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Probability Distributions
Chapter 3
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Probability Distribution of Discrete
Variable
Definition: The probability distribution of adiscrete variable is a table, graph, formula,
or other device used to specify all possible
values of a discrete randam variable along
with their respective probabilities
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Example 3.2.1
A public health nurse has a case load of50 families.
The goal is to construct the probabilitydistribution of X, the number of children
per family for the population
Table 3.2.1 gives the probabilitydistribution on the next slide
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Table 3.2.1Probability distribution of number of
children per family in a population of 50 families
x Frequency of occurance of x P(X) = x
0 1 1/50
1 4 4/50
2 6 6/50
3 4 4/50
4 9 9/50
5 10 10/50
6 7 7/50
7 4 4/50
8 2 2/50
9 2 2/50
10 1 1/50
50 50/50
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Figure 3.2.1Graphical representation of Probability
distribution of number of children per family in a
population of 50 families
0
0.05
0.1
0.15
0.2
0.25
0 1 2 3 4 5 6 7 8 9 10
x
Probability
P(X) = x
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Probability Distribution of Discrete
Variable
So There are two essential properties ofprobability distribution of a discrete
variable
1. 0 P(X = x) 1
2. P(X = x) = 1
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Example 3.2.1
With this probability distribution, it ispossible to make probability statements
regarding the random variable X
Question: For any one of the familyselected, what is the probability that the
family will have three children?
The answer: P(X=3) = 4/50 = 0.08
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Example 3.2.1
What is the probability that a familychosen at random will have either three or
four children?
The answer: From the addition rule
P(X=3 or X=4) = P(X=3) + P(X=4)
P(X=3 or X=4) = 0.08 + 0.18 = 0.26
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Cumulative Distribution It is obained by succesive
addition of probabilities
Cumulative probabilitydistribution can help us tuanswer several questions
1. What is the probability that a
family picked at random fromthe 50 will have fewer than fivechildren?
Look at the excel example1
x Frequency of
occurance of x
P(X= x) P(X x)
0 1 0.02 0.02
1 4 0.08 0.1
2 6 0.12 0.22
3 4 0.08 0.3
4 9 0.18 0.48
5 10 0.2 0.68
6 7 0.14 0.82
7 4 0.08 0.9
8 2 0.04 0.949 2 0.04 0.98
10 1 0.02 1
sum 50 1
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Cumulative Distribution 1. What is the probability that a
family picked at random from
the 50 will have fewer than fivechildren?
For X = 0 to X = 4 P(X
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Cumulative Distribution 2. What is the probability that a
ranomly picked family will have
five or more children? Here, the set of families with
five or more children is thecompliment set of the et offamilies with fewer than fivechildren
The sum always equal to 1 As a result
P(X 5) = 1 P(X
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Cumulative Distribution 3. What is the probability that a
ranomly picked family will have
between three and six children,inclusive?
Here, we need to find out:P(3 X 6)
Now, this equals to:
P(3 X 6) = P(X 6) P (X< 3) P(3 X 6) = 0.82 0.22 P(3 X 6) = 0.60
x Frequency of
occurance ofx
P(X= x) P(X
x)
0 1 0.02 0.02
1 4 0.08 0.1
2 6 0.12 0.22
3 4 0.08 0.3
4 9 0.18 0.48
5 10 0.2 0.68
6 7 0.14 0.82
7 4 0.08 0.9
8 2 0.04 0.94
9 2 0.04 0.98
10 1 0.02 1
su
m
50 1
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Theoretical Probability Distributions
Binominal distribution
Poisson Distribution
Normal Distrbution
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Binominal distribution
One of the most encountered distribution
Originated from Bernoulli trialnamed afterin the honor of the Swiss mathematician
James Bernoulli
When a single trial of some process orexperiment can result in only one of two
mutually exclusive outcomes (i.e. Male or
female) the trial is said to be binominal
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Binominal distribution
Bernoulli trials forms the benoulli process on thefolowing conditions
1. each trial result in one of two possible
mutually exclusive outcomes (one success andother failure)
2. The probability of success denoted asp andthe probability of failure with q as q=1-p
3. The trials are independent means that theoutcome of any particular trial is not effected bythe outcome of any other trial
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Example 3.3.1
The task is to calculate the probability of xsuccesses in n benoulli trial
Let say, in a certain population 52 percent of all
recorded birts are male Then we say that the probability of a recorded
male birth is 0.52
If we randomly selct five birth records from thispopulation, what is the probability that exaclythree of records will be for male births?
