Probability Distribution

Some Basic Probability Concept
and Probability Distribution
and Probability Distribution
Lecture Series on
Lecture Series on
Biostatistics
Biostatistics
No. Bio-Stat_9
No. Bio-Stat_9
Date – 01.02.2009
Date – 01.02.2009
By
By
Dr. Bijaya Bhusan Nanda,
Dr. Bijaya Bhusan Nanda,
M.Sc. (Gold Medalist), Ph. D. (Stat.)
M.Sc. (Gold Medalist), Ph. D. (Stat.)

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CONTENT
CONTENT
 Some basic probability concept
Some basic probability concept
 Screening test, sensitivity,
Screening test, sensitivity,
specificity, predictive value
specificity, predictive value
positive and negative
positive and negative
 Probability distribution
Probability distribution
• Introduction
Introduction
• Probability distribution of discrete variable
Probability distribution of discrete variable
 The Binomial Distribution
The Binomial Distribution
 The Poisson Distribution
The Poisson Distribution
 Continuous Probability distribution
Continuous Probability distribution
• The Normal Distribution
The Normal Distribution

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 The theory of probability provides the
The theory of probability provides the
foundation for statistical inference.
foundation for statistical inference.
 Some Common probability parlance in
Some Common probability parlance in
medical:
medical:
• A patient has a 50-50 chance of surviving a
A patient has a 50-50 chance of surviving a
certain operation.
certain operation.
• Another physician may say that she is 95%
Another physician may say that she is 95%
certain that a patient has a particular disease.
certain that a patient has a particular disease.
• A public health nurse may say that nine times
A public health nurse may say that nine times
out of ten a certain client will break an
out of ten a certain client will break an
appointment.
appointment.
 We measure the probability of occurrence of
We measure the probability of occurrence of
some event by a number between zero and
some event by a number between zero and
one
one.
.

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 The more likely the event, the closer the prob. is to one and
The more likely the event, the closer the prob. is to one and
more unlikely it is the closer the prob. to ‘0’.
more unlikely it is the closer the prob. to ‘0’.
 An event that can’t occur has a probability zero, and an
An event that can’t occur has a probability zero, and an
event that is certain to occur has a probability of one.
event that is certain to occur has a probability of one.
 Health sciences researchers continually ask themselves if
Health sciences researchers continually ask themselves if
the result of their efforts could have occurred by chance
the result of their efforts could have occurred by chance
alone or if some other force was operating to produce the
alone or if some other force was operating to produce the
observed effect.
observed effect.
 For example, suppose six out of ten patients suffering from
For example, suppose six out of ten patients suffering from
some disease are cured after receiving a certain treatment.
some disease are cured after receiving a certain treatment.
 Is such a cure rate likely to have occurred if the patients had
Is such a cure rate likely to have occurred if the patients had
not received the treatment, or is it evidence of a true curative
not received the treatment, or is it evidence of a true curative
effect on the part of the treatment?
effect on the part of the treatment?
 Question such as these can be answered through the
Question such as these can be answered through the
application of the concepts and laws of probability.
application of the concepts and laws of probability.

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Some Basic Probability Concept Contd..
 TWO VIEWS OF PROBABILITY-
TWO VIEWS OF PROBABILITY-
OBJECTIVE AND SUBJECTIVE
OBJECTIVE AND SUBJECTIVE
 Objective Probability
Objective Probability
• Classical or a priori probability
Classical or a priori probability
• The relative frequency , or a posteriori,
The relative frequency , or a posteriori,
 Classical Probability
Classical Probability
 Definition- If an event can occur in N
Definition- If an event can occur in N
mutually exclusive and equally likely ways,
mutually exclusive and equally likely ways,
and if ‘m’ of these possess a trait, E, the
and if ‘m’ of these possess a trait, E, the
probability of the occurrence of E is equal to
probability of the occurrence of E is equal to
m/N.
m/N.
 Symbolically-
Symbolically-
• P(E) – Prob. of E
P(E) – Prob. of E
P(E)=
m
N

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 Relative frequency probability
Relative frequency probability
• If the process is repeated a large number of times, n ,
If the process is repeated a large number of times, n ,
and if some resulting event with the characteristic E
and if some resulting event with the characteristic E
occurs ‘m’ times, the relative frequency of occurrence
occurs ‘m’ times, the relative frequency of occurrence
of E, m/n, will be approximately equal to the
of E, m/n, will be approximately equal to the
probability of E.
probability of E. Symbolically P(E) = m/n
Symbolically P(E) = m/n
 Subjective Probability
Subjective Probability (In the early 1950s, L.J. Savage)
(In the early 1950s, L.J. Savage)
• Probability measures the confidence that a particular
Probability measures the confidence that a particular
individual has in the truth of a particular proposition.
individual has in the truth of a particular proposition.
This concept doesn’t rely on the repeatability of any
This concept doesn’t rely on the repeatability of any
process. In fact, by applying this concept of
process. In fact, by applying this concept of
probability, one may evaluate the probability of an
probability, one may evaluate the probability of an
event that can only happen once, for example, the
event that can only happen once, for example, the
probability that a cure for cancer will be discovered
probability that a cure for cancer will be discovered
within the next 10 years.
within the next 10 years.

