Transcript ST3905 - Mathematical Sciences| |UCC

ST3905
Lecturer : Supratik Roy
Email : [email protected]
(Unix) : [email protected]
Phone: ext. 3626
What do we want to do?
1. What is statistics?
2. Describing Information :
3. Summarization, Visual and non-Visual representation
4. Drawing conclusion from information :
5. Managing uncertainty and incompleteness of information
Resources
1. Recommended textbook : Probability and Statistics for
Engineering and the Sciences : Jay L. Devore. International
Thomson Publishing.
2. Software : R : homepage www.r-project.home
Describing Information
1. Why summarization of information?
2. Visual representation (aka graphical Descriptive Statistics)
3. Non-visual representation (numerical measures)
4. Classical techniques vs modern IT
Stem and Leaf Plot
Decimal point is 2 places to the right of the colon
0:8
1 : 000011122233333333333344444
1 : 55555566666677777778888888899999999999
2 : 0000000111111111111222222233333333444444444
2 : 555556666666666777778889999999999999999
3 : 000000001111112222333333333444
3 : 55555555666667777777888888899999999
4 : 0122234
4 : 55555678888889
5 : 111111134
5 : 555667
6 : 44
6:7
com
ple
x
c om
ple
x
Pie-Chart
diffgeom
algebra
ics
reals
reals
diffgeom
algebra
sta
tist
sta
tist
ics
DotChart
Child Care
Old Suburb
Coast County
New Suburb
o
o
o
Health Services
Old Suburb
Coast County
New Suburb
o
o
o
Community Centers
Old Suburb
Coast County
New Suburb
o
o
Family & Youth
Old Suburb
Coast County
New Suburb
Other
Old Suburb
Coast County
New Suburb
o
o
o
o
o
10
o
20
30
o
Histogram
0
5
10
15
50 samples from a t distribution with 5 d.f.
-4
-3
-2
-1
my.sample
0
1
2
0
5
10
15
Histogram-Categorical
Northeast
South
North Central
state.region
West
Rules for Histograms
1. Height of Rectangle proportional to frequency of class
2. No. of classes proportional to sqrt(total no. of observations)
[not a hard and fast rule]
3. In case of categorical data, keep rectangle widths identical,
and base of rectangles separate.
4. Best, if possible, let the software do it.
Data
-0.053626486 -0.828128399 0.214910482 0.346570399
[5] -0.849316517 0.001077376 0.736191791 1.417540397
[9] -2.382332275 -2.699019949 -0.111907192 1.384903284
[13] 2.113286699 -1.828108272 -1.108280724 0.131883612
[17] -0.394494473 0.829806888 0.023178033 0.019839537
[21] -0.346280222 -0.251981108 1.159853307 -0.249501904
[25] -1.342704742 -2.012653224 -1.535503208 0.869806233
[29] -1.313495887 -0.244408426 -0.998886998 -1.446769605
[33] 1.224528053 -0.410163230 0.032230907 -0.137297112
[37] -2.717620031 -0.728570438 0.034697116 2.202863874
[41] -0.170794163 0.353651680 -0.673296374 3.136364814
[45] -1.260108638 -0.367334893 -0.652217259 -0.301847039
[49] 0.315180215 0.190766333
Tabulation
Class
-3,-2
-2,-1
-1,0
0,1
1,2
2,3
3,4
////
////
////
////
////
//
/
//
////
////
////
////
freq
4
7
///
18
14
4
2
1
Total 50
200
400
600
800
Box-Plot - I
1
2
3
4
5
6
7
Box Plot – II
18-24
25-34
35-44
45-54
55-64
65+
Box Plot – III
200
400
Payoff
600
800
NJ Pick-it Lottery (5/22/75-3/16/76)
0
1
2
3
4
5
6
Leading Digit of Winning Numbers
7
8
9
Non-Visual (numerical measures)
1. Pictures vs. quantitative measures
2. Criteria for selection of a measure – purpose of study
3. Qualities that a measure should have
4. We live in an uncertain world – chances of error
Measures of Location
1. Mean :
2. Mode
3. Median
Location : mean, median
algebra test scores
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
43 50 41 69 52 38 51 54 43 47 54 51 70 58 44 54 52 32 42 70
21 22 23 24 25 50 49 56 59 38
Mean = 50.68
10% trimmed mean of scores = 50.33333
Median = 51
Location : Non-classical
An M-estimate of location is a solution mu of the equation:
sum(psi( (y-mu)/s )) = 0.
