complete business statistics
DESCRIPTION
COMPLETE BUSINESS STATISTICS. by AMIR D. ACZEL & JAYAVEL SOUNDERPANDIAN 6 th edition (SIE). Chapter 4. The Normal Distribution. Using Statistics Properties of the Normal Distribution The Template The Standard Normal Distribution The Transformation of Normal Random Variables - PowerPoint PPT PresentationTRANSCRIPT
4-1
COMPLETE COMPLETE BUSINESS BUSINESS
STATISTICSSTATISTICSbyby
AMIR D. ACZELAMIR D. ACZEL
&&
JAYAVEL SOUNDERPANDIANJAYAVEL SOUNDERPANDIAN
66thth edition (SIE) edition (SIE)
4-2
Chapter 4 Chapter 4
The Normal DistributionThe Normal Distribution
4-3
Using StatisticsProperties of the Normal DistributionThe TemplateThe Standard Normal DistributionThe Transformation of Normal Random
VariablesThe Inverse TransformationThe Normal Approximation of Binomial
Distributions
The Normal DistributionThe Normal Distribution44
4-4
Identify when a random variable will be normally distributed
Use the properties of normal distributionsExplain the significance of the standard normal distributionUse normal distribution tables to compute probabilitiesTransform a normal distribution into a standard normal
distributionConvert a binomial distribution into an approximate normal
distributionUse spreadsheet templates to solve normal distribution
problems
LEARNING OBJECTIVESLEARNING OBJECTIVES44
After studying this chapter, you should be able to:After studying this chapter, you should be able to:
4-5
As n increases, the binomial distribution approaches a ...
n = 6 n = 14n = 10
Normal Probability Density Function:
6543210
0.3
0.2
0.1
0.0
x
P(x
)
Binomial Distribution: n=6, p=.5
109876543210
0.3
0.2
0.1
0.0
x
P(x
)
Binomial Distribution: n=10, p=.5
14131211109876543210
0.3
0.2
0.1
0.0
x
P(x
)
Binomial Distribution: n=14, p=.5
50-5
0.4
0.3
0.2
0.1
0.0
x
f(x)
Normal Distribution: = 0, = 1
4-1 Introduction4-1 Introduction
...14159265.3 and ...7182818.2 where
for 22
2
22
1)(
e
x
x
exf
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The normal probability density function:
50-5
0.4
0.3
0.2
0.1
0.0
x
f(x)
Normal Distribution: = 0, = 1
The Normal Probability DistributionThe Normal Probability Distribution
f x e
x
x
e
( )
. ... . ...
1
2 2
2
2 2
2 7182818 314159265
for
where and
4-7
• The normal is a family ofBell-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.
4-2 Properties of the Normal 4-2 Properties of the Normal DistributionDistribution
4-8
• 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).
4-2 Properties of the Normal 4-2 Properties of the Normal Distribution (continued)Distribution (continued)
4-9
• 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.
4-2 Properties of the Normal 4-2 Properties of the Normal Distribution (continued)Distribution (continued)
4-10
Example 4.1: Let X1, X2, and X3 be independent random variables that are normally distributed with means and variances as shown.
4-2 Properties of the Normal 4-2 Properties of the Normal Distribution – Example 4-1Distribution – Example 4-1
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 = 2.45.
6
4-11
• 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) = a1
2 V(X1) + a22 V(X2) + … + an
2 V(Xn)• Note: It is the variances that can be added above
and not the standard deviations.
4-2 Properties of the Normal 4-2 Properties of the Normal Distribution (continued)Distribution (continued)
4-12
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
4-2 Properties of the Normal 4-2 Properties of the Normal Distribution – Example 4-3Distribution – Example 4-3
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) = 544.873
4-13Computing the Mean, Variance and Standard Computing the Mean, Variance and Standard Deviation for the Sum of Independent Random Deviation for the Sum of Independent Random Variables Using the TemplateVariables Using the Template
4-14
All of these are normal probability density functions, though each has a different mean and variance.
Z~N(0,1)
50-5
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0.3
0.2
0.1
0.0
z
f(z)
Normal Distribution: =0, =1
W~N(40,1) X~N(30,25)
454035
0.4
0.3
0.2
0.1
0.0
w
f(w)
Normal Distribution: =40, =1
6050403020100
0.2
0.1
0.0
x
f(x)
Normal Distribution: =30, =5
Y~N(50,9)
65554535
0.2
0.1
0.0
y
f(y)
Normal Distribution: =50, =3
50
Consider:
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.
Normal Probability DistributionsNormal Probability Distributions
4-15
4-3 Computing Normal Probabilities 4-3 Computing Normal Probabilities Using the TemplateUsing the Template
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The standard normal random variable, Z, is the normal random variable with mean = 0 and standard deviation = 1: Z~N(0,12).
