4-1
COMPLETE
BUSINESS
STATISTICS
by
AMIR D. ACZEL
&
JAYAVEL SOUNDERPANDIAN
6th edition.
4-2
Pertemuan 11 dan 12
The Normal Distribution
4-3
4 The Normal Distribution
Using Statistics
 Properties of the Normal Distribution
 The Template
 The Standard Normal Distribution
 The Transformation of Normal Random
Variables
 The Inverse Transformation
 The Normal Approximation of Binomial
Distributions

4-4
4
LEARNING OBJECTIVES
After studying this chapter, you should be able to:







Identify when a random variable will be normally
distributed
Use the properties of normal distributions
Explain the significance of the standard normal distribution
Use normal distribution tables to compute probabilities
Transform a normal distribution into a standard normal
distribution
Convert a binomial distribution into an approximate normal
distribution
Use spreadsheet templates to solve normal distribution
problems
4-5
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
x
1
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:
1
0.4




0.3
for
- < x<
2p 2
where e = 2 . 7182818 ... and p = 3 . 14159265 ...
f(x)
f ( x) =
- 2
x
e 2 2




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
x
5
4-6
The Normal Probability Distribution
The normal probability density function:
- 2
x
e 2  2 for
f (x) =
1
0.4




0.3
-< x<
2 p 2
where e = 2 .7182818 ... and p = 3.14159265 ...
f(x)




Normal Distribution:  = 0, = 1
0.2
0.1
0.0
-5
0
x
5
4-7
4-2 Properties of the Normal
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.
4-8
4-2 Properties of the Normal
Distribution (continued)
• If several independent random variables are
normally distributed then their sum will also be
normally distributed.
• The mean of the sum will be the sum of all the
individual means.
• The variance of the sum will be the sum of all
the individual variances (by virtue of the
independence).
4-9
4-2 Properties of the Normal
Distribution (continued)
• If X1, X2, …, Xn are independent normal random
variable, then their sum S will also be normally
distributed with
• E(S) = E(X1) + E(X2) + … + E(Xn)
• V(S) = V(X1) + V(X2) + … + V(Xn)
• Note: It is the variances that can be added
above and not the standard deviations.
4-10
4-2 Properties of the Normal
Distribution – Example 4-1
Example 4.1: Let X1, X2, and X3 be independent random
variables that are normally distributed with means and
variances as shown.
Mean
Variance
X1
10
1
X2
20
2
X3
30
3
Let S = X1 + X2 + X3. Then E(S) = 10 + 20 + 30 = 60 and
V(S) = 1 + 2 + 3 = 6. The standard deviation of S is 6
= 2.45.
4-11
4-2 Properties of the Normal
Distribution (continued)
• If X1, X2, …, Xn are independent normal random
variable, then the random variable Q defined as
Q = a1X1 + a2X2 + … + anXn + b will also be
normally distributed with
• E(Q) = a1E(X1) + a2E(X2) + … + anE(Xn) + b
• V(Q) = a12 V(X1) + a22 V(X2) + … + an2 V(Xn)
• Note: It is the variances that can be added
above and not the standard deviations.
4-12
4-2 Properties of the Normal
Distribution – Example 4-3
Example 4.3: Let X1 , X2 , X3 and X4 be independent random variables
that are normally distributed with means and variances as shown.
Find the mean and variance of Q = X1 - 2X2 + 3X2 - 4X4 + 5
Mean
Variance
X1
12
4
X2
-5
2
X3
8
5
X4
10
1
E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11
V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73
SD(Q) = 73 = 8.544
4-13
Computing the Mean, Variance and Standard
Deviation for the Sum of Independent Random
Variables Using the Template
4-14
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
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
Z~N(0,1)
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.
65
4-15
4-3 Computing Normal Probabilities
Using the Template
4-16
4-4 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
=0
Z
1
2
3
4
5
4-17
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
5
Z
Look in row labeled 1.5
and column labeled .06 to
find P(0  z  1.56) =
0.4406
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
.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
4-18
Finding Probabilities of the Standard
Normal Distribution: P(Z < -2.47)
z ...
.
.
.
2.3 ...
2.4 ...
2.5 ...
.
.
.
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
0.4909
0.4931
0.4948
.06
.
.
.
0.4911
0.4932
0.4949
.07
.
.
.
0.4913
0.4934
0.4951
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
Z
1
2
3
4
5
.08
.
.
.
4-19
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
.
.
.
= .9772 - .8413 = 0.1359
Standard Normal Distribution
0.4
Area between 1 and 2
P(1  Z  2) = .9772 - .8413 = 0.1359
f(z)
0.3
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
.00
.
.
.
0.3159
0.3413
0.3643
.
.
.
0.4713
0.4772
0.4821
.
.
.
...
...
...
...
...
...
...
4-20
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
.
.
.
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
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
Z = 1.28
.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
.
.
.
4-21
99% Interval around the Mean
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
Area in center left = .495
to find that:
.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
.
.
.
Total area in center = .99
0.4
 z.005  
z.005  
P(-.2575 Z  ) = .99
Area in center right = .495
f(z)
0.3
0.2
Area in right tail = .005
Area in left tail = .005
0.1
0.0
-5
-4
-3
-2
-z.005
-2.575
-1
0
Z
1
2
3
z.005
2.575
4
5
4-22
4-5 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 =

