Download Chapter 4 - The Normal Distribution

COMPLETE
BUSINESS STATISTICS
4









4-1
fourth edi tion
The Normal Distribution
Using Statistics
The Normal Probability Distribution
The Standard Normal Distribution
The Transformation of Normal Random Variables
The Inverse Transformation
More Complex Problems
The Normal Distribution as an Approximation to Other
Probability Distributions
Using the Computer
Summary and Review of Terms
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COMPLETE
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4-1 Introduction
As n increases, the binomial distribution approaches a ...
n=6
n = 10
Binomial Distribution: n=10, p=.5
Binomial Distribution: n=14, p=.5
0.3
0.3
0.2
0.2
0.2
0.1
P(x)
0.3
P(x)
P(x)
Binomial Distribution: n=6, p=.5
n = 14
0.1
0.0
0.1
0.0
0
1
2
3
4
5
6
0.0
0
1
x
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
x
x
Normal Probability Density Function:




0.3
 x 
2
2
where e  2.7182818... and   314159265
.
...
Irwin/McGraw-Hill
Aczel
f(x)
f ( x) 
1
0.4
x  2

e 2 2 for




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
5
x
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COMPLETE
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BUSINESS STATISTICS
fourth edi tion
4-2 The Normal Probability Distribution
The normal probability density function:
1
e
0.4
x  2




2 2
0.3
for
 x 
2 2
where e  2.7182818... and   314159265
.
...
f(x)
f ( x) 





Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
5
x
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COMPLETE
BUSINESS STATISTICS
4-4
fourth edi tion
The Normal Probability Distribution
• The normal is a family of
– Bell-shaped and symmetric distributions. because the
–
–
–
distribution is symmetric, one-half (.50 or 50%) lies
on either side of the mean.
Each is characterized by a different pair of mean, ,
and variance, . That is: [X~N()].
Each is asymptotic to the horizontal axis.
The area under any normal probability density
function within k of  is the same for any normal
distribution, regardless of the mean and variance.
Irwin/McGraw-Hill
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Normal Probability Distributions
All of these are normal probability density functions, though each has a different mean and variance.
Normal Distribution:  =40, =1
Normal Distribution:  =30, =5
0.4
Normal Distribution:  =50, =3
0.2
0.2
0.2
f(y)
f(x)
f(w)
0.3
0.1
0.1
0.1
0.0
0.0
35
40
45
0.0
0
10
20
30
w
40
x
W~N(40,1)
X~N(30,25)
50
60
35
45
50
55
65
y
Y~N(50,9)
Normal Distribution:  =0, =1
Consider:
0.4
f(z)
0.3
0.2
0.1
0.0
-5
0
z
5
P(39  W  41)
P(25  X  35)
P(47  Y  53)
P(-1  Z  1)
The probability in each
case is an area under a
normal probability density
function.
Z~N(0,1)
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fourth edi tion
4-3 The Standard Normal Distribution
The standard normal random variable, Z, is the normal random
variable with mean  = 0 and standard deviation  = 1: Z~N(0,12).
Standard Normal Distribution
0 .4
=1
{
f( z)
0 .3
0 .2
0 .1
0 .0
-5
-4
-3
-2
-1
0
1
2
3
4
5
=0
Z
Irwin/McGraw-Hill
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COMPLETE
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BUSINESS STATISTICS
fourth edi tion
Finding Probabilities of the Standard
Normal Distribution: P(0 < Z < 1.56)
Standard Normal Probabilities
Standard Normal Distribution
0.4
f(z)
0.3
0.2
0.1
{
1.56
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
Z
Look in row labeled 1.5
and column labeled .06
to find
P(0  z  1.56) = .4406
Irwin/McGraw-Hill
5
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
0.4192
0.4332
0.4452
0.4554
0.4641
0.4713
0.4772
0.4821
0.4861
0.4893
0.4918
0.4938
0.4953
0.4965
0.4974
0.4981
0.4987
.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
0.4207
0.4345
0.4463
0.4564
0.4649
0.4719
0.4778
0.4826
0.4864
0.4896
0.4920
0.4940
0.4955
0.4966
0.4975
0.4982
0.4987
Aczel
.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
0.4222
0.4357
0.4474
0.4573
0.4656
0.4726
0.4783
0.4830
0.4868
0.4898
0.4922
0.4941
0.4956
0.4967
0.4976
0.4982
0.4987
.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
0.4236
0.4370
0.4484
0.4582
0.4664
0.4732
0.4788
0.4834
0.4871
0.4901
0.4925
0.4943
0.4957
0.4968
0.4977
0.4983
0.4988
.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
0.4251
0.4382
0.4495
0.4591
0.4671
0.4738
0.4793
0.4838
0.4875
0.4904
0.4927
0.4945
0.4959
0.4969
0.4977
0.4984
0.4988
.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
0.4265
0.4394
0.4505
0.4599
0.4678
0.4744
0.4798
0.4842
0.4878
0.4906
0.4929
0.4946
0.4960
0.4970
0.4978
0.4984
0.4989
.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
0.4279
0.4406
0.4515
0.4608
0.4686
0.4750
0.4803
0.4846
0.4881
0.4909
0.4931
0.4948
0.4961
0.4971
0.4979
0.4985
0.4989
.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
0.4292
0.4418
0.4525
0.4616
0.4693
0.4756
0.4808
0.4850
0.4884
0.4911
0.4932
0.4949
0.4962
0.4972
0.4979
0.4985
0.4989
.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
0.4306
0.4429
0.4535
0.4625
0.4699
0.4761
0.4812
0.4854
0.4887
0.4913
0.4934
0.4951
0.4963
0.4973
0.4980
0.4986
0.4990
.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
0.4319
0.4441
0.4545
0.4633
0.4706
0.4767
0.4817
0.4857
0.4890
0.4916
0.4936
0.4952
0.4964
0.4974
0.4981
0.4986
0.4990
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-8
BUSINESS STATISTICS
fourth edi tion
Finding Probabilities of the Standard
Normal Distribution: P(Z < -2.47)
To find P(Z<-2.47):
Find table area for 2.47
P(0 < Z < 2.47) = .4932
P(Z < -2.47) = .5 - P(0 < Z < 2.47)
= .5 - .4932 = 0.0068
z ...
.
.
.
2.3 ...
2.4 ...
2.5 ...
.
.
.
0.4909
0.4931
0.4948
.06
.
.
.
0.4911
0.4932
0.4949
.07
.
.
.
0.4913
0.4934
0.4951
.08
.
.
.
Standard Normal Distribution
Area to the left of -2.47
P(Z < -2.47) = .5 - 0.4932
= 0.0068
0.4
Table area for 2.47
P(0 < Z < 2.47) = 0.4932
