Transcript Slide 1
Statistics for Business and Economics
Module 1:Probability Theory and Statistical Inference Spring 2010 Lecture 2: Probability and discrete random variables Priyantha Wijayatunga, Department of Statistics, Umeå University [email protected]
These materials are altered ones from copyrighted lecture slides (
© 2009 W.H. Freeman and Company
) from the homepage of the book: The Practice of Business Statistics Using Data for Decisions :Second Edition by Moore, McCabe, Duckworth and Alwan.
Randomness and probability
A phenomenon is
random
if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions.
The
probability
of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions.
Randomness and Probability Models
Randomness and probability Random variables and probability distributions Independence and rules of probability Assigning probabilities: finite number of outcomes (discrete variables) Conditional probability and Baye’s theorem Mean and variance of a discrete random variable Rules for means and variances Discrete probability models: binomial model, Poisson model, etc.
Coin toss
The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are
independent
(i.e., the outcome of a new coin flip is not influenced by the result of the previous flip).
The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials.
First series of tosses Second series
Two events are
independent
if the probability that one event occurs on any given trial of an experiment is not affected or changed by the occurrence of the other event.
When are trials not independent?
Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0.5.
The trials are independent only when you put the coin back each time. It is called
sampling with replacement.
Probability models
Probability models
describe mathematically the outcome of random processes and consist of two parts: 1)
S = Sample Space :
This is a set, or list, of all possible outcomes of a random process. An
event
is a subset of the sample space.
2) A
probability
for each possible event in the sample space
S
.
Example: Probability Model for a Coin Toss
:
S
= {Head, Tail} Probability of heads = 0.5
Probability of tails = 0.5
Sample spaces
It ’s the question that determines the sample space.
A.
A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)?
H M
…
H M H HHH M HHM H HMH M HMM
… B.
A basketball player shoots three free throws. What is the number of baskets made?
S
= { 0, 1, 2, 3 }
S
= { HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } Note: 8 elements, 2 3
Probability rules
1)
Probabilities range from 0 (
no chance of the event
) to 1 (
the event has to happen
). For any event A, 0 ≤
P
(A) ≤ 1 Coin Toss Example:
S =
{Head, Tail} Probability of heads = 0.5
Probability of tails = 0.5
Probability of getting a Head = 0.5
We write this as:
P
(Head) = 0.5
P
(neither Head nor Tail) = 0
P
(getting either a Head or a Tail) = 1
2)
Because some outcome must occur on every trial, the sum of the probabilities for all possible outcomes (the sample space) must be exactly 1.
Coin toss:
S
= {Head, Tail}
P
(head) +
P
(tail) = 0.5 + 0.5 =1
P
(sample space) = 1
P
(sample space) = 1
Probability rules ( cont d )
Coin Toss Example:
S
= {Head, Tail} Probability of heads = 0.5
Probability of tails = 0.5
3)
The complement of any event A is the event that A does not occur, written as A c .
The complement rule states that the probability of an event not occurring is 1 minus the probability that is does occur.
P
(not A) =
P
(A c ) = 1 −
P
(A) Venn diagram: Sample space made up of an event A and its complementary A c , i.e., everything that is not A.
Tail c = not Tail = Head
P
(Tail c ) = 1 −
P
(Head) = 0.5
Probability rules ( cont d )
Venn diagrams: A and B disjoint
4)
Two events A and B are
disjoint
if they have no outcomes in common and can never happen together
.
The probability that A or B occurs is then the sum of their individual probabilities.
P
(A or B) = “
P
(A U B)” =
P
(A) +
P
(B) This is the
addition rule for disjoint events .
A and B not disjoint Example: If you flip two coins, and the first flip does not affect the second flip:
S
= {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25.
The probability that you obtain “only heads or only tails” is:
P
(HH or TT) =
P
(HH) +
P
(TT) = 0.25 + 0.25 = 0.50
Probability rules ( cont d )
The
joint probability
of happening events A and B is written as
P
(A and B).
For
disjoint events A and B,
P(A and B)=0
Venn diagrams: A and B disjoint A and B not disjoint
General addition rule
General addition rule
for any two events A and B: The probability that A occurs, or B occurs, or both events occur is:
P
(
A
or
B
) =
P
(
A
) +
P
(
B
) –
P
(
A
and
B
) What is the probability of randomly drawing either an ace or a heart from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts. However, 1 card is both an ace and a heart. Thus:
P
(ace or heart) =
P
(ace) +
P
(heart) –
P
(ace and heart) = 4/52 + 13/52 1/52 = 16/52 ≈ .3
Multiplication Rule for Independent Events
Two events A and B are independent when knowing that one occurs does not change the probability that the other occurs. If A and B are independent,
P
(A and B) =
P
(A)
P
(B) This is the
multiplication rule for independent events .
