Random%20Variables%20and%20Discrete%20probability%20Distributions

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Random Variables and Discrete probability
Distributions

• Chapter 7

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7.2 Random Variables and Probability Distributions

• A random variable is a function or rule that
  assigns a numerical value to each simple event in
  a sample space.
• A random variable reflects the aspect of a random
  experiment that is of interest for us.
• There are two types of random variables
• Discrete random variable
• Continuous random variable.

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Discrete and Continuous Random Variables

• A random variable is discrete if it can assume a
  countable number of values.
• A random variable is continuous if it can assume
  an uncountable number of values.

Discrete random variable
Continuous random variable
After the first value is defined the second
value, and any value thereafter are known.
After the first value is defined, any number can
be the next one
0
1
1/2
1/4
1/16
Therefore, the number of values is countable
Therefore, the number of values is uncountable
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Discrete Probability Distribution

• A table, formula, or graph that lists all
  possible values a discrete random variable can
  assume, together with associated probabilities,
  is called a discrete probability distribution.
• To calculate the probability that the random
  variable X assumes the value x, P(X x),
• add the probabilities of all the simple events
  for which X is equal to x, or
• Use probability calculation tools (tree diagram),
• Apply probability definitions

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Requirements for a Discrete Distribution

• If a random variable can assume values xi, then
  the following must be true

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Distribution and Relative Frequencies

• In practice, often probability distributions are
  estimated from relative frequencies.
• Example 7.1
• A survey reveals the following frequencies
  (1,000s) for the number of color TVs per
  household.
• Number of TVs Number of Households x p(x) 0
  1,218 0 1218/Total .012
  1 32,379 1 .319 2 37,961
  2 .374 3 19,387 3 .191 4 7,714 4
  .076 5 2,842 5 .028 Total 101,501
  1.000

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Determining Probability of Events

• The probability distribution can be used to
  calculate the probability of different events
• Example 7.1 continued
• Calculate the probability of the following
  events
• P(The number of color TVs is 3) P(X3) .191
• P(The number of color TVs is two or more)
  P(X³2)P(X2)P(X3)P(X4)P(X5).374 .191
  .076 .028 .669

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Developing a Probability Distribution

• Probability calculation techniques can be used to
  develop probability distributions
• Example 7.2
• A mutual fund sales person knows that there is
  20 chance of closing a sale on each call she
  makes.
• What is the probability distribution of the
  number of sales if she plans to call three
  customers?

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Developing a Probability Distribution
Probability Distribution
Finding Probability/ Probability Distribution

• Solution
• Use probability rules and trees
• Define event S A sale is made.

(.2)(.2)(.8) .032
S S S S S SC S SC S S SC SC SC S S SC S SC SC
SC S SC SC SC

• X P(x)
• .23 .008
• 3(.032).096
• 3(.128).384
• 0 .83 .512

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7.3 Describing the Population/ Probability
Distribution

• The probability distribution represents a
  population
• Were interested in describing the population by
  computing various parameters.
• Specifically, we calculate the population mean
  and population variance.

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Population Mean (Expected Value)

• Given a discrete random variable X with values
  xi, that occur with probabilities p(xi), the
  population mean of X is.

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Population Variance

• Let X be a discrete random variable with possible
  values xi that occur with probabilities p(xi),
  and let E(xi) m. The variance of X is defined
  by

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The Mean and the Variance

• Example 7.3
• Find the mean the variance and the standard
  deviation for the population of the number of
  color television per household in example 7.1
• Solution
• E(X) m Sxip(xi) 0p(0)1p(1)2p(2) 0(.01
  2)1(.319)2(.374) 2.084
• V(X) s2 S(xi - m)2p(xi) (0-2.084)2p(0)(1-2.
  084)2p(1) (2-2.084)2 1.107
• s 1.1071/2 1.052

Using a shortcut formula for the variance
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The Mean and the Variance

• Solution continued
• The variance can also be calculated as follows

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Laws of Expected Value and Variance

• Laws of Expected Value
• E(c) c
• E(X c) E(X) c
• E(cX) cE(X)

• Laws of Variance
• V(c) 0
• V(X c) V(X)
• V(cX) c2V(X)

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Laws of Expected Value Variance

• Example 7.4
• The monthly sales at a computer store have a mean
  of 25,000 and a standard deviation of 4,000.
• Profits are 30 of the sales less fixed costs of
  6,000.
• Find the mean and standard deviation of the
  monthly profit.

