Transcript Ch5 - YSU

Chapter 5
Discrete Probability Distributions
1
Chapter Outline




Random Variables
Discrete Probability Distributions
Expected Value and Variance
Binomial Probability Distribution
2
Random Variables
 A random variable is a numerical description of
the outcome of an experiment.
 A discrete random variable assumes numerical
values that have gaps or jumps between them.
 A continuous random variable assumes numerical
values that have NO gaps or jumps between them.
3
Random Variables
Experiment
Take a quiz with 10
Ture/False
questions
Run 5K
Weigh a sample of
36 cans of coffee
(labeled as 3lbs
Random Variable x
x = Number of correct
Type
Discrete
answers
x = time to finish a 5k run
Continuous
x = the average weight of a
Continuous
sample of 36 cans of coffee
4
Discrete Probability Distributions
 The probability distribution for a random variable
describes how probabilities are distributed over the values
of the random variable.
 We can describe a discrete probability distribution with a
table, graph, or formula.
5
Discrete Probability Distributions
 The probability distribution is defined by a probability
function, denoted by f(x), which provides the probability
for each value of the random variable.
 The required conditions for a discrete probability function
are:
f(x) > 0
f(x) = 1
6
Discrete Probability Distributions
 Example: Probabilities of the # of correct answers to a quiz of 4
True/False questions.
 We can use a table to represent the probability distribution.
 The random variable x represents the number of correct answers.
x
0
1
2
3
4
f(x)
.10
.25
.35
.20
.10
1.00
7
Discrete Uniform Probability Distribution
The discrete uniform probability distribution is the
simplest example of a discrete probability distribution
given by a formula.
The discrete uniform probability function is
f(x) = 1/n
the values of the
random variable
are equally likely
where:
n = the number of values the random
variable may assume
8
Expected Value
 The expected value, or mean, of a random variable is a
measure of its central location.
E(x) =  = xf(x)
 The expected value is a weighted average of the values the
random variable may assume. The weights are the
probabilities.
 The arithmetic mean introduced in Chapter 3 can be
viewed as a special case of weighted average, where the
weights for all the values are the same, i.e. 1/n.
9
Variance and Standard Deviation
 The variance summarizes the variability in the values of a
random variable.
Var(x) =  2 = (x - )2f(x)
 The variance is a weighted average of the squared
deviations of a random variable from its mean. The
weights are the probabilities.
 The standard deviation, , is defined as the positive square
root of the variance.
10
Expected Value
 Example: Take a quiz (# of correct answers)
x
0
1
2
3
4
f(x)
xf(x)
.10
.00
.25
.25
.35
.70
.20
.60
.10
.40
E(x) = 1.95
Expected
number of
correct
answers
11
Variance
 Example: Take a quiz (# of correct answers)
x
0
1
2
3
4
x-
-1.95
-0.95
0.05
1.05
2.05
(x - )2
f(x)
(x - )2f(x)
3.8025
0.9025
0.0025
1.1025
4.2025
.10
.25
.35
.20
.10
.3803
.2256
.0009
.2205
.4203
Variance of daily sales =  2 = 1.2476
Standard deviation of daily sales =  = 1.11 correct answers
12
Binomial Probability Distribution
Properties of a Binominal Experiment:
1. The experiment consists of a sequence of n identical
trials;
2. Only two outcomes, success and failure, are possible on
each trial;
3. The probability of a success, denoted by p, does not
change from trial to trial;
4. The trials are independent from one another.
13
Binomial Probability Distribution



Our interest is in the number of successes occurring in the
n trials.
We let x denote the number of successes occurring in the
n trials.
Either outcome can be named as ‘Success’. We need to
make sure that in the calculation, the probability p is
matched with the definition of the random variable x.
14
Binomial Probability Distribution
 Binomial Probability Function
n!
f (x) 
p x (1  p)( n  x )
x !(n  x )!
where:
x = the number of successes
p = the probability of a success on one trial
n = the number of trials
f(x) = the probability of x successes in n trials
n! = n(n – 1)(n – 2) ….. (2)(1)
15
Binomial Probability Distribution
 Binomial Probability Function
n!
f (x) 
p x (1  p)( n  x )
x !(n  x )!
Number of experimental
outcomes providing exactly
x successes in n trials
Probability of a particular
sequence of trial outcomes
with x successes in n trials
16
Binomial Probability Distribution
 Example: Purchasing a pair of shoes
Based on recent sales data, a shoe store manager
estimates that the probability a customer makes a purchase
is 30%. For the next three customers, what is the
probability that exactly 1 of them will make a purchase?
Analysis: Is this example a binomial experiment? If so,
which outcome is to be named ‘Success’? And what is the
probability of Success?
17
Binomial Probability Distribution
 Example: Purchasing a pair of shoes
Does the example satisfy the properties of a binomial
distribution?

N trials? – Yes, 3 trials ( 3 customers)
 Two outcomes for each trial? – Yes, purchase or not
 Probability of success stays the same – 30% chance for
making a purchase can be assumed to be the same for all
the customers.
 Independent trials – Assume the three customers are
independent in their decision on making a purchase.
18
Binomial Probability Distribution
 Example: Purchasing a pair of shoes
How many favorable outcomes are there where exactly ONE of the
next three customers makes a purchase? With the success
representing ‘making a purchase’ and the three customers assumed to
be independent, we should have the following outcomes and their
probabilities:
Experimental
Outcome
Probability of
Experimental Outcome
(S, F, F)
(F, S, F)
(F, F, S)
p(1 – p)(1 – p) = (.3)(.7)(.7) = .147
(1 – p)p(1 – p) = (.7)(.3)(.7) = .147
(1 – p)(1 – p)p = (.7)(.7)(.3) = .147
Total = .441
19
Binomial Probability Distribution
 Example: Purchasing a pair of shoes
Let: p = .30, n = 3, x = 1
Using the
probability
function
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
3!
0.31 0.72  30.30.49  0.441
f (1) 
1!3  1!
20
Binomial Probability Distribution
Example: Purchasing a pair of shoes
1st Customer
2nd Customer
P (.3)
Purchase
(.3)
Using a tree diagram
3rd Customer
P (.3)
x
3
Prob.
.027
NP (.7)
P (.3)
2
.063
2
.063
1
.147
P (.3)
2
.063
NP (.7)
P (.3)
1
.147
1
.147
0
.343
NP (.7)
NP(.7)
P (.3)
Not
Purchase
(.7)
NP (.7)
NP (.7)
21
Binomial Probability Table
 Statisticians have developed tables that give probabilities
and cumulative probabilities for a binomial random
variable. In the appendix of our textbook, you can locate
the binomial probability tables.
 For our example, the table is presented as below (where x
represents the number of success):
x
0
1
2
3
f(x)
.343
.441
.189
.027
1.00
22
Binomial Probability Distribution
 We can apply the formulas of expected value and variance
for a binomial probability distribution. However, those
formulas can be further simplified as follows:
 Expected Value
E(x) =  = np

Variance
Var(x) =  2 = np(1  p)

Standard Deviation
  np(1  p)
23
Binomial Probability Distribution
 Example: Purchasing a pair of shoes

Expected Value
E(x) = np = 3(.3) = .9 customers out of 3

Variance
Var(x) = np(1 – p) = 3(.3)(.7) = .63

Standard Deviation
  3  0.3  0.7  0.63  0.79 customers
24