Binomial, Poisson, and Hypergeometric Distributions12.7

Show that for a symmetric distribution (whose third central moment exists) the skewnessis zero.

(e) Find the skewness of the distribution with density f(x) = xe-z when x > 0 and f(x) = 0otherwise. Sketch f(x).

(f) Calculate the skewness of a few simple discrete distributions of your own choice.(g) Find a nonsymmetric discrete distribution with 3 possible values, mean 0, and skewness O.

22.7 Binomial, Poisson, andHypergeometric DistributionsThese 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 ofemployees favoring certain schedule changes, etc.), medicine (e.g., number of patientsrecovered 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. Ineach trial the event A has the same probability P(A) = p. Then in a trial, A will not occurwith probability q = I - p. In n trials the random variable that interests us is

X can assume the values 0, I,..', n, and we want to determine the correspondingprobabilities. Now X = x means that A occurs in x trials and in n - x it does not occur.

This may look as follows.

(1)

Here B = A C is the complement of A, meaning that A does not 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-)

Now (1) is just one order of arranging x A's and n - x B's. We now use Theorem 1(b)

in Sec. 22.4, which gives the number of permutations of n things (the n outcomes of then trials) consisting of 2 classes, class 1 containing the nl = x A's and class 2 containing

."-the n - nl = ~ - x B's. This number is

n! (n);1(" - x)i = x .

1079

x = Number of times A occurs in n trials.

BB...B.-~~

n -xtimes

AA"'A- -~xtimes

qq . . . q = pZq"-Z.- ~~

n-xtimes

pp" 'p- -~

.\: times

Data Analysis. Probability Theory1080

Accordingly, (I *) multiplied by this number gives the probability P(X = x) of X = x,that is, of obtaining precisely x A's in n trials. Hence X has the probability function

(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. Numericalvalues are given in Table AS in Appendix 5.

The mean of the binomial distribution is (see Team Project 14)

(3)

and the variance is (see Team

(4)

For the symmetric case of equal chance of success and failure (p = q = 1/2) this givesthe mean nl2, the variance nl4, and the probability function

(2.)

Binomial distributionCompute the probability of obtaini.MIe8t two "SU" in rolling a fair die 4 times.

Solution. p = P(A) = P("Six") = 116. q = 5/6." - 4. Answer:

EXAMPLE 1

P = /(2) + /(3) + /(4) - (:) (i)l(i)l+ (~ (i)8(i) + (:) (i).

',-.,.. ;... . .0.5

,

5s---

p.O.!Fig. 485.

