binomial normal distribution

7/30/2019 Binomial Normal Distribution

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Normal Distribution


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Normal Distribution

Why are normal distributions so important?

Many dependent variables are commonly assumed

to be normally distributed in the population

If a variable is approximately normally distributed wecan make inferences about values of that variable

Example: Sampling distribution of the mean


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TYPES OF HISTOGRAM


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Normal Distribution

Symmetrical, bell-shaped curve

Also known as Gaussian distribution

Point of inflection = 1 standard deviationfrom mean

Mathematical formula

f(X) 1

2

(e)

(X)2

22


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Since we know the shape of the curve, we can

calculate the area under the curve

The percentage of that area can be used to

determine the probability that a given value could bepulled from a given distribution

The area under the curve tells us about the probability- in

other words we can obtain a p-value for our result (data)

by treating it as a normally distributed data set.


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Problem:

Each normal distribution with its own values of

and would need its own calculation of the area

under various points on the curve


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The normalBell shaped curve:

=100,2=10

90 95 100 105 110

x

0.

00

0.

02

0.0

4

0.

06

0.

08

0.

10

0.

12

fx


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Normal curves:

(=0,2=1) and (=5,2=1)

-2 0 2 4 6 8

x

0.

0

0.

1

0.

2

0.

3

0.

4

fx1


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Normal curves:(=0,2=1) and (=0,2=2)

-3 -2 -1 0 1 2 3

x

0.

0

0.

1

0.

2

0.

3

0.

4

y


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Normal curves:(=0,2=1) and (=2,2=0.25)

-2 0 2 4 6 8

0.

0

0.

2

0.

4

0.

6

0.

8

1.

0

fx1


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The standard normal curve:

=0, and2=1

-3 -2 -1 0 1 2 3

x

0.

0

0.

1

0.

2

0.

3

0.

4

y


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Normal Probability DistributionsStandard Normal Distribution N(0,1)

We agree to use thestandard normaldistribution

Bell shaped

=0

=1

Note: not all bellshaped distributionsare normaldistributions


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Normal Distribution

The standard normal distribution will allow us

to make claims about the probabilities of

values related to our own data

How do we apply the standard normal

distribution to our data?


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Z-score

If we know the population mean and

population standard deviation, for any value

of X we can compute a z-score by subtracting

the population mean and dividing the resultby the population standard deviation

zX


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Probabilities are depicted by areas under the curve

Total area under the curve is1

The area in red is equal top(z > 1)

The area in blue is equal top(-1< z


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Binomial Probability Distribution

A fixed number of observations (trials), n

e.g., 15 tosses of a coin; 20 patients; 1000 people surveyed

A binary random variable

e.g., head or tail in each toss of a coin; defective or notdefective light bulb

Generally called success and failure

Probability of success is p, probability of failure is 1 p

Constant probability for each observation

e.g., Probability of getting a tail is the same each time we

toss the coin


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Binomial example

Take the example of 5 coin tosses. Whats the

probability that you flip exactly 3 heads in 5

coin tosses?


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Binomial distribution

Solution:

One way to get exactly 3 heads: HHHTT

Whats the probability of this exact arrangement?P(heads)xP(heads) xP(heads)xP(tails)xP(tails) =(1/2)3x

(1/2)2

Another way to get exactly 3 heads: THHHT

Probability of this exact outcome = (1/2)1x (1/2)3x(1/2)1 = (1/2)3x(1/2)2


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Binomial distribution

In fact, (1/2)3x(1/2)2 is the probability of each

unique outcome that has exactly 3 heads and 2

tails.

So, the overall probability of 3 heads and 2 tails is:

(1/2)3x(1/2)2 + (1/2)3x(1/2)2+ (1/2)3x(1/2)2 +

.. for as many unique arrangements as there arebut how many are there??


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Outcome Probability

THHHT (1/2)3x(1/2)2

HHHTT (1/2)3x(1/2)2

TTHHH (1/2)3x(1/2)2

HTTHH (1/2)3x(1/2)2

HHTTH (1/2)3x(1/2)2

THTHH (1/2)3x(1/2)2

HTHTH (1/2)3x(1/2)2

HHTHT (1/2)3x(1/2)2

THHTH (1/2)3x(1/2)2

HTHHT (1/2)3x(1/2)210 arrangementsx (1/2)3x(1/2)2

The probabilityof each unique

outcome (note:

they are all

equal)

ways toarrange 3

heads in

5 trials

5

3

5C3 = 5!/3!2! = 10


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P(3 heads and 2 tails) = x P(heads)3 x P(tails)2 =

10 x ()5=

31.25%

5

3


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x

p(x)

0 3 4 51 2

Binomial distribution function:X= the number of heads tossed in 5 coin tosses

number of heads

p(x)

number of heads


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Example 2

As voters exit the polls, you ask a representative

random sample of 6 voters if they voted for

proposition 100. If the true percentage of voters who

vote for the proposition is 55.1%, what is theprobability that, in your sample, exactly 2 voted for

the proposition and 4 did not?


