Applied Business Statistics ,ken black , ch 5

Copyright 2010 John Wiley & Sons, Inc. 1
Copyright 2010 John Wiley & Sons, Inc.
Business Statistics, 6th ed.
by Ken Black
Chapter 5
Discrete
Probability
Distributions

Learning Objectives
Distinguish between discrete random variables and
continuous random variables.
Know how to determine the mean and variance of a
discrete distribution.
Identify the type of statistical experiments that can
be described by the binomial distribution, and know
how to work such problems.

Discrete vs. Continuous Distributions
Discrete distributions – constructed from discrete
(individually distinct) random variables
Continuous distributions – based on continuous
random variables
Random Variable - a variable which contains the
outcomes of a chance experiment

Discrete vs. Continuous Distributions
Categories of Random Variables
Discrete Random Variable - the set of all possible values is
at most a finite or a countable infinite number of possible
values
Continuous Random Variable - takes on values at every
point over a given interval

Describing a Discrete Distribution
A discrete distribution can be described by
constructing a graph of the distribution
Measures of central tendency and variability can be
applied to discrete distributions
Discrete values of outcomes are used to represent
themselves

Mean of discrete distribution – is the long run
average
If the process is repeated long enough, the average of the
outcomes will approach the long run average (mean)
Requires the process to eventually have a number which is
the product of many processes
Mean of a discrete distribution
µ = ∑ (X * P(X))
where (X) is the long run average;
X = outcome, P = Probability of X

Variance and Standard Deviation of a discrete
distribution are solved by using the outcomes (X) and
probabilities of outcomes (P(X)) in a manner similar
to computing a mean
Standard Deviation is computed by taking the square
root of the variance

Some Special Distributions
Discrete
binomial
Poisson
Hypergeometric
Continuous
normal
uniform
exponential
t
chi-square
F

Discrete Distribution -- Example
Observe the discrete distribution in the following
table.
An executive is considering out-of-town business
travel for a given Friday. At least one crisis could occur
on the day that the executive is gone. The distribution
contains the number of crises that could occur during
the day the executive is gone and the probability that
each number will occur. For example, there is a .37
probability that no crisis will occur, a .31 probability of
one crisis, and so on.

0
1
2
3
4
5
0.37
0.31
0.18
0.09
0.04
0.01
Number of
Crises
Probability
Distribution of Daily Crises
0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5
P
r
o
b
a
b
i
l
i
t
y
Number of Crises
Discrete Distribution -- Example

( ) 2.1)(
22
==  − XPX   = = 
2
12 110. .
Variance and Standard Deviation
of a Discrete Distribution
X
-1
0
1
2
3
P(X)
.1
.2
.4
.2
.1
-2
-1
0
1
2
X − 
4
1
0
1
4
.4
.2
.0
.2
.4
1.2
)(
2
−X
2
( ) ( )X P X− 

Requirements for a Discrete
Probability Function -- Examples
X P(X)
-1
0
1
2
3
.1
.2
.4
.2
.1
1.0
X P(X)
-1
0
1
2
3
-.1
.3
.4
.3
.1
1.0
X P(X)
-1
0
1
2
3
.1
.3
.4
.3
.1
1.2
: YES NO NO

Mean of a Discrete Distribution
( ) = = E X X P X( )
X
-1
0
1
2
3
P(X)
.1
.2
.4
.2
.1
-.1
.0
.4
.4
.3
1.0
X P X ( )
 = 1.0

( ) = =  =E X X P X( ) .115
Mean of the Crises Data Example
X P(X) X•P(X)
0 .37 .00
1 .31 .31
2 .18 .36
3 .09 .27
4 .04 .16
5 .01 .05
1.15
0
0.1
0.2
0.3
0.4
0.5
0 1 2 3 4 5
P
r
o
b
a
b
i
l
i
t
y
Number of Crises

( )2
2
141 =  =− X P X( ) .  = = =
2
141 119. .
Variance and Standard Deviation
of Crises Data Example
X P(X) (X-) (X-)2 • P(X)
0 .37 -1.15 1.32 .49
1 .31 -0.15 0.02 .01
2 .18 0.85 0.72 .13
3 .09 1.85 3.42 .31
4 .04 2.85 8.12 .32
5 .01 3.85 14.82 .15
1.41
(X-)2

Binomial Distribution
( )
nX
XnX
n
XP qp
XnX

−
=
−
0for
!!
!
)(
 = n p
2
2

 
=  
= =  
n p q
n p q
Probability
function
Mean value
Variance
and
Standard
Deviation

According to the U.S. Census Bureau,
approximately 6% of all workers in Jackson,
Mississippi, are unemployed. In conducting a
random telephone survey in Jackson, what is the
probability of getting two or fewer unemployed
workers in a sample of 20?
Binomial Distribution:
Demonstration Problem 5.3