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ELEMENTARY PROPERTIES OF PROBABILITY
(Axiomatic approach to probability--Russian
Mathematician A. N. Kolmogorov in 1933 )
The basic of this approach is embodied in three
properties:
Axiom of non-negativity: Given some process (or
experiment) with n mutually exclusive outcomes (called
events), E1, E2, ……En, the probability of any event Ei is
assigned a nonnegative number. That is P(Ei)≥ 0

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Axiom of exhaustiveness: The sum of the probabilities
of the all mutually exclusive and exhaustive outcomes is
equal to 1. i.e.P(E1)+ P (E2) +………P(En) = 1
This is the property of exhaustiveness.
Axiom of additive-ness: Consider any two mutually
exclusive event, Ei and Ej.The probability of the
occurrence of either Ei or Ej is equal to the sum of their
individual probabilities.
P(Ei or Ej) = P(Ei)+ P(Ej)
Suppose the two events were not mutually exclusive; in
that case the probability:
P(Ei or Ej) = P(Ei)+ P(Ej) – P(Ei E
∩ j)

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CALCULATING THE PROBABILITY OF AN
EVENT
Example:
The subjects in the study consisted of a sample of
75 men and 36 women. The subjects are a fairly
representative sample of “typical” adult users who
were neither in treatment nor in jail. Table 1 shows
the life time frequency of cocaine use and the
gender of these subjects. Suppose we pick a person
at random from this sample. What is the
probability that this person will be a male?

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 We define the desired probability as the
We define the desired probability as the
number of subjects with the characteristic of
number of subjects with the characteristic of
interest (male) divided by the total number
interest (male) divided by the total number
of subjects. Symbolically
of subjects. Symbolically
• P(M)= number of males/ total number of
P(M)= number of males/ total number of
subjects = 75/111 = .6757
subjects = 75/111 = .6757
Life frequency of
cocaine use
Male(M) Female(F) total
1-19 times(A) 32 7 39
20-99 times(B) 18 20 38
100+ times (C ) 25 9 34
Total 75 36 111

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 Conditional Probability
Conditional Probability
 When probabilities are calculated with a subset of the
When probabilities are calculated with a subset of the
total group as the denominator, the result is a
total group as the denominator, the result is a
conditional probability.
conditional probability.
 Example:-
Example:-
 Suppose we pick a subject at random from the 111
Suppose we pick a subject at random from the 111
subjects and find that he is a male (M). What is the
subjects and find that he is a male (M). What is the
probability that this male will be one who has used
probability that this male will be one who has used
cocaine 100 times or more during his lifetime?
cocaine 100 times or more during his lifetime?
 Solution:
Solution:
 This is a conditional probability and is written as
This is a conditional probability and is written as
P(C|M). The 75 males become the denominator of
P(C|M). The 75 males become the denominator of
this conditional probability, and 25 the number of
this conditional probability, and 25 the number of
males who have used cocaine 100 times or more
males who have used cocaine 100 times or more
during their lifetime, becomes the numerator. Our
during their lifetime, becomes the numerator. Our
desired probability, then is P(C|M) = 25/75 = 0.33
desired probability, then is P(C|M) = 25/75 = 0.33

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 Joint Probability
Joint Probability
 The probability that a subject picked at a random from
The probability that a subject picked at a random from
a group of subjects possesses two characteristics at the
a group of subjects possesses two characteristics at the
same time.
same time.
 Example:
Example:Ref. (Table 1) what is the probability that a
Ref. (Table 1) what is the probability that a
person picked at a random from the 111 subjects will be
person picked at a random from the 111 subjects will be
a male (M) and be person who has used cocaine 100
a male (M) and be person who has used cocaine 100
times or more during his lifetime (C)?
times or more during his lifetime (C)?
 Solution:
Solution:The probability is written as P (M∩C)
The probability is written as P (M∩C)
 (M∩C) indicates the joint occurrence of conditions M
(M∩C) indicates the joint occurrence of conditions M
and C.
and C.
 No. of subjects satisfying both of the desired conditions
No. of subjects satisfying both of the desired conditions
is found in Table-1 at the intersection of the column
is found in Table-1 at the intersection of the column
labeled M and the row labeled C i.e. 25 since the
labeled M and the row labeled C i.e. 25 since the
selection will be made from the total set of subjects, the
selection will be made from the total set of subjects, the
denominator is 111.
denominator is 111.
 P(M∩C) = 25/111 = 0.2252
P(M∩C) = 25/111 = 0.2252

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 Multiplication Rule
Multiplication Rule
 A probability may be computed from other
A probability may be computed from other
probability. For example, a joint probability may be
probability. For example, a joint probability may be
computed as the product of an appropriate marginal
computed as the product of an appropriate marginal
probability and an appropriate conditional
probability and an appropriate conditional
probability. This relationship is known as the
probability. This relationship is known as the
multiplication rule of probability.
multiplication rule of probability.
 We may state the multiplication rule in general
We may state the multiplication rule in general
terms as follows for any two events A & B:
terms as follows for any two events A & B:
 P(A ∩ B) = P(B) P(A|B), if P(B) ≠ 0 or
P(A ∩ B) = P(B) P(A|B), if P(B) ≠ 0 or
 P(A ∩ B) = P(A) P(B|A), if P(A) ≠ 0
P(A ∩ B) = P(A) P(B|A), if P(A) ≠ 0

