Transcript probability distribution

Pharmaceutical Statistics
Probability
Lectures 6-10
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Elementary properties of probability
• Foundation of statistics because of the concept of
sampling and the concept of variation or dispersion and
how likely an observed difference is due to chance.
• Probability statements used frequently in statistics – e.g.,
we say that we are 90% sure that an observed treatment
effect in a study is real.
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Elementary properties of probability
1. Given some process (or experiment) with n mutually
exclusive outcomes (called events), E1, E2, E3, …, En,
the probability f any event Ei is assigned a nonnegative
number. That is:
P( Ei )  0
• In other words, all events must have a probability of more
than or equal to zero (no negative probability)
• A key concept in the statement of this property is the
concept of mutually exclusive outcome. Two events are
said to be mutually exclusive if they can not occur
simultaneously.
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Elementary properties of probability
2. The sum of probabilities of the mutually exclusive
outcomes is equal to 1:
P( E1 )  P( E2 )  P( E3 )  ...............P( En )  1
•
This is the property of exhaustiveness and refers to the
fact that the observer of a probabilistic process must
allow for all possible events, and when all are taken
together, their total probability is 1.
•
Mutually exclusive events E1, E2, ….., En do not overlap,
that is no two of them can occur at the same time.
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Elementary properties of probability
3. Consider any two mutually exclusive events Ei and Ej.
The probability of the occurrence of Ei and Ej is equal to
the sum of their individual probabilities:
P( Ei  E j )  P( Ei )  P( E j )
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Calculating the probability of an event
• Probability of an event is usually referred to as relative
frequency.
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Calculating the probability of an event
• In an article published in the The American Journal of
Drug and Alcohol Abuse, the authors state that women
have been identified as a group at a particular risk for
cocaine addiction and that it has been suggested that
their problems with cocaine are greater than those of
men.
• Based on their review of scientific literature and their
analysis of the results of an original research study, the
authors argue that there is no evidence for the above
assumption.
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Calculating the probability of an event
• The subjects in the authors study were 75 men and 36 women. The
authors state that the subjects are representative sample of typical
adult users who were neither in treatment nor in jail.
Frequency of cocaine use by gender among adult cocaine users
Lifetime frequency of
cocaine use
Male (M)
Female (F)
Total
1-19 times (A)
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7
39
20-99 times (B)
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20
38
100+ times (c)
25
9
34
Total
75
36
111
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Calculating the probability of an event
• Suppose we pick a person at random from this sample,
what is the probability that this person is a male?
– 111 subjects are our population.
– Male and female are mutually exclusive categories
– The likelihood of selecting any one person is equal to the
likelihood of selecting any other
– The desired probability is the number of subjects with the
characteristic of interest (M) divided by the total number of
subjects
P(M) = Number of males / Total number of subjects =
75/111 = 0.6757
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Calculating the probability of an event
• Conditional probability:
– The size of the group of interest may be reduced by conditions not
applicable to the total group.
– Encountered when the set of “all possible outcomes” may constitute a
subset of the total group.
– When probabilities are calculated with a subset of the total group as the
denominator, the result is a conditional probability.
• Suppose we pick a subject at random from the 111
subjects and find that he is a male (M), what is the
probability that this male will be one who has used
cocaine 100 times or more during his lifetime?
– What is the probability that a subject has used cocaine 100 times or
more given he is a male?
– P(C\M), the verical line is read: given.
– P(C\M) = 25/75 = .33
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Calculating the probability of an event
• Joint probability:
– What is the probability that a person picked at random
from the 111 subjects will be male and be a person
who had used cocaine 100 times or more?
– The joint probability is given the symbolic notation:
P(M∩C) in which the symbol “∩” is read either as
“intersection” or “and”.
– The statement M∩C indicates the joint occurrence of
conditions M and C.
P(M∩C) = 25/111 = 0.2252
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Calculating the probability of an event
• The multiplication rule:
– A probability may be calculated from other
probabilities.
