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Transcript of # 6 Distribusi Probabilitas Diskrit
Chapter 5
Discrete Probability Distributions
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-1
Business Statistics: A Decision-Making Approach
8th Edition
Chapter Goals
After completing this chapter, you should be able to:
Calculate and interpret the expected value of a discrete
probability distribution
Apply the binomial distribution to business problems
Compute probabilities for the Poisson and
hypergeometric distributions
Recognize when to apply discrete probability
distributions to decision making situations
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-2
Introduction to Probability Distributions
Probability is the way decision makers express their
uncertainty about outcomes and events.
Random Variables A variable that takes on different numerical values based on chance.
Represents a possible numerical value from a random event.
Can vary from trial to trial.
Contoh : Inspektor memeriksa 3 TV plasma dan menentukan ‘diterima’ atau
‘tidak diterima’.
Outcome/hasil dari eksperimen yang mungkin adalah :
x = {0, 1, 2, 3} random variable
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-3
Introduction to Probability Distributions
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-4
Random
Variables
Discrete
Random Variable
Continuous
Random VariableCh. 5 Ch. 6
Discrete Random Variable
A discrete random variable is a variable that can assume only
a countable number of values
Many possible outcomes:
number of complaints per day
number of TV’s in a household
number of rings before the phone is answered
Only two possible outcomes:
gender: male or female
defective: yes or no
spreads peanut butter first vs. spreads jelly first
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-5
Contoh :
Kuis Statistik Industri 1 berupa soal pilihan ganda. Apabila x
merepresentasikan jumlah pertanyaan yang dijawab dengan
benar. Maka x adalah discrete random variable. Nilai yang
mungkin untuk x adalah :
x = {0, 1, 2, 3, 4, 5, 6}
Asuransi Prudential melakukan survey pada konsumennya untuk
menentukan jumlah anak berusia di bawah 22 tahun yang tinggal
di rumah.
Maka random variable adalah x yaitu jumlah anak yang tinggal di
rumah.
x = {0, 1, 2, 3, 4, ...} discrete random variable
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-6
Continuous Random Variable
A continuous random variable is a variable that can assume
any value on a continuum (can assume an uncountable
number of values)
thickness of an item
time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any value, depending only on
the ability to measure accurately.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-7
Discrete Random Variables
Can only assume a countable number of values
Examples:
Roll a die twice
Let x be the number of times 4 comes up
x = {0, 1, or 2 times}
Toss a coin 5 times
Let x be the number of heads
x = {0, 1, 2, 3, 4, or 5}
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-8
Discrete Probability Distribution
x Value Probability
0 1/4 = 0.25
1 2/4 = 0.50
2 1/4 = 0.25
5-9
Experiment: Toss 2 Coins. Let x = # heads.
T
T
4 possible outcomes
T
T
H
H
H H
Probability Distribution
0 1 2 x
0.50
0.25 P
rob
ab
ilit
y
Discrete Probability Distribution
A list of all possible [ xi , P(xi) ] pairs
xi = Value of Random Variable (Outcome)
P(xi) = Probability Associated with Value
xi’s are mutually exclusive (no overlap)
xi’s are collectively exhaustive (nothing left out)
0 ≤ P(xi) ≤ 1 for each xi
▪ The probability of xi is between 0 and 1
S P(xi) = 1
▪ The sum of all probabilities in the sample space = 1
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-10
Discrete Probability Distribution
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-11
3 possible outcomes 21 possible outcomes
The number of possible outcomes increase, the distribution besome smoother dan the individual probability of any particular
value tends to be reduce
Discrete Random Variable Mean
Expected Value (or mean) of a discrete distribution
(Weighted Average) the weight are probability
assigned ti the values.