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 Conditional Probability
Conditional Probability
 The conditional probability of A given B is equal to the
The conditional probability of A given B is equal to the
probability of A∩B divided by the probability of B,
probability of A∩B divided by the probability of B,
provided the probability of B is not zero.
provided the probability of B is not zero.
 That is =>
That is => P (A|B) = P (A ∩ B)/ P (B), P (B) ≠ 0 and
P (A|B) = P (A ∩ B)/ P (B), P (B) ≠ 0 and
same way P (B|A) = P (A ∩ B)/ P (A), P (A) ≠ 0
same way P (B|A) = P (A ∩ B)/ P (A), P (A) ≠ 0
 Example
Example:
:
 Using the equation and the date of Table-1 to find the
Using the equation and the date of Table-1 to find the
conditional probability, P(C|M).
conditional probability, P(C|M).
 Solution:
Solution:
 According to the equation
According to the equation
 P(C|M) = P(C∩M)/ P(M)
P(C|M) = P(C∩M)/ P(M)
 Earlier we found P(C∩M) = P(M∩C) = 25/111 =
Earlier we found P(C∩M) = P(M∩C) = 25/111 =
0.2252.We have also determined that P(M) = 75/111
0.2252.We have also determined that P(M) = 75/111
=0.6757.
=0.6757.
 Using these results we are able to compute
Using these results we are able to compute
 P(C|M) = 0.2252/ .6757 = 0.3333
P(C|M) = 0.2252/ .6757 = 0.3333

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 Addition Rule:
Addition Rule:
 Given two events A and B, the probability that event A, or event
Given two events A and B, the probability that event A, or event
B, or both occur is equal to the probability that event A occurs,
B, or both occur is equal to the probability that event A occurs,
plus the probability that event B occurs, minus the probability
plus the probability that event B occurs, minus the probability
that the events occur simultaneously.
that the events occur simultaneously.
 The addition rule may be written P(AUB) = P(A) + P(B) – P(A∩B)
The addition rule may be written P(AUB) = P(A) + P(B) – P(A∩B)
 Example:
Example:
 If we select a person at random from the 111 subjects represented in
If we select a person at random from the 111 subjects represented in
Table-1 what is the probability that this person will be a male (M) or
Table-1 what is the probability that this person will be a male (M) or
will have used cocaine 100 times or more during his lifetime(C) or
will have used cocaine 100 times or more during his lifetime(C) or
both?
both?
 Solution:
Solution:
 The probability is P(MUC).By the addition rule as expressed this
The probability is P(MUC).By the addition rule as expressed this
probability may be written as P(MUC) = P(M) + P(C) - P(M∩C)
probability may be written as P(MUC) = P(M) + P(C) - P(M∩C)
 We have already found that P(M)=75/111=0.6757, and P(M∩C) =
We have already found that P(M)=75/111=0.6757, and P(M∩C) =
25/111 = 0.2252
25/111 = 0.2252
 P(C) = 34/111= 0.3063
P(C) = 34/111= 0.3063
 Substituting these results in to the equation for P(MUC)=
Substituting these results in to the equation for P(MUC)=
0.6757+0.3063-0.2252 = 0.7568.
0.6757+0.3063-0.2252 = 0.7568.

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 Independence of Events
Independence of Events
 Suppose that the probability of event A is the same
Suppose that the probability of event A is the same
regardless of whether or not B occurs. In this
regardless of whether or not B occurs. In this
situation, P(A|B) = P(A). In such cases we say that
situation, P(A|B) = P(A). In such cases we say that
A and B are independent events. The
A and B are independent events. The
multiplication rule for two independent events,
multiplication rule for two independent events,
then may be written as:
then may be written as:
 P(A∩B) = P(B)P(A); P(A) ≠ 0, P(B) ≠ 0
P(A∩B) = P(B)P(A); P(A) ≠ 0, P(B) ≠ 0
 Thus we see that if two events are independent, the
Thus we see that if two events are independent, the
probability of their joint occurrence is equal to the
probability of their joint occurrence is equal to the
product of the probabilities of their individual
product of the probabilities of their individual
occurrences.
occurrences.
 When two events with non zero probabilities are
When two events with non zero probabilities are
independent each of the following statements is
independent each of the following statements is
true:
true:
 P(A|B) = P(A), P(B|A) =P(B), P(A∩B) = P(A) P(B)
P(A|B) = P(A), P(B|A) =P(B), P(A∩B) = P(A) P(B)

