3+ +Probability+Distributions
-
Upload
prashant-bhanu-sharma -
Category
Documents
-
view
216 -
download
0
Transcript of 3+ +Probability+Distributions
-
8/8/2019 3+ +Probability+Distributions
1/32
Quantitative Methods 2010
3
Probability Distributions
1QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
2/32
Frequency Distributions
When data is very voluminous, we may find it
useful to view it in compressed form. One
way to compress is to show data as a
frequency table or a frequency
distribution.
Divide the entire range of data into groups or
classes
Show how many data points fall into each class.
QM 2010 - Kingston 2
-
8/8/2019 3+ +Probability+Distributions
3/32
QM 2010 - Kingston 3
Store Stock, Days
1 2.02 3.8
3 4.1
4 4.7
5 5.5
6 3.4
7 4.08 4.2
9 4.8
10 5.5
11 3.4
12 4.1
13 4.3
14 4.9
15 5.5
16 3.8
17 4.1
18 4.7
19 4.9
20 5.5
0
1
2
3
4
5
6
7
8
9
2.0 to
2.5
2.6 to
3.1
3.2 to
3.7
3.8 to
4.3
4.4 to
4.9
5.0 to
5.5
More
Raw Data Frequency Distribution
As Table
Frequency Distribution
As Histogram
Stock, Days Frequency
2 0 2 5 1
2 6 3 1 0
3 2 3 7 2
3 8 4 3 84 4 4 9 5
5 0 5 5 4
0
If ins ad f c un s, y u sh w as
f ac i n p c n ag f al
numb f bs va i ns, y u g a
R la iv Frequency Dis ribu i n
-
8/8/2019 3+ +Probability+Distributions
4/32
Probability Distributions
They are related to Frequency Distributions.
Frequency Distributions show observedfrequencies of outcomes in an experiment.
Probability Distributions indicate expectedfrequencies of all possible outcomes in theexperiment.
Discrete Probability Distributions indicate expectedfrequencies directly
Continuous Probability Distributions indicate expectedfrequencies indirectly, by area under the curve.
QM 2010 - Kingston 4
-
8/8/2019 3+ +Probability+Distributions
5/32
Probability Distributions
Discrete Continuous
QM 2010 - Kingston 5
0 1 2 3 4 5 ------- 500
P[ X ]
X = No Of Parts Bad X = Weight Of Students, kg
f(x)
45 50 55 60 65 70 75
P[x] directly indicates probability
of particular x
f[x] does not directly indicate probability
of particular x.
-
8/8/2019 3+ +Probability+Distributions
6/32
Discrete Probability Distributions
QM 2010 - Kingston 6
-
8/8/2019 3+ +Probability+Distributions
7/32
Counts Of Events
Many situations, where we are interested in counts:how many times something happened. We receive 500 bolts every week.
How many bolts were defective each week?
Each possible no can be thought of as event, and it will have itsown probability.
Zero defective ------ P[0]
1 defective ------- P[1]
2 defective -------- P[2]
500 defective ------ P[500]
Counts are one example of discrete variables.
7QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
8/32
Discrete Probability Distributions
A table or picture
(graph) showing
probabilities for all
possible values of a
discrete variable is
called a discrete
distribution.
8
Value P[ ]
0 0.02
1 0.15
2 0.18
3 0.25
500 0.0005
Total 1.0000
0 1 2 3 4 5 ------- 500
P[ ]
ValueQM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
9/32
Discrete Cumulative Probability Distributions
9
A table or picture
(graph) showing for all
possible values of a
discrete variable, thesum of probabilities is
called a cumulative
discrete distribution.
Value P[ ]
0 0.02
1 0.17
2 0.35
3 0.6
500 1.0000
0 1 2 3 4 5 ------- 500
P[ ]
1.00
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
10/32
Binomial Distribution
Describes discrete data resulting from aBernoulli Process
Bernoulli Process is a process that has only 2
outcomes (example: success, failure) withfixed probability p and q = (1 p), andindependent trials.
Gives the probability of r successes in n trials: P[r/n] =
Binomial tables are available to look up probabilities
Mean, = np, Variance 2 = npq
10
rnrqp
rnr
n
)!(!
!