– For example, a joint probability may be computed as
the product of an appropriate marginal probability and
an appropriate conditional probability. This
relationship is known as the multiplication rule of
probability.
– We wish to compute the joint probability of male (M)
and a lifetime frequency of cocaine use of 100 times
or more (C) from a knowledge of an appropriate
marginal probability and an appropriate conditional
probability.
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Calculating the probability of an event
• The multiplication rule:
– The probability we seek is P(M∩C).
– We have already computed a marginal probability
P(M)=75/111=0.6757 and a conditional probability
P(C\M)=25/75=0.3333
– We may now compute
P(M∩C)=P(M)*P(C\M)=0.6757*0.3333=0.2252
– The multiplication rule may be stated in general terms
as:
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
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Calculating the probability of an event
• The addition rule:
– In the previous example, if we picked a person at
random from the 111 subject sample, what is the
probability that this person will be a male (M) or
female (F)?
– We state this probability as P(M∪F) were the symbol
∪ is read either “union” or “or”.
– Since the two genders are mutually exclusive, P(M∪F)
= P(M) + P(F) =(75/111)+(36/111)=0.6757+0.3243=1
– Now if the two events are not mutually exclusive:
• 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, plus the probability that event B occurs minus the
probability that events occur simultaneously.
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Calculating the probability of an event
• The addition rule:
– P(A∪B)=P(A)+P(B)-P(A∩B)
– If we select a person at random from the 111
subjects, what is the probability that this person will
be male (M) or will have used cocaine 100 times or
more during his lifetime (C) or both?
P(M∪C)=P(M)+P(C)-P(M∩C)
P(M∪C)=0.6757+0.3063-0.2252=0.7568
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Calculating the probability of an event
• Independent events:
– If the probability of event A is the same regardless of
whether or not event B occurs, we can say that A and
B are independent events.
P(A\B)=P(A)
– And the multiplication rule:
P(A∩B)=P(B)P(A\B), if P(B) ≠0
– May be written as:
P(A∩B)=P(B)P(A), if P(B)≠0, P(A)≠0
For independent events:
P(A\B)=P(A), P(B\A)=P(B), P(A∩B)=P(A)P(B)
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Calculating the probability of an event
• Independent events:
– In a class consisting of 60 girls and 40 boys, it is
observed that 24 girls and 16 boys wear eye glasses.
If a student is picked at random from this class, the
probability that the student wears eye glasses is
40/100=0.4
– What is the probability that the student wears eye
glasses given that the student is a boy?
– P(E\B)=(P(E∩B)/(P(B))=(16/100)/(40/100)=0.4
– This indicates that the additional information that a
student is a boy does not alter the probability that a
student wears glasses.
– The events of being a boy and wearing glasses for
this group are independent.
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Calculating the probability of an event
• Complimentary events:
– The probability of an event A is equal to 1 minus the
probability of its compliment which is written as Ā, and
P(Ā)=1-P(A)
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Calculating the probability of an event
• Marginal probability:
– A marginal probability is a probability in which the
numerator of the probability is a marginal total from a
table.
– The probability of a person picked at random from the
111 persons in the cocaine use study is a male
– P(M)=75/111=0.6757
– The numerator is the total number of males, 75 and
we refer to the probability as a marginal one.
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Probability distributions
• The probability distribution of a discrete random variable
is a table, graph, formula or other device used to specify
all possible values of a discrete random variable along
with the respective probabilities.
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Probability distributions
• In an article in the American Journal of Obstetrics and
Gynecology, Buitendijk and Bracken assessed the use of
medication in a population of women who were delivered
of infants at a large Eastern hospital in a 2 years period.
• The following table shows the prevalence of prescription
and nonprescription drug use in pregnancy among the
study subjects.
• We wish to construct the probability distribution of
the discrete variable X = number of prescription and
nonprescription drugs used by the study subjects.
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Probability distributions
• The values of X are x0=0, x1=1, ….. x10=10,x11=12.