543210- 1- 2- 3- 4- 5
0 . 4
0 . 3
0 . 2
0 . 1
0 . 0
Z
f(z )
Standard Normal Distribution
= 0
=1{
4-4 The Standard Normal 4-4 The Standard Normal DistributionDistribution
4-17
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.41771.4 0.4192 0.4207 0.4222 0.4236 0.4251 0.4265 0.4279 0.4292 0.4306 0.43191.5 0.4332 0.4345 0.4357 0.4370 0.4382 0.4394 0.4406 0.4418 0.4429 0.44411.6 0.4452 0.4463 0.4474 0.4484 0.4495 0.4505 0.4515 0.4525 0.4535 0.45451.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.46331.8 0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.47061.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.47672.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.48172.1 0.4821 0.4826 0.4830 0.4834 0.4838 0.4842 0.4846 0.4850 0.4854 0.48572.2 0.4861 0.4864 0.4868 0.4871 0.4875 0.4878 0.4881 0.4884 0.4887 0.48902.3 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909 0.4911 0.4913 0.49162.4 0.4918 0.4920 0.4922 0.4925 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 0.4938 0.4940 0.4941 0.4943 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 0.4953 0.4955 0.4956 0.4957 0.4959 0.4960 0.4961 0.4962 0.4963 0.49642.7 0.4965 0.4966 0.4967 0.4968 0.4969 0.4970 0.4971 0.4972 0.4973 0.49742.8 0.4974 0.4975 0.4976 0.4977 0.4977 0.4978 0.4979 0.4979 0.4980 0.49812.9 0.4981 0.4982 0.4982 0.4983 0.4984 0.4984 0.4985 0.4985 0.4986 0.49863.0 0.4987 0.4987 0.4987 0.4988 0.4988 0.4989 0.4989 0.4989 0.4990 0.4990
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f( z)
Standard Normal Distribution
1.56{
Standard Normal Probabilities
Look in row labeled 1.5 and column labeled .06 to find P(0 z 1.56) = 0.4406
Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(0 < Z < 1.56)P(0 < Z < 1.56)
4-18
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
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Normal Distribution
Table area for 2.47P(0 < Z < 2.47) = 0.4932
Area to the left of -2.47P(Z < -2.47) = .5 - 0.4932= 0.0068
Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(Z < -2.47)P(Z < -2.47)
z ... .06 .07 .08. . . .. . . .. . . .2.3 ... 0.4909 0.4911 0.49132.4 ... 0.4931 0.4932 0.49342.5 ... 0.4948 0.4949 0.4951...
4-19
Finding Probabilities of the Standard Finding Probabilities of the Standard Normal Distribution: Normal Distribution: P(1< Z < 2)P(1< Z < 2)
z .00 ... . . . . . .0.9 0.3159 ...1.0 0.3413 ...1.1 0.3643 ... . . . . . .1.9 0.4713 ...2.0 0.4772 ...2.1 0.4821 ... . . . . . .
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)
= .9772 - .8413 = 0.1359
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Normal Distribution
Area between 1 and 2P(1 Z 2) = .9772 - .8413 = 0.1359
4-20
Finding Values of the Standard Normal Finding Values of the Standard Normal Random Variable: Random Variable: P(0 < Z < z) = 0.40P(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.
2. Then determine the value of z from the corresponding row and column.
P(0 Z 1.28) .40
Also, since P(Z 0) = .50
P(Z 1.28) .90543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Normal Distribution
Area = .40 (.3997)
Z = 1.28
Area to the left of 0 = .50P(z 0) = .50
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.1950 0.1985 0.2019 0.2054 0.2088 0.2123 0.2157 0.2190 0.22240.6 0.2257 0.2291 0.2324 0.2357 0.2389 0.2422 0.2454 0.2486 0.2517 0.25490.7 0.2580 0.2611 0.2642 0.2673 0.2704 0.2734 0.2764 0.2794 0.2823 0.28520.8 0.2881 0.2910 0.2939 0.2967 0.2995 0.3023 0.3051 0.3078 0.3106 0.31330.9 0.3159 0.3186 0.3212 0.3238 0.3264 0.3289 0.3315 0.3340 0.3365 0.33891.0 0.3413 0.3438 0.3461 0.3485 0.3508 0.3531 0.3554 0.3577 0.3599 0.36211.1 0.3643 0.3665 0.3686 0.3708 0.3729 0.3749 0.3770 0.3790 0.3810 0.38301.2 0.3849 0.3869 0.3888 0.3907 0.3925 0.3944 0.3962 0.3980 0.3997 0.40151.3 0.4032 0.4049 0.4066 0.4082 0.4099 0.4115 0.4131 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4-21
z .04 .05 .06 .07 .08 .09. . . . . . . . . . . . . .. . . . . . .2.4 ... 0.4927 0.4929 0.4931 0.4932 0.4934 0.49362.5 ... 0.4945 0.4946 0.4948 0.4949 0.4951 0.49522.6 ... 0.4959 0.4960 0.4961 0.4962 0.4963 0.4964. . . . . . .. . . . . . .. . . . . . .
