Random Variables and Discrete probability Distributions

1
Random Variables and Discrete probability
Distributions

• SESSION 2

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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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Requirements for a Discrete Distribution

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

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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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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

• Example
• 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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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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Calculating the Binomial Probability
In general, The binomial probability is
calculated by
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Calculating the Binomial Probability

• Example
• 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
• 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
• 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.