Data set : car.miles
(bisquare) 204.5395
(Huber’s ) 204.2571
Tabular method of computing
Class freq
-3,-2
-2,-1
-1,0
0,1
1,2
2,3
3,4
4
7
18
14
4
2
1
50
Classmidpt
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
Rel.
freq
0.08
0.14
0.36
0.28
0.08
0.04
0.02
r.f X
midpt
-0.20
-0.21
-0.18
0.14
0.12
0.10
0.07
-0.16
Tabular method of computing
Class freq Classmidpt(x)
-3,-2
-2,-1
-1,0
0,1
1,2
2,3
3,4
4
7
18
14
4
2
1
50
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
A=-0.5
x-A/d
-2
-1
0
1
2
3
4
Rel.
freq
r.f X x
0.08
0.14
0.36
0.28
0.08
0.04
0.02
-0.16
-0.14
0
0.28
0.16
0.12
0.08
0.34
Measures of Scale (aka
Dispersion)
1. Variance (unbiased) : sum((x-mean(x))^2)/(N-1)
2. Variance (biased) : sum((x-mean(x))^2)/(N)
3. Standard Deviation : sqrt( variance)
Tabular method of computing
Class Classmidpt(x)
-3,-2
-2,-1
-1,0
0,1
1,2
2,3
3,4
-2.5
-1.5
-0.5
0.5
1.5
2.5
3.5
A=-0.5
x’=(xA)/d
-2
-1
0
1
2
3
4
x^2
Rel.
freq
r.f X
x^2
4
1
0
1
4
9
16
0.08
0.14
0.36
0.28
0.08
0.04
0.02
0.32
0.14
0
0.28
0.32
0.36
0.32
1.74
Robust measures of scale
1. The MAD scale estimate generally has very small bias
compared with other scale estimators when there is
"contamination" in the data.
2. Tau-estimates and A-estimates also have 50% breakdown,
but are more efficient for Gaussian data.
3. The A-estimate that scale.a computes is redescending, so it
is inappropriate if it necessary that the scale estimate always
be increasing as the size of a datapoint is increased.
However, the A-estimate is very good if all of the
contamination is far from the "good" data.
Comparison of scale measures
MAD(corn.yield) =4.15128
scale.tau(corn.yield) = 4.027753
scale.a(corn.yield) = 4.040902
var(corn.yield) = 19.04191
sqrt(var(corn.yield)) = 4.363703
N.B. To really compare you have to compare for various
probability distributions as well as various sample sizes.
Probability
1. Concept of an Experiment on Random observables
2. Sets and Events, Random variables, Probability
(a).Set of all basic outcomes = Sample space = S
(b).An element of S or union of elements in S = An event
(Asingleton event = simple event, else compound)
(c) A numerical function that associates an event with a
number(s) = Random Variable
(d) A map from E onto [0,1] obeying certain rules =
probability
Examples of Probability
Consider toss of single coin :
1. A single throw : Only two possible outcomes – Head or Tail
2. Two consecutive throws : Four possible outcomes – (Head,
Head), (Head, Tail), (Tail, Head), (Tail, Tail)
3. Unbiased coin : P(Head turns up) = 0.5
4. Define R.V. X to be X(Head)=1, X(Tail)=0. P(X=1)=0.5,
P(X=0)=0.5.
Axioms of Probability
1. 0 <= P(A) <= 1 for any event A
2. P[A  B] = P[A]+P[B] if A,B are disjoint sets/events
3. P[S] =1
Basic Formulae-I
1. P[A’] = 1- P[A]
2. P[A  B] = 0 if A,B are disjoint
3. P[A  B] = P[A]+P[B]-P[A  B]
4. P[A  B  C] = P[A]+P[B]+ P[C]
-P[A  B] –P[A  C] – P[B  C]
+P[A  B  C]
Basic Formulae-I-Examples-1
Consider the coin tossing experiment with three consecutive
tosses, and Head or Tail being equally likely in any throw.