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
Look to the table of standard normal probabilities to find that:
z.005 z.005
P(-.2575 Z ) = .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
Look to the table of standard normal probabilities to find that:
z.005 z.005
P(-.2575 Z ) = .99
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
-z.005 z.005
Area in right tail = .005Area in left tail = .005
Area in center right = .495
Area in center left = .495
2.575-2.575
Total area in center = .99
99% Interval around the Mean99% Interval around the Mean
4-22
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 sinceand
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 sinceand
1009080706050403020100
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0.00
X
f(x)
Normal Distribution: =50, =10
=10{
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Normal Distribution
1.0
{
Transformation
(2) Division by x)
The transformation of X to Z:The transformation of X to Z:
The inverse transformation of Z to X:The inverse transformation of Z to X:
4-5 The Transformation of Normal 4-5 The Transformation of Normal Random VariablesRandom Variables
(1) Subtraction: (X - x)
ZX x
x
X x Z x
4-23
Example 4-9Example 4-9
X~N(160,302)
Example 4-9Example 4-9
X~N(160,302)
Example 4-10Example 4-10
X~N(127,222)
Example 4-10Example 4-10
X~N(127,222)
Using the Normal TransformationUsing the Normal Transformation
P X
PX
P Z
P Z
( )
.. . .
100 180100 180
100 160
30
180 160
302 6667
0 4772 0 2475 0 7247
P X
PX
P Z
P Z
( )
.. . .
150150
150 127
22
1 0450 5 0 3520 0 8520
4-24
Example 4-11Example 4-11
X~N(383,122)
Example 4-11Example 4-11
X~N(383,122)
440390340
0.05
0.04
0.03
0.02
0.01
0.00
X
f( X)
Normal Distribution: = 383, = 12
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Normal Distribution
Equivalent areas
Using the Normal Transformation - Using the Normal Transformation - Example 4-11Example 4-11
Template solutionTemplate solution
P X
PX
P Z
P Z
( )
. .
. . .
394 399
394 399
394 383
12
399 383
12
0 9166 1 333
0 4088 0 3203 0 0885
4-25
The transformation of X to Z:The transformation of X to Z: The inverse transformation of Z to X:The inverse transformation of Z to X:
The transformation of X to Z, where a and b are numbers::The transformation of X to Z, where a and b are numbers::
The Transformation of Normal The Transformation of Normal Random VariablesRandom Variables
P X a P Za
P X b P Zb
P a X b Pa
Zb
( )
( )
( )
ZX x
x
X
xZ
x
4-26
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
S tand ard N o rm al D is trib utio n• The probability that a normal 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.
Normal Probabilities (Empirical Rule)Normal Probabilities (Empirical Rule)
4-27
z .07 .08 .09 . . . . . . . . . . . . . . .1.1 . . . 0.3790 0.3810 0.38301.2 . . . 0.3980 0.3997 0.40151.3 . . . 0.4147 0.4162 0.4177 . . . . . . . . . . . . . . .
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),
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.
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),
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.
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
18013080
0.04
0.03
0.02
0.01
0.00
X
f(x)
Normal Distribution: = 124, = 12
4-6 The Inverse Transformation4-6 The Inverse Transformation
0.01
139.36
P X Px
P Z P Z( ) ( )
70
70 70 50
102
4-28
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
Template Solution for Example 4-12Template Solution for Example 4-12
4-29
Example 4-13 X~N(5.7,0.52)P(X > x)=0.01 and P(Z > 2.33) 0.01x = + z = 5.7 + (2.33)(0.5) = 6.865
Example 4-13 X~N(5.7,0.52)P(X > x)=0.01 and P(Z > 2.33) 0.01x = + z = 5.7 + (2.33)(0.5) = 6.865
Example 4-14 X~N(2450,4002)P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)P(1666 < X < 3234) = 0.95
Example 4-14 X~N(2450,4002)P(a<X<b)=0.95 and P(-1.96<Z<1.96)0.95x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)P(1666 < X < 3234) = 0.95
z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .
4000300020001000
0.0015
0.0010
0.0005
0.0000
X
f(x)
Normal Distribution: = 2450 = 400
4000300020001000
0.0015
0.0010
0.0005
0.0000
543210-1-2-3-4-5
Z
.4750.4750
.0250.0250
-1.96 1.96
The Inverse Transformation (Continued)The Inverse Transformation (Continued)
z .02 .03 .04 . . . . . . . . . . . . . . .2.2 . . . 0.4868 0.4871 0.48752.3 . . . 0.4898 0.4901 0.49042.4 . . . 0.4922 0.4925 0.4927 . . . . . . . . . . . . . . .