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Example 3.3.1
Let say, male birth is a success (arbitrary) Suppose the five birth records selected
resulted in this squence of sexes:
MFMMF f code this as 10110 Now, with p and q notation, the probability
of above sequence is found by means ofmultiplication rule as
P(1.0,1,1,0) = pqppq = q2p3
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Example 3.3.1
The three male andtwo female squences
could occur as
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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Example 3.3.1
So, the question:What is the
probability, in a
random sample ofsize 5, drawn from the
specified population,
of observing three
success and twofailure?
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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Example 3.3.1
Since in the population, p =0.52 and
q = 1 p = 1 0.52 = 0.48
The the answer to thequestion: 10*q
2*p
3
= 10*(0.48)2
*(0.52)3
=10*0.2304*0.140608=0.32 So, where does the 10 came
from?
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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Example 3.3.1
The list of number of sequenceswill become difficult as the sample
size increases
Thus, if whe have n things andxofwhich one type and the remainder
are of another type we use the
equation:
nCx= n!/x!(n-x)!
This equation gives the number ofcombinations ofn things takenxat
a time
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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Example 3.3.1
Thus, for our example
nCx= n!/x!(n-x)!
nCx= 5!/2!(5-2)! = 120/12 =10
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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The Binominal Distribution
Then, the probability of obtainingexactly x successes in n trials as
f(x) = nCxqn-xpx
f(x) = nCxqxpn-x
for x = 0,1,2,...,n
Number Sequence
1 111OO
2 11O1O
3 11OO1
4 1O11O
5 1OO11
6 1O1O1
7 O111O
8 OO111
9 O1O11
10 O11O1
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The Binominal Distribution
Number of Success Probablity, f(x)
0 nC0qn-0p0
1 nC1qn-1p1
2 nC2qn-2p2
... ...
x nCxqn-xpx
... ...
n nC0qn-npn
Total 1
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Example 3.3.2
30 % of a certain population are immune to somedisease.
If a random sample of size 10 is selected from thispopulation, what is the probability that it will contain
exactly 4 immune persons?
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Example 3.3.2
Here we can set the probability of an immune person tobe 0.3
Then
f(4) =10
C4(0.7)6(0.3)4 = 10! /4!6! (0.117649)(0.0081)
=0.2001
This prblem can be solved with Excel
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Example 3.3.2
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For the birth record example
Here, the objectve was the probabilty of observing exactly3male birth records when n=5 and p=0.52
Hawever there is no =0.52 in Table A in Appendix II
But, if we ask the question as the probability of observing 2female birth for n=2 and p=1-0.52=0.48, we can solve the
problem
Since from the table: P(X2)=0.5373 andP(X1)=0.2135
Then P(X=2) = P(X2) - P(X1)= 0.5373 0.2135 = 0.324
This number is same wth our previous example
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Example 3.3.3
Let say for a certain pupation 10 % of thepopulation is color blind. If a randam sample of 25
people is drawn from this population.
a. Find the probability that five or fewer will be colorblind
From the Table A for n=25 and p=0.10
P(X5) = 0.9666
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Example 3.3.3
Let say for a certain pupation 10 % of thepopulation is color blind. If a randam sample of 25
people is drawn from this population.
b. Find the probability that six or more will be colorblind
This is the compliment of the set specified in part a.