• We compute the probabilities of these values by dividing
their respective frequencies by the total 4185.
• For example:P(X=x0)=1425/4185=0.3405
• We can display the results in the following table.
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Probability distributions
• Alternatively we can present this probability distribution
in the form of a graph.
0.40
0.35
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
10
12
x (number of drugs)
Graphical presentation of the probability distribution
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Probability distributions
• We have two essential properties of a probability
distribution of a discrete variable:
0≤P(X=x)≤1
ΣP(X=x)=1
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Probability distributions
• What is the probability that a randomly selected woman
will be one who used three prescription and
nonprescription drugs?
P(X=3)=0.0832
• What is the probability that a randomly selected woman
used wither one or two drugs?
We use the addition rule for mutually exclusive events
P(1∪2)=P(1)+P(2)=0.3228+0.1895=0.5123
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Probability distributions
• Sometimes we may want to work with a cumulative
probability distribution of a random variable.
• The cumulative probability distribution is obtained by
successively adding the probabilities, P(X=x) given in the
last column of the previous table.
• The cumulative probability for xi is written as
F(x)=P(X≤xi)
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Probability distributions
• What is the probability that a woman picked at random
will be one who used two or fewer drugs?
The probability of interest may be found directly by reading
the cumulative probability opposite to x=2 which is
0.8528
• What is the probability that a randomly selected woman
will be one who used fewer than two drugs?
Since a woman who used fewer than two drugs used either
one or no drugs, the answer is cumulative probability for
1. P(X≤1)= P(X2)=0.6633
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Probability distributions
• What is the probability that a randomly selected woman
is one who used between three and five drugs,
inclusive?
P(x≤5)=0.9872 is the probability that a woman used
between zero and five drugs, inclusive.
P(X≤2)= 0.8528 is the probability that a woman used
between zero and two drugs, inclusive.
P(3≤x≤5)= 0.9872- 0.8528=0.1344
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Probability distributions
• The probability distribution of the drug use during
pregnancy described earlier was developed out of actual
experience, so to find another variable following this
distribution would be coincidental.
• However, the probability distributions of many variables
of interest may be determined or assumed on the basis
of theoretical considerations.
– Binomial distribution
– Poisson distribution
– Normal distribution
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Binomial
Distribution
Discrete
Poisson
Distribution
Probability
Distribution
Continuous
Normal
Distribution
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Probability distributions
Binomial Distribution
• One of the most widely encountered distributions in applied
statistics.
• It is derived from a process known as Bernoulli trial (James
Bernoulli, 1654-1705).
• When a random process or experiment, called a trial, can result in
only one of two mutually exclusive outcomes, such as dead or alive,
sick or well, male or female, the trial is called a Bernoulli trial.
Head
2 mutually exclusive
events
Tail
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Probability distributions
Binomial Distribution
• Repeated identical trials are called Bernoulli trials if:
– There are two possible mutually exclusive outcomes
for each trial, denoted arbitrarily as success (s) and
failure (f).
– The trials are independent; that is, the outcome of any
particular trial is not affected by the outcome of any
other trial.
– The probability of a success, denoted by p, remains
the same from trial to trial. The probability of failure, 1p, is denoted by q.
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Binomial Distribution
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Probability distributions
Binomial Distribution
• A drug is known to be 80% effective in curing a certain
disease. Suppose that four patients with the disease are
to be given the drug and the cure - no cure results
recorded.
a) Formulate this process as a sequence of four
Bernoulli trials.
b) Determine the possible outcomes of the four Bernoulli
trials.
c) Determine the probability of each outcome in part (b).
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Probability distributions
Binomial Distribution
• Formulate this process as a sequence of four Bernoulli
trials.
– Each trial consists of administering the drug to one of
the patients.
– There are two possible outcomes for each trial: cure
or no cure.
– The trials are independent (why?).
– If we let a success, s, correspond to a cure, then the
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success probability is p = 0.8 (80%).
Probability distributions
Binomial Distribution
• Determine the possible outcomes of the four Bernoulli
trials.