8.27.26.25.24.23.2
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
8.27.26.25.24.23.2
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
X
f(x)
Normal Distribution: = 5.7 = 0.5
543210-1-2-3-4-5
z Z.01 = 2.33
Area = 0.49
Area = 0.01
X.01 = +z = 5.7 + (2.33)(0.5) = 6.865
4-30
4000300020001000
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
0.0000
X
f( x)
Normal Distribution: = 2450, = 400
.
.
.
.
.
.
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Norm al D istribution
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability
4-31
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
2. Shade the area corresponding to the desired probability.
2. Shade the area corresponding to the desired probability.
Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability
4000300020001000
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
0.0000
X
f( x)
Normal Distribution: = 2450, = 400
.
.
.
.
.
.
.4750.4750
.9500
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Norm al D istribution
.4750.4750
.9500
4-32
z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .
3. From the table of the standard normal distribution, find the z value or values.
3. From the table of the standard normal distribution, find the z value or values.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
2. Shade the area corresponding to the desired probability.
2. Shade the area corresponding to the desired probability.
Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability
4000300020001000
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
0.0000
X
f( x)
Normal Distribution: = 2450, = 400
.
.
.
.
.
.
.4750.4750
.9500
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Norm al D istribution
.4750.4750
.9500
-1.96 1.96
4-33
4. Use the transformation from z to x to get value(s) of the original random variable.
4. Use the transformation from z to x to get value(s) of the original random variable.
x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)
x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)
Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability
z .05 .06 .07 . . . . . . . . . . . . . . .1.8 . . . 0.4678 0.4686 0.46931.9 . . . 0.4744 0.4750 0.47562.0 . . . 0.4798 0.4803 0.4808 . . . . . . . . . .
3. From the table of the standard normal distribution, find the z value or values.
3. From the table of the standard normal distribution, find the z value or values.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
2. Shade the area corresponding to the desired probability.
2. Shade the area corresponding to the desired probability.
4000300020001000
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
0.0000
X
f( x)
Normal Distribution: = 2450, = 400
.
.
.
.
.
.
.4750.4750
.9500
543210-1-2-3-4-5
0.4
0.3
0.2
0.1
0.0
Z
f(z)
Standard Norm al D istribution
.4750.4750
.9500
-1.96 1.96
4-34
1050
0.3
0.2
0.1
0.0
X
f(x)
Normal Distribution: = 3.5, = 1.323
76543210
0.3
0.2
0.1
0.0
X
P(x
)
Binomial Distribution: n = 7, p = 0.50
The normal distribution with = 3.5 and = 1.323 is a close approximation to the binomial with n = 7 and p = 0.50.
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
MTB > cdf 4.5;SUBC> normal 3.5 1.323.Cumulative Distribution Function
Normal with mean = 3.50000 and standard deviation = 1.32300
x P( X <= x) 4.5000 0.7751
MTB > cdf 4.5;SUBC> normal 3.5 1.323.Cumulative Distribution Function
Normal with mean = 3.50000 and standard deviation = 1.32300
x P( X <= x) 4.5000 0.7751
MTB > cdf 4;SUBC> binomial 7,.5.Cumulative Distribution Function
Binomial with n = 7 and p = 0.500000
x P( X <= x) 4.00 0.7734
MTB > cdf 4;SUBC> binomial 7,.5.Cumulative Distribution Function
Binomial with n = 7 and p = 0.500000
x P( X <= x) 4.00 0.7734
P( x 4) = 0.7734
Finding Values of a Normal Random Finding Values of a Normal Random Variable, Given a ProbabilityVariable, Given a Probability
4-35
4-7 The Normal Approximation of Binomial 4-7 The Normal Approximation of Binomial DistributionDistribution
1050
0.3
0.2
0.1
0.0
X
f(x)
Normal Distribution: = 5.5, = 1.6583
11109876543210
0.2
0.1
0.0
X
P(x
)
Binomial Distribution: n = 11, p = 0.50
The normal distribution with = 5.5 and = 1.6583 is a closer approximation to the binomial with n = 11 and p = 0.50.
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
P(x 4) = 0.2744
4-36
or:
NOTE: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.
NOTE: If p is either small (close to 0) or large (close to 1), use the Poisson approximation.
Approximating a Binomial Probability Approximating a Binomial Probability Using the Normal Distribution Using the Normal Distribution
P a X b Pa np
np pZ
b npnp p
( ) ( ) ( )
1 1
for large (n 50) and not too close to 0 or 1.00n p
P a X b Pa np
np pZ
b npnp p
( ) .( )
.( )
0 51
0 51
for moderately large (20 n < 50).n
4-37
Using the Template for Normal Approximation Using the Template for Normal Approximation of the Binomial Distributionof the Binomial Distribution