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 Complementary Events:
Complementary Events:
 Earlier using the data in Table 1 we computed the
Earlier using the data in Table 1 we computed the
probabilities:
probabilities:
 P(M) =75/111=.6757. P(F) = 36/111=.3243.The sum of these
P(M) =75/111=.6757. P(F) = 36/111=.3243.The sum of these
two probabilities = 1. This is because the events being male
two probabilities = 1. This is because the events being male
and being female are complementary events.
and being female are complementary events.
 The probability of an event A is equal to 1 minus the
The probability of an event A is equal to 1 minus the
probability of its complement, which is written as A´, and
probability of its complement, which is written as A´, and
 P(A
P(A´
´) = 1- P(A)
) = 1- P(A)
 This follows from the third property of probability since
This follows from the third property of probability since
the event, A and its complement A’, are mutually exclusive
the event, A and its complement A’, are mutually exclusive
and exhaustive
and exhaustive
 Example:
Example:
 Suppose that of 1200 Admissions to a general hospital
Suppose that of 1200 Admissions to a general hospital
during a certain period of time, 750 are private
during a certain period of time, 750 are private
admissions. If we designate these as set A´, then A is equal
admissions. If we designate these as set A´, then A is equal
to 1200 minus 750, or 450. we may compute
to 1200 minus 750, or 450. we may compute
 P(A)=750/1200=0.625 and P(A´) = 450/1200 =0.375 and see
P(A)=750/1200=0.625 and P(A´) = 450/1200 =0.375 and see
that P(A) + P(A´) =1
that P(A) + P(A´) =1

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 Marginal Probability
Marginal Probability
 Given some variable that can be broken down in to ‘m’
Given some variable that can be broken down in to ‘m’
categories as A
categories as A1
1, A
, A2
2…..A
…..Ai
i,……A
,……Am
m and another jointly occurring
and another jointly occurring
variable into ‘n’ categories as B
variable into ‘n’ categories as B1
1, B
, B2
2,…..B
,…..Bj
j,….B
,….Bn
n the marginal
the marginal
probability of Ai , P(Ai), is equal to the sum of the joint
probability of Ai , P(Ai), is equal to the sum of the joint
probabilities of Ai with all the categories of B.
probabilities of Ai with all the categories of B.
 That is P(A
That is P(Ai
i) = ∑ P (A
) = ∑ P (Ai
i∩B
∩Bj
j), for all values of j.
), for all values of j.
 Example:
Example:
 We use the equation and the data in Table-1 to compute the
We use the equation and the data in Table-1 to compute the
marginal probability P(M).
marginal probability P(M).
 Solution:
Solution:
 The variable gender is broken down in to two categories male
The variable gender is broken down in to two categories male
and female and the variable ‘frequency of cocaine use’ in to 3
and female and the variable ‘frequency of cocaine use’ in to 3
categories, 1-19 times (A), 20-99 times (B), 100+times (C). The
categories, 1-19 times (A), 20-99 times (B), 100+times (C). The
category male occurs jointly with all 3 categories of the variable
category male occurs jointly with all 3 categories of the variable
‘frequency of cocaine use’. The three joint probabilities are:
‘frequency of cocaine use’. The three joint probabilities are:
 P(M∩A) = 32/111 = 0.2883, P(M∩B) = 18/111 = 0.1662,
P(M∩A) = 32/111 = 0.2883, P(M∩B) = 18/111 = 0.1662,
 P(M∩C) = 25/111 = 0.2252
P(M∩C) = 25/111 = 0.2252
 We obtain the marginal probability: P(M)= P(M∩A) +P(M∩B)+
We obtain the marginal probability: P(M)= P(M∩A) +P(M∩B)+
P(M∩C) = 0.2883 + 0.1622 + 0.2252 = 0.6757
P(M∩C) = 0.2883 + 0.1622 + 0.2252 = 0.6757

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Screening Test, Sensitivity, Specificity, Predictive
Screening Test, Sensitivity, Specificity, Predictive
Value Positive And Negative
Value Positive And Negative
 Screening tests are not always infallible.
Screening tests are not always infallible.
That is a testing procedure may yield a false
That is a testing procedure may yield a false
positive or a false negative.
positive or a false negative.
 A false positive results when a test indicates
A false positive results when a test indicates
a positive status when the true status is
a positive status when the true status is
negative.
negative.
 A false negative results when a test indicates
A false negative results when a test indicates
a negative status when the true status is
a negative status when the true status is
positive.
positive.

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 The following questions must be answered in order to
The following questions must be answered in order to
evaluate the usefulness of test results.
evaluate the usefulness of test results.
• Given that a subject has the disease, what is the
Given that a subject has the disease, what is the
probability of a positive test result (or the
probability of a positive test result (or the
presence of a symptom)?
presence of a symptom)?
• Given that a subject doesn’t have the disease, what
Given that a subject doesn’t have the disease, what
is the probability of a negative test result (or the
is the probability of a negative test result (or the
absence of a symptom)?
absence of a symptom)?
• Given a positive screening test (or the presence of
Given a positive screening test (or the presence of
a symptom), what is the probability that the subject
has the disease?
has the disease?
• Given a negative screening test (or the absence of
Given a negative screening test (or the absence of
doesn’t have the disease?
doesn’t have the disease?