– For this part we are to obtain the possible outcomes
of the four Bernoulli trials; that is, the possible cure-no
cure results for the four patients.
– The possible outcomes are shown in the following
table.
ssss
sfss
fsss
ffss
sssf
sfsf
fssf
ffsf
ssfs
sffs
fsfs
fffs
ssff
fsff
fsff
ffff
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Probability distributions
Binomial Distribution
Note that:
1. Sum of all outcome’s probability is 1, why?
2. The probability of each outcome is not necessary the same, why?
3. p(sfss)=p(fsss)=p(sssf)=p(sfsf), why?
4. Probability of exactly 3 patients will be cured
P(X=3)=P(sfss or fsss or sssf or sfsf) = P(sfss)+P(fsss)+p(sssf)+p(sfsf)
P(X=3)=0.1024+0.1024+0.1024+0.1024 = 4 * 0.1024= 0.4096
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Probability distributions
Binomial Distribution
Number of
success
P(X=x)
0
1
0.0016
0.0256
2
0.1536
3
0.4096
4
0.4096
1.0
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Probability distributions
Binomial Distribution
• Binomial Distribution: Is the probability distribution for the
number of successes in a sequence of Bernoulli Trials
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Probability distributions
Binomial Distribution
• As the size of the sample increases, listing the number
of sequences becomes more and more difficult and
tedious.
• An easy method for counting the number of sequences
is provided by means of a counting formula that allows
us to determine quickly how many subsets can be
produced as a result of Bernoulli trials.
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Probability distributions
Binomial Distribution
•
•
•
•
STEP 1 Identify a success.
STEP 2 Determine p, the success probability.
STEP 3 Determine n, the number of trials.
STEP 4 The binomial probability formula for the number of
successes, x, is
P( x) n Cx p x (1  p)n  x
•
Where nCx is the number of combinations of n objects that can be
formed by taking x of them at a time:
n!
n Cx 
x!(n  x)!
•
Where x! is read as x factorial, it is the product of all the whole
numbers from x down to 1 (x!=x(x-1)(x-2)……(1))
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Probability distributions
Binomial Distribution
• Suppose that 30% of a certain population are immune to
a certain disease. If a random sample of size 10 is to be
selected from that population.
What is the probability that it will contain exactly 4
immune persons?
P(4)=10C4(0.7)6(0.3)4
P(4)=(10!/4!6!)0.117649)(0081)
P(4)=0.2001
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Probability distributions
Binomial Distribution
• The calculation of a probability using the previous
equation can be a tedious undertaking if the sample size
is large.
• Fortunately, probabilities for different values of n, p, and
x have been tabulated, so that we need only to consult
an appropriate table to obtain the desired probability.
• The table gives the probability that x is less than or
equal to some specified value. That is, the table gives
the cumulative probabilities from x = 0 up through
some specified value.
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This table shows the probability of x successes in n independent Bernoulli trials, each
with probability of success P
Probability distributions
Binomial Distribution
• In the previous example, we wanted to find the
probability that x = 4 when n = 10 and p = 0.3
• Drawing on our knowledge of cumulative probability
distributions, we know that P(x = 4) may be found by
subtracting P(X≤3) from P(X≤4).
• If in Binomial distribution table we locate P = 0.3 for n =
10, we find that P(X≤4) = 0.8497 and P(X≤3) = 0.6496.
Subtracting the latter from the former gives
0.8497 - 0.6496= 0.2001, which agrees with our hand
calculation.
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Probability distributions
Binomial Distribution
•
Suppose it is known that in a certain population 10
percent of the population is color blind. If a random
sample of 25 people is drawn from this population, use
the Binomial table in the to find the probability that:
1. Five or fewer are color blind.
This probability is an entry in the table. No addition or
subtraction is necessary. P(X≤5) = 0.9666.
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Probability distributions
Binomial Distribution
2. Six or more will be color blind.
We cannot find this probability directly in the table.