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Solution:

Outcome Probability

YYNNNN = (.551)2x(.449)4

NYYNNN(.449)1x (.551)2x(.449)3 = (.551)2x(.449)4

NNYYNN(.449)2x (.551)2x(.449)2 = (.551)2x(.449)4

NNNYYN(.449)3x (.551)2x(.449)1 = (.551)2x(.449)4

NNNNYY(.449)4x (.551)2 = (.551)2x(.449)4

.

.

ways to

arrange 2

Obama votes

among 6

voters

6

2

15 arrangementsx(.551)2x(.449)4

6

2

P(2 yes votes exactly) = x (.551)2x(.449)4 = 18.5%


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Binomial distribution, generally

XnXn

X

pp

)1(

1-p =probabilityof failure

p =

probability of

success

X = #

successes

out ofntrials

n = number of trials

Note the general pattern emerging if you have only two possibleoutcomes (call them 1/0 or yes/no or success/failure) in n independent

trials, then the probability of exactlyXsuccesses=


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Definitions: Binomial

Binomial: Suppose that n independent experiments, or trials,

are performed, where n is a fixed number, and that each

experiment results in a success with probabilityp and a

failure with probability 1-p. The total number of successes,

X, is a binomial random variable with parameters n andp.

We write:X ~ Bin (n, p) {reads: X is distributed binomially

with parameters n and p}

And the probability thatX=r(i.e., that there are exactly r

successes) is:

rnrn

r

pprXP

)1()(


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Definitions: Bernouilli

Bernouilli trial: If there is only 1 trial with

probability of successp and probability of

failure 1-p, this is called a Bernouilli

distribution. (special case of the binomial withn=1)

Probability of success:

Probability of failure:pppXP

111

1

1

)1()1(

pppXP

1)1()0( 010

1

0


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Binomial distribution: example

If I toss a coin 20 times, whats the

probability of getting exactly 10 heads?

176.)5(.)5(. 101020

10


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Binomial distribution: example

If I toss a coin 20 times, whats the

probability of getting of getting 2 or fewer

heads?

4

472018220

2

572019120

1

72020020

0

108.1

108.1105.9190)5(.!2!18

!20)5(.)5(.

109.1105.920)5(.!1!19

!20)5(.)5(.

105.9)5(.!0!20

!20)5(.)5(.

x

xxx

xxx

x


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Things that follow a binomial

distribution

Cohort study (or cross-sectional):

The number of exposed individuals in your sample that

develop the disease

The number of unexposed individuals in your sample that

develop the disease

Case-control study:

The number of cases that have had the exposure The number of controls that have had the exposure


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Answer

1. You are performing a cohort study. If the probability of developing disease in the

exposed group is .05 for the study duration, then if you sample (randomly) 500

exposed people, how many do you expect to develop the disease? Give a margin

of error (+/- 1 standard deviation) for your estimate.

X ~ binomial (500, .05)E(X) = 500 (.05) = 25

Var(X) = 500 (.05) (.95) = 23.75

StdDev(X) = square root (23.75) = 4.8725 4.87


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Answer

2. Whats the probability that at most 10 exposed subjects

develop the disease?

01.)95(.)05(....)95(.)05(.)95(.)05(.)95(.)05(.49010

500

10

4982500

2

4991500

1

5000500

0

This is asking for a CUMULATIVE PROBABILITY: the probability of 0 getting the

disease or 1 or 2 or 3 or 4 or up to 10.

P(X10) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4)+.+ P(X=10)=

(well learn how to approximate this long sum next week)


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Youll rarely calculate the binomial by hand. However, it is good to know how to

Pascals Triangle Trick for calculating binomial coefficients

Recall from math in your past that Pascals Triangle is used to get thecoefficients for binomial expansion

For example, to expand: (p + q)5

The powers follow a set pattern: p5 + p4q1 + p3q2 + p2q3+ p1q4+ q5

But what are the coefficients?

Use Pascals Magic Triangle

A brief distraction: Pascals Triangle

Trick


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Pascals Triangle

1

1 11 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

Edges are all 1s

Add the twonumbers in the row

above to get the

number below, e.g.:

3+1=4; 5+10=15

To get the

coefficient forexpanding to the

5th power, use the

row that starts

with 5.

