Discrete & Continuous Probability Distributions

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Discrete & Continuous Probability Distributions Sunu Wibirama Basic Probability and Statistics Department of Electrical Engineering and Information Technology Faculty of Engineering, Universitas Gadjah Mada

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Transcript of Discrete & Continuous Probability Distributions

OUTLINE
SOME IMPORTANT THINGS
Please read Walpole 8th Chapter 5 (for Binomial and Poisson) and Chapter 6 (for Normal Distribution)
You can use Table A.1, A.2, and A.3 from Walpole book (p. 751)
Some exercises from these chapters (5 and 6) will be used in Midterm Exam
Binomial Distribution
CASE 1
Suppose that 80% of the jobs submitted to a data-processing center are of a statistical nature.
Selecting a random sample of 10 submitted jobs would be analogous to tossing an unbalanced coin 10 times, with the probability of observing a head (drawing a statistical job) on a single trial equal to 0.80.
CASE 2
Test for impurities commonly found in drinking water from private wells showed that 30% of all wells in a particular country have impurity A.
If 20 wells are selected at random then it would be analogous to tossing an unbalanced coin 20 times, with the probability of observing a head (selecting a well with impurity A) on a single trial equal to 0.30.
BINOMIAL EXPERIMENT
Inspection of public opinian (approve or not)
Inspection of digging wells (success or failed)
Etc….
BERNOULLI PROCESS
An experiment often consists of repeated trials, each with two possible outcome that may be labeled success and failed
The process is called Bernoulli Process:
Consists of n repeated trials
The outcome can be classified as success of failed (i.e., only 2 possible outcomes)
The probability of success (p) remains constant from trial to trial
The repeated trials are independent
BINOMIAL DISTRIBUTION
The number X of successes in n Bernoulli trials is called binomial random variable
The probability of success is denoted by p
The probability distribution of binomial random variable is called binomial distribution
b(x; n, p)
Each failure has probability, q q = 1 - p
x = number of successes
Since independent, combination probabilities are multiplied
b(x ;n , p ) = n ! x !(n ! x )!
" #$
b(x;n,p)
6 5 8 7 10 9 12 11 2 1 4 3 13 0
BINOMIAL CUMULATIVE DISTRIBUTION
0
1),;(),;(
• As with all other distributions, the sum of all probabilities must equal unity:
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Table A.1: Binomial Cumulative Distribution Proportion, p
Trials Success 0.1 0.2 0.25 0.3 0.4 0.5 0.6 0.7 0.8 0.9
1 0 0.9000 0.8000 0.7500 0.7000 0.6000 0.5000 0.4000 0.3000 0.2000 0.1000
1 1 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
2 0 0.8100 0.6400 0.5625 0.4900 0.3600 0.2500 0.1600 0.0900 0.0400 0.0100
2 1 0.9900 0.9600 0.9375 0.9100 0.8400 0.7500 0.6400 0.5100 0.3600 0.1900
2 2 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
3 0 0.7290 0.5120 0.4219 0.3430 0.2160 0.1250 0.0640 0.0270 0.0080 0.0010
3 1 0.9720 0.8960 0.8438 0.7840 0.6480 0.5000 0.3520 0.2160 0.1040 0.0280
3 2 0.9990 0.9920 0.9844 0.9730 0.9360 0.8750 0.7840 0.6570 0.4880 0.2710
3 3 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
4 0 0.6561 0.4096 0.3164 0.2401 0.1296 0.0625 0.0256 0.0081 0.0016 0.0001
4 1 0.9477 0.8192 0.7383 0.6517 0.4752 0.3125 0.1792 0.0837 0.0272 0.0037
4 2 0.9963 0.9728 0.9492 0.9163 0.8208 0.6875 0.5248 0.3483 0.1808 0.0523
4 3 0.9999 0.9984 0.9961 0.9919 0.9744 0.9375 0.8704 0.7599 0.5904 0.3439
4 4 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
BINOMIAL DISTRIBUTION
EXAMPLE
Test for impurities commonly found in drinking water from private wells showed that 30% of all wells in a particular country have impurity A.
If a random sample of 5 wells is selected from the large number of wells in the country, what is the probability that: Exactly 3 will have impurity A?
At least 3?
Fewer than 3?
Confirm that this experiment is a binomial experiment.
This experiment consists of n = 5 trials, one corresponding to each random selected well.
Each trial results in an S (the well contains impurity A) or an F (the well does not contain impurity A).
Since the total number of wells in the country is large, the probability of drawing a single well and finding that it contains impurity A is equal to 0.30 and this probability will remain the same for each of the 5 selected wells.