Sample space = { HHH,HHT,HTH,HTT,THT,THH,TTH,TTT }
Define A = {there are at least 2 Heads }; P(A)=0.5
Define B = {there are at least 1 Tail };
P(B)=0.875
•
A  B = {HHT,THH,HTH}; P[A  B ]=3/8
•
P[A  B] = P[A]+P[B]-P[A  B] = 1
Basic Formulae-I-Examples-2
Venn Diagrams:
A
AB
B
Basic Formulae-I-Examples-3
Venn Diagrams:
A
AB
B
Basic Formulae-I-Examples-4
Venn Diagrams:
A
B’ or complement of B
B
Basic Formulae – I-Examples
1. A family that owns 2 cars is selected, and for both the older
car and the newer car we note whether the car was
manufactured in America, Europe or Asia. (a) what are the
possible outcomes of this experiment (b) which outcomes are
contained in the event that one car is American and the other
is non-American? ( c) which outcomes are contained in the
event that at least one car is non-American?
2. In a certain residential suburb, 60% of all households
subscribe to the metropolitan newspaper published in a
nearby city, 80% subscribe to the local afternoon paper, and
50% of all households subscribe to both papers. If a
household is selected at random, what is the probability that
it subscribes to (1) at least one of the two (2) exactly one
newspaper?
Basic Formulae - II
1. Counting Principle : For an ordered sequence to be formed
from N groups G1,G2,….GN with sizes k1,k2,….kN, the total
no. of sequences that can be formed are k1 x k2 x ….kN.
2. For any positive integer m, m! is read as “m-factorial” and
defined by m!=m(m-1)(m-2)…3.2.1
3. An ordered sequence of k objects taken from a set of n
distinct objects is called a Permutation of size k of the
objects, and is denoted by Pk,n = n(n-1)…(n-k+1) = n!/(n-k)!
4. Any unordered subset of size k from a set of n distinct
objects is called a Combination, denoted Ck,n. = Pk,n /k!
=n!/[k! (n-k)!]
Basic Formulae – II-Example
1. A student wishes to commute first to a junior college for two
years and then to a state college campus. Within commuting
range there are four junior colleges and three state colleges.
How many choices of junior college and state college are
available to her? I f junior colleges are denoted by 1,2,3,4
and state colleges by a,b,c, choices are (1,a),(1,b),…,(4,c), a
total of 12 choices. With n1 =4 and n2=3, N=n1n2=12 without
a list.
2. There are 8 teaching assistants available for grading papers
in a particular course. The first exam consists of 4 questions,
and the professor wishes to select a different assistant to
grade each question (only 1 assistant per question). In how
many ways can assistants be chosen to grade the exam? Ans.
P4,8 =(8)(7)(6)(5)=1680.
Basic Formulae – II-Examples
1. Consider the set {A,B,C,D,E}. We know that there are
5!/(5-3)! =60 permutations of size 3. There are 6 permutations of
size 3 consisting of the elements {A,B,C} since these 3 can
be ordered 3.2.1 = 3! = 6 ways: (A,B,C), (A,C,B),
(B,A,C),(B,C,A), (C,A,B) and (C,B,A). These 6
permutations are equivalent to the single combination
{A,B,C}. Similarly for any other combination of size 3, there
are 3! Permutations, each obtained by ordering the 3 objects.
Thus,
60 = P3,5 = C3,5 .3! So C3,5 =60 / 3! =10.
These 10 combinations are
{A,B,C},{A,B,D},{A,B,E},{A,C,D},{A,C,E},{A,D,E},{B,
C,D},{B,C,E},{B,D,E},{C,D,E}.
Basic Formulae – II-Example
1. The student Engineers council at a certain college has one
student representative from each of the 6 engineering majors
(civil, food, electrical, industrial, materials, and mechanical).
In how many ways can (a) Both a council president and a
vice president be selected? (b) A president, a vice-president,
and a secretary be selected? ( c) Two members be selected
for the Presidents Council?
2. A real estate agent is showing homes to a prospective buyer.
There are 10 homes in the desired price range listed in the
area. The buyer has time to visit only 3 of them. (a) In how
many ways could the 3 homes be chosen if the order of
visiting is considered? (b) how many ways could the 3 homes
chosen if the order is unimportant? If 4 of the homes are new
and 6 been previously occupied and if 3 homes to visit are
randomly chosen, what is the prob. That all 3 are new?
Basic Formulae-III
1. Pk,n = n!/(n-k)!
2. Ck,n = n!/[k!(n-k)!]
3. For any two events A and B with P(B)>0, the Conditional
Probability of A given (that ) B (has occurred)is defined by
P(A|B) = P(A  B)/P(B) [=0 if P(B)=0]
4. Let A,B be disjoint and C be any event with P[C]>0. Then
P(C)=P(C|A)P(A)+P(C|B)P(B) [Law of Total Probability]
5.