The average value when the experiment that generates
values for the random varianle is repeated over the long
run. = E(x) = xi P(xi)
Where :E(x) = Expected value of xx = values of the random variableP(x) = probability of the randim variable taking on the value x
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-12
Discrete Random Variable Mean
Example: Toss 2 coins, x = # of heads,
compute expected value of x:
E(x) = (0 x 0.25) + (1 x 0.50) + (2 x 0.25) = 1.0
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-13
x P(x)
0 0.25
1 0.50
2 0.25
Discrete Random Variable Standard Deviation
Variace of a discrete random variable
Standard Deviation of a discrete distribution
where:E(x) = Expected value of the random variable (x)
x = Values of the random variable
P(x) = Probability of the random variable having the value of x
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-14
)P(xE(x)}{xσ i
2
ix
)P(XE(X)][Xσ i
2
i
2
Discrete Random Variable Standard Deviation
Example: Toss 2 coins, x = # heads,
compute standard deviation (recall E(x) = 1)
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-15
)P(xE(x)}{xσ i
2
ix
.7070.50(0.25)1)(2(0.50)1)(1(0.25)1)(0σ 222
x
(continued)
Possible number of heads = 0, 1, or 2
Contoh
Karena cuaca buruk, jumlah hari pada minggu depan untuk
pelayaran menjadi tidak tentu. Misalkan x adalah hari
pelayaran terjadi. Berikut ini adalah distribusi probabilitas
dari variabel, x ditentukan berdasarkan data historis saat
cuaca buruk :
Berdasarkan distribusi probabilitas, tentukan expected dari
hari per minggu yang memungkinkan untuk pelayaran?
Tentukan pula standar deviasinya!
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-16
X 0 1 2 3 4 5 6 7
P(x) 0,05 0,1 0,1 0,2 0,2 0,15 0,15 0,05
Expected value :
Standar deviasi :
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-17
x 0 1 2 3 4 5 6 7
P(x) 0,05 0,1 0,1 0,2 0,2 0,15 0,15 0,05 Sum
x.P(x) 0 0,1 0,2 0,6 0,8 0,75 0,9 0,35 3,7
Probability Distributions
5-18
Binomial*
DiscreteProbability
Distributions
Normal
Uniform
Exponential
Ch. 5 Ch. 6
Binomial Negative
Hypergeometric*
Poisson*
ContinuousProbability
Distributions
Multinomial
Geometric
The Binomial Distribution
A distribution that gives the probability of x successes in n
trial in a process that meets the following condition
(characteristics the Binomial Distribution) :
A trial has only two possible outcomes – “success” or “failure”
There is a fixed number, n (finite), of identical trials
The trials of the experiment are independent of each other
The probability of a success, p, remains constant from trial to
trial
If p represents the probability of a success, then
(1-p) = q is the probability of a failure
The sampling is performed with replacement from a finite
population.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-19
Binomial Distribution Examples
A manufacturing plant labels items as either defective or
acceptable
A firm bidding for a contract will either get the contract
or not
A marketing research firm receives survey responses of
“yes I will buy” or “no I will not”
New job applicants either accept the offer or reject it
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-20
Counting Rule for Combinations
A combination is an outcome of an experiment where x
objects are selected from a group of n objects.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-21
)!xn(!x
!nCn
x
where:
Cx = number of combinations of x objects selected from n objects
n! =n(n - 1)(n - 2) . . . (2)(1)
x! = x(x - 1)(x - 2) . . . (2)(1)
n
NOTE: 0! = 1 (by definition)
Order does not matter
i.e. ABC = CBA only one outcome
Contoh
Dilakukan pemeriksaan terhadap 4 kelompok sepatu untuk
menentukan kualitas “good” dan “defect”.
Maka kombinasi cara untuk mendapatkan kemungkinan
“defect” bersifat discrete dengan hasil sebagai berikut :
x = {0, 1, 2, 3, 4}
x = 0
x = 1
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-22
1)1234(1
1234
)!04(!0
!44
0
xxx
xxxC
4)123(1
1234
)!14(!1
!44
1
xx
xxxC
G, G, G, G
G, G, G, D
G, G, D, G
G, D, G, G
D, G, G, G
x = 2
x = 3
x = 4
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-23
4)1(123
1234
)!34(!3
!44
3
xx
xxxC
1)1(1234
1234
)!44(!4
!44
4
xxx
xxxC
6)12(12
1234
)!24(!2
!44
2
xx
xxxC
G, G, D, D G, D, G, D
G, D, D, G D, G, D, G
D, D, G, G D, G, G, D
G, D, D, D
D, G, D, D
D, D, G, D
D, D, D, G
D, D, D, D
Binomial Distribution Formula
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-24
P(x) = probability of x successes in n trials, with probability of success p on each trial
x = number of successes in sample, (x = 0, 1, 2, ..., n)
p = probability of “success” per trialq = probability of “failure” = (1 – p)n = number of trials (sample size)
P(x)n
x ! n xp q
x n x!
