. Figure 485 shows typical examples. Numerical
values are given in Table AS in Appendix 5.
The mean of the binomial distribution is (see Team Project 14)
(3)
~~~J
and the varianceis (seeTeam
14)
[~~~]
(4)
For the symmetriccaseof equal chanceof successand failure (p
the meannl2, the variancenl4, and the probability function
= q = 1/2) this gives
(x- 0.I, . . . ,It).
(2.)
EXAMPLE1
Binomial distribution
Computethe probability of obtaini.MIe8t
P
= /(2)
two "SU" in rolling a fair die 4 times.
= 116.q = 5/6."-
Solution. p = P(A) = P("Six")
+ /(3) + /(4)
4. Answer:
-(:) (i)l(i)l+ (~ (i)8(i) +(:) (i).
1
171
= "6i (6 . 2.S+ 4 . ~ + I) - ""iiii - 13.2411.
',-.,.. ;... .
.
0.5
,
5
s---
0
%--p.O.2
5
p.O.!
p=O.5
p.O.8
p.O.9
Fig.485. Probability function (2) of the binomial distribution
for n
= S and various values of p
...
22.1
Sec.22.7
Binomial, Poisson,and HypergeometricDistributions
1081
Poisson Distribution
The discretedistribution with infinitely many possiblevaluesand probability function
EE"-
(5)
f(x)
=t
x.
(x = 0, I, . . .)
e-"
is called the Poisson distribution, named after S. D. Poisson(Sec. 16.S).Figure 486
shows(5) for some valuesof IJ..It can be proved that this distribution is obtained as a
limiting caseof the binomial distribution, if we let p - 0 and n - 00so that the mean
IJ. = np approachesa finite value. (For instance,IJ. = np may be kept constant.)The
Poissondistribution hasthe meanIJ.and the variance(seeTeam Project 14)
0"2 --
(6)
II
roo
Figure486 givesdie impressiondlat widl increasingmeandie spreadof die distribution
increases,dlereby illusb'ating formula (6), and that the distribution becomesmore and
more (approximately)symmetric.
0.5
II = 0.5
50,,-1 5~ ,,-2
Fig. 486.
,,-s
Probability function (5) of the Poissondistribution
for variousvaluesof JJ.
EXAMPLE2
-:'
.
Poisson distribution
J,f die probability of producing a defective screw is p
'.,.;r. ;")I! ~tain
- 0.01, what is die probability
!bat a lot of 100 screws
m~ than 2 defectives?
Solution. The complementaryevent is Ac: Not mon ,hall 2 "«ti~s.
binomial
distributioo
with- " - np = 1 die Yal~ (lee (2»)
P(AC)
-r~)
0.99100+
For its probability we get from die
r~) 0.01 .0.99- + r~ 0.011.0.99-,
SirM:eP is very small. we can 8PIXUximate this by the much more convenientPoissondistribution with mean
.~ z lip - 100' 0.01 - 1. obIaiDiDI[see(5)J
P(A°)
11101P(A)
= 8.03"'.
is quite Jood.
- e-l (I + I + j) - 91."".
Show that the binomial distribution gives P(A)
c
7.94"',10 thII: the Poissoo
..
1082
Data Analysis.Probability Theory
EXAMPLE 3
Chap. 22
Parking problems. Poisson distribution
If on d)Caverage,2 cars enter a cenain P8fking lot per minute. what is the probability that during any given
minute 4 or more cars will enter the lot?
Solution. To undentandthat the PoissondistriOOti<lllis a m<xIelof d)Csituation.we imaginethe minute to be
divided into very many shorttime intervals,let p be the (constant)probability that a car will enterthe lot during
any suchshort interval. and assumeinck.peIIdeDce
of d)Ceventsthat happenduring thOle intervals.Then we ~
dealing with a binomial distribution with Vely 1arge" and very amaIIp. which we can approximateby the
Poiuon distribution with #£ =
lip
= 2. Thus (be ~emeatary
JXob8bi\ity
1(0) + 1(1) + 1(2) + 1(3)
(~
21
event "3 can or I~r
~
~
ellUr 1M lot" baa (be
)
= e-2 O! + "Ii" + 21 +""3"i"- 0.857. Answer:
14.3%.
...
Sampling with Replacement
This meansthat we draw things from a given set one by one, and after each trial we
replacethe thing drawn (put it back to the given set and mix) before we draw the next
thing. This guaranteesindependenceof trials and leads to the binomial distribution.
Indeed,if a box containsN things, for example,screws,M of which ~ defective, the
probability of drawing a defectivescrewin a trial is p = MIN. Hencethe probability of
drawing a nondefectivescrew is q = I
P = I - MIN, and(2) givesthe probability
-
of drawing x defectivesin II trials in the fonn
(or- O. 1. . . . . 11).
(7)
Sampling Without Replacement. Hypergeometric Distribution
Sampling without replacement meansthat we return no screwto the box. Then we no
longer have independenceof trials (why?), and insteadof (7) the probability of drawing
x defectivesin n trials is
(x
(8)
~
..
- 0, I,
. . . ,II).
, /'
..~..,..~.,'
The distribution with this probability function is called the
6
Derivation of (8). By (4a) in Sec.22.4 there are
(a)
(~)
different ways of picking n things from N.
(~)
different ways of picking x defectivesfrom M.
'"
~)
can be expressed
Sec.
Sec. 22.7
(c)
(N-
J<1ltterentwaysor plcKmgn - x nofluelecuvesrrom JV-
IYI,
n-x
and each way in (b) combined with each way in (c) gives the total number of mutually
exclusive ways of obtaining x defectives in n drawings without replacement. Since (a) is
the total number of outcomes and we draw at random. each such way has the probability
I/(~).
From this, (8) follows.
...
The hypergeometric distribution has the mean (Team Project 14)
M
(9)
j
p,=nN
and the variance
(10)
EXAMPLE4
;"[,
nM(N- MXN- n)
001=
N2(N- 1)
Samplingwith and without replacement
We want to draw random santples of two gaskets from a box containing 10 gaskets, three of which are defective.
Find the probability function of the random variable X = Numberof defectivesin the sample.
Solution,
We have N = 10,M = 3, N - M = 7, n = 2. For samplingwith replacement,(7) yields
).(2.)
(xl1\ (..'[.
1910
1-%
f(x)-
'
1(0)
= 0.49. /(1) - 0.42. /(2) = 0.09.
For sampling without replacement we have to use (8), finding
/(x).
x
2
7
~
i
%
JO
2'
/(0)
- /(1)
-
21
~
-.s- 0.47. /(2) a ':4;" -0.07.
...
If N, M, and N - M are large comparedwith n, then it does not matter too much
whether we sample with or without replacement,and in this case the hypergeometric
distribution may be approximatedby the binomial distribution (with p = M/N), which is
somewhatsimpler.
Hence in samplingfrom an indefinitely large population ("infinite population") we
may use the binomial distribution, regardlessof whether we sample with or without
replacement.
~.
, . ,
i
"4"."-.
so: ..
..
(3)( )~'~( )
" . .
;'.
1. Five fair coins are tossedsimultaneously.Find the probability function of the randomvariable
X = Number of headsand computethe probabilitiesof obtaining no heads.precisely I head,
at least I head,not more than 4 heads.
2. If the probability of hitting a target is 25% and 4 shots are fired independently,what is the
probability that the target will be hit at leastonce?
3. In Prob.2, if the probability of hitting would be 5% andwe fired 20 shots,would the probability
of hitting at leastonce be less than, equal to, or greaterthan that in Prob. 2? Guessfirst.
~j
:~
1
[';
I