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Irwin/McGraw-Hill
Aczel
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COMPLETE
4-9
BUSINESS STATISTICS
fourth edi tion
Finding Probabilities of the Standard
Normal Distribution: P(1< Z < 2)
To find P(1  Z  2):
1. Find table area for 2.00
F(2) = P(Z  2.00) = .5 + .4772 =.9772
2. Find table area for 1.00
F(1) = P(Z  1.00) = .5 + .3413 = .8413
3. P(1  Z  2.00) = P(Z  2.00) - P(Z  1.00)
z
.
.
.
0.9
1.0
1.1
.
.
.
1.9
2.0
2.1
.
.
.
.00
.
.
.
0.3159
0.3413
0.3643
.
.
.
0.4713
0.4772
0.4821
.
.
.
...
...
...
...
...
...
...
= .9772 - .8413 = .1359
Standard Normal Distribution
0.4
Area between 1 and 2
P(1  Z  2) = .4772 - .8413 = 0.1359
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
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COMPLETE
4-10
BUSINESS STATISTICS
fourth edi tion
Finding Values of the Standard Normal
Random Variable: P(0 < Z < z) = 0.40
To find z such that
P(0  Z  z) = .40:
1. Find a probability as close as
possible to .40 in the table of
standard normal probabilities.
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
.
.
.
.00
0.0000
0.0398
0.0793
0.1179
0.1554
0.1915
0.2257
0.2580
0.2881
0.3159
0.3413
0.3643
0.3849
0.4032
.
.
.
.01
0.0040
0.0438
0.0832
0.1217
0.1591
0.1950
0.2291
0.2611
0.2910
0.3186
0.3438
0.3665
0.3869
0.4049
.
.
.
2. Then determine the value of z
from the corresponding row
and column.
Area to the left of 0 = .50
Also, since P(Z  0) = .50
.03
0.0120
0.0517
0.0910
0.1293
0.1664
0.2019
0.2357
0.2673
0.2967
0.3238
0.3485
0.3708
0.3907
0.4082
.
.
.
.04
0.0160
0.0557
0.0948
0.1331
0.1700
0.2054
0.2389
0.2704
0.2995
0.3264
0.3508
0.3729
0.3925
0.4099
.
.
.
.05
0.0199
0.0596
0.0987
0.1368
0.1736
0.2088
0.2422
0.2734
0.3023
0.3289
0.3531
0.3749
0.3944
0.4115
.
.
.
.06
0.0239
0.0636
0.1026
0.1406
0.1772
0.2123
0.2454
0.2764
0.3051
0.3315
0.3554
0.3770
0.3962
0.4131
.
.
.
.07
0.0279
0.0675
0.1064
0.1443
0.1808
0.2157
0.2486
0.2794
0.3078
0.3340
0.3577
0.3790
0.3980
0.4147
.
.
.
.08
0.0319
0.0714
0.1103
0.1480
0.1844
0.2190
0.2517
0.2823
0.3106
0.3365
0.3599
0.3810
0.3997
0.4162
.
.
.
.09
0.0359
0.0753
0.1141
0.1517
0.1879
0.2224
0.2549
0.2852
0.3133
0.3389
0.3621
0.3830
0.4015
0.4177
.
.
.
Standard Normal Distribution
0.4
P(z  0) = .50
Area = .40 (.3997)
0.3
f(z)
P(0  Z  1.28)  .40
.02
0.0080
0.0478
0.0871
0.1255
0.1628
0.1985
0.2324
0.2642
0.2939
0.3212
0.3461
0.3686
0.3888
0.4066
.
.
.
0.2
0.1
0.0
P(Z  1.28)  .90
Irwin/McGraw-Hill
-5
-4
-3
-2
-1
0
Z
Aczel
1
2
3
4
5
Z = 1.28
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-11
BUSINESS STATISTICS
fourth edi tion
99% Interval around the Mean:
P(-z.005< Z < z.005) = 0.99
To have .99 in the center of the distribution, there
should be (1/2)(1-.99) = (1/2)(.01) = .005 in each
tail of the distribution, and (1/2)(.99) = .495 in
each half of the .99 interval. That is:
P(0  Z  z.005) = .495
z
.
.
.
2.4 ...
2.5 ...
2.6 ...
.
.
.
.04
.
.
.
0.4927
0.4945
0.4959
.
.
.
.05
.
.
.
0.4929
0.4946
0.4960
.
.
.
Look to the table of standard normal probabilities
to find that:
Area in center left = .495
.06
.
.
.
0.4931
0.4948
0.4961
.
.
.
.07
.
.
.
0.4932
0.4949
0.4962
.
.
.
.08
.
.
.
0.4934
0.4951
0.4963
.
.
.
.09
.
.
.
0.4936
0.4952
0.4964
.
.
.
Area in center = .99
0.4
 z.005  
z.005  
Area in center right = .495
f(z)
0.3
P(-.2575 Z  ) = .99
0.2
Area in right tail = .005
Area in left tail = .005
0.1
0.0
-5
-4
-3
-2
-z.005
-2.575
Irwin/McGraw-Hill
Aczel
-1
0
Z
1
2
3
4
5
z.005
2.575
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-12
BUSINESS STATISTICS
fourth edi tion
4-4 The Transformation of Normal
Random Variables
The area within k of the mean is the same for all normal random variables. So an area
under any normal distribution is equivalent to an area under the standard normal. In this
example: P(40  X  P(-1  Z     since and 
The transformation of X to Z:
X x
Z
x
Normal Distribution:  =50, =10
0.07
0.06
Transformation
f(x)
(1) Subtraction: (X - x)
0.05
0.04
0.03
=10
{
0.02
Standard Normal Distribution
0.01
0.00
0.4
0
20
30
40
50
60
70
80
90 100
X
0.3
0.2
(2) Division by x)
{
f(z)
10
1.0
0.1
X  x  Z x
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Irwin/McGraw-Hill
The inverse transformation of Z to X:
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-13
fourth edi tion
Using the Normal Transformation
Example 4-1
X~N(160,302)
Example 4-2
X~N(127,222)
P (100  X  180)
 100    X    180   
 P

