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EC 421 STATISTICAL
COMMUNICATION THEORY
Instructor: Dr. Heba A. Shaban
Lecture # 2
BINOMIAL RANDOM VARIABLES
Class
of
discrete
random
variables
Binomial -- results from a binomial experiment.
=
A binomial random variable is defined as X=number
of successes in the n trials of a binomial experiment.
Conditions for a binomial experiment:
1. There are n “trials” where n is determined in advance
and is not a RANDOM value.
2. Two possible outcomes on each trial, called “success”
and “failure” and denoted S and F
F.
3. Outcomes are independent from one trial to the next.
4. Probability of a “success”, denoted by p, remains
same from one trial to the next. Probability of “failure” is
1 – p.
2
1
FINDING BINOMIAL PROBABILITIES
P X  k  
n!
nk
p k 1  p 
k!n  k !
for k = 0, 1, 2, …, n
E
Example:
l Probability
P b bili off Two
T
Wins
Wi in
i Three
Th
Plays
Pl
p = probability win = 0.2; plays of game are
independent.
X = number of wins in three plays.
What is P(X = 2)?
P X  2 
3!
3 2
.2 2 1  .2 
2!3  2 !
 3(.2) 2 (.8)1  0.096
3
EXPECTED VALUE AND STANDARD DEVIATION
FOR A BINOMIAL RANDOM VARIABLE
For a binomial random variable X based
on n trials and success probability p,
Mean
  E  X   np
Standard deviation   np 1  p 
4
2
EXAMPLE: CRV - TIME SPENT WAITING FOR BUS
Bus arrives at stop every 10 minutes. Person arrives at
stop at a random time, how long will s/he have to wait?
X = waiting time until next bus arrives.
X is a continuous random variable over 0 to 10
minutes.
Note: Height is 0.10
so total area under the
curve is (0.10)(10) = 1
This is an example of a
Uniform random
variable density
curve
This
is
a
uniform
distribution, thus area of
rectangle = w x h (there is
a correspondence between
area and probability
5
EXAMPLE: WAITING FOR BUS (CONT)
What is the probability of waiting time X was in the
interval from 5 to 7 minutes?
Probability = area under curve between 5 and 7
= (base)(height) = (2)(.1) = .2
P(a ≤ X ≤ b) is the
area
under
the
density curve over
th interval
the
i t
l between
b t
values a and b
6
3
NORMAL RANDOM VARIABLES
If a population of measurements follows a
normal curve,
curve and if X is the measurement for
a randomly selected individual from that
population, then
X is said to be a normal random variable
X is also said to have a normal distribution
Any normal random variable can be completely
characterized by its mean, m, and standard
deviation, s.
7
NORMAL DISTRIBUTIONS (OF NORMAL RANDOM
VARIABLES)
Just as there are many
y
different uniform
distributions (with different ranges of values), there
are also many different normal distributions, with
each one depending on 2 parameters: the
population mean and population SD.
The standard normal distribution is a normal
probability distribution that has a mean = 0
and a SD = 1.0, and the total area under its
density curve = 1.0
8
4
EXAMPLE NORMAL CURVES (HEIGHTS; MALES;
FEMALES)
Women:
µ = 63.6
 = 2.5
Men:
µ = 69.0
 = 2.8
63.6
69.0
Height (inches)
9
STANDARD SCORES
The formula for converting any value x
to a zz-score
score is
z
Value  Mean
x

Standard deviation

A z-score
z score measures the number of standard
deviations that a value falls from the mean.
10
5
EXAMPLE STANDARD SCORES FOR HEIGHT
For a population of college women, the zscore corresponding to a height of 62 inches is
z
Value  Mean
62  65

 1.11
Standard deviation
2 .7
This z-score tells
Thi
t ll us that
th t 62 inches
i h
i 1.11
is
1 11 standard
t d d
deviations below the mean height for this population.
11
EXAMPLE: PROBABILITY Z > 1.31
P(Z > 1.31) = 1 – P(Z  1.31)
= 1 – .9049 = .0951
12
6
EXAMPLE: PROBABILITY Z IS BETWEEN
–2.59 AND 1.31
P(-2.59  Z  1.31)
= P(Z
(  1.31)) – P(Z
(  -2.59))
= .9049 – .0048 = .9001
13
STANDARD NORMAL (Z) DISTRIBUTION
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
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
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
14
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
7
EXAMPLE:

If thermometers have an average (mean) reading of 0
degrees and a standard deviation of 1 degree for
freezing water and if one thermometer is randomly
selected, find the probability that it reads freezing
water between 0 degrees and 1.58 degrees.
P ( 0 < x < 1.58 ) =
0
1.58
STANDARD NORMAL (Z) DISTRIBUTION
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
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
.0000
.0398
.0793
.1179
.1554
.1915
.2257
.2580
.2881
.3159
.3413
.3643
.3849
.4032
.4192
.4332
.4452
.4554
.4641
.4713
.4772
.4821
.4861
.4893
.4918
.4938
.4953
.4965
.4974
.4981
.4987
.0040
.0438
.0832
.1217
.1591
.1950
.2291
.2611
.2910
.3186
.3438
.3665
.3869
.4049
.4207
.4345
.4463
.4564
.4649
.4719
.4778
.4826
.4864
.4896
.4920
.4940
.4955
.4966
.4975
.4982
.4987
.0080
.0478
.0871
.1255
.1628
.1985
.2324
.2642
.2939
.3212
.3461
.3686
.3888
.4066
.4222
.4357
.4474
.4573
.4656
.4726
.4783
.4830
.4868
.4898
.4922
.4941
.4956
.4967
.4976
.4982
.4987
.0120
.0517
.0910
.1293
.1664
.2019
.2357
.2673
.2967
.3238
.3485
.3708
.3907
.4082
.4236
.4370
.4484
.4582
.4664
.4732
.4788
.4834
.4871
.4901
.4925
.4943
.4957
.4968
.4977
.4983
.4988
.0160
.0557
.0948
.1331
.1700
.2054
.2389
.2704
.2995
.3264
.3508
.3729
.3925
.4099
.4251
.4382
.4495
.4591
.4671
.4738
.4793
.4838
.4875
.4904
.4927
.4945
.4959
.4969
.4977
.4984
.4988
.0199
.0596
.0987
.1368
.1736
.2088
.2422
.2734
.3023
.3289
.3531
.3749
.3944
.4115
.4265
.4394
.4505
.4599
.4678
.4744
.4798
.4842
.4878
.4906
.4929
.4946
.4960
.4970
.4978
.4984
.4989
.0239
.0636
.1026
.1406
.1772
.2123
.2454
.2764
.3051
.3315
.3554
.3770
.3962
.4131
.4279
.4406
.4515
.4608
.4686
.4750
.4803
.4846
.4881
.4909
.4931
.4948
.4961
.4971
.4979
.4985
.4989
.0279
.0675
.1064
.1443
.1808
.2157
.2486
.2794
.3078
.3340
.3577
.3790
.3980
.4147
.4292
.4418
.4525
.4616
.4693
.4756
.4808
.4850
.4884
.4911
.4932
.4949
.4962
.4972
.4979
.4985
.4989
.0319
.0714
.1103
.1480
.1844
.2190
.2517
.2823
.3106
.3365
.3599
.3810
.3997
.4162
.4306
.4429
.4535
.4625
.4699
.4761
.4812
.4854
.4887
.4913
.4934
.4951
.4963
.4973
.4980
.4986
.4990
16
.0359
.0753
.1141
.1517
.1879
.2224
.2549
.2852
.3133
.3389
.3621
.3830
.4015
.4177
.4319
.4441
.4545
.4633
.4706
.4767
.4817
.4857
.4890
.4916
.4936
.4952
.4964
.4974
.4981
.4986
.4990
*
*
8
USING SYMMETRY TO FIND THE AREA
TO THE LEFT OF THE MEAN
Because of symmetry, these areas are equal.
(a)
(b)
0.4925
0.4925
0
z = - 2.43
0
Equal distance away from 0
z = 2.43
NOTE: Although a z-score can be negative, the area under the
curve (or the corresponding probability) can never be negative.
THE EMPIRICAL RULE
STANDARD NORMAL DISTRIBUTION: µ = 0 AND S = 1
99.7% of data are within 3 standard deviations of the mean
95% within
2 standard deviations
68% within
1 standard deviation
34%
34%
2.4%
2.4%
0.1%
0.1%
13.5%
x - 3s
x - 2s
13.5%
x-s
x
x+s
x + 2s
x + 3s
9
PROBABILITY OF HALF OF A DISTRIBUTION
0.5
0
FINDING THE AREA TO THE RIGHT OF Z = 1.27
Value found
in Table
0.3980
0
This area is
0.5 - 0.3980 = 0.1020
z = 1.27
10
FINDING THE AREA BETWEEN Z = 1.20
AND Z
= 2.30
0.4893 (from Table with z = 2.30)
Area A is 0.4893 - 0.3849 =
0.1044
0.3849
A
0
z = 1.20 z = 2.30
CUMULATIVE DISTRIBUTION FUNCTION
11
PROPERTIES OF Q(.) FUNCTION
2
12