(p + q)5 = 1p5 + 5p4q1 + 10p3q2 + 10p2q3+ 5p1q4+ 1q5


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50

5

0

)5(.)5(.

415

1

)5(.)5(.

325

2

)5(.)5(.

235

3

)5(.)5(.

145

4

)5(.)5(.

055

5

)5(.)5(.

X P(X)

0

1

2

3

4

5

Same coefficients for X~Bin(5,p)

X P(X)

0 5)5(.1

1 5)5(.5

2 5)5(.10 3 5)5(.10

4 5)5(.5

5 5)5(.1

32(.5)5=

1.0

5

0

=5!/0!5!=1

5

1

=5!/1!4! = 5

5

2

= 5!/2!3!=5x4/2=10

5

3

=5!/3!2!=10

5

4

=5!/4!1!= 5

5

5

=5! /5!1!=1 (Note the symmetry!)

For example, X=# heads in 5 coin tosses:

From line 5

of Pascalstriangle!


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Relationship between binomial probability

distribution and binomial expansion

If p + q = 1 (which is the case if they are binomial probabilities)

then: (p + q)5= (1)

5= 1 or, equivalently:

1p5 + 5p4q1 + 10p3q2 + 10p2q3+ 5p1q4+ 1q5 = 1

(the probabilities sum to 1, making it a

probability distribution!)

P(X=0) P(X=1) P(X=2) P(X=3) P(X=4) P(X=5)


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Introduction to the Poisson Distribution

Poisson distribution is for countsif events happen ata constant rate over time, the Poisson distributiongives the probability of X number of events occurring in

time T.


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Poisson Mean and Variance

Mean

Variance and Standard Deviation

2

where = expected number of hits in a given

time period

For a Poisson

random variable, the

variance and mean

are the same!


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Poisson Distribution, example

The Poisson distribution models counts, such as the number of new cases

of SARS that occur in women in New England next month.

The distribution tells you the probability of all possible numbers of new

cases, from 0 to infinity.

If X= # of new cases next month andX~ Poisson (), then the probabilitythatX=k(a particular count) is:

!

)(

k

ekXp

k


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Example

For example, if new cases of West Nile Virus in

New England are occurring at a rate of about 2

per month, then these are the probabilities

that: 0,1, 2, 3, 4, 5, 6, to 1000 to 1 million tocases will occur in New England in the next

month:


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Poisson Probability table

!0

220

e

!1

2 21

e

!2

222

e

!3

223

e

X P(X)

0 =.135

1 =.27

2 =.27

3 =.18

4 =.09

5


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Example: Poisson distribution

Suppose that a rare disease has an incidence of 1 in 1000 person-

years. Assuming that members of the population are affected

independently, find the probability of k cases in a population of

10,000 (followed over 1 year) for k=0,1,2.

The expected value (mean) = = .001*10,000 = 10

10 new cases expected in this population per year

00227.!2

)10()2(

000454.!1

)10()1(

0000454.!0

)10()0(

)10(2

)10(1

)10(0

eXP

eXP

eXP


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more on Poisson

Poisson Process (rates)

Note that the Poisson parameter can be given as themean number of events that occur in a defined time

period OR, equivalently, can be given as a rate, such as=2/month (2 events per 1 month) that must bemultiplied by t=time (called a Poisson Process)

X ~ Poisson ()

!

)()(

k

etkXP

tk

E(X) = t

Var(X) = t


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Example

For example, if new cases of West Nile in NewEngland are occurring at a rate of about 2 permonth, then whats the probability that exactly 4

cases will occur in the next 3 months?

X ~ Poisson (=2/month)

%4.13!4

6

!4

)3*2(months)3in4P(X

)6(4)3*2(4

ee

Exactly 6 cases?

%16!6

6

!6

)3*2(months)3in6P(X

)6(6)3*2(6

ee


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Practice problems

1a. If calls to your cell phone are a Poisson process

with a constant rate =2 calls per hour, whats the

probability that, if you forget to turn your phone off

in a 1.5 hour movie, your phone rings during thattime?

1b. How many phone calls do you expect to get

during the movie?


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Answer

1a. If calls to your cell phone are a Poisson process with a constantrate =2 calls per hour, whats the probability that, if you forget toturn your phone off in a 1.5 hour movie, your phone rings during thattime?

X ~ Poisson (=2 calls/hour)P(X1)=1 P(X=0)

05.!0

)3(

!0

)5.1*2()0( 3

30)5.1(20

eee

XP

P(X1)=1 .05 = 95% chance

1b. How many phone calls do you expect to get during the movie?

E(X) = t = 2(1.5) = 3

binomial normal distribution

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