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Table-2 Sample of n subjects (where n is large) cross-
classified according to disease status and screening test
result.
Test
result
Present
(D)
Absent
(D´)
Total
Positive(T) a b a+b
Negetive
(T´)
c d c+d
Total a+c b+d n

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Sensitivity of a Test (Probability of True positive)
It is the probability of a positive test result ( or presence
of the system) given the presence of the disease.
Symbolically this is P(T|D) = a/(a+c)
Specificity of a Test (Probability of True Negative)
It is the probability of a negative test result (absence of
the system) given the absence of the disease.
Symbolically estimate of Specificity is given by the
conditional probability P(T´|D´) = d/(b+d)

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The Predictive Value of a Positive Test
The predictive value positive of a screening test (or
symptom) is the probability that a subject has the disease
given that the subject has a positive screening result (or has
the symptom).
Symbolically, this is given by the conditional probability
P(D|T) = a/(a+b)
Predictive Value of a Negative test
The predictive value negative of a screening test (or
symptom) is the probability that a subject doesn’t have the
disease given that the subject has a negative screening test
result (or doesn’t have the symptom).
Symbolically, this is given by the conditional probability
P(D´|T´) = d/(c+d)

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 Probability of false negative
Probability of false negative: It is the
: It is the
probability that the screening test is negative
probability that the screening test is negative
given that the subject has disease.
given that the subject has disease.
• P(T´|D) = c/ (a + c).
P(T´|D) = c/ (a + c).
 Probability of false positive
Probability of false positive: It is the
: It is the
probability that the screening test is positive
probability that the screening test is positive
given that the subject does not have disease.
given that the subject does not have disease.
• P(T|D´) = b/(b+d).
P(T|D´) = b/(b+d).

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Some Theoretical Probability Distribution
Some Theoretical Probability Distribution
 The relationship between the values of a random
The relationship between the values of a random
variable and the probability of their occurrence
variable and the probability of their occurrence
summarized by means of a device called a
summarized by means of a device called a
probability distribution.
probability distribution.
 It may be expressed in the form of a table, a graph or
It may be expressed in the form of a table, a graph or
a formula.
a formula.
 Knowledge of probability distribution of a random
Knowledge of probability distribution of a random
variable provides the researchers with a powerful tool
variable provides the researchers with a powerful tool
for summarizing and describing a set of data and for
for summarizing and describing a set of data and for
reaching conclusion about a population on the basis
reaching conclusion about a population on the basis
of sample.
of sample.

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Definition
Definition
 The probability distribution of a discrete random
The probability distribution of a discrete random
variable is a table, graph, formula, or other device
variable is a table, graph, formula, or other device
used to specify all possible values of a discrete
used to specify all possible values of a discrete
random variable along with their respective
random variable along with their respective
probabilities.
probabilities.
Example: Prevalence of prescription and
Example: Prevalence of prescription and
nonprescription drug use in pregnancy among
nonprescription drug use in pregnancy among
women delivered at a large eastern hospital. N=4185
women delivered at a large eastern hospital. N=4185
No of
No of
drugs
drugs
0
0 1
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
10 12
12
f
f 1425
1425 1351
1351 793
793 348
348 156
156 58
58 28
28 15
15 6
6 3
3 1
1 1
1

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Construct a probability distribution
table.
What is the probability that a
randomly selected women who use 3
drugs?
What is the probability that a woman
picked at random who used 2 or fewer
drugs?
What is the probability that a woman
use 5 or more drugs?

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 Most widely encountered probability
Most widely encountered probability
distribution.
distribution.
 Derived from a process known as a Bernoulli
Derived from a process known as a Bernoulli
trial.
trial.
 When a random process or experiment, called
When a random process or experiment, called
a trial can result only one of two mutually
a trial can result only one of two mutually
exclusive outcomes, such as dead or alive, sick
exclusive outcomes, such as dead or alive, sick
or well, male or female, the trial is called a
or well, male or female, the trial is called a
Bernoulli trial.
Bernoulli trial.

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29
Characteristics of a binomial random variable
Characteristics of a binomial random variable
 The experiment consists of
The experiment consists of n
n identical trials
identical trials
 There are only 2 possible outcomes on each trial.
There are only 2 possible outcomes on each trial.
We will denote one outcome by
We will denote one outcome by S
S (for Success)
(for Success)
and the other by
and the other by F
F (for Failure).
(for Failure).
 The probability of
The probability of S
S remains the same from trial
remains the same from trial
to trial. This probability will be denoted by
to trial. This probability will be denoted by p
p,
,
and the probability of
and the probability of F
F will be denoted by
will be denoted by q
q (
( q
q
=
= 1-
1-p
p).
).
 The trials are independent.
The trials are independent.
 The binomial random variable
The binomial random variable x
x is the number
is the number
of
of S
S’ in
’ in n
n trials.
trials.

30
30
The probability distribution:
The probability distribution:
(
(x =
x = 0, 1, 2, ...,
0, 1, 2, ..., n
n),
),
Where,
Where,
p
p = probability of a success on a single trial,
= probability of a success on a single trial, q=1-p
q=1-p
n
n = number of trials,
= number of trials, x
x= number of successes in
= number of successes in n
n
trials
trials
= combination of
= combination of x
x from
from n
n
Px = n
Cx px
qn-x
n
Cx =
n!
x!(n-x)!

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31
Binomial Table
Binomial Table
 The calculation of probability using the
The calculation of probability using the
probability mass function is tedious if the sample
probability mass function is tedious if the sample
size is large. There are binomial probability tables
size is large. There are binomial probability tables
are available where probabilities for different
are available where probabilities for different
values of n, p and x have been tabulated.
values of n, p and x have been tabulated.

