Transcript Probability and discrete Probability distributions

Chapter 7
Random Variables and
Discrete probability
Distributions
1
7.1 Random Variables and Discrete
Probability Distribution



Many times simple events in the sample space can be assigned
numerical values. A random variable can then be defined that takes
on these values.
The random variable reflects a certain characteristic of interest of
the experimental outcome.
Examples:




The number of cars entering a gas station in the next five minutes.
The amount of gas filled in a gas tank by a driver.
The grade in a physics test.
The number of new employees to be hired next month.
2
7.1 Random Variables and Discrete
Probability Distribution

There are two types of random variables:
Discrete random variable
 Continuous random variable.
A random variable is discrete if it can assume only a
countable number of values.
A random variable is continuous if it can assume an
uncountable number of values.



3
A short demonstration:
Discrete and Continuous Random Variables
Discrete random variable Continuous random variable
Only certain values can be
assigned to the random
variable
0
1
2
Therefore, the number of
values is countable
3 ...
Any value within the range where the
variable is defined can be selected
0
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 the
probabilities associated with each value, is called a
discrete probability distribution.

To calculate the probability that the random variable X
assumes the value x, P(X = a),


add the probabilities of all the simple events for which X is
equal to ‘a’, or
Use probability calculation tools (such as tree diagram),
5
Discrete Probability Distribution

Example 1


Find the probability distribution of the random variable
describing the number of heads that turn-up when a coin is
flipped twice.
Solution
H
HH (½)(½)=1/4
H
T
T
T
Simple event
TT
HT
TH
HH
x
0
1
1
2
H
HT (½)(½)=1/4
TH (½)(½)=1/4
TT (½)(½)=1/4
Probability
1/4
1/4
1/4
1/4
x
p(x)
0
1
2
1/4
1/2
1/4
6
Requirements for a Discrete Distribution

If a random variable can assume values xi, then
the following must be true:
1. 0  f(x i )  1 for all x i
2.  f ( x i )  1
all x i

With the probability distribution we can more
conveniently calculate probabilities (see example 2)
next.
7
Distribution and Relative Frequencies

Example 2

A survey reveals the following frequencies for the
number of colored TV per household.
Number of TVs
0
1
2
3
4
5
Total
Number of Households
1,218
32,379
37,961
19,387
7,714
2,842
101,501
x
p(x)
0 1218/Total = .012
1 32379/Total = .319
2
.374
3
.191
4
.076
5
.028
1.000
8
Determining Probability of Events
Example 2 – continued
Calculate the probability of the following events:


P(The number of colored TVs is 3) = P(X=3) =.191
P(The number of colored TVs is two or more) =
P(X2)=P(X=2)+P(X=3)+P(X=4)+P(X=5)=
.319+.374+…= .669
9
Determining the Probability Distribution
and the Probability of Events

Example 3

The number of cars a dealer is selling daily were
recorded in the last 100 days. This data was
summarized as follows:
Daily sales Frequency
0
5
1
15
2
35
3
25
4
20
100


Estimate the probability
distribution.
State the probability of
selling more than 2 cars
a day.
10
Determining the Probability Distribution
and the Probability of Events

Example 3 - Solution
From the table of frequencies we can calculate the
relative frequencies, which becomes .35
.25
our estimated probability distribution
.20
.15
Daily sales Relative Frequency
0
5/100=.05
1
15/100=.15
2
35/100=.35
3
25/100=.25
4
20/100=.20
1.00
.05
0

1
2
3
4
X
The probability of selling
more than 2 cars a day is
P(X>2) = P(X=3) + P(X=4)
= .25 + .20 = .45
11
Developing a Probability Distribution

Example 4

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?
Solution


Let us use probability rules and probability tree
Define event Ai = {a sale is made in the iit phone call}.
12
Developing a Probability Distribution