Two consecutive coin tosses:
P
(first Tail and second Tail) =
P
(first Tail) *
P
(second Tail) = 0.5 * 0.5 = 0.25
Venn diagram: Event A and event B. The intersection represents the event {A and B} and outcomes common to both A and B.
Applying the Multiplication Rule
If two events A and B are independent, the event that A does not occur is also independent of B. The Multiplication Rule extends to collections of more than two events, provided that all are independent.
Example : A transatlantic data cable contains repeaters to amplify the signal. Each repeater has probability 0.999 of functioning without failure for 25 years. Repeaters fail independently of each other. Let A 1 denote the event that the first repeater operates without failure for 25 years, A 2 denote the event that the second repeater operates without failure for 25 years, and so on. The last transatlantic cable had 662 repeaters. The probability that all 662 will work for 25 years is: P(A 1 and A 2 and…and A 662 ) = 0.999
662 = 0.516
Independent vs. disjoint events
Disjoint events are not independent.
If A and B are disjoint, then the fact that A occurs tells us that B cannot occur. So A and B are not independent.
Independence cannot be pictured in a Venn Diagram.
Assigning probabilities: finite number of outcomes
Finite sample spaces
deal with
discrete data
— data that can only take on a limited number of values. These values are often integers or whole numbers.
Throwing a die:
S
= {1, 2, 3, 4, 5, 6} The individual outcomes of a random phenomenon are always disjoint. The probability of any event is the sum of the probabilities of the outcomes making up the event (addition rule).
M&M candies
If you draw an M&M candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here:
Color Probability Brown 0.3
Red 0.2
Yellow 0.2
Green 0.1
Orange 0.1
Blue ?
What is the probability that an M&M chosen at random is blue?
S
= {brown, red, yellow, green, orange, blue}
P
(S) =
P
(brown) +
P
(red) +
P
(yellow) +
P
(green) +
P
(orange) +
P
(blue) = 1
P
(blue) = 1 – [
P
(brown) +
P
(red) +
P
(yellow) +
P
(green) +
P
(orange)] = 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1
What is the probability that a random M&M is any of red, yellow, or orange?
P
(red or yellow or orange) =
P
(red) +
P
(yellow) +
P
(orange) = 0.2 + 0.2 + 0.1 = 0.5
Probabilities: equally likely outcomes
We can assign probabilities either:
empirically
from our knowledge of numerous similar past events Mendel discovered the probabilities of inheritance of a given trait from experiments on peas without knowing about genes or DNA.
or theoretically
from our understanding the phenomenon and symmetries in the problem A 6-sided fair die: each side has the same chance of turning up Genetic laws of inheritance based on meiosis process If a random phenomenon has
k
equally likely possible outcomes, then each individual outcome has probability 1/
k
. And, for any event A:
P
( A ) count of outcomes in A count of outcomes in S
Dice
You toss two dice. What is the probability of the outcomes summing to 5?
This is
S
: {(1,1), (1,2), (1,3), ……etc.} There are 36 possible outcomes in
S
, all equally likely (given fair dice). Thus, the probability of any one of them is 1/36.
P
(the roll of two dice sums to 5) =
P
(1,4) +
P
(2,3) +
P
(3,2) +
P
(4,1) = 4 / 36 = 0.111
Example: A couple wants three children. What are the arrangements of boys (B) and girls (G)?
Genetics tell us that the probability that a baby is a boy or a girl is the same, 0.5. Sample space: {BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG} All eight outcomes in the sample space are
equally likely .
The probability of each is thus 1/8.
Each birth is independent of the next, so we can use the
multiplication rule
. Example:
P
(BBB) =
P
(B)*
P
(B)*
P
(B) = (1/2)*(1/2)*(1/2) = 1/8
Conditional probability
Conditional probabilities
reflect how the probability of an event can change if we know that some other event has occurred/is occurring. Example: The probability that a cloudy day will result in rain is different if you live in Los Angeles than if you live in Seattle.
Our brains effortlessly calculate conditional probabilities, updating our “degree of belief” with each new piece of evidence.
