Transcript Chapter5a--Discrete probability distributions

Discrete Probability
Distributions
•
•
•
•
Random variables
Discrete probability distributions
Expected value and variance
Binomial probability distribution
Random Variables
A random variable is a
numerical description of the
outcome of an experiment
Discrete Random Variables
A discrete random variable may assume a
finite number of numerical values or an infinite
sequence of values such as 0, 1, 2, . . .
Example:
1 = passed driving test
2 = failed driving test
This is a discrete
variable because it is
either pass or fail—you
can’t score 1½, for
example. Notice also it
assumes a finite
number of values
Example: JSL Appliances
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Discrete random variable with an infinite sequence
of values
Let x = number of customers arriving in one day,
where x can take on the values 0, 1, 2, . . .
We can count the customers arriving, but there is no
finite upper limit on the number that might arrive.
Continuous Random
Variables
Continuous random variables can assume an
infinite number of values within a defined interval.
The liquid in these
bottles (x) must be
between 0 and 32
ounces. But x could
be 2 oz., 2.1 oz., 2.01
oz, 2.001 oz., . . .
0  x  32
Random Variables
Question
Family
size
Random Variable x
x = Number of dependents in
family reported on tax return
Distance from x = Distance in miles from
home to store home to the store site
Own dog
or cat
x = 1 if own no pet;
= 2 if own dog(s) only;
= 3 if own cat(s) only;
= 4 if own dog(s) and cat(s)
Type
Discrete
Continuous
Discrete
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
Example: JSL Appliances
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Using past data on TV sales, …
a tabular representation of the probability
distribution for TV sales was developed.
Units Sold
0
1
2
3
4
Number
of Days
80
50
40
10
20
200
x
0
1
2
3
4
f(x)
.40
.25
.20
.05
.10
1.00
80/200
Example: JSL Appliances
• Graphical Representation of the Probability
Distribution
Probability
.50
.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
Discrete Uniform Probability Distribution
This is the simplest probability
distribution described by a
formula. It assumes that
possible values of random
variables are equally likely
f ( x)  1 / n
n = number of values the random variable may
assume.
Example: Rolling a Die
x
1
2
3
4
5
6
f(x)
1/6
1/6
1/6
1/6
1/6
1/6
Note that n = 6
Expected Value E(x)
The expected value, or mean, of a random
variable, is a measure of central location for the
random variable
For a discrete random variable x we have
E ( x)     xf ( x)
E ( x)     xf ( x)
Notice that this is a
weighted average of
the values a random
variable can assume.
The “weights” are the
probabilities.
Example: JSL Appliances
• Expected Value of a Discrete Random Variable
x
0
1
2
3
4
f(x)
xf(x)
.40
.00
.25
.25
.20
.40
.05
.15
.10
.40
E(x) = 1.20
expected number of
TVs sold in a day
Variance of a Discrete Random
Variable
The variance of a random variable x is a weighted
average of the squared deviations of a random
variable from its mean (expected) value. The weights
are the probabilities.
Var ( x)   2  ( xi   ) 2 f ( x)
Standard Deviation of a
Random Variable (σ)
StDev( x)    ( xi   ) f ( x)
2
Example: JSL Appliances
• Variance and Standard Deviation
of a Discrete Random Variable
x
x-
0
1
2
3
4
-1.2
-0.2
0.8
1.8
2.8
(x - )2
f(x)
(x - )2f(x)
1.44
0.04
0.64
3.24
7.84
.40
.25
.20
.05
.10
.576
.010
.128
.162
.784
Variance of daily sales =  2 = 1.660
TVs
squared
Standard deviation of daily sales = 1.2884 TVs
Using Excel to Compute the Expected
Value, Variance, and Standard Deviation
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2
3
4
5
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9
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Formula Worksheet
A
Sales
0
1
2
3
4
B
Probability
0.40
0.25
0.20
0.05
0.10
Mean =SUMPRODUCT(A2:A6,B2:B6)
Variance =SUMPRODUCT(C2:C6,B2:B6)
Std.Dev. =SQRT(B9)
C
Sq.Dev.from Mean
=(A2-$B$8)^2
=(A3-$B$8)^2
=(A4-$B$8)^2
=(A5-$B$8)^2
=(A6-$B$8)^2
Using Excel to Compute the Expected
Value, Variance, and Standard Deviation
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Value Worksheet
A
Sales
0
1
2
3
4
Mean 1.2
Variance 1.66
Std.Dev. 1.2884
B
Probability
0.40
0.25
0.20
0.05
0.10
C
Sq.Dev.from Mean
1.44
0.04
0.64
3.24
7.84
The Binomial Distribution
This is a very useful tool for
multi-step experiments where
each step has 2 outcomes—
hence the term binomial.
Properties of Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
2. Two outcomes are possible on each trial. We
refer to one outcome as a success and the other
outcome as a failure.
3. The probability of success, denoted by ρ, does
not change from trial to trial. Consequently, the
probability of failure, denoted by 1 – ρ , does not
change from trial to trial.
4. The trials are independent.
Stationarity
assumption
We are interested in
computing the number of
successes (x) for n number
of trials
Binomial Probability Distribution
The binomial distribution is given by:
n!
x
(n x)
f ( x) 
 (1   )
x!(n  x)!
Where:
f(x) = probability of success in n trials
n = number of trials
p = probability of success in any one trial.
Example: Evans Electronics
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Binomial Probability Distribution
Evans is concerned about a low retention rate for
employees. In recent years, management has seen a
turnover of 10% of the hourly employees annually.
Thus, for any hourly employee chosen at random,
management estimates a probability of 0.1 that the
person will not be with the company next year.
Evans electronics
If we selected three (3)
employees at random, what is
the probability that one(1) will
leave the company within the
year?
Notice that:
•The experiment has three identical trials—that is, n = 3.
•There are two outcomes for each trial—the employee leaves (S)
or the employee stays (F).
•The probability that an employee will leave is .1—that is, ρ = .1
•The decision of each employee to leave is independent of the
decisions made by the other employees.
Counting the Number of Outcomes
First
Employee
Second
Employee
Third
Experimental
Employee
Outcome
(S,S,S)
S
S
F
F
S
S
F
F
S
S
F
S
F
3
(S,S,F)
2
(S,F,S)
2
1
(S,F,F)
(F,S,S)
F
Value of x
(F,S,F)
(F,F,S)
(F,F,F)
2
1
1
0
Number of Experimental Outcomes
Providing Exactly x Successes in n trials
n
n!
  