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Laws of Expected Value and Variance
Laws of Expected Value and Variance

• Solution

• Profit .30(Sales) 6,000
• E(Profit) E.30(Sales) 6,000
  E.30(Sales) 6,000 .30E(Sales) 6,000
  .(30)(25,000) 6,000 1,500

• V(Profit) V(.30(Sales) 6,000
  V(.30)(Sales) (.30)2V(Sales) 1,440,000

• s 1,440,0001/2 1,200

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7.4 Bivariate Distributions

• The bivariate (or joint) distribution is used
  when the relationship between two random
  variables is studied.
• The probability that X assumes the value x, and Y
  assumes the value y is denoted
• p(x,y) P(Xx and Y y)

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Bivariate Distributions
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Bivariate Distributions

• Example 7.5
• Xavier and Yvette are two real estate agents.
  Let X and Y denote the number of houses that
  Xavier and Yvette will sell next week,
  respectively.
• The bivariate probability distribution is
  presented next.

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Bivariate Distributions
0.42
Example 7.5 continued
p(x,y)
X Y 0 1
2 0 .12 .42 .06 1 .21 .06 .03 2 .07 .02 .01
0.21
0.12
0.06
X
y0
0.06
0.03
0.07
0.02
y1
0.01
Y
y2
X0
X2
X1
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Marginal Probabilities

• Example 7.5 continued
• Sum across rows and down columns

p(0,0)
p(0,1)
p(0,2)
The marginal probability P(X0)
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Describing the Bivariate Distribution

• The joint distribution can be described by the
  mean, variance, and standard deviation of each
  variable.
• This is done using the marginal distributions.

x p(x) y p(y) 0
.4 0 .6 1 .5
1 .3 2 .1 2
.1 E(X) .7 E(Y) .5 V(X) .41
V(Y) .45
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Describing the Bivariate Distribution

• To describe the relationship between the two
  variables we compute the covariance and the
  coefficient of correlation
• Covariance COV(X,Y) S(X mx)(Y-
  my)p(x,y)
• Coefficient of Correlation COV(X,Y) sxsy

r
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Describing the Bivariate Distribution

• Example 7.6
• Calculate the covariance and coefficient of
  correlation between the number of houses sold by
  the two agents in Example 7.5
• Solution
• COV(X,Y) S(x-mx)(y-my)p(x,y)
  (0-.7)(0-.5)p(0,0)(2-.7)(2-.5)p(2,2) -.15
• rCOV(X,Y)/sxsy - .15/(.64)(.67) -.35

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Conditional Probability (Optional)
Example 7.5 - continued
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Conditions for Independence (optional)

• Two random variables are said to be independent
  when
• This leads to the following relationship for
  independent variables
• Example 7.5 - continued
• Since P(X0Y1).7 but P(X0).4, The variables
  X and Y are not independent.

P(XxYy)P(Xx) or P(YyXx)P(Yy).
P(Xx and Yy) P(Xx)P(Yy)
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Sum of Two Variables

• The probability distribution of X Y is
  determined by
• Determining all the possible values that XY can
  assume
• For every possible value C of XY, adding the
  probabilities of all the combinations of X and Y
  for which XY C
• Example 7.5 - continued
• Find the probability distribution of the total
  number of houses sold per week by Xavier and
  Yvette.
• Solution
• XY is the total number of houses sold. XY can
  have the values 0, 1, 2, 3, 4.

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The Probability Distribution of XY
The probabilities P(XY)3 and P(XY) 4 are
calculated the same way. The distribution follows
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The Expected Value and Variance of XY

• The distribution of XY
• The expected value and variance of XY can be
  calculated from the distribution of XY.
• E(XY)0(.12) 1(63)2(.19)3(.05)4(.01)1.2
• V(XY)(0-1.2)2(.12)(1-1.2)2(.63) .56

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The Expected Value and Variance of XY

• The following relationship can assist in
  calculating E(XY) and V(XY)
• E(XY) E(X) E(Y)
• V(XY) V(X) V(Y) 2COV(X,Y)
• When X and Y are independent COV(X,Y) 0, and
  V(XY) V(X)V(Y).

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7.6 The Binomial Distribution

• The binomial experiment can result in only one of
  two possible outcomes.
• Typical cases where the binomial experiment
  applies
• A coin flipped results in heads or tails
• An election candidate wins or loses
• An employee is male or female
• A car uses 87octane gasoline, or another gasoline.

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Binomial Experiment

• There are n trials (n is finite and fixed).
• Each trial can result in a success or a failure.
• The probability p of success is the same for all
  the trials.
• All the trials of the experiment are independent.
• Binomial Random Variable
• The binomial random variable counts the number of
  successes in n trials of the binomial experiment.
• By definition, this is a discrete random variable.