Thus
P(X6) = 1 - P(X5) = 1 - 0.9666 = 0.0334
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Example 3.3.3
Let say for a certain pupation 10 % of thepopulation is color blind. If a randam sample of 25people is drawn from this population.
c. Find the probability that between six and nineinclusve will be color blind
This is the compliment of the set specified in part a. Thus
P(6X9) = P(X9) - P(X5) P(6X9) = - 0.9999 0.9666 = 0.0333
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Example 3.3.3
Let say for a certain pupation 10 % of thepopulation is color blind. If a randam sample of 25people is drawn from this population.
c. Find the probability that two, three, or four will becolor blind
This is the compliment of the set specified in part a. Thus
P(2X4) = P(X4) - P(X1) P(6X9) = - 0.9020 0.2712 = 0.6308
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When we have p > 0.50
The probability thatXis equal to some specifiedvalue given the sample size and probability of
success greather than 0.50 is equal to the
probability thatXis equal to nxgiven the samplesize and the probability of a failure of 1 p.
This statement is given as
P(X=x|n, p>0.50) = P(X=n - x|n, 1 p)
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When we have p > 0.50
If we are looking for cumulative probability whenp >0.50
P(Xx|n, p>0.50) = P(Xn - x|n, 1 p) And to find the probability that X is grather than or
equal to some x when p>0.50 we use the following
equation P(Xx|n, p>0.50) = P(Xn - x|n, 1 p)
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Example 3.3.4
In a certain community, on a given evening,someone is at home in 85 % of hoseholds. A healthcare research team conducting a telephone surveyselects a random sample of 12 households.
a. Find the probability that the team will findsomeone at home in exactly 7 households
We can look at this problem as the probability thatthe team conducting the survey gets no answerfrom exaxtly 5 calls out of 12, if no one is at home in15 % of households
Then the answer
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Example 3.3.4
We can look at this problem as the probability thatthe team conducting the survey gets no answer
from exaxtly 5 calls out of 12, if no one is at home in
15 % of households Then the answer
P(X=5|12, p>0.15) = P(X5) P(X4)
P(X=5|12, p>0.15) = 0.9954 0.9761 = 0.0193
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Example 3.3.4
In a certain community, on a given evening,someone is at home in 85 % of hoseholds. A health
care research team conducting a telephone survey
selects a random sample of 12 households. b. Find the probability that the team will find
someone at home in 5 or fewer hoseholds
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Example 3.3.4
b. Find the probability that the team will findsomeone at home in 5 or fewer hoseholds
P(X5|12, p>0.85) = P(X12 - 5|n=12, p=1 0.85) P(X5|12, p>0.85) = P(X7|n=12, p=0.15)
P(X5|12, p>0.85) = 1 - P(X6|n=12, p=0.15)
P(X5|12, p>0.85) = 1 0.9993 = 0.0007
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Example 3.3.4
c. Find the probability that the team will findsomeone at home in 8 or more hoseholds
P(X8|n=12, p=0.85) = P(X4 |n=12, p=0.15)
P(X8|n=12, p=0.85) = 0.9761
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The poisson Distribution
Poisson distribution named for the Frenchmathematician Simoen Denis Poisson
It has been used in biology and medicine
If x is the number of occurances of some randomevent in an interval of time or space, the probabiity
that x will occur is given by
,...2,1,0
!
x
x
exf
x
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The poisson Distribution
Ifxis the number of occurances of some randomevent in an interval of time or space, the probabiitythatxwill occur is given by
The is called the parameter of the distribution and
is the average number of occurances of the randamevent in the interval
The symbol e is the constant as 2.7183
,...2,1,0
!
