Bernoulli

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Bernoulli

• Bernoulli population is a population in which
  each element is one of two possibilities. The two
  possibilities are usually designed as success and
  failure.
• A Bernoulli trial is observing one element in a
  Bernoulli population.

• The probability of success is ? which means that
  the
• probability of failure is 1- ?.

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Example 4.12

• Ex1. The toss of a coin results in a Bernoulli
  population.
• Ex2. The homes in Dallas either heat with gas or
  not.
• Ex3. In a local election, the voters either
  favor a candidate for mayor or do not. The
  choices of voters result in a Bernoulli
  population.

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Binomial experiment

• A Binomial experiment is an experiment that
  consists of n repeated independent Bernoulli
  trials in which the probability of success on
  each trial is ? and the probability of failure
  on each trial is 1- ? .

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Criteria for a Binomial Setting

• 1.consists of n Bernoulli trials (each trial
  yields either S or F).
• 2.For each trial, P(S) ? ,and P(F)1- ? .
• 3.The trials are independent.
• Ex1. The toss of a coin 3 times

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binomial random variable.

• The random variable x, which gives the number of
  successes in the n trials of Binomial experiment,
  is called a binomial random variable. The sample
  space of x is
• Sx 0, 1, 2, , n.
• Ex1. The number of head after tossing a coin 10
  times.

• The binomial random variable is a discrete random
  variable.

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Question4.3 Are the following random variables
binomial random variable?

• The number of getting head after flipping a coin
  10 times.
• 40 of all airline pilot are over 40 years of
  age. The number of pilots who are over 40 out of
  15 randomly chosen pilots.
• Suppose a salesperson makes sales to 20 of her
  customers, the number of customers until her
  first sale.
• A room contains 6 women and four men, Three
  people are selected to form a committee. The
  number of women on the committee.

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Note

• If we looked at less than 10 of the
  population,we can pretend that the trials are
  still independent.

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Recall Example 4.9

• Suppose 30 of homes in Dallas heats with gas. We
  select 3 homes randomly from Dallas and observe
  whether or not they are heated with gas.
• Event A Exactly 2 of the three heat with gas.
• Event B Only one of the three heats with
  gas.
• Event C All 3 of the three heat with gas.
  Event D None of them heat with gas.
• Whats the probability of A,B,C,D

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Recall Example 4.9
Binomial R.V.

• Solution Let X be number of homes out of three
  that are heated with gas,then
• AX2,BX1,CX3,DX0

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Example 4.9

• X2ggnorgngorngg
• P(X2)P(ggn)P(gng)P(ngg)
• 3(0.3)2(0.7)
• 3 ?2(1- ?)

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Example 4.9

• Assign the probability to each outcome!

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• Similarly, Similarly, P(X1) 3(0.3) (0.7) 2
• Clearly P(X3) 1(0.3)3 P(X0)
  1(0.7)3
• Now we can complete the probability distribution
  of X.

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Binomial distribution
Probability of k successes out of n trials is
given by
k
where
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Exercise 4.7

• 40 of homes in Dallas are heated with gas. Let
  Y of homes out of 5 that are heated with gas.
• a) What is the probability of Ylt3?
• b) What are the mean and SD of Y

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• 0 0.07776
• 1 0.25920
• 2 0.34560
• P(Ylt3)0.68256
• 2
• 1.09545

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Remark

• We can use computer or binomial table to find out
  the probabilities for certain values of n, k and
  ? .
• Look at table B.1 on Appendix B.

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Exercise

• Suppose a couple plans to have 4 children. The
  chance they have a boy is 0.2.The gender of one
  child is independent of the gender of another
  child. What is the probability that they have
  exactly 2 boys out of 4 children?

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How the two parameters (n, ?) affect binomial
distribution?
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How the two parameters (n, ?) affect binomial
distribution?

• When the number of independent trials n is fixed,
  if ? is close to .5, the distribution is near to
  a bell curve. For small values of ?,the
  distribution is heavily concentrated on the lower
  values of k. For large values of ?, the
  distribution is heavily concentrated on the
  higher values of k.
• When the success probability ? on each trial is
  fixed and n is larger, the distribution is more
  close to a bell curve.

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mean and standard deviation

• For binomial distribution
• Meanµ n ?
• Variances 2 n? (1- ? )
• Standard deviation s

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Exercise 4.8

• Suppose that the probability of successfully
  rehabilitating a convicted criminal in a penal
  institution is .4. Let r represent the number
  successfully rehabilitated out of a random sample
  of ten convicted criminal, find out the mean and
  the standard deviation of the variable r.

