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Introduction Discrete random variables take on only a finite
or countable number of values. Three discrete probability distributions serve as
models for a large number of practical applications:
The binomial random variable
The Poisson random variable
The hypergeometric random variable
The binomial random variable
The Poisson random variable
The hypergeometric random variable
Chapter 5: Several Useful Discrete Distributions
The Binomial Experiment
Ex: A coin-tossing experiment is a simple example of a binomial random variable, with n tosses and x number of heads
• The experiment consists of n identical trials.• Each trial results in one of two outcomes, success (S) or
failure (F).• The probability of success on a single trial is p and remains
constant from trial to trial. The probability of failure is q = 1 – p.
• The trials are independent.• We are interested in x, the number of successes in n
trials.
The Binomial Probability Distribution
For a binomial experiment with n trials and probability p of success on a given trial, the probability of k successes in n trials is
.1!0 and )1)(2)...(2)(1(!with
)!(!
! Recall
.,...2,1,0for )!(!
!)(
nnnn
knk
nC
nkqpknk
nqpCkxP
nk
knkknknk
.1!0 and )1)(2)...(2)(1(!with
)!(!
! Recall
.,...2,1,0for )!(!
!)(
nnnn
knk
nC
nkqpknk
nqpCkxP
nk
knkknknk
The Mean and Standard Deviation
For a binomial experiment with n trials and probability p of success on a given trial, the measures of centre and spread are:
npq
npq
np
:deviation Standard
:Variance
:Mean2
npq
npq
np
:deviation Standard
:Variance
:Mean2
n = p = x =success =
ExampleA cancerous tumour that is irradiated will die 80% of the time. A doctor treats 5 patients by irradiating their tumours. What is the probability that exactly 3 patients will have their tumours disappear?
333)3( nn qpCxP
5 .8cure # of cures
353 )2(.)8(.!2!3
!5
2048.)2(.)8(.10 23
AppletApplet
Example
What is the probability that more than 3 patients are cured?
55555
45454)3( qpCqpCxP
0514 )2(.)8(.!0!5
!5)2(.)8(.
!1!4
!5
7373.)8(.)2(.)8(.5 54
AppletApplet
Cumulative Probability Tables
You can use the cumulative probability tables to find probabilities for selected binomial distributions.
Find the table for the correct value of n.
Find the column for the correct value of p.
The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)
Find the table for the correct value of n.
Find the column for the correct value of p.
The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)
Examplek p = .80
0 .000
1 .007
2 .058
3 .263
4 .672
5 1.000
What is the probability that exactly 3 patients are cured?
P(x = 3) = P(x 3) – P(x 2)= .263 - .058= .205
P(x = 3) = P(x 3) – P(x 2)= .263 - .058= .205 Check from formula:
P(x = 3) = .2048
AppletApplet
Examplek p = .80
0 .000
1 .007
2 .058
3 .263
4 .672
5 1.000
What is the probability that more than 3 patients are cured?
P(x > 3) = 1 - P(x 3)= 1 - .263 = .737
P(x > 3) = 1 - P(x 3)= 1 - .263 = .737
Check from formula: P(x > 3) = .7373
AppletApplet
Example Would it be unusual
to find that none of the patients are cured?
• The value x = 0 lies
49.489.
40
x
z 49.489.
40
x
z
more than 4 standard deviations below the mean. Very unusual.
AppletApplet
89.)2)(.8(.5
4)8(.5
npq
np
:deviation Standard
:Mean
89.)2)(.8(.5
4)8(.5
npq
np
:deviation Standard
:Mean
The Poisson Random Variable The Poisson random variable x is a model for data
that represents the number of occurrences of a specified event in a given unit of time or space.
It is a special approximation to the Binomial Distribution for which n is large and the probability of success (p) is small. ( and ) 50n
Examples:
• The number of traffic accidents at a given intersection during a given time period.
• The number of radioactive decays in a certain time.