To find the answer, we use the concept of
complementary probabilities. The probability that six
or more are color blind is the complement of the
probability that five or fewer are not color blind.
P(X≥5) = 1- P(X≤5) = 0.963 = 1 - 0.9666 = 0.0334
3. Between six and nine inclusive will be color blind.
We find this by subtracting the probability that X is less than or
equal to 5 from the probability that X is less than or equal to 9.
P(6≤X≤9) = P(X≤9) - P(X≤5) = 1 - 0.967 = 0.033
4. Two, three, or four will be color blind.
This is the probability that is between 2 and 4 inclusive.
P(2 ≤ X ≤ 4) = P(X ≤ 4) - P(X ≤ 1) = 0.902 - 0.271 = 0.631
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Probability distributions
Binomial Distribution
•
Many binomial tables do not give probabilities for values
of p greater than 0.5.
•
We may obtain probabilities from Binomial tables by
restating the problem in terms of the probability of a
failure, 1 - p, rather than in terms of the probability of a
success, p.
•
As part of the restatement, we must also think in terms
of the number of failures, n - x, rather than the number
of successes, x.
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•
Probability distributions
Binomial
Distribution
We may summarize this idea as follows:
P(X = x|n, p > 0.50) = P(X=(n - x)|n, 1 - p)
•
In words, "The probability that X is equal to some
specified value given the sample size and a probability
of success greater than .5 is equal to the probability that
X is equal to n - x given the sample size and the
probability of a failure of 1 -p.“
•
For purposes of using the binomial table we treat the
probability of a failure as though it were the probability
of a success. When p is greater than 0.5, we may
obtain cumulative probabilities from Binomial tables by
using the following relationship:
P(X≤ x|n, p >0.5) = P(X≥(n - x)\ n, 1 - p)
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•
Probability distributions
Binomial Distribution
Finally, to use Binomial tables to find the probability that
X is greater than or equal to some x when P > 0.5, we
use the following relationship:
P(X≥x|n, p >0.5) = P(X≤ n - x| n, 1 - p)
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•
Probability distributions
Binomial Distribution
In a certain community, on a given evening, someone is
at home in 85 percent of the households. A health
research team conducting a telephone survey selects a
random sample of 12 households. Use binomial tables
to find the probability that:
1. The team will find someone at home in exactly 7
households.
We restate the problem as follows: What is the
probability that the team conducting the survey gets no
answer from exactly 5 calls out of 12, if no one is at
home in 15 percent of the households?
We find the answer as follows:
P(X=5|n=12, p =0.15) = P(X≤5) - P(X≤4)= 0.9954 - 0.9761 = 0.0193
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Probability distributions
Binomial Distribution
2. The team will find someone at home in 5 or fewer
households.
Solution: The probability we want is
P(X≤5|n = 12, p =0.85) = P(X≥12 -5|n = 12, p =0.15)
= P(X≥7|n =12, p =0.15)
= 1 - P(X≤61n =12, p =0.15)
= 1 - 0.9993 = 0.0007
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Probability distributions
Binomial Distribution
3. The team will find someone at home in 8 or more
households.
Solution: The probability we desire is
P(X≥8|n =12, p =0.85) = P(X≤4|n=12, p =0.15) =0.9761
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Probability distributions
Binomial Distribution
•
The binomial distribution has two parameters, n and p.
•
They are parameters in the sense that they are
sufficient to specify a binomial distribution.
•
The binomial distribution is really a family of
distributions with each possible value of n and p
designating a different member of the family.
•
The mean and variance of the binomial distribution are
µ = np
2
 = np(1— p)
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p = 0.1 (blue)
p = 0.8 (red)
p = 0.5 (green)
Binomial distribution for n = 20
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Probability distributions
Poisson Distribution
•
Another discrete distribution is the Poisson distribution,
named for the French mathematician Simeon Denis
Poisson (1781-1840), who is generally credited for
publishing its derivation in 1837.