32
32
Binomial Parameters
Binomial Parameters
 The binomial distribution has two parameters n
The binomial distribution has two parameters n
and p which are sufficient to specify a binomial
and p which are sufficient to specify a binomial
distribution.
distribution.
 The mean of the binomial distribution is µ=np
The mean of the binomial distribution is µ=np
 The variance of the binomial distribution
The variance of the binomial distribution σ
σ =
=
np(1-p)
np(1-p)
 Strictly speaking the binomial distribution is
Strictly speaking the binomial distribution is
applicable where the sampling is from an infinite
applicable where the sampling is from an infinite
population or finite population with replacement.
population or finite population with replacement.
But in practice the sampling is mostly without
But in practice the sampling is mostly without
replacement. Hence question arises on the
replacement. Hence question arises on the
appropriateness of this distribution under these
appropriateness of this distribution under these
circumstances.
circumstances.

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33
Example:
Example:
In a certain population 10% of the population is color
In a certain population 10% of the population is color
blind. If a random sample of 25 people is drawn from
blind. If a random sample of 25 people is drawn from
this population. Find the probability that:
this population. Find the probability that:
a) Five or fewer will be color blind
a) Five or fewer will be color blind
P(X
P(X≤ 5) =.9666
≤ 5) =.9666
b) Six or more will be color blind
b) Six or more will be color blind
The probability that six or more are color blind is the
The probability that six or more are color blind is the
complement of the probability that five or fewer are not
complement of the probability that five or fewer are not
color blind.
color blind.
P( X ≥6)=1- P(X ≤ 5)=1-.9666=.0334
P( X ≥6)=1- P(X ≤ 5)=1-.9666=.0334
c) Between six and nine inclusive will be color blind
c) Between six and nine inclusive will be color blind
P(6 ≤ X ≤ 9)=P(X ≤ 9)-P(X ≤ 5)=.9999-.9666=.0333
P(6 ≤ X ≤ 9)=P(X ≤ 9)-P(X ≤ 5)=.9999-.9666=.0333
d) Two, three or four will be color blind
d) Two, three or four will be color blind
P(2 ≤ X ≤ 4)=P(X ≤) – P(X ≤ 1)=.9020-.2712=.6308
P(2 ≤ X ≤ 4)=P(X ≤) – P(X ≤ 1)=.9020-.2712=.6308

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34
 This distribution has been used extensively as a
This distribution has been used extensively as a
probability model in biology and medicine.
probability model in biology and medicine.
If x is the number of occurrences of some random events
If x is the number of occurrences of some random events
in an interval of time or space, the probability that x will
in an interval of time or space, the probability that x will
occur is:
occur is:
x=0,1,2……,
x=0,1,2……, λ
λ>0
>0
λ
λ = parameter (average number of occurrence of the
= parameter (average number of occurrence of the
random event in the interval
random event in the interval
e= constant =2.7183
e= constant =2.7183
f(x) =
e-λ
λx
X !

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35
Characteristics defining a Poisson
Characteristics defining a Poisson
random variable
random variable
 Poisson distribution occurs when there are events
Poisson distribution occurs when there are events
which don’t occur as outcome of a definite
which don’t occur as outcome of a definite
numbers of trials of an experiment but which
numbers of trials of an experiment but which
occur at random point of time and space and our
occur at random point of time and space and our
interest lies in the occurrence of the event only.
interest lies in the occurrence of the event only.
 The experiment consists of counting the number
The experiment consists of counting the number x
x
of times a particular event occurs during a given
of times a particular event occurs during a given
unit of time
unit of time
 The probability that an event occurs in a given
The probability that an event occurs in a given
unit of time is the same for all units.
unit of time is the same for all units.
 The number of events that occur in one unit of
The number of events that occur in one unit of
time is independent of the number that occur in
time is independent of the number that occur in
other units of time.
other units of time.

36
36
 Theoretically an infinite number of
Theoretically an infinite number of
occurrence of the event must be possible in
occurrence of the event must be possible in
the interval.
the interval.
 The poisson event is an infrequently occurring
The poisson event is an infrequently occurring
events with very small probability of
events with very small probability of
occurrence.
occurrence.
 Some situations where the poisson process can
Some situations where the poisson process can
be employed successfully are
be employed successfully are
No. of deaths from a disease such as heart attack
No. of deaths from a disease such as heart attack
or cancer or due to snake bite
or cancer or due to snake bite
No. of suicide reported in a particular city
No. of suicide reported in a particular city

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37
Example:
Example:
In a study of suicide it was found that monthly
In a study of suicide it was found that monthly
distribution of suicide in Cook country, between
distribution of suicide in Cook country, between
1977 and 1987 closely followed a poisson
1977 and 1987 closely followed a poisson
distribution with parameter
distribution with parameter λ
λ= 2.75. Find the
= 2.75. Find the
probability that a randomly selected month will be
probability that a randomly selected month will be
one in which three adolescent suicide occurs.
one in which three adolescent suicide occurs.
Solution
Solution
P(X=3) =
P(X=3) =
e
e -2.75
-2.75
2.75
2.753
3
3!
(.063928) (20.796875)
(.063928) (20.796875)
=
6
= .221584