Assuming each phone call is independent of all other phone calls,
we can assign the following probabilities to each branch on the tree:
A1 and A2 and A3
A1 and A2 and A3C
A1 and A2C and A3
A1c and A2 and A3
A1 and A1C and A3C
A1C and A2 and A3C
A1C and A2C and A3
A1 and A2 and
C
C
AC
(.2)(.2)(.8)= .032
X P(x)
3 = .008
3 If X.2represents
the
2 number
3(.032)=.096
of sales
1 made,
3(.128)=.384
then.. (click)
0 .83 = .512
(.2)(.8)(.8)= .128
13
Expected value and Variance
Describing the Population
Probability Distribution
14
The Expected Value (Mean)

Given a discrete random variable X with values xi,
that occurs with probabilities p(xi), the population
mean of X is.
E( X )     x i  p( x i )
all x i
15
The Population Variance
 Let
X be a discrete random variable with possible
values xi that occur with probabilities p(xi), and let
E(xi) = . The variance of X is defined by
V(X)    E(X  )    ( x i  ) p( x i )
2
2
2
all xi
The s tan dard deviation is
  2
16
The Mean and the Variance


Example 5
Find the mean the variance and the standard deviation for
the population of the number of colored TV per household
in example 2
Solution



E(X) =  = Sxip(xi) = 0p(0)+1p(1)+2p(2)+…=
0(.012)+1(.319)+2(.374)+… = 2.084
V(X) = 2 = S(xi - )2p(xi) = (0-2.084)2p(0)+(1-2.084)2p(1) +
(2-2.084)2+… =1.107
Using a shortcut formula
for the variance
 = 1.1071/2 = 1.052
17
The Mean and the Variance

Solution – continued

The variance can also be calculated as follows:
V(X)  2  E(X 2 )   2   x i2p( x i )   2 
all xi
0 p(0)  1 p(1)  2 p(2)  ... 2.084
2
2
2
2
 1.107
18
7.4 The Binomial Distribution
 The
binomial experiment has the following
characteristics:
 There
are n trials (n is finite and fixed).
 Each trial can result in one out of two outcomes,
success or a failure.
 The probability p of a success is the same for all the
trials.
 All the trials of the experiment are independent.
19
Binomial Random Variable
The binomial random variable counts the
number of successes in n trials of the binomial
experiment.
 The possible values of this count are 0,1, 2, …,n,
and therefore the binomial variable is discrete.

20
The Binomial Probability Distribution
The binomial probability distribution is described by
the following closed form formula:
P(X  x)  p( x)   p (1  p)
n
x
x
n x
n!
w here  
x! (n  x)!
n
x
21
Developing the Binomial Probability
Distribution (n = 3)
S2
S1
F2
S2
F1
F2
22
Developing the Binomial Probability
Distribution (n = 3)
P(SSS)=p
S2
S1
S3
3
F3 P(SSF)=p2(1-p)
S3 P(SFS)=p(1-p)p
F2
S2
Since the outcome of each
F3 P(SFF)=p(1-p)
trial is independent
of the 2
previous outcomes, we can
3 P(FSS)=(1-p)p2
replace theSconditional
probabilities with the
unconditional
F probabilities.
P(FSF)=(1-p)P(1-p)
3
S3 P(FFS)=(1-p)2p
F1
F2
F3 P(FFF)=(1-p)3
23
Developing the Binomial Probability
Distribution (n = 3)
P(X = 3) = P(Three successes) = P(SSS) = p3
P(X = 2) = P(Two successes, one Failure) = P(SSF) + P(SFS) +
P(FSS) = p2(1-p) + p(1-p)p + (1-p)pp =
p2(1-p) + p2(1-p) + p2(1-p) =
3 p2(1-p).
P(X = 1) = P(One success, two failures) = P(SFF) + P(FSF) +
P(FFS) = p(1-p)2 + (1-p)p(1-p) + (1-p)2p =
p(1-p)2 + p(1-p)2 + p(1-p)2 =
3 p(1-p)2
P(X = 0) = P(Three failures) = P(FFF) = (1-p)3
24
Calculating the Binomial Probability

