probability distributionsa

7/30/2019 Probability Distributionsa

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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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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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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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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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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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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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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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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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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


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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

probability distributionsa

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