Slide 1
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Shakeel NoumanM.Phil Statistics
The Normal Distribution
Slide 2
Using Statistics The Normal Probability Distribution The Standard Normal Distribution The Transformation of Normal Random
Variables The Inverse Transformation The Normal Distribution as an
Approximation to Other Probability Distributions
Summary and Review of Terms
The Normal Distribution4
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 3
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 Dis tribution: 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 Dis trib ution: n=14, p=.5
50-5
0.4
0.3
0.2
0.1
0.0
x
f( x)
Normal Distribution: = 0, = 1
f x ex
x
e
( )
. ... . ...
12 2
2
2 2
2 7182818 314159265
for
where and
4-1 Introduction
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 4
The normal probability density function:
50-5
0.4
0.3
0.2
0.1
0.0
x
f( x)
Normal Distribution: = 0, = 1
f x ex
x
e
( )
. ... . ...
12 2
2
2 2
2 7182818 314159265
for
where and
The Normal Probability Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 5
• 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.
Properties of the Normal Probability Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 6
• 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).
Properties of the Normal Probability Distribution (continued)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 7
• 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.
Properties of the Normal Probability Distribution
(continued)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 8
Example 4.1: Let X1, X2, and X3 be independent random variables that are normally distributed with means and variances as shown.
Properties of the Normal Probability Distribution – Example
4-1
Mean VarianceX1 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
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 9
• 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.
Properties of the Normal Probability Distribution (continued)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 10
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
Properties of the Normal Probability Distribution – Example 4-3
Mean VarianceX1 12 4
X2 -5 2
X3 8 5
X4 10 1
E(Q) = 12 – 2(-5) + 3(8) – 4(10) + 5 = 11V(Q) = 4 + (-2)2(2) + 32(5) + (-4)2(1) = 73SD(Q) =
544.873
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 11Computing the Mean, Variance and Standard Deviation for the Sum of Independent Random Variables Using the Template
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 12
All of these are normal probability density functions, though each has a different mean and variance.
Z~N(0,1)
50-5
0.4
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 Distributions
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 13Computing Normal Probabilities
Using the Template
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 14
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{
4-3 The Standard Normal Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 15
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) = .4406
Finding Probabilities of the Standard Normal Distribution: P(0
< Z < 1.56)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 16
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 Normal Distribution: 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
.
.
.
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 17
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 2.00F(2) P(Z 2.00) .5 + .4772 .97722. Find table area for 1.00F(1) P(Z 1.00) .5 + .3413 .8413
3. P(1 Z 2.00) P(Z 2.00) P(Z 1.00) .9772 .8413 .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
Finding Probabilities of the Standard Normal Distribution:
P(1< Z < 2)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 18
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
Finding Values of the Standard Normal Random Variable: P(0 < Z
< z) = 0.40z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
0.0 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
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 19
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
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 Mean
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 20
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 m = 50 and s = 10.
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:
ZX x
x
The inverse transformation of Z to X:
X x Z x +
4-4 The Transformation of Normal Random Variables
(1) Subtraction: (X - x)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 21
Example 4-9 X~N(160,302)
( )
P X
PX
P Z
P Z
( )
.. . .
100 180100 180
100 160
30
180 160
302 6667
0 4772 0 2475 0 7247
+
Example 4-10X~N(127,222)
( )
P X
PX
P Z
P Z
( )
.. . .
+
150150
150 127
221 045
0 5 0 3520 0 8520
Using the Normal Transformation
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 22
Example 4-11 X~N(383,122)
( )
P X
PX
P Z
P Z
( )
. .. . .
394 399394 399
394 383
12
399 383
120 9166 1 333
0 4088 0 3203 0 0885
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 - Example 4-11
Template solution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 23
The transformation of X to Z:
ZX x
x
The inverse transformation of Z to X:X
xZ
x +
The transformation of X to Z, where a and b are numbers::
P X a P Z a
P X b P Z b
P a X b P a Z b
( )
( )
( )
The Transformation of Normal Random Variables
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 24
543210-1-2-3-4-5
0 .4
0 .3
0 .2
0 .1
0 .0
Z
f(z)
S t a n d a rd N o rm a l D is trib u tio 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)
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 25
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.
P X Px
P Z P Z( ) ( )
70
70 70 5010
2
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
4-5 The Inverse Transformation
0.01
139.36
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 26
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 4-12
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 27
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
. . . . . . . . . . . . . . .
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
. . . . . . . . . .
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)0.95
x = z = 2450 ± (1.96)(400) = 2450 ±784=(1666,3234)
P(1666 < X < 3234) = 0.95
8.27.26.25.24.23.2
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.08.27.26.25.24.23.2
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
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)
X.01 = +z = 5.7 + (2.33)(0.5) = 6.865
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 28
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)
S ta nd ard Norm al Dis trib utio n
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
Finding Values of a Normal Random Variable, Given a
Probability
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 29
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
2. Shade the area corresponding to the desired probability.
Finding Values of a Normal Random Variable, 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)
S ta nd ard Norm al Dis trib utio n
.4750.4750
.9500
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 30
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.
1. Draw pictures of the normal
distribution in question and of the standard normal
distribution.
2. Shade the area corresponding to the desired probability.
Finding Values of a Normal Random Variable, Given a
Probability
4000300020001000
0.0012
0.0010
0.0008
0.0006
0.0004
0.0002
0.0000
X
f( x)
Norm al 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)
S ta nd ard Norm al Dis trib utio n
.4750.4750
.9500
-1.96 1.96
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 31
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)
Finding Values of a Normal Random Variable, 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.
1. Draw pictures of the normal distribution in question and of the standard normal distribution.
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)
S ta nd ard Norm al Dis trib utio n
.4750.4750
.9500
-1.96 1.96
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 32
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.
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;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 Variable, Given a
Probability
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 33
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.
P(x < 4.5) = 0.2732P(x 4) = 0.2744
MTB > cdf 4.5;SUBC> normal 5.5 1.6583.
Cumulative Distribution Function
Normal with mean = 5.50000 and standard deviation = 1.65830
x P( X <= x) 4.5000 0.2732
MTB > cdf 4;SUBC> binomial 11,.5.
Cumulative Distribution Function
Binomial with n = 11 and p = 0.500000
x P( X <= x) 4.00 0.2744
4-6 The Normal Approximation of Binomial Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 34
P a X b P a npnp p
Z b npnp p
( ) ( ) ( )
1 1
for large (n 50) and not too close to 0 or 1.00n p
P a X b P a npnp p
Z b npnp p
( ) .( )
.( )
+
0 51
0 51
for moderately large (20 n < 50).n
or:
If p is either small (close to 0) or large (close to 1), use the Poisson approximation.
Approximating a Binomial Probability Using the Normal
Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 35
Using the Template for Normal Approximation of the Binomial
Distribution
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer
Slide 36
M.Phil (Statistics)
GC University, . (Degree awarded by GC University)
M.Sc (Statistics) GC University, . (Degree awarded by GC University)
Statitical Officer (BS-17) (Economics & Marketing Division)
Livestock Production Research Institute Bahadurnagar (Okara), Livestock & Dairy Development
Department, Govt. of Punjab
Name Shakeel NoumanReligion ChristianDomicile Punjab (Lahore)Contact # 0332-4462527. 0321-9898767E.Mail [email protected] [email protected]
The Normal Distribution By Shakeel Nouman M.Phil Statistics Govt. College University Lahore, Statistical Officer