•
If x is the number of occurrences of some random event
in an interval of time or space (or some volume of
matter), the probability that x will occur is given by:
e  x
f ( x) 
, x  0, 1, 2, ....................
x!
The Greek letter λ (Lambda) is called the parameter of
the distribution and is the average number of
occurrences of the random event in the interval (or
volume).
The symbol e is the constant (to four decimals) 2.7183.
•
•
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Probability distributions
Poisson Distribution
• The Poisson process:
1. The occurrences of the events are independent. The
occurrence of an event in an interval of space or time
has no effect on the probability of a second occurrence
of the event in the same, or any other, interval.
2. Theoretically, an infinite number of occurrences of the
event must be possible in the interval.
3. The probability of the single occurrence of the event in a
given interval is proportional to the length of the interval.
4. In any infinitesimally small portion of the interval, the
probability of more than one occurrence of the event is
negligible.
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Number of accidents at this intersection/day:
1. Probability of second accident is independent
on the first one
2. Infinite number of accidents can occur (not
like the binomial model)
3. Probability of accidents in two days more than
one day
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Probability distributions
Poisson Distribution
•
In a study of suicides, it was found that the monthly
distribution of adolescent suicides in one county in Illinois
between 1977 and 1987 closely followed a Poisson
distribution with a parameter of λ=2.75.
•
What is the probability that a randomly selected month
will be one in which three adolescent suicides occurred?
e 2.75  2.753 (0.063928)  (20.796875)
P( X  3) 

 0.221584
3!
6
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Probability distributions
Poisson Distribution
•
Assume that future adolescent suicides in the studies
population will follow a Poisson distribution. What is the
probability that a randomly selected future month will be
one in which either three or four suicides will occur.
e 2.75  2.754
P( X  3)  P( X  4)  (
)  0.221584  0.373922
4!
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Probability distributions
Poisson Distribution
•
Special Poisson tables give cumulative probabilities for
various values of λ and X.
•
In a study of a certain aquatic organism, a large number
of samples were taken from a pond, and the number of
organisms in each sample was counted. The average
number of organisms per sample was found to be two.
Assuming that the number of organisms follows a
Poisson distribution, find the probability that the next
sample taken will contain one or fewer organisms.
•
From the table when λ=2, the probability that X≤1 is
0.406, P(X≤1|2)=0.406
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67
68
• Only one parameter
(λ)
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Probability distributions
Normal distribution
•
A continuous variable is one that can assume any value
within a specified interval of values assumed by the
variable.
•
Consequently, between any two values assumed by a
continuous variable, there exist an infinite number of
values.
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Normal distribution
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Normal distribution
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Normal distribution
•
In such distributions, as the number of observations, n,
approaches infinity, and the width of the class intervals
approaches zero, the frequency polygon approaches a
smooth curve.
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Normal distribution
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Normal distribution
•
The total area under the curve is equal to one.
•
The relative frequency of occurrence of values between
any two points on the x-axis is equal to the total area
bounded by the curve, the x-axis and the perpendicular
lines erected at the two points on the x-axis.
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Normal distribution
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Normal distribution
•
The most important of continuous probability
distributions is the normal distribution (some times
referred to as Gaussian distribution, Carl Friedrich
Gauss, 1777-1855).
•
The equation describing the normal distribution was first
published by Abraham De Moivre (1667-1754):
•
•
π=3.14159, e=2.71828
The parameters of distribution are μ, the mean and σ,
the standard deviation.
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Normal distribution
•
Characteristics of the normal distribution:
1. It is symmetrical about its mean, µ,. As is shown in
the folowing figure, the curve on either side of µ is a
mirror image of the other side.
2. The mean, the median, and the mode are all equal.
3. The total area under the curve above the x-axis is
one square unit. This characteristic follows from the
fact that the normal distribution is a probability
distribution.
Because of the symmetry already mentioned, 50
percent of the area is to the right of a perpendicular
erected at the mean, and 50 percent is to the left.