The conditional probability of event B given event A is: (provided that
P
(A) ≠ 0)
P
(
B
|
A
)
P
(
A and B
)
P
(
A
)
General Multiplication Rule
Conditional probability gives the probability of one event under the condition that we know another event.
General multiplication rule: The probability that any two events, A and B, happen together is:
P
(A and B) =
P
(A)
P
(B|A)
Here P(B|A) is the conditional probability that B occurs, given the information that A occurs.
Independent Events
Recall: A and B are independent when they have no influence on each other’s occurrence.
Two events A and B that both have positive probability are independent if
P
(B|A) =
P
(B) The general multiplication rule then becomes:
P
(A and B) =
P
(A)
P
(B) What is the probability of randomly drawing an ace of hearts from a deck of 52 playing cards? There are 4 aces in the pack and 13 hearts.
P
(heart|ace) = 1/4
P
(ace) = 4/52
P
(ace and heart) =
P
(ace)*
P
(heart|ace) = (4/52)*(1/4) = 1/52 Notice that heart and ace are independent events.
Tree diagrams
Conditional probabilities can get complex, and it is often a good strategy to build a
probability tree
that represents all possible outcomes graphically and assigns conditional probabilities to subsets of events.
Tree diagram for chat room habits for three adult age groups A 1 , A 2 & A 3 .
Internet user
P(A 1 |C ) = 0.47 P(A 1 and C) = P(C|A 1 )P(A 1 ) = 0.136
.
0.47
P
(chatting) =
P(C and A 1 ) + P(C and A 2 ) + P(C and A 3 )
= 0.136 + 0.099 + 0.017 = 0.252
About 25% of all adult Internet users visit chat rooms.
Breast cancer screening
If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer? Disease incidence
0.0004
Mammography
0.9996
Incidence of breast cancer among women ages 20 –30
Diagnosis sensitivity
0.8
Positive Cancer Negative
False negative
0.2
0.1
Positive
False positive
No cancer
0.9
Diagnosis specificity Negative
Mammography performance
She could either have a positive test and have breast cancer or have a positive test but not have cancer (false positive).
Disease incidence
0.0004
Mammography
0.9996
Incidence of breast cancer among women ages 20 –30
Diagnosis sensitivity
0.8
Positive Cancer
0.2
Negative
False negative
0.1
Positive
False positive
No cancer
0.9
Diagnosis specificity Negative
Mammography performance
Possible outcomes given the positive diagnosis: positive test and breast cancer or positive test but no cancer (false positive).
) | ) 0.0004*0.8
) 0.3% ) This value is called the positive predictive value, or
PV
+. It is an important piece of information but, unfortunately, is rarely communicated to patients.
Bayes’s rule
An important application of conditional probabilities is Bayes’s rule. It is the foundation of many modern statistical applications beyond the scope of this textbook.
* If a sample space is decomposed in
k
disjoint events, A 1 , A 2 , … , A k — none with a null probability but
P
(A 1 ) +
P
(A 2 ) + … +
P
(A k ) = 1, * And if
C
is any other event such that
P
(
C
) is not 0 or 1, then: However, it is often intuitively much easier to work out answers with a probability tree than with these lengthy formulas.
If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer?
Disease incidence
0.0004
Mammography
0.9996
Incidence of breast cancer among women ages 20 –30
Diagnosis sensitivity
0.8
Positive Cancer
0.2
Negative
False negative
0.1
Positive
False positive
No cancer
0.9
Diagnosis specificity Negative
Mammography performance
This time, we use Bayes’s rule: A1 is cancer, A2 is no cancer,
C
is a positive test result. | ) | ) ( 0.8*0.0004
| ) ) ( | 0.3% ) ) ( )
Random variable
A
random variable
is a variable whose value is a numerical outcome of a random phenomenon.
A couple want three children. We define the random variable
X
as the number of girls they may get (or equivalently number of boys that they may get).
A
discrete random variable
X
has a finite number of possible values.
A
continuous random variable
X
takes all values in an
interval
.
Probability distributions
The probability distribution values
X
of a random variable
X
tells us what can take and how to assign probabilities to those values.
Because of the differences in the nature of sample spaces for discrete and continuous sample random variables, we describe probability distributions for the two types of random variables separately.
Define a discrete random variable
A couple wants three children. What are the numbers of girls they could have?
