Download Chapter 6

Business Statistics, 5th ed.
by Ken Black
Chapter 6
Discrete Distributions
Continuous
Distributions
PowerPoint presentations prepared by Lloyd Jaisingh,
Morehead State University
Learning Objectives
• Understand concepts of the uniform
distribution.
• Appreciate the importance of the normal
distribution.
• Recognize normal distribution problems, and
know how to solve them.
• Decide when to use the normal distribution to
approximate binomial distribution problems,
and know how to work them.
• Decide when to use the exponential distribution
to solve problems in business, and know how to
work them.
Uniform Distribution
 1
b  a

f ( x)  
 0

for
a xb
for
all other values
1
ba
f (x)
Area = 1
a
x
b
Uniform Distribution of Lot Weights
 1
 47  41

f ( x)  
 0

for
for
41  x  47
all other values
1
1

47  41 6
f (x)
Area = 1
41
47
x
Uniform Distribution Probability
P( x1  X 
x
2
)
 x1
ba
x
2
45 42 1
P(42  X  45) 

47 41 2
45  42
1

47  41
2
f (x)
Area
= 0.5
41
42
45 47 x
Uniform Distribution
Mean and Standard Deviation
Mean

=
Mean
a +
2
b
Standard Deviation
ba

12
41 + 47
88
 =

 44
2
2
Standard Deviation
47  41
6


 1. 732
12
3. 464
Uniform Distribution of Assembly of
Plastic Modules
1
 1
39  27  12

f ( x)  

0


for
27  x  39
for all other values
1
1

39  27 12
f ( x)
Area = 1
27
39 x
Uniform Distribution
Mean and Standard Deviation
Mean

=
Mean
a +
2
b
39 + 27
 33
 =
2
Standard Deviation
Standard Deviation
ba

12
39  27
12

 3 .464
 
3.464
12
Uniform Distribution of Assembly of
Plastic Modules
P(30  X  35) 
35 30

5
39 27 12
 0.4167
35  30
 0 . 4167
39  27
f (x)
42
27
30
35
39 x
45
Uniform Distribution of Assembly of
Plastic Modules
P ( X  30 ) 
30  27
39  27

3
 0 . 2500
12
35  30
 0 . 4167
39  27
f (x)
27
30
39 x
Normal Distribution
• Probably the most widely known and used of
all distributions is the normal distribution.
• It fits many human characteristics, such as
height, weight, length, speed, IQ scores,
scholastic achievements, and years of life
expectancy, among others.
• Many things in nature such as trees, animals,
insects, and others have many characteristics
that are normally distributed.
Normal Distribution
• Many variables in business and industry are
also normally distributed. For example
variables such as the annual cost of household
insurance, the cost per square foot of renting
warehouse space, and managers’ satisfaction
with support from ownership on a five-point
scale, amount of fill in soda cans, etc.
• Because of the many applications, the normal
distribution is an extremely important
distribution.
Normal Distribution
• Discovery of the normal curve of errors is
generally credited to mathematician and
astronomer Karl Gauss (1777 – 1855), who
recognized that the errors of repeated
measurement of objects are often normally
distributed.
• Thus the normal distribution is sometimes
referred to as the Gaussian distribution or the
normal curve of errors.
• In addition, some credit were also given to
Pierre-Simon de Laplace (1749 – 1827) and
Abraham de Moivre (1667 – 1754) for the
discovery of the normal distribution.
Properties of the Normal Distribution
• The normal distribution exhibits the following
characteristics:
• It is a continuous distribution.
• It is symmetric about the mean.
• It is asymptotic to the horizontal axis.
• It is unimodal.
• It is a family of curves.
• Area under the curve is 1.
• It is bell-shaped.
Graphic Representation of the Normal
Distribution
Probability Density of the Normal
Distribution
This image cannot currently be display ed.
Family of Normal Curves
Standardized Normal Distribution
• Since there is an infinite number of
combinations for  and , then we can generate
an infinite family of curves.
• Because of this, it would be impractical to deal
with all of these normal distributions.
• Fortunately, a mechanism was developed by
which all normal distributions can be
converted into a single distribution called the
z distribution.
• This process yields the standardized normal
distribution (or curve).
Standardized Normal Distribution
• The conversion formula for any x value of a
given normal distribution is given below. It is
called the z-score.
x