Let A,B be disjoint and C be any event with P[C]>0. Then
P(A|C)=P(C|A)P(A)/[P(C|A)P(A)+P(C|B)P(B)]. [Bayes
Theorem]
Basic Formulae-III-examples
1. Suppose that of all individuals buying a certain PC, 60%
include a word processing program in their purchase, 40%
include a spreadsheet program, and 30% include both types of
programs. Consider randomly selecting a purchaser and let
A={word processing program included} and B={spreadsheet
program included}. Then P(A)=0.6, p(B)=0.4, and P(both
included)=P(AB)=0.30. Given that the selected individual
included a spreadsheet program, the probability that a word
program was also included is P(A|B) = P(A  B)/P(B)
=0.30/0.40 =0.75.
Basic Formulae-III-examples
1. A chain of video stores sells 3 different brands of VCR’s. Of
its VCR sales, 50% are brand 1 (the least expensive), 30% are
brand 2, and 20% are brand 3. Each manufacturer offers a 1year warranty on parts and labour. It is known that 25% of
brand 1’s VCR’s require warranty repair work, whereas the
corresponding percentages for brands 2 and 3 are 20% and
10%, respectively. (a) What is the probability that a randomly
selected purchaser has a VCR that will need repair while
under warranty?
2. Let Ai ={brand I is purchased} for i=1,2,3. Let B={needs
repair}, the given data implies p(B|A1)=0.25,
P(B|A2)=0.20,P(B|A3)=0.10.
Basic Formulae-III-examples
1. Only 1 in 100 adults is afflicted with a rare disease for which a
diagnostic test has been developed. The test is such that when
an individual actually has the disease, a positive result will
occur 99% of the time, while an individual without the disease
will show a positive test result only 2% of the time. If a
randomly selected individual is tested and the result is
positive, what is the probability that the individual has the
disease? Let A1 ={individual has the disease}, A2
={individual does not have disease}, B={positive test result}.
Then P(A1)=0.001, P(A2)=0.999, P(B|A1)=0.99, and
P(B|A2)=0.02. P(B)=0.02097, P(A1|B)=P(A1B)/P(B)=0.047
Basic Formulae-IV
1. Two events A and B are independent if P(A|B) = P(A 
B)/P(B) = P(A) and are dependent otherwise.
2. Two events A and B are independent if and only if P(AB) =
P(A)P(B).
Random Variables - Discrete
1. A discrete set is a set such that either it is finite or there exists a
map from each element of the set into a subset of the set of
Natural numbers.
2. A discrete random variable is a r.v. which takes values in a
discrete set consisting of numbers.
3. The probability distribution or probability mass function (pmf)
of a discrete r.v. X is defined for every number x by
p(x)=P(X=x)=P(all s  S: X(s)=x)
[P[X=x] is read “the probability that the r.v. X assumes the value
x”. Note, p(x) >= 0, sum of p(x) over all possible x is 1
Random Variables - Discrete
1. Bernoulli trials : (Coin toss is a particular example). The
random variable X takes two values 1, and 0.
2. Notation : P[X=1]=p, 0<p<1 (Note that this automatically
implies P[X=0]=1-p)
3. A general (arbitrary) discrete random variable can be denoted
by an uppercase letter, say, X
4. The discrete values that can be taken by X are x1,x2,x3,…xn
(assuming that total no. of values possible is n)
5. Typically, the corresponding probability masses are denoted
by p1,p2,…,pn
Cumulative Distribution Function
1. The probability distribution or probability mass function of a
discrete r.v. is defined for every number x by p(x) =P(X=x) =
P(all s  S: X(s)=x).
2. The Cumulative distribution function (cdf) F(x) of a discrete
r.v. X with pmf (probability mass function) p(x) is defined for
every number x by F(x)=P(Xx)={y : y  x} p(y)
3. For any number x, F(x) is the probability that the observed
value of X will be at most x.
4. For any two numbers a,b with a  b, P(a  X  b) = F(b)-F(a-)
where a- represents the largest possible X value that is strictly
less than a.
Discrete R.V.-illustration
1. Consider the Bernoulli r.v. X with P[X=1]=p, 0<p<1. The
probability mass function can be given by px(1-p)1-x
2. The Cumulative distribution function (cdf) F(x) = P(Xx)
={y : y  x} p(y) = (1-p)1[x<1] .1[x>0]
3.