( )!=
-
-
Example: Flip a coin four times, let x = #
heads:n = 4
p = 0.51 - p = (1 - 0.5) = 0.5
x = 0, 1, 2, 3, 4
Dengan menggunakan persoalan pada pemeriksaan sepatu
diketahui n = 4, p = 0,1 dan x = 2 defects
Maka :
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-25
0486,0)9,0)(1,0(!2!2
!4)2(
)!(!
!)(
22
P
qpxnx
nxP xnx
Distribution
x = # of defects
P(x)
0 0,6561
1 0,2916
2 0,0486
3 0,0036
4 0,0001
Total 1,0000
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-26
Perhitungan dengan menggunakan rumus seringkali memerlukan waktu.
Untuk itu dapat digunakan tabel Binomial.
Tabel Binomial
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-27
Example: Calculating a Binomial Probability
Business Statistics, A First Course (4e) © 2006 Prentice-Hall, Inc. 5-28
What is the probability of one success in five observations if the probability of success is .1?
x = 1, n = 5, and p = 0.1
0.32805
.9)(5)(0.1)(0
0.1)(1(0.1)1)!(51!
5!
p)(1px)!(nx!
n!1)P(x
4
151
nx
x
Binomial Distribution Summary Measures
Mean
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-29
Variance and Standard Deviation
npE(x)μ
npqσ2
npqσ
Where n = sample size
p = probability of success
q = (1 – p) = probability of failure
Binomial Distribution
The shape of the binomial distribution depends on the
values of p and n
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-30
n = 5 p = 0.1
n = 5 p = 0.5
Mean
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
Here, n = 5 and p = 0.1
(Binomial distribution skewed. The skeweness will most pronoinced when n is small and p approaches 0 or 1)
Here, n = 5 and p = 0.5
(bell-shapes distribution)
Binomial Characteristics
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-31
n = 5 p = 0.1
n = 5 p = 0.5
Mean
0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
.2
.4
.6
0 1 2 3 4 5
X
P(X)
0
0.5(5)(0.1)npμ
0.6708
.1)(5)(0.1)(1npqσ
2.5(5)(0.5)npμ
1.118
.5)0(5)(0.5)(1npqσ
Examples
Binomial Distribution Example
Example: 35% of all voters support Proposition A. If a
random sample of 10 voters is polled, what is the probability
that exactly three of them support the proposition?
i.e., find P(x = 3) if n = 10 and p = 0.35 :
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-32
.25220(0.65)(0.35)3!7!
10!qp
x)!(nx!
n!3)P(x 73xnx
There is a 25.22% chance that exactly 3 out of the 10 voters will support Proposition A
n = 10
x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50
012345678910
0.19690.54430.82020.95000.99010.99860.99991.00001.00001.00001.0000
0.10740.37580.67780.87910.96720.99360.99910.99991.00001.00001.0000
0.05630.24400.52560.77590.92190.98030.99650.99961.00001.00001.0000
0.02820.14930.38280.64960.84970.95270.98940.99840.99991.00001.0000
0.01350.08600.26160.51380.75150.90510.97400.99520.99951.00001.0000
0.00600.04640.16730.38230.63310.83380.94520.98770.99830.99991.0000
0.00250.02330.09960.26600.50440.73840.89800.97260.99550.99971.0000
0.00100.01070.05470.17190.37700.62300.82810.94530.98930.99901.0000
Using Cumulative Binomial Distribution Tables
5-33Example: n = 10, p = 0.35, x = 3: P(x = 3, n =10, p = 0.35) = 0.5138
If a random sample of 10 voters is polled, what is the probability that three or fewer of them support the proposition?
Contoh
Manager PT Coca Cola Amatil mengirimkan 300 email kepada konsumen untuk mengukur kepuasan terhadap pelayanan tahunan. Selama ini tingkat pengembalian email (membalas email) adalah p = 0,11.