Continuous Probability Distribution
Continuous Probability Distribution
Continuous variable
Continuous variable:
:
 Assumes any value within a specified
Assumes any value within a specified
interval/range.
interval/range.
 Consequently any two values within a specified
Consequently any two values within a specified
interval, there exists an infinite number of values.
interval, there exists an infinite number of values.
 As the number of observation, n, approaches
As the number of observation, n, approaches
infinite and the width of the class interval
infinite and the width of the class interval
approaches zero, the frequency polygon approaches
approaches zero, the frequency polygon approaches
smooth curve.
smooth curve.
 Such smooth curves are used to represent
Such smooth curves are used to represent
graphically the distribution of continuous random
graphically the distribution of continuous random
variable
variable

39
39
 This has some important consequences when we deal
This has some important consequences when we deal
with probability distributions.
with probability distributions.
 The total area under the curve is equal to 1 as in the
The total area under the curve is equal to 1 as in the
case of the histogram.
case of the histogram.
 The relative frequency of occurrence of values
The relative frequency of occurrence of values
between any two points on the x-axis is equal to the
between any two points on the x-axis is equal to the
total area bounded by the curve, the x-axis and the
total area bounded by the curve, the x-axis and the
perpendicular lines erected at the two points.
perpendicular lines erected at the two points.
Graph of a continuous
distribution showing
area between a and b
f (x)
a b x

40
40
Definition of probability distribution:
Definition of probability distribution:
 A density function is a formula used to represent
A density function is a formula used to represent
the probability distribution of a continuous
the probability distribution of a continuous
random variable.
random variable.
 This is a nonnegative function f (x) of the
This is a nonnegative function f (x) of the
continuous r.v, x if the total area bounded by its
continuous r.v, x if the total area bounded by its
curve and the x-axis is equal to 1 and if the sub
curve and the x-axis is equal to 1 and if the sub
area under the curve bounded by the curve, the x-
area under the curve bounded by the curve, the x-
axis and perpendiculars erected at any two points
axis and perpendiculars erected at any two points
a and b gives the probability that x is between the
a and b gives the probability that x is between the
point a and b.
point a and b.

41
41
 The distribution is frequently called the Gaussian
The distribution is frequently called the Gaussian
distribution.
distribution.
 It is a relative frequency distribution of errors, such
It is a relative frequency distribution of errors, such
errors of measurement. This curve provides an
errors of measurement. This curve provides an
adequate model for the relative frequency
adequate model for the relative frequency
distributions of data collected from many different
distributions of data collected from many different
scientific areas.
scientific areas.
 The density function for a normal random variable
The density function for a normal random variable
 The parameters
The parameters µ
µ and
and σ
σ2
2
are the mean and the
are the mean and the
variance , respectively, of the normal random variable
variance , respectively, of the normal random variable
Normal Distribution( C.F.Gauss, 1777-1855)
2
2
2
/
)
(
2
1
)
( σ
µ
π
σ
−
−
= x
e
x
f

42
42
 It is symmetrical about its mean,
It is symmetrical about its mean, µ
µ.
.
 The mean, the median, and the mode are all
The mean, the median, and the mode are all
equal.
equal.
 The total area under the curve above the x-axis is
The total area under the curve above the x-axis is
one square unit.
one square unit.
 This characteristic follows that the normal
This characteristic follows that the normal
distribution is a probability distribution.
distribution is a probability distribution.
 Because of the symmetry already mentioned,
Because of the symmetry already mentioned,
50% of the area is to the right of a perpendicular
50% of the area is to the right of a perpendicular
erected at the mean, and 50% is to the left.
erected at the mean, and 50% is to the left.
Characteristics Of The Normal Distribution

43
43
0
0.1
0.2
0.3
0.4
0.5
-3.4
-2.8
-2.2
-1.6
-1
-0.4
0.2
0.8
1.4
2
2.6
3.2
 If µ = 0 and σ =1 then . The distribution with this
density function is called the standardized normal
distribution. The graph of the standardized normal
density distribution is shown in Figure

44
44
 If ‘x’ is a normal random variable with the mean
If ‘x’ is a normal random variable with the mean µ
µ
and variance
and variance σ
σ then
then
1) the variable
1) the variable
is the standardized normal random variable.
is the standardized normal random variable.
The equation of pdf for standard normal distribution
The equation of pdf for standard normal distribution
 Area properties of normal distribution
Area properties of normal distribution
σ
µ
−
=
x
z
f (z) =
1
√(2∏)
e –z2 / 2
, -∞ < z < ∞
680
.
0
)
1
(
P =
≤
− σ
µ
x
9544
.
0
)
2
(
P =
≤
− σ
µ
x
9973
.
0
)
3
(
P =
≤
− σ
µ
x

45
45
0.16
0.16
µ - 1σ µ + 1σ
(Fig. a )
0.68
0.025
0.025
µ - 2σ µ + 2σ
(Fig. b )
0.95
0.0015
0.0015
µ - 3σ µ + 3σ
(Fig. c )
0.997

46
46
 Namely, if a population of measurements has
Namely, if a population of measurements has
approximately a normal distribution the
approximately a normal distribution the
probability that a random selected observation
probability that a random selected observation
falls within the intervals (
falls within the intervals (µ
µ -
- σ
σ,
, µ
µ +
+ σ
σ), (
), (µ
µ -
-
2
2σ
σ,
, µ
µ +2
+2σ
σ), and (
), and (µ
µ - 3
- 3σ
σ,
, µ
µ + 3
+ 3σ
σ), is
), is
approximately 0.6826, 0.9544 and 0.9973,
approximately 0.6826, 0.9544 and 0.9973,
respectively.
respectively.