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Developing the Binomial Probability Distribution
(n 3)
P(SSS)p3
S2
P(S3)p
P(S2)p
P(S2S1)
S1
P(F3)1-p
P(SSF)p2(1-p)
P(S3)p
P(SFS)p(1-p)p
P(F2S1)
P(S1)p
P(F2)1-p
P(F3)1-p
Since the outcome of each trial is independent
of the previous outcomes, we can replace the
conditional probabilities with the marginal
probabilities.
F2
P(SFF)p(1-p)2
P(FSS)(1-p)p2
P(S3)p
S2
P(F1)1-p
P(S2F1)
P(F3)1-p
P(FSF)(1-p)p(1-p)
P(FFS)(1-p)2p
F1
P(S3)p
P(F2F1)
F2
P(F3)1-p
P(FFF)(1-p)3
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Developing the Binomial Probability Distribution
(n 3)
P(SSS)p3
SSS
Let X be the number of successes in three
trials. Then,
P(SSF)p2(1-p)
SS
X 3 X 2 X 1 X 0
P(X 3) p3
P(SFS)p(1-p)p
S S
P(X 2) 3p2(1-p)
P(SFF)p(1-p)2
P(X 1) 3p(1-p)2
P(FSS)(1-p)p2
SS
P(X 0) (1- p)3
P(FSF)(1-p)p(1-p)
P(FFS)(1-p)2p
This multiplier is calculated in the following
formula
P(FFF)(1-p)3
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Calculating the Binomial Probability
In general, The binomial probability is
calculated by
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Calculating the Binomial Probability

• Example 7.9 7.10
• Pat Statsdud is registered in a statistics course
  and intends to rely on luck to pass the next
  quiz.
• The quiz consists on 10 multiple choice questions
  with 5 possible choices for each question, only
  one of which is the correct answer.
• Pat will guess the answer to each question
• Find the following probabilities
• Pat gets no answer correct
• Pat gets two answer correct?
• Pat fails the quiz

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Calculating the Binomial Probability

• Solution
• Checking the conditions
• An answer can be either correct or incorrect.
• There is a fixed finite number of trials (n10)
• Each answer is independent of the others.
• The probability p of a correct answer (.20) does
  not change from question to question.

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Calculating the Binomial Probability

• Solution Continued
• Determining the binomial probabilities
• Let X the number of correct answers

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Calculating the Binomial Probability

• Solution Continued
• Determining the binomial probabilities
• Pat fails the test if the number of correct
  answers is less than 5, which means less than or
  equal to 4.

P(X4)
p(0) p(1) p(2) p(3) p(4) .1074
.2684 .3020 .2013 .0881 .9672
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Mean and Variance of Binomial Variable
Binomial Distribution- summary

• E(X) m np
• V(X) s2 np(1-p)
• Example 7.11
• If all the students in Pats class intend to
  guess the answers to the quiz, what is the mean
  and the standard deviation of the quiz mark?
• Solution
• m np 10(.2) 2.
• s np(1-p)1/2 10(.2)(.8)1/2 1.26.

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7.7 Poisson Distribution

• The Poisson experiment typically fits cases of
  rare events that occur over a fixed amount of
  time or within a specified region
• Typical cases
• The number of errors a typist makes per page
• The number of customers entering a service
  station per hour
• The number of telephone calls received by a
  switchboard per hour.

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Properties of the Poisson Experiment

• The number of successes (events) that occur in a
  certain time interval is independent of the
  number of successes that occur in another time
  interval.
• The probability of a success in a certain time
  interval is
• the same for all time intervals of the same size,
• proportional to the length of the interval.
• The probability that two or more successes will
  occur in an interval approaches zero as the
  interval becomes smaller.

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The Poisson Variable and Distribution

• The Poisson Random Variable
• The Poisson variable indicates the number of
  successes that occur during a given time interval
  or in a specific region in a Poisson experiment
• Probability Distribution of the Poisson Random
  Variable.

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Poisson Distributions (Graphs)
0 1 2 3 4 5
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Poisson Distributions (Graphs)
Poisson probability distribution with m 2
0 1 2 3 4 5 6
Poisson probability distribution with m 5
0 1 2 3 4 5 6 7 8 9
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Poisson probability distribution with m 7
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
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Poisson Distribution

• Example 7.12
• The number of Typographical errors in new
  editions of textbooks is Poisson distributed with
  a mean of 1.5 per 100 pages.
• 100 pages of a new book are randomly selected.
• What is the probability that there are no typos?
• Solution
• P(X0)

.2231


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Poisson Distribution
Finding Poisson Probabilities

• Example 7.13
• For a 400 page book calculate the following
  probabilities
• There are no typos
• There are five or fewer typos
• Solution
• P(X0)
• P(X5)ltuse the formula to find p(0),
  p(1),,p(5), then calculate p(0)p(1)p(5)
  .4457

Important! A mean of 1.5 typos per100 pages, is
equivalent to 6 typos per 400 pages.