Sehingga expected value (mean) pengembalian email adalah : E(x) = n.p = 300 x 0,11 = 33
Pada tahun ini ternyata da 44 konsumen yang membalas email sebanyak 44 orang. Pengembalian ini melebihi 33 orang. Selanjutnya manager melakukan pengujian terhadap data tersebut : P(x≥44) = 1 – P(x≤43)
Tabel binomial memiliki keterbatasan untuk sample > 300, sehingga perhitungan dilakukan dengan menggunakan software excel ataupun minitab.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-34
Using EXCEL
1. Buka sheet kosong
2. Pilih Formulas
3. Klik pada fx (Function
wizard)
4. Pilih Statistical Category
5. Pilih BINOMDIST
function
6. Lengkapi informasi yang
diminta
7. True mengindikasikan
probabilitas kumulatif
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-35
P(x≤43) = 0,97P(x≥44) = 1 – 0,97 = 0,03Hanya 3% peluang dari 44 atau lebih konsumen
yang membalas email dari peluang 11%.
The Poisson Distribution
Difficult to count the number of failures, so diffivult to
know possible outcomes (successes + failures).
Characteristics of the Poisson Distribution:
The outcomes of interest are rare relative to the possible
outcomes
The average number of outcomes of interest per time or
space interval is
The number of outcomes of interest are random, and the
occurrence of one outcome does not influence the chances
of another outcome of interest
The probability that an outcome of interest occurs in a
given segment is the same for all segmentsCopyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-36
Contoh :
Mendeskripsikan suatu situasi :
Jumlah pasien yang datang pada unit IGD per jam.
Jumlah panggilan pada customer service Telkom setiap
30 menit.
Mendeskripsikan random variable :
Jumlah ‘dot’ pada setuap lembar kertas.
Jumlah kontaminan dalam segalon air
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-37
Poisson Distribution Summary Measures
Mean
λ and t must be in compatibe unit.the average number in t segment, is not necessarily the number we eill see if we observe the process for t segments.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-38
Variance and Standard Deviation
λtμ E(x) x
λtσ2
λtσ
where = number of successes in a segment of unit size
t = the size of the segment of interest
Graph of Poisson Probabilities
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-39
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P(x
)X
t =
0.50
0
1
2
3
4
5
6
7
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000P(x = 2) = 0.0758
Graphically:
= .05 and t = 100
Poisson Distribution Shape
The shape of the Poisson Distribution depends on the
parameters and t:
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-40
0.00
0.05
0.10
0.15
0.20
0.25
1 2 3 4 5 6 7 8 9 10 11 12
x
P(x
)
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0 1 2 3 4 5 6 7
x
P(x
)
t = 0.50 t = 3.0
Poisson Distribution Formula
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-41
where:
t = size of the segment of interest
x = number of successes in segment of interest
= expected number of successes in a segment of unit size
e = base of the natural logarithm system (2.71828...)
!
)()(
x
etxP
tx
Using Poisson Tables
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-42
Example: Find P(x = 2) if = 0.05 and t = 100
.075802!
e(0.50)
!
)()2(
0.502
x
etxP
tx
X
t
0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90
01234567
0.90480.09050.00450.00020.00000.00000.00000.0000
0.81870.16370.01640.00110.00010.00000.00000.0000
0.74080.22220.03330.00330.00030.00000.00000.0000
0.67030.26810.05360.00720.00070.00010.00000.0000
0.60650.30330.07580.01260.00160.00020.00000.0000
0.54880.32930.09880.01980.00300.00040.00000.0000
0.49660.34760.12170.02840.00500.00070.00010.0000
0.44930.35950.14380.03830.00770.00120.00020.0000
0.40660.36590.16470.04940.01110.00200.00030.0000
Contoh
Figur kedatangn dari konsumen Mc. D disajikan dlam
distribusi Poissson dengan λ = 16 konsume per jam.
Berdasarkan data didapatkan waktu pelayanan cukup
konstan pada 6 menit. Di restoran terdapat 3 kasir, sehingga
dalam 6 menit 3 konsumen dapat terlayani. Manager
bermaksud mengetahui probabilitas 1 atau 2 konsumen
tidak terlayani dalam 6 menit.
Convert 6 menit menjadi 0,1 jam sehingga t = 0,1. sehingga
λt = 16 (0,1) = 1,6 konsumen.
P(1 atau lebih konsumen menunggu) = P(4) + P(5) + P(6) + ...