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Normal distribution
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Normal distribution
•
Characteristics of the normal distribution:
5. If we erect perpendiculars a distance of 1 standard
deviation from the mean in both directions, the area
enclosed by these perpendiculars, the x-axis, and the
curve will be approximately 68 percent of the total
area.
If we extend these lateral boundaries a distance of 2
standard deviations on either side of the mean,
approximately 95 percent of the area will be
enclosed, and extending them a distance of 3
standard deviations will cause approximately 99.7
percent of the total area to be enclosed. These
approximate areas are illustrated in the following
figure.
80
Normal distribution
81
Normal distribution
82
Normal distribution
83
Normal distribution
•
Characteristics of the normal distribution:
5. The normal distribution is completely determined by
the parameters µ and .
In other words, a different normal distribution is
specified for each different value of µ and . Different
values of µ, shift the graph of the distribution along
the x-axis as is shown in the following figure. Different
values of  determine the degree of flatness or
peakedness of the graph of the distribution as is
shown in the figures.
84
Normal distribution
85
Normal distribution
86
The Standard Normal Distribution
•
The last-mentioned characteristic of the normal
distribution implies that the normal distribution is really a
family of distributions in which one member is
distinguished from another on the basis of the values of
µ and .
•
The most important member of this family is the
standard normal distribution (SND) or unit normal
distribution, as it is sometimes called, because it has a
mean of 0 and a standard deviation of 1.
87
The Standard Normal Distribution
88
The Standard Normal Distribution
•
The SND equation may be obtained from normal
distribution equation by creating a random variable z =
(x — µ)/.
•
The equation for the standard normal distribution is:
•
Which is derived from the normal distribution equation:
89
The Standard Normal Distribution
•
To find the probability that z takes on a value between
any two points on the z-axis (z0 and z1), we must find the
area bounded by the perpendiculars erected at these
points, the curve and the horizontal z-axis.
•
The area can be found by integrating the f(z) function
between two values of the variable. In the standard
normal distribution, the integral (area) is given by the
equation:
z1
1
z0
2

e
z2 / 2
dz
90
The Standard Normal Distribution
•
There are tables that provide the results of all such
integrations.
•
These tables provide the areas under the curve between
- and z.
91
92
93
Application of Normal Distribution
• As a part of a study of Alzheimer’s disease, data
reported were compatible with the hypothesis that brain
weights of victims of the disease are normally
distributed. From the reported data we may compute a
mean of 1076.80 grams and a standard deviation of
105.76 grams. If we assume that these results are
applicable to all victims of Alzheimer, find the probability
that a randomly selected victim will have a brain that
weighs less than or equal to 800 grams.
94
Application of Normal Distribution
800  1076.80
z
 2.62
105.76
• The distance from the mean, 1076.80 to the x value of
interest (800) is
• 800 - 1076.80 = -276.80
• which is the distance of 2.62 standard deviations.
• When we transform brain weight values to z values, the
distance of the z value of interest from its mean 0 is
equal to the distance of the corresponding x value from
its mean (1076.80), in standard deviation units.
• From the table:P(z≤-2.62)=P(X≤800)=0.0044
95
Application of Normal Distribution
• Suppose it is known that the heights of a certain
population of individuals are approximately normally
distributed with a mean of 70 inches and a standard
deviation of 3 inches. What is the probability that a
person picked at random from this group will be between
65 and 74 inches tall?
65  70
   1.67
3
74  70
z 74 
 1.33
3
P(65  x  74)  P(1.67  z  1.33)  P(  z  1.33)  P(  z  1.67)  0.9082  0.0475  0.8607
z 65 
96
Application of Normal Distribution
• In a population of 10,000 of the people described in the
previous example, how many would you expect to be 6
feet 5 inches tall or taller?
77  70
 2.33
3
P( x  77)  P( z  2.33)  1  0.9901  0.0099
z 65 
• Out of 10,000 people, we would expect
10,000*0.0099=99 to be 6 feet 5 inches tall (77 inches) or
taller.
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