0
1
Operations on RV’s
1. Expectation of a RV
2. Expectations of functions of RV’s
3. Special Cases : Moments, Covariance
Expected Values of Random
Variables
1. Let X be a discrete r.v. with set of possible values D and pmf
p(x). The expected value or mean value of X, denoted by E(X)
or X , is E(X) = X ={xD} x.p(x)
2. Note that E(X) may not always exists. Consider p(x)=k/x2
3. For Bernoulli X, E(X)=p.1+(1-p).0 = p
4. E(a +bX) = a+bE(X) [linearity property of expectation]
Expected Values of functions of
Random Variables
1. Let X be a discrete r.v. with set of possible values D and pmf
p(x). The expected value or mean value of f(X), denoted by
E(f(X)) or  f(X) , is E(f(X)) ={xD} f(x).p(x)
2. Example : Variance.
E(X)]2=E(X2)-[E(X)]2
Var(X)=V(X)=E[X-
3. Variance of Bernoulli X : E(X-p)2 =E(X2)-p2 = 1.p –p2 =
p(1-p)
4. Classical expression of variance of n numbers x1,x2,…xn is
simply the variance of a r.v. X that takes the values x1,x2,…xn ,
each with probability 1/n.
Expected Values of functions of
Random Variables
1. E[a+bX]=a+bE[X]; Var(a+bX)=b2Var(X)
2. Standard deviation; aka s.d. is Var(X)
3. Let X be the r.v. with pmf
x
3
4
5
p(x)
.3
.4
.3
E(X)=30.3 +40.4 +50.3=4.0
Var(X) = (3-4)2 0.3 + (4-4)2 0.4 +(5-4)2 0.3 =0.6
s.d. (X) = 0.77
Expected Values of functions of
Random Variables
1. Let X be the r.v. with pmf
x
0
1
2
3
4
p(x)
.08
.15
.45
0.27
0.05
Find E(X), Var(X), s.d.(X)
R.V.D - Binomial
1. Binomial experiment : total number of a particular outcome
in a sequence of trials with only two possible outcomes.
2. The Binomial r.v. X, with parameters, (n,p) denoted for
short by BIN(n,p) is defined by
P[X=x]=Ck,n px(1-p)n-x , x=0,1,2,…,n
3. X=X1+X2+…+Xn, where Xk’s are independent Bernoulli
r.v.s.
4. E[X] = np (Exercise!) ; Var[X] = np(1-p)
R.V.D – Binomial-2
Consider the outcome for a binomial experiment with 4 trials
R .V . D - B in o m ial
O ut com e
SSSS
SSSF
SSF S
SSF F
SF SS
SF SF
SF F S
SF F F
x
4
3
3
2
3
2
2
1
P rob.
p4
p3 ( 1 ¡ p)
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
O ut com e
F SSS
F SSF
F SF S
F SF F
F F SS
F F SF
FFFS
FFFF
x
3
2
2
1
2
1
1
0
P rob
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
p( 1 ¡ p) 3
( 1 ¡ p) 4
R.V.D – Binomial-3
Each of six r andomly select ed cola dr inker s is given a
glass cont aining cola S and one cont aining cola F. T he
glasses ar e ident ical in appear ance except for a code
on t he bot t om t o ident ify t he cola. Suppose t her e is
act ually no t endency among cola dr inker s t o pr efer one
cola t o t he ot her . T hen p = P( a select ed individual
pr efer s S)= 0:5, so wit h X = t he number among t he six
who pr efer S, X » B I N (6; 0:5). T hus,
P(X = 3) =
0
1
B
B
B
@
C
C
C
A
6
3
(0:5) 3(0:5) 3 = 20(0:5) 6 = 0:313
t he pr obabilit y t hat at least t hr ee pr efer S is
0
P(X ¸ 3) =
X6 B
B
B
@
x= 3
1
6
x
C
C
C
A
(0:5) x (0:5) 6¡
x
= 20(0:5) 6 = 0:0:656
R.V.D – Binomial-4
E(X ) = np. Pr oof:
Fir st M et hod:
0
Xn
1
n
x @ A px (1 ¡ p) n¡ x =
E(X ) =
x
x= 0
Xn
x= 0
x
n!
px (1 ¡ p) n¡ x
x!(n ¡ x)!