SOLUTION (a)
Therefore, the sampling process represents a binomial experiment with n = 5 and p = 0.30.
a) The probability of drawing exactly x = 3 wells containing impurity A, with n = 5, p = 0.30 and x = 3
b(3;5, 0.30) = 5! 3!2!
SOLUTION (b)
b) The probability of observing at least 3 wells containing impurity A is:
P(x ≥ 3) = p(3)+p(4)+p(5).
We have calculated p(3) = 0.1323 p(4) = 0.02835, p (5) = 0.00243. In result,
P(x ≥ 3) = 0.1323+0.02835+0.00243 = 0.16380.
SOLUTION (c)
c) Probability of observing less than 3 wells containing impurity A is P(x<3) = p(0)+p(1)+p(2). We can avoid calculating 3 probabilities by using the complementary relationship P(x<3) = 1-P(x ≥ 3) = 1-0.16380 = 0.83692.
Poisson Distribution
POISSON DISTRIBUTION
The experiment consists of counting the number x of times a particular event occurs during a given unit of time
The probability that an event occurs in a given unit of time is the same for all units
The number of events that occur in one unit of time is independent of the number that occur in other units.
The mean number of events in each unit will be denoted by the Greek letter !
POISSON DISTRIBUTION
EXAMPLE
Suppose that we are investigating the safety of a dangerous intersection.
Past police records indicate a mean of 5 accidents per month at this intersection.
Suppose the number of accidents is distributed according to a Poisson distribution. Calculate the probability in any month of exactly 0, 1, 2, 3 or 4 accidents.
SOLUTION
Since the number of accidents is distributed according to a Poisson distribution and the mean number of accidents per month is 5, we have the probability of happening accidents in any month
By this formula we can calculate:
p(x) = 5 x e!5
p(0) = 0.00674, p(1) = 0.03370, p(2) = 0.08425, p(3) = 0.14042, p(4) = 0.17552.
POISSON CUMULATIVE DISTRIBUTION
Consider the previous example. Suppose we want to know cumulative distribution of probability less than 3 accidents per month
Than we have p(x < 3) = p(0) + p(1) + p(2) p(x < 3) = 0.00674+0.03370+0.08425 = 0.12469
Simplify your computation using Table A.2
P(r;!) = p( x=0
! x;!) see Table A.2
!
Most important continuous probability distribution
Graph called the “normal curve” (bell-shaped). Total area under the curve = 1
Derived by DeMoivre and Gauss. Called the “Gaussian” distribution.
Describes many phenomena in nature, industry and research
Random variable, X
0.0000
0.0500
0.1000
0.1500
0.2000
0.2500
0.3000
0.3500
0.4000
0.4500
f( x)
NORMAL DISTRIBUTION
Definition: Density function of the normal random variable, X, with mean and variance such that:
f (x) = n(x;µ,! ) = e !0.5 (x!µ )
! " #$
Different means and _______
Area under the curve between x= x1 and x= x2
equals P(x1 < x < x2)
STANDARD NORMAL DISTRIBUTION
• General density function:
% &' 2
2"
! " #$
STANDARD NORMAL DISTRIBUTION  
It’s  also  called  z-­distribu*on   It  has  a  mean  of  0  and  standard  [email protected]  of  1  
Transforming  normal  random  variable  x  to  standard  normal   random  variable  z   xz µ
σ −=
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THE ORIGIN OF Z-DISTRIBUTION
To enable use of z-table (Table A.3), transform to unit values Mean = 0
Variance =1 σ µ−= XZ
{ }
e dz
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EXAMPLE
Given the standard normal distribution, find the area under the curve that lies to the LEFT of z = 1.84
Solution: (use Table A.3)
f(x)
0
z 0.0000 0.0100 0.0200 0.0300 0.0400 0.0500 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678
SOLVING CASE WITH Z-DISTRIBUTION
Suppose we have a problem finding probability of a case in X domain p(X)
We can use Z-Distribution to solve the problem
We should transform X Z
“Finding X from Z”
Finding  x  from  z  
TUGAS INDIVIDU (sekaligus latihan untuk Ujian MIDTERM)
Kerjakan  dari  ebook  Walpole  (8th  [email protected]):   Exercise  5.9  (page  151,  kasus  truck)   Exercise  5.58  (page  165,  kasus  traffic  accidents)   Exercise  6.11  (page  186,  kasus  lawyer)   Exercise  6.14  (page  187,  kasus  students)  
Kerjakan  dengan  tulisan  yang  jelas,  bila  diperlukan,   tambahkan  ilustrasi  pada  jawaban  Anda.  Jangan  lupa   cantumkan  NAMA  dan  NIM  
Batas  waktu  pengumpulan,  hari  Senin  04  April  2011,       pkl.  12.00  WIB  (siang),  di  lab  SE  
THANK YOU and