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Exercise 4.9

• Calculate the unknown probability of random
  variable y with n4 and ?.2
• P(y2)
• P(y lt 2)
• P(y 2)
• P(ygt3)

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The diagram is the graph of binomial
distribution B (n, 0.5) for n 20, 40, 60, 80,
100
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Probability density function
A Probability Density Function , f(x), describes
the probability distribution of a continuous
random variable with properties 1.f(x) 0 2.The
total area under the probability density curve
1.0, which corresponds to 100 3.P(a x b)
area under the probability density curve between
a and b.
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Note

• P(a x b) P(a lt x lt b).
• Histogram and relative polygons can approximate
  the probability density curve.
• From the shape of the probability density curve,
  we can know the shape of the distribution such
  as skew to the right, skew to the left, bell
  shaped, multimodal or bimodal.

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normal random variable
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normal random variable

• The pdf of a normal random variable is continuous
  and bell-shaped.

• We denote X N (?, ?) when X is normal random
• variable with parameters ? and ?. (s,m,.5)

• Key Properties
• It is characterized by its mean ? and its
  standard deviation ?.
• The probability density function is symmetric
  about ?.
• The maximum value occurs at ?.
• The area to the left of ? is .5.

?
?
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Standard Normal R.V.

• The normal distribution N(0,1) with ? 0, ? 1
  is called the standard normal distribution.
• Given X N (?, ?), the z-score
• is a standard normal random variable, or say
• Z N(0,1)

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Example 4.14

• SAT scores have a normal distribution with mean
  500 and SD 100,suppose you earned a 600 on the
  SAT test, where do you stand among all students
  who took the SAT. (What percent of test scores
  were better than 600)

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Normal Curve
Approximate percentage of area within given
standard deviations (empirical rule).
99.7
95
68
500
400
600
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• Solution
• Let X be the SAT scores, then X N (500, 100)

Sketch the normal density, specify ? ,?, ? ?,?
2?, ?3?
Note that 600 is one SD above the
mean, P(Xgt600)(1-.68)/20.16 16 of scores were
better than 600.
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Example 4.15

• Let X be the SAT score of a randomly chosen
  student, X N (500, 100),
• a) What is the chance that Xgt 680.
• b) What is the probability that 450X600

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Solution of a)
1.8

• Step1. Determine ? and ? and expression for
  desired probability, here to derive
• P(Xgt 680)
• Step2.Standadize X by converting X to z-score.

Step3. Sketch a Standard normal density and shade
the desired region
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(How to compute probabilities)

• Step4. Use Standard Normal Table (out -gtinside)
  to find probabilities related.

P(X680)P(Z1.8)0.9641

• Step5.Adjust to compute the probability of the
  shaded region

P(Xgt680)1-0.96410.0359 The probability that
Xgt680 is 3.59
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Solution of b)

• X N (500, 100),

Area(Zlt1.0)P(Zlt1.0)0.8413 Area(Zlt-.5)P(Zlt-0.5)
0.3085 P(450X600)P(-.5Z1.0)0.5328
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Example 4.16

• Suppose a college says it admits only people with
  SAT scores among the top 10,what SAT score does
  this correspond to?( i.e. find score b such that
  P(Xgtb).1)

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Solution
zb

• Step1. Sketch a Standard normal density and shade
  the given region.
• Step2.Adjust to compute z-score to the
  probability of the region that agrees with the
  table

Find z such that P(Zz).9

• Step3. Use Standard Normal Table (inside -gt out)
  to find the desired percentile (z-value).

P(Zlt1.28)0.8997 0.9
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• Step4. Convert to X with X ? ? Z.

b5001.28100628 The cutoff score is
628 Remark How to compute percentiles for
Normal random variables is given by these steps
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100P-th Percentiles

• Definition For any continuous random variable,
  X, the 100p-th percentile is a number b such
  that
• P(X b) p
• Useful Formula Convert Z to X
• X ? ? Z

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Exercise 4.10

• Let X be N (8, 3) random variable, ie ? 8 and
  ? 3. 1)Find the following probabilities
• a) P(X 5 ) (area directly above point x 5 )
• b) P(X 5 ) (area to the left of x 5 )
• c) P(X gt9 ) (area to the right of x 9 )
• d)P(5 X 10 )
• 2)Find the 90th percentile of X

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• 0
• .1587
• .3707
• .5899
• 11.84

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Exercise 4.11

• A public health department closed the beach when
  the contamination level index is in the 80th
  percentile. From research we know that the index
  level is normally distributed N(160,20)
• a) What is the probability the index level
  exceeds 190?
• b) What is the probability the index level is
  between 150 and 170?
• c) What is the index level when the beach is
  closed?

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• .0668
• .3830
• 176.8