5np
The Poisson Probability Distribution x is the number of events that occur in a period
of time or space during which an average of such events can be expected to occur. The probability of k occurrences of this event is
For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are
Mean:
Standard deviation:
For values of k = 0, 1, 2, … The mean and standard deviation of the Poisson random variable are
Mean:
Standard deviation:
!)(
k
ekxP
k
ExampleThe average number of traffic accidents on a certain section of highway is two per week. Find the probability of exactly one accident during a one-week period.
!)1(
k
exP
k
2707.2!1
2 221
ee
Cumulative Probability Tables
You can use the cumulative probability tables to find probabilities for selected Poisson distributions.
Find the column for the correct value of .
The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)
Find the column for the correct value of .
The row marked “k” gives the cumulative probability, P(x k) = P(x = 0) +…+ P(x = k)
Example
k = 2
0 .135
1 .406
2 .677
3 .857
4 .947
5 .983
6 .995
7 .999
8 1.000
What is the probability that there is exactly 1 accident? P(x = 1) = P(x 1) – P(x 0)
= .406 - .135= .271
P(x = 1) = P(x 1) – P(x 0)= .406 - .135= .271
Check from formula: P(x = 1) = .2707
ExampleWhat is the probability that 8 or more accidents happen?
P(x 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001
P(x 8) = 1 - P(x < 8)= 1 – P(x 7) = 1 - .999 = .001
k = 2
0 .135
1 .406
2 .677
3 .857
4 .947
5 .983
6 .995
7 .999
8 1.000
This would be very unusual (small probability) since x = 8 lies
standard deviations above the mean.
This would be very unusual (small probability) since x = 8 lies
standard deviations above the mean.
24.4414.1
28
x
z
The Hypergeometric Probability Distribution Example: a bowl contains M red candies and N-
M blue candies. Select n candies from the bowl and record x, the number of red candies selected, where red candies are a success.
M= successes N-M = failures n= total number x= number selected
The probability of exactly k successes in n trials is
Nn
NMkn
Mk
C
CCkxP
)(
The Mean and VarianceThe mean and variance of the hypergeometric random variable x resemble the mean and variance of the binomial random variable:
1 :Variance
:Mean
2
N
nN
N
MN
N
Mn
N
Mn
1 :Variance
:Mean
2
N
nN
N
MN
N
Mn
N
Mn
ExampleA group of 8 drugs used to treat Alzheimer’s disease contains 2 drugs that are not effective. A researcher randomly selects four drugs to test on her subject. What is the probability that all four drugs work? What is the mean and variance for the number of drugs that work?
84
20
64)4(C
CCxP
N = 8 M = 6 n = 4
70
15
)1)(2)(3(4/)5)(6)(7(8
)1(2/)5(6
Success = effective drug
38
64
N
Mn
4286.7
4
8
2
8
64
12
N
nN
N
MN
N
Mn
Key ConceptsI. The Binomial Random Variable
1. Five characteristics: n identical independent trials, each resulting in either success S or failure F; probability of success is p and remains constant from trial to trial; and x is the number of successes in n trials.
2. Calculating binomial probabilities
a. Formula:
b. Cumulative binomial tables
3. Mean of the binomial random variable: np
4. Variance and standard deviation: 2 npq and
knknk qpCkxP )(
npq npq
Key ConceptsII. The Poisson Random Variable
1. The number of events that occur in a period of time or space, during which an average of such events are expected to occur
2. Calculating Poisson probabilities
a. Formula:b. Cumulative Poisson tables
c. Individual and cumulative probabilities using Minitab
3. Mean of the Poisson random variable: E(x) 4. Variance and standard deviation: 2 and
5. Binomial probabilities can be approximated with Poisson probabilities when np 7, using np.
!)(
k
ekxP
k
!)(
k
ekxP
k
Key ConceptsIII. The Hypergeometric Random Variable
1. The number of successes in a sample of size n from a finite population containing M successes and N M failures2. Formula for the probability of k successes in n trials:
3. Mean of the hypergeometric random variable:
4. Variance and standard deviation:
N
Mn
N
Mn
12
N
nN
N
MN
N
Mn
12
N
nN
N
MN
N
Mn
Nn
NMkn
Mk
C
CCkxP
)( N
n
NMkn
Mk
C
CCkxP
)(