P ( X  150)
 X    150   
 P



 

 
100  160
180  160

P
Z

 30
30 




 P Z  1.045
 0.5  0.3520  0.8520
 P 2  Z  .6667
 0.4772  0.2475  0.7247
Irwin/McGraw-Hill
 
 
150  127 

P Z 


22 
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-14
fourth edi tion
Using the Normal Transformation Minitab Solutions for Examples 4-1 & 4-2
MTB > cdf 100;
SUBC> normal 160,30.
Cumulative Distribution Function
Normal with mean = 160.000 and standard
deviation = 30.0000
MTB > cdf 150;
SUBC> normal 127,22.
Cumulative Distribution Function
Normal with  = 127.000 and  = 22.0000
x P( X <= x)
100.0000
0.0228
x P( X <= x)
150.0000
0.8521
MTB > cdf 180;
SUBC> normal 160,30.
Cumulative Distribution Function
Normal with mean = 160.000 and standard
deviation = 30.0000
x P( X <= x)
180.0000
0.7475
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-15
BUSINESS STATISTICS
fourth edi tion
Using the Normal Transformation Example 4-3
Normal Distribution:  = 383,  = 12
Example 4-3
X~N(383,122)
0.05
P ( 394  X  399)
 394   X   399   
 P



f(X)
0.03
0.02
 

 
399  383
 394  383
P
Z 

 12

12

0.01
Standard Normal Distribution
0.00
340
0.4
390
440
X
0.3
f(z)

0.04

 P 0.9166  Z  1.333
 0.4088  0.3203  0.0885
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
MTB > cdf 394;
SUBC> normal 383,12.
MTB > cdf 399;
SUBC> normal 383,12.
Cumulative Distribution Function
Cumulative Distribution Function
Normal with mean = 383.000 and standard deviation = 12.0000
Normal with mean = 383.000 and standard deviation = 12.0000
x P( X <= x)
394.0000
0.8203
Irwin/McGraw-Hill
x P( X <= x)
399.0000
0.9088
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-16
fourth edi tion
Using the Normal Transformation Excel Solution for Example 4-3
Do the same with X = 394 and subtract the two values to get 0.088447.
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-17
fourth edi tion
The Transformation of Normal Random
Variables
The inverse transformation of Z to X:
The transformation of X to Z:
Z 
X  x
X  
 Z
x
x
x
The transformation of X to Z, where a and b are numbers::
a  

P( X  a)  P Z 


 
b  

P( X  b)  P Z 


 
b  
a
P(a  X  b)  P
Z

 
 
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-18
fourth edi tion
Normal Probabilities
S ta n d a rd N o rm a l D is trib utio n
• The probability that a normal
•
•
Irwin/McGraw-Hill
Aczel
0.4
0.3
f(z)
random variable will be within 1
standard deviation from its mean
(on either side) is 0.6826, or
approximately 0.68.
The probability that a normal
random variable will be within 2
standard deviations from its mean
is 0.9544, or approximately 0.95.
The probability that a normal
random variable will be within 3
standard deviation from its mean is
0.9974.
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-19
fourth edi tion
4-5 The Inverse Transformation
The area within k of the mean is the same for all normal random variables. To find a
probability associated with any interval of values for any normal random variable, all that
is needed is to express the interval in terms of numbers of standard deviations from the
mean. That is the purpose of the standard normal transformation. If X~N(50,102),
70  50 
 x   70   