In the following example,
6% are unemployed => p
The sample size is 20 => n
94% are employed => q
x is the number of successes desired
What is the probability of getting 2 or fewer unemployed
workers in the sample of 20?
The hard part of this problem is identifying p, n, and x –
emphasis this when studying the problems.

n
p
q
P X P X P X P X
=
=
=
 = = + = + =
= + + =
20
06
94
2 0 1 2
2901 3703 2246 8850
.
.
( ) ( ) ( ) ( )
. . . .
( )( ) 2901.)2901)(.1)(1(
)!020(!0
!20
)0( 94.06.
0200
==
−
==
−
XP
( ) ( )P X( )
!( )!
( )(. )(. ) .. .= =
−
= =
−
1
20!
1 20 1
20 06 3086 3703
1 20 1
06 94
( ) ( )P X( )
!( )!
( )(. )(. ) .. .= =
−
= =
−
2
20!
2 20 2
190 0036 3283 2246
2 20 2
06 94

Binomial Distribution Table:
n
p
q
P X P X P X P X
=
=
=
 = = + = + =
= + + =
20
06
94
2 0 1 2
2901 3703 2246 8850
.
.
( ) ( ) ( ) ( )
. . . .
P X P X( ) ( ) . . = −  = − =2 1 2 1 8850 1150
 =  = =n p ( )(. ) .20 06 1 20
2
2
20 06 94 1 128
1 128 1 062

 
=   = =
= = =
n p q ( )(. )(. ) .
. .
n = 20 PROBABILITY
X 0.05 0.06 0.07
0 0.3585 0.2901 0.2342
1 0.3774 0.3703 0.3526
2 0.1887 0.2246 0.2521
3 0.0596 0.0860 0.1139
4 0.0133 0.0233 0.0364
5 0.0022 0.0048 0.0088
6 0.0003 0.0008 0.0017
7 0.0000 0.0001 0.0002
8 0.0000 0.0000 0.0000
… … …
20 0.0000 0.0000 0.0000
…

Excel’s Binomial Function
n = 20
p = 0.06
X P(X)
0 =BINOMDIST(A5,B$1,B$2,FALSE)

X P(X =x)
0 0.000000
1 0.000000
2 0.000000
3 0.000001
4 0.000006
5 0.000037
6 0.000199
7 0.000858
8 0.003051
9 0.009040
10 0.022500
11 0.047273
12 0.084041
13 0.126420
14 0.160533
15 0.171236
16 0.152209
17 0.111421
18 0.066027
19 0.030890
20 0.010983
21 0.002789
22 0.000451
23 0.000035
Binomial with n = 23 and p = 0.64
Minitab’s Binomial Function

Mean and Std Dev of Binomial Distribution
Binomial distribution has an expected value or a long
run average denoted by µ (mu)
If n items are sampled over and over for a long time and if
p is the probability of success in one trial, the average long
run of successes per sample is expected to be np
=> Mean µ = np
=> Std Dev = √(npq)

Poisson Distribution
The Poisson distribution focuses only on the number
of discrete occurrences over some interval or
continuum
Poisson does not have a given number of trials (n)
as a binomial experiment does
Occurrences are independent of other occurrences
Occurrences occur over an interval

If Poisson distribution is studied over a long period
of time, a long run average can be determined
The average is denoted by lambda (λ)
Each Poisson problem contains a lambda value from which
the probabilities are determined
A Poisson distribution can be described by λ alone

)logarithmsnaturalofbase(the...718282.2
:
,...3,2,1,0for
!
)(
=
−=
==
−
e
averagerunlong
where
X
X
XP e
X



Probability function

Mean value

Standard deviationVariance


Poisson Distribution:
Bank customers arrive randomly on weekday
afternoons at an average of 3.2 customers
every 4 minutes. What is the probability of
having more than 7 customers in a 4-minute
interval on a weekday afternoon?

Solution
λ = 3.2 customers>minutes X > 7 customers/4 minutes
The solution requires obtaining the values of x = 8, 9, 10, 11, 12,
13, 14, . . . . Each x value is determined until the values are so
far away from λ = 3.2 that the probabilities approach zero. The
exact probabilities are summed to find x 7. If the bank has been
averaging 3.2 customers every 4 minutes on weekday
afternoons, it is unlikely that more than 7 people would
randomly arrive in any one 4-minute period. This answer
indicates that more than 7 people would randomly arrive in a
4-minute period only 1.69% of the time. Bank officers could use
these results to help them make staffing decisions.