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
11/32
Example Binomial Probabilities
There are 15 members of a certain political party in
the Bombay Municipal Corporation (BMC), who are
notorious for not being present when BMC is in
session. It has been found that the probability of amember of this party being absent is 0.35. What is
the probability that for an upcoming crucial budget
session, 10 of them will be absent?
Ans: P[10/15] = = 0.0096
11
510)65.0()35.0(
!5!10
!15
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
12/32
Extract From Binomial Tables
12
n r
1
2
21
22
2
2
2
2
2
2
2
1
2
1
1
2
1
2
11
11
al es
QM 2
1
- Kingston
-
8/8/2019 3+ +Probability+Distributions
13/32
Poisson Distribution
Occurs in situations where
Probability of an event in a time-interval is verysmall, and the same for all intervals
Probability of 2 or more events in that interval ~ 0 No of events in any interval is independent of where
the interval occurs
No of events in any interval is independent of no. in
any other interval P(x) = --- Tables available
Mean, = = Variance, 2
13
!/ xex PP
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
14/32
Example Poisson Probabilities
The Vikhroli intersection on Eastern ExpressHighway is notorious for accidents. Recordsshow a mean of 5 accidents per month, and
Poisson distribution. What is the probability ofmore than 3 accidents in a given month?
P[x > 3] = 1 P[x 3]
= 1 {P[0] + P[1} + P[2] + P[3]}
= 1 {0.00674 + 0.03370 + 0.08425 + 0.14042}
= 1 0.26511
= 0.73 or 73%
14QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
15/32
Extract From Poisson Tables
15
x 1 5 5
1
5
1
1 55
QM
1 - Kingston
-
8/8/2019 3+ +Probability+Distributions
16/32
Poisson
As An Approximation To
Binomial In a Binomial situation, if n is large ( 20) and p
is small ( .05), you can use a Poisson with =np.
Example You are an investment banker, advising onIPO of 20 clients. Probability of any one of them beingunder-subscribed is .02. What is the probability of 3clients being under-subscribed?
Situation is really Binomial: n= 20, p = .02
P[3 out of 20] = .0065 ---- (from Binomial tables)
Using Poisson, with = np = .40 :
P[3] = .0072 ---- (from Poisson tables)
QM 2010 - Kingston 16
-
8/8/2019 3+ +Probability+Distributions
17/32
Continuous Probability Distributions
QM 2010 - Kingston 17
-
8/8/2019 3+ +Probability+Distributions
18/32
Normal Distribution N(,)
18
Roughly 1
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
19/32
Uniform Distribution U(a, b)
QM 2010 - Kingston 19
a b
= 95 = 100
X = mm of rainfall
f(x)
-
8/8/2019 3+ +Probability+Distributions
20/32
Exponential Distribution
QM 2010 - Kingston 20
X = Time Between Arrivals At Toll-booth, mins
f(x)
-
8/8/2019 3+ +Probability+Distributions
21/32
Expected Value Of A Random Variable
QM 2010 - Kingston 21
-
8/8/2019 3+ +Probability+Distributions
22/32
Random Variables
Random variable
Takes different values as the result of a random
experiment
Only limited number of values discrete random
variable
Any values within a range continuous random
variable
22QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
23/32
Expected Values For
Discrete Random Variables It is basically a weighted average of the value of each
possible outcome, weighted by its probability.
Sum of cross-products of variable value and itsprobability E(x) =
Example The no. of defective parts in boxes
of 1000 has the following
probability distribution. What is
the expected value of no. of
defective parts?
Ans: 4.13
23
!
n
i
ii xpx1
)(Variabl
cti
! art"
i # boxof $ % % %
Variabl
Val & "
' xi ( ) xi 0 1 ross- ! rod
0 0.01 0
1 0.07 0.07
2 0.22 0.44
3 0.18 0.54
4 0.14 0.56
5 0.11 0.55
6 0.09 0.54
7 0.08 0.56
8 0.05 0.4
9 0.03 0.27
10 0.02 0.2
E ) xi 0 4.13
CALCULATI 2
E3
(ECTE 4 VALUE 5 F
4 ISCRETE RANDOMVARIABLE
QM2010 - Ki 6 gston
-
8/8/2019 3+ +Probability+Distributions
24/32
Expected Values For
Continuous Random Variables
It is equivalent to that of discrete random
variables
Example - Rainfall in June is uniformly distributed between 95
and 100 mm. What is the expected rainfall in June?
f(x) = 1/(b a)
QM 2010 - Kingston 24
! dxxxfxE )()(
5.97
2
95100
2
)()(
2)(
1
2)(
1
)(
1
)(
1
)()(
222
!
