Title: Random%20Variables%20and%20Discrete%20probability%20Distributions
1Random Variables and Discrete probability
Distributions
27.2 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.
3Discrete 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
4Discrete 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
5Requirements for a Discrete Distribution
- If a random variable can assume values xi, then
the following must be true
6Distribution 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
7Determining 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
8Developing 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?
9Developing 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
107.3 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.
11Population Mean (Expected Value)
- Given a discrete random variable X with values
xi, that occur with probabilities p(xi), the
population mean of X is.
12Population 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
13The 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
14The Mean and the Variance
- Solution continued
- The variance can also be calculated as follows
15Laws 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)
16Laws 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.
17Laws of Expected Value and Variance
Laws of Expected Value and Variance
- 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
187.4 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)
19Bivariate Distributions
20Bivariate 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.
21Bivariate 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
22Marginal 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)
23Describing 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
24Describing 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
25Describing 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
26Conditional Probability (Optional)
Example 7.5 - continued
27Conditions 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)
28Sum 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.
29The Probability Distribution of XY
The probabilities P(XY)3 and P(XY) 4 are
calculated the same way. The distribution follows
30The 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
31The 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).
327.6 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.
33Binomial 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.
34Developing 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
35Developing 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
36Calculating the Binomial Probability
In general, The binomial probability is
calculated by
37Calculating 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
38Calculating 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.
39Calculating the Binomial Probability
- Solution Continued
- Determining the binomial probabilities
- Let X the number of correct answers
40Calculating 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
41Mean 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.
427.7 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.
43Properties 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.
44The 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.
45Poisson Distributions (Graphs)
0 1 2 3 4 5
46Poisson 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
47Poisson 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
48Poisson 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.