P( X  70)  P

  P Z 
  P( Z  2)
 

 
10 
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean
of X: 70 =  + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal
distribution.
Normal Distribution:  = 124,  = 12
Example 4-4
X~N(124,122)
P(X > x) = 0.10 and P(Z > 1.28) 0.10
x =  + z = 124 + (1.28)(12) = 139.36
0.04
0.03
.
.
.
...
...
...
.
.
.
.07
.
.
.
0.3790
0.3980
0.4147
.
.
.
Irwin/McGraw-Hill
.08
.
.
.
0.3810
0.3997
0.4162
.
.
.
.09
.
.
.
0.3830
0.4015
0.4177
.
.
.
f(x)
z
.
.
.
1.1
1.2
1.3
.
.
.
0.02
0.01
0.00
80
130
180
X
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-20
BUSINESS STATISTICS
fourth edi tion
The Inverse Transformation (2)
Example 4-5
X~N(5.7,0.52)
P(X > x)=0.01 and P(Z > 2.33) 0.01
x =  + z = 5.7 + (2.33)(0.5) = 6.865
z
.
.
.
2.2
2.3
2.4
.
.
.
.02
.
.
.
0.4868
0.4898
0.4922
.
.
.
.
.
.
...
...
...
.
.
.
.03
.
.
.
0.4871
0.4901
0.4925
.
.
.
Example 4-6
X~N(2450,4002)
P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95
x =   z = 2450 ± (1.96)(400) = 2450
±784=(1666,3234)
P(1666 < X < 3234) = 0.95
.04
.
.
.
0.4875
0.4904
0.4927
.
.
.
z
.
.
.
1.8
1.9
2.0
.
.
Normal Distribution:  = 5.7 = 0.5
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.0015
Area = 0.49
0.6
.4750
.4750
0.0010
f(x)
0.5
f(x)
.06
.
.
.
0.4686
0.4750
0.4803
.
.
Normal Distribution:  = 2450  = 400
0.8
0.7
.05
.
.
.
0.4678
0.4744
0.4798
.
.
.
.
.
...
...
...
.
.
0.4
X.01 = +z = 5.7 + (2.33)(0.5) = 6.865
0.3
0.0005
0.2
.0250
.0250
Area = 0.01
0.1
0.0
0.0000
3.2
4.2
5.2
6.2
7.2
8.2
1000
2000
X
-5
-4
-3
-2
-1
0
z
Irwin/McGraw-Hill
3000
4000
X
1
2
3
4
5
-5
-4
-3
-2
-1.96
Z.01 = 2.33
Aczel
-1
0
Z
1
2
3
4
5
1.96
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-21
BUSINESS STATISTICS
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
0.0012
.
0.0010
.
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-22
BUSINESS STATISTICS
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
.4750
.4750
0.3
f(z)
2. Shade the area
corresponding to
the desired
probability.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-23
BUSINESS STATISTICS
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
S tand ard Norm al D istrib utio n
0.4
.4750
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
.
.
.
...
...
...
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
Irwin/McGraw-Hill
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
3
4
5
Z
-1.96
Aczel
1.96
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-24
BUSINESS STATISTICS
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
3. From the table
of the standard
normal
distribution,
find the z value
or values.
0.0012
.
.4750
0.0010
.
.4750
0.0008
.
f(x)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.0006
.
0.0004
.
0.0002
.
.9500
0.0000
1000
2000
3000
4000
X
2. Shade the area
corresponding
to the desired
probability.
0.4
.4750
.
.
.
...
...
...
.
.
.05
.
.
.
0.4678
0.4744
0.4798
.
.
Irwin/McGraw-Hill
.06
.
.
.
0.4686
0.4750
0.4803
.
.
.4750
0.3
f(z)
z
.
.
.
1.8
1.9
2.0
.
.
4. Use the
transformation
from z to x to get
value(s) of the
original random