 x  x!(n  x)!
In our Evans Electronics example, n = 3 and x = 1.
Thus:
 3  (3)( 2)(1) 6
  
 3
1  (1)( 2)(1) 2
Refer to the tree diagram to verify this is
right
What is the probability the first
employee selected will leave
and second and third will stay?
Note this is outcome (S, F, F)
Because these events are independent, we can
multiply probabilities. . Thus the probability of
(S, S, F) is given by:
 (1   )(1   )   (1   ) 2
Thus we have:
.1(1  .1)  .1(.81)  .081
2
Trial Outcomes
Experimental
Outcome
Probability of
Experimental
Outcome
(S,F,F)
  (1   )(1   )   (1   ) 2
 .1(1  .9) 2  .081
(F,S,F)
  (1   )(1   )   (1   ) 2
 .1(1  .9) 2  .081
(F,F,S)
 (1   )(1   )    (1   ) 2
 .1(1  .9) 2  .081
Binomial Probability Distribution
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Binomial Probability Function
n!
f (x) 
p x (1  p)( n  x )
x !(n  x )!
n!
x !(n  x )!
Number of experimental
outcomes providing exactly
x successes in n trials
(nx )
p (1  p)
x
Probability of a particular
sequence of trial outcomes
with x successes in n trials
Probability Distribution for the
Number of Employees Leaving Within
the Year
x
f(x)
0
3!
(.1) 0 (.9) 3  .729
0!3!
1
3!
(.1)1 (.9) 2  .243
1!2!
2
3!
(.1) 2 (.9)1  .027
2!1!
3! 3
(.1) (.9) 0  .001
3!
3
Using Excel to Compute
Binomial Probabilities
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Formula Worksheet
A
1
2
3
4
5
6
7
8
9
x
0
1
2
3
B
3 = Number of Trials (n )
0.1 = Probability of Success (p )
f (x )
=BINOMDIST(A5,$A$1,$A$2,FALSE)
=BINOMDIST(A6,$A$1,$A$2,FALSE)
=BINOMDIST(A7,$A$1,$A$2,FALSE)
=BINOMDIST(A8,$A$1,$A$2,FALSE)
Using Excel to Compute
Binomial Probabilities
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Value Worksheet
A
1
2
3
4
5
6
7
8
9
x
0
1
2
3
B
3 = Number of Trials (n )
0.1 = Probability of Success (p )
f (x )
0.729
0.243
0.027
0.001
Using Excel to Compute
Cumulative Binomial Probabilities
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Formula Worksheet
A
1
2
3
4
5
6
7
8
9
x
0
1
2
3
B
3 = Number of Trials (n )
0.1 = Probability of Success (p )
Cumulative Probability
=BINOMDIST(A5,$A$1,$A$2,TRUE)
=BINOMDIST(A6,$A$1,$A$2,TRUE)
=BINOMDIST(A7,$A$1,$A$2,TRUE)
=BINOMDIST(A8,$A$1,$A$2,TRUE)
Using Excel to Compute
Cumulative Binomial Probabilities
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Value Worksheet
A
1
2
3
4
5
6
7
8
9
x
0
1
2
3
B
3 = Number of Trials (n )
0.1 = Probability of Success (p )
Cumulative Probability
0.729
0.972
0.999
1.000
Expected Value and Variance
for a Binominal Distribution
The expected value is computed by:
E ( x)    n
The variance is computed by:
Var( x)   2  n (1   )
Evans Electronics Example
Remember that n = 3 and ρ = .1. Thus:
E ( x)  3(.1)  .3
Var ( x)  3(.1)(.9)  .27
Note also that:
  .27  .52