z

• A z-score gives the number of standard
deviations that a value x, is above or below the
mean.
Standardized Normal Distribution
• If x is normally distributed with a mean of 
and a standard deviation of , then the z-score
will also be normally distributed with a mean
of 0 and a standard deviation of 1.
• Since we can convert to the standard normal
distribution, tables have been generated for this
standard normal distribution which will enable
us to determine probabilities for normal
variables.
• The tables in the text are set up to give the
probabilities between z = 0 and some other z
value, z0 say, which is depicted on the next
slide.
Standardized Normal Distribution
Z Table
Second Decimal Place in Z
Z 0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.00
0.10
0.20
0.30
0.0000
0.0398
0.0793
0.1179
0.0040
0.0438
0.0832
0.1217
0.0080
0.0478
0.0871
0.1255
0.0120
0.0517
0.0910
0.1293
0.0160
0.0557
0.0948
0.1331
0.0199
0.0596
0.0987
0.1368
0.0239
0.0636
0.1026
0.1406
0.0279
0.0675
0.1064
0.1443
0.0319
0.0714
0.1103
0.1480
0.0359
0.0753
0.1141
0.1517
0.90
1.00
1.10
1.20
0.3159
0.3413
0.3643
0.3849
0.3186
0.3438
0.3665
0.3869
0.3212
0.3461
0.3686
0.3888
0.3238
0.3485
0.3708
0.3907
0.3264
0.3508
0.3729
0.3925
0.3289
0.3531
0.3749
0.3944
0.3315
0.3554
0.3770
0.3962
0.3340
0.3577
0.3790
0.3980
0.3365
0.3599
0.3810
0.3997
0.3389
0.3621
0.3830
0.4015
2.00
0.4772
0.4778
0.4783
0.4788
0.4793
0.4798
0.4803
0.4808
0.4812
0.4817
3.00
3.40
3.50
0.4987
0.4997
0.4998
0.4987
0.4997
0.4998
0.4987
0.4997
0.4998
0.4988
0.4997
0.4998
0.4988
0.4997
0.4998
0.4989
0.4997
0.4998
0.4989
0.4997
0.4998
0.4989
0.4997
0.4998
0.4990
0.4997
0.4998
0.4990
0.4998
0.4998
Applying the Z Formula
X is normallydistributed with  = 485, and  = 105
P(485  X  600)  P(0  Z  1.10)  .3643
For X = 485,
X-
485  485
Z=

0
105

Z
0.00
0.01
0.02
0.00
0.10
0.0000 0.0040 0.0080
0.0398 0.0438 0.0478
For X = 600,
1.00
0.3413 0.3438 0.3461
X- 
1.10
0.3643 0.3665 0.3686
1.20
0.3849 0.3869 0.3888
600 485
Z=

 1.10
105

Applying the Z Formula
X is normallydistributed with  = 494, and  = 100
P( X  550)  P(Z  0.56)  .7123
For X = 550
X -  550  494
Z=

 0.56

100
0.5 + 0.2123 = 0.7123
Applying the Z Formula
X is normallydistributed with  = 494, and  = 100
P( X  700)  P(Z  2.06)  .0197
For X = 700
X -  700  494
Z=

 2.06

100
0.5 – 0.4803 = 0.0197
Applying the Z Formula
X is normallydistributed with  = 494, and  = 100
P(300  X  600)  P(1.94  Z  1.06)  .8292
For X = 300
X -  300  494
Z=

 1.94

100
For X = 600
X -  600  494
Z=

 1.06

100
0.4738+ 0.3554 = 0.8292
Demonstration Problem 6.9
• These types of problems can be solved quite
easily with the appropriate technology. The
output shows the MINITAB solution.
Normal Approximation
of the Binomial Distribution
• The normal distribution can be used to
approximate binomial probabilities.
• Procedure
– Convert binomial parameters to normal
parameters.
– Does the interval   3 lie between 0 and n?
If so, continue; otherwise, do not use the
normal approximation.
– Correct for continuity.
– Solve the normal distribution problem.
Normal Approximation of Binomial:
Parameter Conversion
• Conversion equations
  n p
  n pq
• Conversion example:
Given that X has a binomial distribution , find
P( X  25| n  60 and p . 30 ).
  n  p  (60 )(. 30 )  18
  n  p  q  (60 )(. 30 )(. 70 )  3. 55
Normal Approximation of Binomial:
Interval Check
. )  18  1065
.
  3  18  3(355
  3  7.35
.
  3  2865
0
10
20
30
40
50
60
n
70
Graph of the Binomial Problem:
n = 60, p = 0.3
0.12
0.10
P(x)
0.08
0.06
0.04
0.02
0.00
10
15
20
x
25
30
Normal Approximation of Binomial:
Correcting for Continuity
Values
Being
Determined
Correction
X
X
X
X
X
X
+.50
-.50
-.50
+.05
-.50 and +.50
+.50 and -.50
The binomial probability,
P( X  25| n  60 and p . 30)
is approximated by the normal probabilit
P(X  24.5|   18 and   3. 55).
Normal Approximation of Binomial:
Computations
X
P(X)
25
26
27
28
29
30
31
32
33
Total
0.0167
0.0096
0.0052
0.0026
0.0012
0.0005
0.0002
0.0001
0.0000
0.0361
The normal approximation,
. )
P(X  24.5|   18 and   355
24.5  18 

 P Z 



.
355
. )
 P( Z  183
. 
.5  P 0  Z  183
.5.4664
.0336
Exponential Distribution
•
•
•
•
•
•
•
Continuous
Family of distributions
Skewed to the right
X varies from 0 to infinity
Apex is always at X = 0
Steadily decreases as X gets larger
Probability function
X
f ( X)   e
for X  0,   0
Different Exponential Distributions
Exponential Distribution:
Probability Computation
1.2

1.0
0.8
X 0
P X  X 0   e
(12
. )(2)
P X  2|   12
. e
.0907
0.6
0.4
0.2
0.0
0
1
2
3
4
5
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