xx
exf
x
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The poisson Distribution
It can be shown that f(x) 0 for every x and thatf(x) = 1
So that the distribution satisfies the requirements fora probability distribution
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Assumtions in poisson Process
1. The occurances of the events are independent
This means that the occurance of an event in an
interval of space or time has no effect on theprobability of a second occurance of the event in
the same or any other interval
An intresting feature of poisson distribution is thatmean and variance are equal
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Assumtions in poisson Process
2. Theoretically, an infinete number of occurancesof the event must be possible in the interval
3. The probability of the single occurance of theevent in a given interval is proportional to the lengthof the interva
In any infinetelly small portion of the interval theprobability of more than one occurance of the eventis negligible
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Example 3.4.1
A hospital administrator, who has been studiyngdaily emergency admissions over a period ofseveral years, has concluded that they aredistributed according to poisson law
Hospital records reveal that emergency admissionshave averaged three per day during this period
If the admissions is correct in assuming a Poissondistribution,
a. Fin the probability that exactly two emergencyadmissions will occur on a given day
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Example 3.4.1
a. Fin the probability that exactly two emergencyadmissions will occur on a given day
Here = 3 and X is a random variable denoting the
number of daily emergency admission f X follows the Poisson distribution
225.0
12
905.0
!2
322
23
e
fXP
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Example 3.4.1
b. Fin the probability that no emergency admissionswill occur on a particular day
05.0
1
105.0
!0
30
03
e
f
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Example 3.4.1
c. Fin the probability that either three or fouremergency admissions will occur on a particular
day
39.0
24
8105.0
6
2705.0
!43
!3343
4333
eeff
All of this values can also be obtained from Table B in Appendix II for the
known and X
The Poisson distribution is defined by one parameter: lambda
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The Poisson distribution is defined by one parameter: lambda.
This parameter equals the mean and variance.
As lambda increases, the Poisson distribution approaches a
normal distribution.
A variable follows a Poisson distribution if the following
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A variable follows a Poisson distribution if the following
conditions are met:
Data are counts of events (non-negative integers with no
upper bound).
All events are independent.
Average rate does not change over the period of interest.
The Poisson distribution is similar to the binomial distribution
because theyboth model counts of events.
However, the Poisson distribution models a finite observation
space with any integer number of events greater than or equal to
zero.The binomial distribution models a fixed number of discrete trials
from 0 to n events.
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The formula for the Poisson cumulative probability
function is
The following is the plot of the Poisson cumulative distributionfunction with the same values of as the pdf plots above.
The following is the plot of the Poisson cumulative distribution
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The following is the plot of the Poisson cumulative distribution
function with the same values of as the pdfplots above.
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Example 3.4.2
In the study of a certain aquatic organism, a largenumber of samples were taken from a pound, andthe number of organism in each sample wascounted.
The average nuber of organisms per sample wasfound to be two.
Assuming that the number of organisms follows apoisson distribution:
a. Find the probability that the next sample takenwill contain one or fewer organisms
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Example 3.4.2
a. Find the probability that the next sample takenwill contain one or fewer organisms
In the Table B, when = 2, the probability that X1is 0.406
That is, P(X1| 2) = 0.406
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Example 3.4.2
b. Find the probability that the next sample takenwill contain exactly three organisms
P(X = 3 | 2) = P(X 3) P(X 2)
P(X = 3 | 2) =0.857 0.677 = 0.180
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Example 3.4.2
c. Find the probability that the next sample takenwill contain more than five organisms
P(X > 5 | 2) = 1P(X 5)
P(X > 5 | 2) = 1 0.983 = 0.0170
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Continuous Probability Distribution
The distributions (Binominal and Poisson) so far we haveseen are distributions of discrete variables
A continus variable is one that can assume any value withina specified interval of values assumed by the variable.
f(x)
xFigure 3.5.2 Graphical Representation of a continuous distribution
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Continuous Probability Distribution
The total area under the curve is equal to 1 The probability of any specific value of a random variable is
zero
f(x)
xFigure 3.5.2 Graphical Representation of a continuous distribution
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Continuous Probability Distribution To find the area of a smooth curve between any two points, a and b, the density
function is integrated from a to b. A density function is a formula used to represent the distribution of a continuousrandom variable
f(x)
xFigure 3.5.2 Graph of a contnuous distribution showing area between a and b
a b
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Continuous Probability Distribution
Defination: A nonnegative function f(x) is called aprobability distribution (probability density function)
of the continuous random variable X if the total area
bounded by its curve and the x-axis is equal to 1and if the subarea under the curve bounded by the
curve, the x-axis, and perpendiculars drawn at any
two points a and b gives the probability that X is
between the points a and b
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The Normal Distribution
The most importan distribution in all of statistics
It is first formulated by Abraham De Moivre in 1733
It is also called Gaussian distribution in the honor of
Carl Friedrich Gauss The normal density function is given as
xfor
exf
x
2
1 22
2
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The Normal Distribution
Here is the meanand is the standard
deviation
xfor
exfx
2
1 22
2
x
f(x)
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Charecteristics of Normal Distribution
It is symmetrical around the mean, The mean, median, and mode are all same The total area under the curve above the x-axis is one
square unit.