47
47
Normal Distribution Application
Normal Distribution Application
Example:1
Example:1 As a part of a study of Alzeheimer’s disease,
As a part of a study of Alzeheimer’s disease,
reported data that are compatible with the hypothesis
reported data that are compatible with the hypothesis
that brain weights of victims of the disease are normally
that brain weights of victims of the disease are normally
distributed. From the reported data, we may compute a
distributed. From the reported data, we may compute a
mean of 1076.80 grams and a standard deviation of
mean of 1076.80 grams and a standard deviation of
105.76 grams. If we assume that these results are
105.76 grams. If we assume that these results are
applicable to all victims of Alzeheimer’s disease, find
applicable to all victims of Alzeheimer’s disease, find
the probability that a randomly selected victim of the
the probability that a randomly selected victim of the
disease will have a brain that weighs less than 800
disease will have a brain that weighs less than 800
grams.
grams.
800 µ = 1076.80
σ = 105.76

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48
Solution:
Solution:
R.V x ‘Brain weights’
R.V x ‘Brain weights’ follows a
follows a Normal distribution
Normal distribution
with µ=1076.80 and
with µ=1076.80 and σ
σ = 105.76)
= 105.76)
The Corresponding Standard Normal Variate
The Corresponding Standard Normal Variate
=
=
σ
µ
−
=
x
z
76
.
105
80
.
1076
−
=
x
z
We have to find out P (x < 800) i.e P (z
< -2.62). This is the area bounded by
the curve, x axis and to the left of the
perpendicular drawn at z = -2.62.
Thus from the standard normal table
this prob., p= .0044. The probability is .
0044 that a randomly selected patient
will have a brain weight of less than
800 grams.
- 2.62 0
σ = 1

49
49
Example: 2
Example: 2
Suppose it is known that the heights of a
Suppose it is known that the heights of a
certain population of individuals are
certain population of individuals are
approximately normally distributed with a
approximately normally distributed with a
mean of 70 inches and a standard deviation of
mean of 70 inches and a standard deviation of
3 inches. What is the probability that a person
3 inches. What is the probability that a person
picked at random from this group will be
picked at random from this group will be
between 65 and 74 inches tall.
between 65 and 74 inches tall.
Solution:
Solution:
X (height) follows a Normal distribution with
X (height) follows a Normal distribution with
mean (
mean (µ)= 70 inches
µ)= 70 inches and
and σ
σ 3 inches.
3 inches.

50
50
z =
65-70
3
= -1.76
-1.67 1.33
0
σ = 1
65 74
70
σ = 3
For x= 65, Z we have For x= 74, Z we have
z =
74-70
3
= 1.33
P ( 65 ≤ x ≤ 74) = P (
65-70
3
≤ z ≤
74-70
3
)
= P (- 1.76 ≤ z ≤ 1.33)
= P ( -∞ ≤ z ≤ 1.33) – P (-∞ ≤ z ≤ -1.67)
= .9082 - .0475
= .8607
The required probability
asked for in our original
question, is 0.8607.

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51
Example: 3
Example: 3
In a population of 10,000 of the people described in
In a population of 10,000 of the people described in
previous example how many would you expect to be
previous example how many would you expect to be
6 feet 5 inches tall or taller?
6 feet 5 inches tall or taller?
Solution:
Solution:
we first find the probability that one person selected
we first find the probability that one person selected
at random from the population would be 6 feet 5
at random from the population would be 6 feet 5
inches tall or taller. That is,
inches tall or taller. That is,
P ( x ≥ 77 ) = P (z ≥
77-70
3 )
= P ( z ≥ 2.33 ) = 1 - .9901= .0099
Out of 10,000 people we would expect 10,000 (.0099) =
99 to be 6 feet 5 inches (77 inches tall or taller).

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52
Exercise:
Exercise:
1.Given the standard normal distribution, find the area
1.Given the standard normal distribution, find the area
under the curve, above the z-axis between z=-∞ and
under the curve, above the z-axis between z=-∞ and
z = 2.
z = 2.
2. What is the probability that a z picked at random
2. What is the probability that a z picked at random
from the population of z’s will have a value
from the population of z’s will have a value
between -2.55 and + 2.55?
between -2.55 and + 2.55?
3. What proportion of z values are between -2.74 and
3. What proportion of z values are between -2.74 and
1.53?
1.53?
4. Given the standard normal distribution, find P ( z
4. Given the standard normal distribution, find P ( z ≥
≥
2.71)
2.71)
5.Given the standard normal distribution, find P(.84 ≤ z
5.Given the standard normal distribution, find P(.84 ≤ z
≤ 2.45).
≤ 2.45).

Probability Distribution

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