variable.
S tand ard Norm al D istrib utio n
.07
.
.
.
0.4693
0.4756
0.4808
.
.
0.2
0.1
.9500
0.0
-5
-4
-3
-2
-1
0
1
2
Z
-1.96
Aczel
3
4
5
x =   z = 2450 ± (1.96)(400)
= 2450 ±784=(1666,3234)
1.96
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-25
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
Using EXCEL to help solve Example 4-4.
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-26
BUSINESS STATISTICS
fourth edi tion
Finding Values of a Normal Random
Variable, Given a Probability
The normal distribution with  = 3.5 and  = 1.323 is a close
approximation to the binomial with n = 7 and p = 0.50.
P(x<4.5) = 0.7749
Normal Distribution:  = 3.5,  = 1.323
Binomial Distribution: n = 7, p = 0.50
0.3
0.3
P( x 4) = 0.7734
0.2
f(x)
P(x)
0.2
0.1
0.1
0.0
0.0
0
5
10
0
1
2
3
X
4
5
6
7
X
MTB > cdf 4.5;
SUBC> normal 3.5 1.323.
Cumulative Distribution Function
MTB > cdf 4;
SUBC> binomial 7,.5.
Cumulative Distribution Function
Normal with mean = 3.50000 and standard deviation = 1.32300
Binomial with n = 7 and p = 0.500000
x P( X <= x)
4.00
0.7734
x P( X <= x)
4.5000
0.7751
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
4-27
BUSINESS STATISTICS
fourth edi tion
The Normal Distribution as an Approximation to
Other Probability Distributions (2)
The normal distribution with  = 5.5 and  = 1.6583 is a closer
approximation to the binomial with n = 11 and p = 0.50.
P(x<4.5) = 0.2732
Normal Distribution:  = 5.5,  = 1.6583
Binomial Distribution: n = 11, p = 0.50
P(x4) = 0.2744
0.3
0.2
f(x)
P(x)
0.2
0.1
0.1
0.0
0.0
0
5
0
10
1
2
3
4
6
7
8
9 10 11
X
X
MTB > cdf 4;
SUBC> binomial 11,.5.
Cumulative Distribution Function
MTB > cdf 4.5;
SUBC> normal 5.5 1.6583.
Cumulative Distribution Function
Binomial with n = 11 and p = 0.500000
Normal with mean = 5.50000 and standard deviation = 1.65830
x P( X <= x)
4.00
0.2744
x P( X <= x)
4.5000
0.2732
Irwin/McGraw-Hill
5
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-28
fourth edi tion
Approximating a Binomial Probability
Using the Normal Distribution
b  np 
 a  np
P( a  X  b)  P
Z

 np(1  p)
np(1  p) 
for n large (n  50) and p not too close to 0 or 1.00
Or:
b  0.5  np 
 a  0.5  np
P(a  X  b)  P
Z

np(1  p) 
 np(1  p)
for n moderately large (20  n < 50).
If p is either small (close to 0) or large (close to 1), use the Poisson
approximation.
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-29
fourth edi tion
4-7 Using the Computer
• In EXCEL, the command
•
NORMSDIST(number) will return
cumulative probability for a standard normal
random variable.
The command NORMDIST(number,
mean, standard deviation) will return the
cumulative probability for a general random
variable.
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999
COMPLETE
BUSINESS STATISTICS
4-30
fourth edi tion
4-7 Using the Computer
• For example, NORMSDIST(1.0) = 0.8413.
• NORMDIST(10.0, 5, 2) = 0.9938.
• The inverse commands are
•
•
NORMSINV(number) and
NORMINV(number, mean, standard
deviation).
NORMSINV(0.975) = 1.96.
NORMINV(0.975, 20, 10) = 39.6.
Irwin/McGraw-Hill
Aczel
© The McGraw-Hill Companies, Inc., 1999