Atau P(1 atau lebih konsumen menunggu) = 1 – P(x≤3)
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-43
P(x≤3) = 0,9212
Sehingga P(4 atau lebih konsumen) = 1 – 0,9212 = 0,0788
Manager bermaksud memenuhi target agar konsumen yang
menunggu tidak lebih dari 0,05 sebagai acuan untuk
penambahan kasir.
P(4 atau lebih konsumen) = 1 – P(x≤?) ≤0,05Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-44
Sehingga diperlukan P(x≤?) ≥0,95
P(4 atau lebih konsumen) = 1 – P(x≤4) = 1 – 0,9763 = 0,0237
Karena 0,0237 < 0,05 maka sebaiknya manager menambah
satu kasir.
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-45
The Hypergeometric Distribution
“n” trials in a sample taken from a finite population of
size N
Sample taken without replacement
Trials are dependent
The probability changes from trial to trial
Concerned with finding the probability of “x” successes
in the sample where there are “X” successes in the
population
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-46
Hypergeometric Distribution Formula
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-47
N
n
X
x
XN
xn
C
CCxP
)(.
Where
N = population size
X = number of successes in the population
n = sample size
x = number of successes in the sample
n – x = number of failures in the sample
(Two possible outcomes per trial: success or failure)
Hypergeometric Distribution Example
■ Example: 3 Light bulbs were selected from 10. Of the 10
there were 4 defective. What is the probability that 2 of
the 3 selected are defective?
N = 10 n = 3
X = 4 x = 2
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-48
0.3120
(6)(6)
C
CC
C
CC2)P(x
10
3
4
2
6
1
N
n
X
x
XN
xn
Contoh
Pada produksi Nokia bulan Juli didapatkan 20 produk dicurigai
defect karena kesalahan proses. Manager segera memerintahkan
untuk memisahkan 20 produk tersebut. Namun bagian QC
terlanjur meloloskan 10 produk ke bagian pengepakan. Peluang
defect dari 20 product adalah 2.
Maka peluang produk tanpa defect yang dikirim adalah (x = 0)?
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-49
0.2368184,756
)(43,758)(10)P(x
C
CC
C
CC0)P(x
20
10
2
0
2-20
0-10
N
n
X
x
XN
xn
Hypergeometric dustribution untuk semua defect yang
mungkin :
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-50
x P(x)
0 0,2368
1 0,5264
2 0,2368
Total 1,0000
Hypergeometric Distribution (k possible outcomes per trial)
N = Population size
n = Total sample size
Xi = Number of items in the
population with outcome i
xi = Number of items in the sample
with outcome i
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-51
N
n
X
x
X
x
X
x
kC
CCCxxxP
K
k...
),...,,(2
2
1
1
21
nx
NX
k
i
i
k
i
i
1
1
Contoh
Studi terhadapt pasar sepeda motor dengan 4 merek yang ada pada suatu dealer. Pada dealer tersebut
tersebut terdapat stok sebagai berikut : 5 brand Honda
4 brand Suzuki
6 brand Yamaha
4 brand Kawasaki
Sales manager melakukan penelitian pilihan konsumen tanpa melihat harga. Sebanyak 6 konsumen dipilih secara acak dan 3 memilih Suzuki, 2 memilih Kawasaki dan 1 memilih Yamaha. Tidak satupun memilih Honda.
Propabilitas pemilihan secara acak dengan asumsi
memilih tanpa pengembalian adalah :
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-52
Tentukan ukuran populasi dan sample :
N = 19 dan n = 6
Tentukan event of interest :
P(x1 = 0, x2 = 3, x3 = 1, x4 =2) = .....
Tentukan jumlah tiap kategori tiap populasi dan sample
:
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-53
Honda Suzuki Yamaha Kawasaki Total
X1 = 5 X2= 4 X3 = 6 X4 = 4 N = 19
x1 = 0 X2 = 3 x3 = 1 x4 = 2 n = 6
Hitung probabilitas
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-54
Chapter Summary
Reviewed key discrete distributions
Binomial
Poisson
Hypergeometric
Found probabilities using formulas and tables
Recognized when to apply different distributions
Applied distributions to decision problems
Copyright ©2011 Pearson Education, Inc. publishing as Prentice Hall 5-55