Example 10




Pat Statsdud takes a course in statistics, 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 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
25
Calculating the Binomial Probability
 Solution

Checking the conditions:
• Only one out of two outcomes can occur
(An answer can be either correct or incorrect).
• There is a fixed finite number of trials
(There are 10 questions in the test, n=10).
• Each answer is independent of the others.
• The probability p of a correct answer does not change
from question to question
(20% chance that an answer is correct).
26
Calculating the Binomial Probability

Solution – Continued

Determining the binomial probabilities:
Let X = the number of correct answers
10!
P(X  0) 
(.20)0 (.80)10  .1074
0! (10  0)!
10!
P(X  2) 
(.20)2 (.80)102  .3020
2! (10  2)!
27
Calculating the Binomial Probability

Solution – Continued

Determining the binomial probabilities:
Pat fails the test if she gets less than 5 correct
answers.
P(X4 = p(0) + p(1) + p(2) + p(3) + p(4)
= .1074 + .2684 + .3020 + .2013 + .0881
=.9672
This is called cumulative probability
28
Mean and Variance - Binomial Variable
E(X) =  = np
V(X) = 2 = np(1-p)
29
Mean and variance of binomial variable

Example 11


If all the students in Pat’s class practice the same
learning behavior like she does, what is the mean
and the standard deviation of the quiz mark?
Solution
 = np = 10(.2) = 2.
  = [np(1-p)]1/2 = [10(.2)(.8)]1/2 = 1.26.

30
Binomial Distribution - summary

Example 12
Records show that 30% of the customers in a shoe
store make their payments using a credit card.
 This morning 20 customers purchased shoes.
 Use the binomial table to answer some questions
stated in the next few slides.

31
Binomial Distribution - summary

Solution

First we verify that this is a binomial experiment:
• There are two possible outcomes of which only one will take
place (paid with the credit card or not)
• There is a finite number of trials (20 customers are
observed)
• Customers pay independently
• Each customer has the same probability to pay with a credit
card (. 30).
32
Binomial Distribution - summary

Solution – continued

Find the probability that at most 11 customers use a
credit card.
n = 20
k
0
.
.
11
p
In what
.01………..
.30 follows
we demonstrate the
usage of the Binomial Table
P(X  11) = .995
.995
33
Binomial Distribution - summary
To calculate P(X  11) when p = 0.3 and n = 20,
type in an empty cell the following:
=binomdist(11,20,0.3,1). Then click outside the
cell.
In
what follows we demonstrate how
Excel P(X
to determine
probabilities
 to
Touse
calculate
= 11) typebinomial
the following:
=binomdist(11,20,0.3,0)
 To calculate P(X  11) type the following:
= 1 - binomdist(10,20,0.3,1)

34
Binomial Distribution - summary

What is the probability that at least 3 but not more
than 6 customers used a credit card?
k
0
2
.
6
At least 3 but
not more than 6
Not
p
.01……….. .30
.035
.608
P(3X6)=
P(X=3 or 4 or 5 or 6) =
P(X6) - P(X2)
=.608 - .035 = .573
Click
.573
0 1 2 3 4 5 6
35
Binomial Distribution - summary
Find the probability that exactly 14 customers did not
use a credit card.
Let Y be the number of customers who did not use a
credit card, while X (as before) the number of those
who did use a credit card.
P(Y=14) = P(X=6) = P(X  6) - P(X  5) = .608 - .416 = .192
 Using Excel to solve the last two questions:

• P(3  X  6) = binomdist(6,20,0.3,1) – binomdist(2,20,0.3,1)
• P(Y = 14) = binomdist(14,20,0.7,0)
36
Binomial Distribution - summary

What is the expected number of customers who
used a credit card?
E(X) = np = 20(.30) = 6
37