The area defined by 1 around the mean is approximately68 % of the total area
The area defined by 2 around the mean is approximately95 % of the total area
The area defined by 3 around the mean is approximately99.7 % of the total area
The normal distribution completely determined byparameters mean and standard deviation
The satandard normal distribution is a special case wheremean equals to zero and standard deviation equals to one
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The satandard Normal Distribution
For a random variable defined as
The the equation for the standard normal
distribution is given as
xz
zezfz
,2
12
2
2
11
0
2
2z
z
z
dze
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The satandard Normal Distribution
x
f(x)
z
f(z)
=0
=1
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Example 3.6.1
Given the standard normal distribution, find the area underthe curve, above the z-axis between
z = and z = 2
From the Table C in Appendix II, the are is given as 0.9772
x
f(x)
0 2
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Example 3.6.1
From the Table C in Appendix II, the are is given as0.9772
This value says that the probability that a z picked atrandom from the population of zs will have a valuebetween - and 2
x
f(x)
0 2
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Example 3.6.1
From the Table C in Appendix II, the are is given as0.9772
We can also say that the relative frequency ofoccurance (or proportion) of values of z betven -and 2.
x
f(x)
0 2
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Example 3.6.1
From the Table C in Appendix II, the area is given as
0.9772
Put another way, 97.72 % of the zs have a value
between - and 2.
x
f(x)
0 2
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Example 3.6.2
What is the probabilty that a z picked at random fromthe population of zs will have value between
-2.55 and +2.55?
-2.55 2.55
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Example 3.6.2
P(-2.55
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Example 3.6.3
What proportion of z values are between
-2.74 and +1.53?
-2.74 1.53
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Example 3.6.3
P(-2.74z1.53) = 0.9370 0.0031= 0.9892
-2.74 1.53
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Example 3.6.4
Given the satandard normal distribution,
find P(z2.71)
2.71
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Example 3.6.4
P(z2.71) = 1 P(z2.71) = 1 0.9966= 0.0034
2.71
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Example 3.6.5
Given the satandard normal distribution,
find P(0.84z2.45)
2.450.84
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Example 3.6.5
P(0.84z2.45)=P(z2.45) P(z0.84)
=0.9929 0.7995= 0.1934
2.450.84
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Example 3.6.6
A physical terapist belivesthat scores on a certain
manual dexterity test are
approximately normally
distributed with a mean of10 and a standard
deviation of 2.5. If a
randomly selected
individual takes the test,
what is the probability
that he or she will make a
sore of 15 or better?
15=10
=2.5
20
=1
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Example 3.6.6
P(x15)=P(z2) = 0.0228
20
=1
2
5.2
101551for x
xz
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Example 3.6.7
Suppose it is known thatthe weights of a certainpopulation of individualsare aproximately normallydistributed with a mean of
140 pounds and astandard deviation of 25pounds.
What is the probability
that a person picked atrandom from this groupwill weight between 100and 170 pounds
170=140
=25
100
1.20
=1
-1.6
Example 3.6.7
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P(100x170)=P(-1.6z1.2) =P(-z1.2) P(-z-1.6) =0.8849 0.0548P(-z1.2) P(-z-1.6) =0.8301
1.20
=1
-1.6
2.1
25
140170170for x
6.125
140100100for x
xz
xz
Example 3 6 8
7/30/2019 Probability Distributionsa
82/82
Example 3.6.8
In a population of 10,000 of the people described in
example 3.6.7, how many would you expect to
weight more than 200 pounds?
P(x>200)=P(z>2.4) =1-0.9918 = 0.0082
So, for 10.000 people
10,000x0.0082=82 will weigh more than 200 pounds