Xn
(n ¡ 1)!
p:px¡ 1(1 ¡ p) n¡ 1¡ (x¡ 1)
n
=
x= 1 (x ¡ 1)!(n ¡ x)!
n¡
X 1
(n ¡ 1)!
px (1 ¡ p) n¡ 1¡ x = p
= np
x= 0 x!(n ¡ 1 ¡ (x ¡ 1))!
P
Second met hod : Recognize t hat X = nk= 1 X k wher e
each X k is a B er noulli r andom var iable. U se t he linear it y
Pn
pr oper t y of expect at ions, i.e., E(X ) = k= 1 E(X k ) = np
R.V.D – Binomial-5
V ar (X ) = np(1 ¡ p). Pr oof:
Fir st M et hod:
E (X 2) = E (X (X ¡ 1) + X ) = E [X (X ¡ 1)] + E [X ]
0
Xn
n
= np+
x(x¡ 1) @
x
x= 0
= np+
Xn
n(n¡ 1)
x= 2
1
A
x
p (1¡ p)
n¡ x
=
Xn
x(x¡ 1)
x= 0
n!
px (1¡ p) n¡ x
x!(n ¡ x)!
(n ¡ 2)!
p2:px¡ 2(1¡ p) n¡ 2¡ (x¡ 1)
(x ¡ 2)!(n ¡ 2 ¡ (x ¡ 2))!
n¡ 2
2 X
= np+ n(n¡ 1)p
(n ¡ 2)!
px (1¡ p) n¡ 2¡ x = np+ n(n¡ 1)p2
x= 0 x!(n ¡ 2 ¡ (x ¡ 2))!
V ar (X ) = np + n2p2 ¡ np2 ¡ (np) 2 = np ¡ np2 = np(1 ¡ p)
P
Second met hod : R ecognize t hat X = nk= 1 X k wher e
each X k is a B er noulli r andom var iable. U se t he linear it y
pr oper t y of expect at ions, as ext ended t o var iance,i.e.,
P
V ar (X ) = nk= 1 V ar (X k ) = np(1 ¡ p)
R.V.D – Poisson
1. Poisson r.v. can be thought of as a limit of Binomial
experiment where n is very large, and np approaches a limit
, say .
2. The Poisson r.v. X, with parameter, , denoted for short by
POI() is defined by
P[X=x]=e- x /x! , x=0,1,2,…
3. E[X] =  (Exercise!) ; Var[X] = 
R.V.D – Poisson-2
L et X denot e t he number of ° aws on t he sur face of
r andomly select ed boiler of a cer t ain t ype. Suppose X
has a Poisson dist r ibut ion wit h ¸ = 5. T hen t he pr obabilit y t hat a r andomly select ed boiler has exact ly t wo
° aws is
e¡ 5(5) 2
P(X = 2) =
= 0:084
2!
T he pr obabilit y t hat a boiler cont ains at most t wo ° aws
is
Ã
!
e¡ 5(5) x
25
¡5
P(X · 2) =
= e 1+ 5+
= 0:125
x!
2
x= 0
X2
R.V.D – Poisson-3
( A s an appr oximat ion t o B inomial)
I f a Publisher of non-t echnical books t akes gr eat pains
t o ensur e t hat it s books ar e fr ee of t ypogr aphical er r or s,
so t hat t he pr obabilit y of any given page cont aining at
least one such er r or is 0.005 and er r or s ar e independent
fr om page t o page, what is t he pr obabilit y t hat one of it s
400 page novels will cont ain exact ly 1 page wit h er r or s?
A t most 3 pages wit h er r or s?
W it h S denot ing a page cont aining at least 1 er r or and
F an er r or fr ee page, t he number X of pages cont aining
at least 1 er r or is a B inomial r .v. wit h n = 400 and
p = 0:005, so np = 2.
e¡ 2(2) 1
P(X = 1) = b(1; 400; 0:005) ¼ p(1; 2) =
= 0:271
1!
R.V.D
–
Poisson-4
E (X ) = ¸ .
e¡ ¸ ¸ x
E (X ) =
x
=
x!
x= 0
X1
e¡ ¸ ¸ x¡ 1
= ¸
= ¸
(x
¡
1)!
x= 1
X1
e¡ ¸ ¸ x
x= 1 (x ¡ 1)!
X1
e¡ ¸ ¸ x
x= 0 x!