0528.0
!10
=)10=(
!
=P(X)
minutes8customers/4.6=
Adjusted
minutes8customers/10=X
minutes4customers/2.3
4.610
X
6.4 =
=
−
−
e
e
XP
X




 




=
=
−
−
3 2
6 4
6
6
0 1586
6 4
.
!
!
.
.
customers/ 4 minutes
X = 6 customers/ 8 minutes
Adjusted
= . customers/ 8 minutes
P(X) =
( = ) =
X
6
6.4
e
e
X
P X

0060.0000.0002.0011.0047.
)9()8()7()6()5(
6.1
=+++=
=+=+=+==
=
XPXPXPXPXP

Using the Poisson Tables

X 0.5 1.5 1.6 3.0
0 0.6065 0.2231 0.2019 0.0498
1 0.3033 0.3347 0.3230 0.1494
2 0.0758 0.2510 0.2584 0.2240
3 0.0126 0.1255 0.1378 0.2240
4 0.0016 0.0471 0.0551 0.1680
5 0.0002 0.0141 0.0176 0.1008
6 0.0000 0.0035 0.0047 0.0504
7 0.0000 0.0008 0.0011 0.0216
8 0.0000 0.0001 0.0002 0.0081
9 0.0000 0.0000 0.0000 0.0027
10 0.0000 0.0000 0.0000 0.0008
11 0.0000 0.0000 0.0000 0.0002
12 0.0000 0.0000 0.0000 0.0001

4751.3230.2019.1
)1()0(1)2(1)2(
6.1
=−−=
=−=−=−=
=
XPXPXPXP

Using the Poisson Tables

X 0.5 1.5 1.6 3.0
0 0.6065 0.2231 0.2019 0.0498
1 0.3033 0.3347 0.3230 0.1494
2 0.0758 0.2510 0.2584 0.2240
3 0.0126 0.1255 0.1378 0.2240
4 0.0016 0.0471 0.0551 0.1680
5 0.0002 0.0141 0.0176 0.1008
6 0.0000 0.0035 0.0047 0.0504
7 0.0000 0.0008 0.0011 0.0216
8 0.0000 0.0001 0.0002 0.0081
9 0.0000 0.0000 0.0000 0.0027
10 0.0000 0.0000 0.0000 0.0008
11 0.0000 0.0000 0.0000 0.0002
12 0.0000 0.0000 0.0000 0.0001

Excel’s Poisson Function
= 1.6
X P(X)
0 =POISSON(D5,E$1,FALSE)

Minitab’s Poisson Function
X P(X =x)
0 0.149569
1 0.284180
2 0.269971
3 0.170982
4 0.081216
5 0.030862
6 0.009773
7 0.002653
8 0.000630
9 0.000133
10 0.000025
Poisson with mean = 1.9

Mean and Std Dev of a Poisson Distribution
Mean of a Poisson Distribution is λ
Understanding the mean of a Poisson distribution
gives a feel for the actual occurrences that are likely
to happen
Variance of a Poisson distribution is also λ
Std Dev = Square root of λ

Poisson Approximation
of the Binomial Distribution
Binomial problems with large sample sizes and small
values of p, which then generate rare events, are
potential candidates for use of the Poisson
Distribution
Rule of thumb, if n > 20 and np < 7, the
approximation is close enough to use the Poisson
distribution for binomial problems

Procedure for Approximating binomial with Poisson
Begin with the computation of the binomial mean
distribution µ = np
Because µ is the expected value of the binomial, it becomes
λ for Poisson distribution
Use µ as the λ, and using the x from the binomial problem
allows for the approximation of the probabilities from the
Poisson table or Poisson formula

If and the approximation is acceptable.n n p  20 7,
Use  = n p.
Binomial probabilities are difficult to calculate when
n is large.
Under certain conditions binomial probabilities may
be approximated by Poisson probabilities.
Poisson approximation

Hypergeometric Distribution
Sampling without replacement from a finite
population
The number of objects in the population is denoted N.
Each trial has exactly two possible outcomes, success
and failure.
Trials are not independent
X is the number of successes in the n trials
The binomial is an acceptable approximation,
if n < 5% N. Otherwise it is not.

Hypergeometric Distribution
 =
A n
N
2
2
2
1

 
=
− −
−
=
A N A n N n
NN
( ) ( )
( )
( )( )
P x
C C
C
A x N A n x
N n
( ) =
− −
Probability function
N is population size
n is sample size
A is number of successes in population
x is number of successes in sample
Mean
Value
Variance and standard
deviation

N = 24
X = 8
n = 5
x
0 0.1028
1 0.3426
2 0.3689
3 0.1581
4 0.0264
5 0.0013
P(x)
( )( )
( )( )
( )( )
P x
C C
C
C C
C
A x N A n x
N n
( )
,
.
= =
=
=
=
− −
− −
3
56 120
42 504
1581
8 3 24 8 5 3
24 5
Hypergeometric Distribution:
Probability Computations

Excel’s Hypergeometric Function
N = 24
A = 8
n = 5
X P(X)
0 =HYPGEOMDIST(A6,B$3,B$2,B$1)
=SUM(B6:B11)

Minitab’s Hypergeometric Function
X P(X =x)
0 0.102767
1 0.342556
2 0.368906
3 0.158103
4 0.026350
5 0.001318
Hypergeometric with N = 24, A = 8, n = 5

Applied Business Statistics ,ken black , ch 5

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