Normal Distribution:  =50, =10
x
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
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
The inverse transformation of Z to X:
X = x + Z x
4-23
Using the Normal Transformation
Example 4-9
Example 4-10
X~N(160,302)
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 -2  Z .6667
= 0.4772 + 0.2475 = 0.7247
(

 
 
150 - 127

= P Z <


22 
(
= P Z < 1.045
= 0.5 + 0.3520 = 0.8520
4-24
Using the Normal Transformation Example 4-11
Normal Distribution:  = 383,  = 12
Example 4-11
0.05
0.04
X~N(383,122)
0.03
(

= P 0.9166  Z  1.333
= 0.4088 - 0.3203 = 0.0885
Template solution
0.02
0.01
Standard Normal Distribution
0.00
340
0.4
390
X
0.3
f(z)
=
 

 
 394 - 383   399 - 383

P
Z
 12

12
f(X)
P ( 394  X  399)
394 -  X -  399 -  

= P



0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
440
4-25
The Transformation of Normal
Random Variables
The transformation of X to Z:
Z =
X - x
x
The inverse transformation of Z to X:
X = 
+ Z
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
 
 
4-26
Normal Probabilities (Empirical Rule)
• The probability that a normal random
S ta n d a rd N o rm a l D is trib u tio n
variable will be within 1 standard
deviation from its mean (on either
side) is 0.6826, or approximately 0.68.
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.
0.3
f(z)
• The probability that a normal random
0.4
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
4-27
4-6 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),
 x -  70 -  

70 - 50 
>
 = P Z >
 = P( Z > 2)
P( X > 70) = P
 

 
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 4-12
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
.
.
.
...
...
...
.
.
.
.07
.
.
.
0.3790
0.3980
0.4147
.
.
.
.08
.
.
.
0.3810
0.3997
0.4162
.
.
.
.09
.
.
.
0.3830
0.4015
0.4177
.
.
.
f(x)
0.03
z
.
.
.
1.1
1.2
1.3
.
.
.
0.02
0.01
0.01
0.00
80
130
X
139.36
180
4-28
Template Solution for Example 4-12
Example 4-12
X~N(124,122)
P(X > x) = 0.10 and P(Z > 1.28) 0.10
x =  + z = 124 + (1.28)(12) = 139.36
4-29
The Inverse Transformation (Continued)
X~N(5.7,0.52)
Example 4-13
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 4-14
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
3000
4000
X
1
2
3
4
5
Z.01 = 2.33
-5
-4
-3
-2
-1.96
-1
0
Z
1
2
1.96
3
4
5
4-30
Finding Values of a Normal Random
Variable, Given a Probability
Normal Distribution:  = 2450,  = 400
0.0012
.
0.0010
.
f(x)
0.0008
.
0.0006
.
0.0004
.
0.0002
.
0.0000
1000
2000
3000
4000
X
S tand ard Norm al D istrib utio n
0.4
0.3
f(z)
1. Draw pictures of
the normal
distribution in
question and of the
standard normal
distribution.
0.2
0.1
0.0
-5
-4
-3
-2
-1
0
Z
1
2
3
4
5
4-31
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
Z
1
2
3
4
5
4-32
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
.
.
.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
Z
-1.96
1.96
3
4
5
4-33
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
.
.
.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
1.96
3
4
5
x =   z = 2450 ± (1.96)(400)
= 2450 ±784=(1666,3234)
4-34
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)
4.5000
0.7751
x P( X <= x)
4.00
0.7734
4-35
4-7 The Normal Approximation of Binomial
Distribution
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
10
X
0
1
2
3
4
5
6
X
7
8
9 10 11
4-36
Approximating a Binomial Probability
Using the Normal Distribution
 a - np
b - np 
Z
P( a  X  b) =& P

 np(1 - p)
np(1 p) 
for n large (n  50) and p not too close to 0 or 1.00
or:
 a - 0.5 - np
b + 0.5 - np 
Z
P(a  X  b) =& P

 np(1 - p)
np(1 p) 
for n moderately large (20  n < 50).
NOTE: If p is either small (close to 0) or large (close to 1), use the
Poisson approximation.
4-37
Using the Template for Normal Approximation
of the Binomial Distribution