!
!
!
!
!
!!
ab
abab
xab
xdxabdxabxdxxxfx
b
a
b
a
b
a
-
8/8/2019 3+ +Probability+Distributions
25/32
Use Of Normal Distribution
Example
A certain type of project activity takes an average
of 22 days to complete, and has a standard
deviation of 4 days. What is the probability that in
the current project it may take upto 30 days?
25QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
26/32
The Required Probability
26
= 22 30
Let x be no of days to complete the activity
We want to know P[x 30]
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
27/32
Standard Normal Distribution
Since there are an infinite number of Normal
distributions, we cannot have table for all of
them
Every Normal x with (, ) can be converted
into equivalent Standard Normal z with (0, 1)
by the conversion
We have table for z, Standard Normal.
QM 2010 - Kingston 27
W
Q)(
! xz
-
8/8/2019 3+ +Probability+Distributions
28/32
28
z F(z) z F(z) z F(z) z F(z)
-4 3.16712 -05 -2 0.02275 0.1 0.539828 2 0.97725
-3.9 4.80963 -05 -1.9 0.028717 0.2 0.57926 2.1 0.982136
-3.8 7.2348 -05 -1.8 0.03593 0.3 0.617911 2.2 0.986097
-3.7 0.0001078 -1.7 0.044565 0.4 0.655422 2.3 0.989276
-3.6 0.000159109 -1.6 0.054799 0.5 0.691462 2.4 0.991802
-3.5 0.000232629 -1.5 0.066807 0.6 0.725747 2.5 0.99379
-3.4 0.000336929 -1.4 0.080757 0.7 0.758036 2.6 0.995339
-3.3 0.000483424 -1.3 0.0968 0.8 0.788145 2.7 0.996533
-3.2 0.000687138 -1.2 0.11507 0.9 0.81594 2.8 0.997445-3.1 0.000967603 -1.1 0.135666 1 0.841345 2.9 0.998134
-3 0.001349898 -1 0.158655 1.1 0.864334 3 0.99865
-2.9 0.001865813 -0.9 0.18406 1.2 0.88493 3.1 0.999032
-2.8 0.00255513 -0.8 0.211855 1.3 0.9032 3.2 0.999313
-2.7 0.003466974 -0.7 0.241964 1.4 0.919243 3.3 0.999517
-2.6 0
.004661188 -0
.6 0
.274253 1
.5 0
.933193 3
.4 0
.999663
-2.5 0.006209665 -0.5 0.308538 1.6 0.945201 3.5 0.999767
-2.4 0.008197536 -0.4 0.344578 1.7 0.955435 3.6 0.999841
-2.3 0.01072411 -0.3 0.382089 1.8 0.96407 3.7 0.999892
-2.2 0.013903448 -0.2 0.42074 1.9 0.971283 3.8 0.999928
-2.1 0.017864421 -0.1 0.460172 2 0.97725 3.9 0.999952
-2 0.022750132 0 0.5 4 0.999968
AREA UNDER STANDARD NORMAL TO THE LEFT OF Z
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
29/32
-
8/8/2019 3+ +Probability+Distributions
30/32
The Standard Normal Distribution
30
= 22 X =30 = 0 Z = 2
= 1
Z = (x - ) /
From table or EXCEL =normsdist(2), P[z 2] = P[x 30] = .97725
QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
31/32
P[x 25 days]
31
= 0
Z = 0.75
= 1
Z = (x - ) /
= 22X =25
= 4
From table or EXCEL =normsdist(2), P[z 0.75] = 0.7734. Then P[x 25] = 1
P[z < 0.75] = 0.2266QM 2010 - Kingston
-
8/8/2019 3+ +Probability+Distributions
32/32
End OfProbability Distributions
32QM 2010 - Kingston