Chap. 22

(x. - 0, 1. . . . .11)

~~~J14)

[~~~]

(x - 0. I, . . . ,It).

1 171= "6i (6 . 2.S + 4 . ~ + I) - ""iiii - 13.2411. ...

0 5%---

p.O.2 p=O.5 p.O.8 p.O.9

Probability function (2) of the binomial distributionfor n = S and various values of p

Sec. 22.7 Binomial, Poisson, and Hypergeometric Distributions 1081Sec. 22.1

Poisson DistributionThe discrete distribution with infinitely many possible values and probability function

EE"- (5) f(x) = t e-" (x = 0, I, . . .)x.

is called the Poisson distribution, named after S. D. Poisson (Sec. 16.S). Figure 486shows (5) for some values of IJ.. It can be proved that this distribution is obtained as alimiting case of the binomial distribution, if we let p - 0 and n - 00 so that the meanIJ. = np approaches a finite value. (For instance, IJ. = np may be kept constant.) ThePoisson distribution has the mean IJ. and the variance (see Team Project 14)

(6)

Figure 486 gives die impression dlat widl increasing mean die spread of die distributionincreases, dlereby illusb'ating formula (6), and that the distribution becomes more andmore (approximately) symmetric.

0.5

50 5 ~II = 0.5

Fig. 486.

Poisson distributionEXAMPLE 2-:' . J,f die probability of producing a defective screw is p - 0.01, what is die probability !bat a lot of 100 screws

'.,.;r. ;")I! ~tain m~ than 2 defectives?

Solution. The complementary event is Ac: Not mon ,hall 2 "«ti~s. For its probability we get from diebinomial distributioo with - " - np = 1 die Yal~ (lee (2»)

P(AC) - r~) 0.99100 + r~) 0.01 . 0.99- + r~ 0.011. 0.99-,

SirM:e P is very small. we can 8PIXUximate this by the much more convenient Poisson distribution with mean.~ z lip - 100' 0.01 - 1. obIaiDiDI [see (5)J

11101 P(A) = 8.03"'. Show that the binomial distribution gives P(A) c 7.94"',10 thII: the Poissoo

is quite Jood.

2 - II0" - roo

,,-1 ,,-2 ,,-sProbability function (5) of the Poisson distribution

for various values of JJ.

P(A°) - e-l (I + I + j) - 91."".

..

Data Analysis. Probability Theory1082

EXAMPLE 3 Parking problems. Poisson distributionIf on d)C average, 2 cars enter a cenain P8fking lot per minute. what is the probability that during any givenminute 4 or more cars will enter the lot?

Solution. To undentand that the Poisson distriOOti<lll is a m<xIel of d)C situation. we imagine the minute to bedivided into very many short time intervals, let p be the (constant) probability that a car will enter the lot duringany such short interval. and assume inck.peIIdeDce of d)C events that happen during thOle intervals. Then we ~dealing with a binomial distribution with Vely 1arge" and very amaII p. which we can approximate by thePoiuon distribution with #£ = lip = 2. Thus (be ~emeatary event "3 can or I~r ellUr 1M lot" baa (be

JXob8bi\ity (~ 21 ~ ~)1(0) + 1(1) + 1(2) + 1(3) = e-2 O! + "Ii" + 21 +""3"i" - 0.857. Answer: 14.3%. ...

Sampling with ReplacementThis means that we draw things from a given set one by one, and after each trial wereplace the thing drawn (put it back to the given set and mix) before we draw the nextthing. This guarantees independence of trials and leads to the binomial distribution.Indeed, if a box contains N things, for example, screws, M of which ~ defective, theprobability of drawing a defective screw in a trial is p = MIN. Hence the probability ofdrawing a nondefective screw is q = I - P = I - MIN, and (2) gives the probabilityof drawing x defectives in II trials in the fonn

(7)

Sampling Without Replacement. Hypergeometric DistributionSampling without replacement means that we return no screw to the box. Then we nolonger have independence of trials (why?), and instead of (7) the probability of drawingx defectives in n trials is

(8)

~ .. , /'

..~..,.. ~.,'The distribution with this probability function is called the

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 defectives from M.

Chap. 22 Sec.

(or - O. 1. . . . . 11).

(x - 0, I, . . . ,II).

6

can be expressed

Sec. 22.7

(N- (c) J<1ltterent ways or plcKmg n - x nofluelecuves rrom JV - IYI,

n-x

and each way in (b) combined with each way in (c) gives the total number of mutuallyexclusive ways of obtaining x defectives in n drawings without replacement. Since (a) isthe 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)

(9)

and the variance

(10)

Sampling with and without replacementWe 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 = Number of defectives in the sample.

Solution, We have N = 10, M = 3, N - M = 7, n = 2. For sampling with replacement, (7) yields

EXAMPLE 4

For sampling without replacement we have to use (8),

(3)( 7 )i ~(JO) 21 ~/(x). x 2 ~ % ~' 2' /(0) - /(1) - -.s- - 0.47. /(2) a ':4;" -0.07.

If N, M, and N - M are large compared with n, then it does not matter too much

whether we sample with or without replacement, and in this case the hypergeometricdistribution may be approximated by the binomial distribution (with p = M/N), which issomewhat simpler.

Hence in sampling from an indefinitely large population ("infinite population") wemay use the binomial distribution, regardless of whether we sample with or withoutreplacement.~. ,, . i

"4"."-. so: .... " . .

1. Five fair coins are tossed simultaneously. Find the probability function of the random variableX = Number of heads and compute the probabilities of 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 theprobability that the target will be hit at least once?

3. In Prob. 2, if the probability of hitting would be 5% and we fired 20 shots, would the probabilityof hitting at least once be less than, equal to, or greater than that in Prob. 2? Guess first.

;'.

Mp,=nN

j

;" [,

~j

:~1[';

nM(N - MXN - n)N2(N - 1)001=

(1\(..'[.).(2.)1-%

xl 1910 'f(x) - 1(0) = 0.49. /(1) - 0.42. /(2) = 0.09.

finding

...

I