X1
= ¸ e¡ ¸ e¸ = ¸
V ar (X ) = E[X (X ¡ 1)] + E [X ] ¡ (E[X ]) 2
e¡ ¸ ¸ x
= ¸ ¡ ¸ +
x(x ¡ 1)
= ¸ ¡ ¸2+
x!
x= 0
2
X1
e¡ ¸ ¸ x
= ¸ ¡ ¸ +¸
x= 0 x!
2
1
2X
= ¸ ¡ ¸ 2 + ¸ 2e¡ ¸ e¸
e¡ ¸ ¸ 2¸ x¡ 2
x= 2 (x ¡ 2)!
X1
Random Variables - Continuous
1. A continuous random variable is a r.v. which takes values
in an interval on the real number line. (If multivariate then
on the two dimensional real plane, etc.).
2. The probability distribution or probability density function
(pdf) of a continuous r.v. X is defined by, a function, say,
f(x) such that P[a X  b] =  a b f(x)dx
3. i.e. the probability that X lies between a and b is given by
the area under the graph of f(x) enclosed on the x-axis by a
and b.
4. If X is a continuous r.v., then for any constant c,
P[X=c]=0.
Cumulative distribution functions
1. The cumulative distribution function (c.d.f) of a
continuous r.v. X with pdf f(x) is defined by
F(x) = P[X  x] = -x f(x)dx
2. The density, f(x) is obtained by differentiating F(x) as a
function of x.
3. The expectation of a continuous r.v. X with pdf f(x) is
defined by E(X) = - xf(x)dx
4. The variance of a continuous r.v. X with pdf f(x), and
expectation  is defined by Var(X) = - [x]2f(x)dx
R.V.C - Uniform
1. The Uniform r.v. X, with parameters, (a,b) denoted for short
by UNIF(a,b), with a<b, is defined by the density
f(x)=1/[b-a] if a<x<b, and 0 otherwise.
2. F(x)= 0 if x<a,
= x/[b-a] if a<x<b,
=1, if x>b
3. E[X] = (b-a)/2 (Exercise!) ; Var[X] = ? (Find out!)
R.V.C - Exponential
1. The Exponential r.v. X, with parameter , denoted for short
by EXP(), with >0, is defined by the density
f(x)=(1/  )exp(- x/) if 0<x, and 0 otherwise.
2. F(x)= 0 if x<0,
= 1- exp(- x/) , 0<x
3. E[X] =  (Exercise!) ; Var[X] = ? (Find out!)
R.V.C – Normal or Gaussian
1. The Normal r.v. X, with parameters (, 2), denoted for
short by N(, 2), with 2 >0, and -<  < , is defined
by the density
2. f(x)=(1/ (2))exp(- (x-  )2/ (22))
3. E[X] =  (Exercise!) ; Var[X] = 2 (Exercise!)
4. It has a symmetric density function (about its mean). All the
measures of central tendency, mean, mode, median are the
same.
5. It occurs as the most common limiting distribution for
averages of random variables, i.e., averages of large no. of
r.v.s can be well approximated by it, for most r.v.s.
R.V.C – Normal or Gaussian-2
1. If X follows N(1, 12) and Y follows N(2, 22) then
aX+bY, where a,b are real constants, follows
N(a1 +b2, a212 +b222)
2. The above result can be extended to any finite number of
independent Normal random variables.
3. If X follows N(0, 1), then X is called a standard normal r.v.,
and the corresponding distribution function is called a
standard normal distribution.
4.
If X follows N(, 2), then Y=(X- )/  follows N(0, 1).
Percentiles of a continuous R.V.
1. Let p be a no. between 0 and 1. The (100p)th percentile of
the distribution of a continuous r.v. X, with density f(x),
denoted by (p), is defined by p = - (p) f(x)dx
R .V . D - B in o m ial
O ut com e
SSSS
SSSF
SSF S
SSF F
SF SS
SF SF
SF F S
SF F F
x
4
3
3
2
3
2
2
1
P rob.
p4
p3 ( 1 ¡ p)
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
O ut com e
F SSS
F SSF
F SF S
F SF F
F F SS
F F SF
FFFS
FFFF
x
3
2
2
1
2
1
1
0
P rob
p3 ( 1 ¡ p)
p2 ( 1 ¡ p) 2
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
p2 ( 1 ¡ p) 2
p( 1 ¡ p) 3
p( 1 ¡ p) 3
( 1 ¡ p) 4
Gaussian or Normal Distribution
Sample as Random Observables
Parametric Inference
Tests of Hypothesis
Hypothesis Tests for Normal
Population