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


1
Random Variables and Discrete probability
Distributions
  • Chapter 7

2
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.

3
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
4
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

5
Requirements for a Discrete Distribution
  • If a random variable can assume values xi, then
    the following must be true

6
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

7
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

8
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?

9
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

10
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.

11
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.

12
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

13
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
14
The Mean and the Variance
  • Solution continued
  • The variance can also be calculated as follows

15
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)

16
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.

17
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

18
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)

19
Bivariate Distributions
20
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.

21
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
22
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)
23
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
24
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
25
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

26
Conditional Probability (Optional)
Example 7.5 - continued
27
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)
28
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.

29
The Probability Distribution of XY
The probabilities P(XY)3 and P(XY) 4 are
calculated the same way. The distribution follows
30
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

31
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).

32
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.

33
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.

34
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
35
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
36
Calculating the Binomial Probability
In general, The binomial probability is
calculated by
37
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

38
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.

39
Calculating the Binomial Probability
  • Solution Continued
  • Determining the binomial probabilities
  • Let X the number of correct answers

40
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
41
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.

42
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.

43
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.

44
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.

45
Poisson Distributions (Graphs)
0 1 2 3 4 5
46
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
10
Poisson probability distribution with m 7
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15
47
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


48
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.
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