Define the
random variable
: X=number of girls
Sample space
of X (possible
values of X
): {0, 1, 2, 3} Then we get the probabilty model:
P
(X = 0) =
p(0)
=
P
(BBB) = 1/8
P
(X = 1)
= p(1)
=
P
(BBG or BGB or GBB) =
P
(BBG) +
P
(BGB) +
P
(GBB) = 3/8 x P(X=x) or p(x) or f(x) 0 1/8 1 2 3 3/8 3/8 1/8 F(x) 1/8 4/8 7/8 8/8 =1 p(x) : probability distribution F(x) : distribution function or cumulative distribution function
The probability distribution of a discrete random variable
X
lists the values and their probabilities: The probabilities
p
i must add up to 1.
A basketball player shoots three free throws. The random variable
X
is the number of baskets successfully made.
H M
…
H M H HHH M HHM H HMH M HMM
… Value of X Probability 0 1/8 1 3/8 2 3/8 MMM HMM MHM MMH HHM HMH MHH 3 1/8 HHH
The probability of any event is the sum of the probabilities
p
i of the values of
X
that make up the event. A basketball player shoots three free throws. The random variable
X
is the number of baskets successfully made.
What is the probability that the player
Value of X
successfully makes at least two baskets (“at least two” means “two or more”)?
P
(
X
≥2) =
P
(
X
=2) +
P
(
X
=3) = 3/8 + 1/8 = 1/2
Probability 0 1/8 MMM 1 3/8 HMM MHM MMH 2 3/8 HHM HMH MHH 3 1/8 HHH
What is the probability that the player successfully makes fewer than three baskets?
P
(
X
<3) =
P
(
X
=0) +
P
(
X
=1) +
P
(
X
=2) = 1/8 + 3/8 + 3/8 = 7/8 or
P
(
X
<3) = 1 –
P
(
X
=3) = 1 – 1/8 = 7/8
Frequency tables for discrete variables
Frequency
: Number of observations in/for each category/value of the variable
Relative Frequency
: proportion/percentage of observations in/for each category/value of the variable
Frequency Distribution
: Listing of frequncies for all values of the variable
Relative Frequency Distribution
: Listing of relative frequencies for all values of the variable
Example
To study the age structure of the people in Umeå, a random sample of 90 people was taken and recorded their : 23 people were less than 21yrs, 30 people were less than 36yrs, 15 people were less than 51yrs, 12 people were less than 71yrs and 10 people were 71yrs or older Age 0 –20 21 –35 36 –50 51 –70 71 – Total 23 30 15 12 10 90 Frequency Relative frequency 0.26
0.33
0.17
0.13
0.11
1.00
Cumulative relative frequency 0.26
0.59
0.76
0.89
1.00
Obtaining a probability model
If our random sample is a good representation of the total population Define a random variable 1, if the person’s age is in 0–20 2, if the person’s age is in 21–35 X= 3, if the person’s age is in 36–50 4, if the person’s age is in 51–70 5, if the person’s age is 71 or older Value of X 1 2 3 4 5 Probability f(x) 0.26
0.33
0.17
0.13
0.11
1.00
Cumulative probability F(x) 0.26
0.59
0.76
0.89
1.00
Mean of a discrete random variable
The mean of a set of observations is their arithmetic average. The mean
µ
of a random variable
X
is a weighted average of the possible values of
X
, reflecting the fact that all outcomes might not be equally likely. A basketball player shoots three free throws. The random variable
X
is the number of baskets successfully made (“H”).
MMM HMM MHM MMH HHM HMH MHH HHH Value of X Probability 0 1/8 1 3/8 2 3/8 3 1/8
The mean of a random variable
X
is also called
expected value
of
X
.
Mean of a discrete random variable
For a discrete random variable
X
with probability distribution the mean
µ
of
X
is found by multiplying each possible value of
X
by its probability, and then adding the products.
A basketball player shoots three free throws. The random variable
X
is the number of baskets successfully made.
Value of X Probability 0 1/8 1 3/8 2 3/8 3 1/8
The mean
µ
of
X
is
µ
= (0*1/8) + (1*3/8) + (2*3/8) + (3*1/8) = 12/8 = 3/2 = 1.5
Variance of a discrete random variable
The variance and the standard deviation are the measures of spread that accompany the choice of the mean to measure center.
The
variance
σ 2 X
of a random variable is a weighted average of the squared deviations (
X
−
µ X
) 2 of the variable
X
from its mean
µ X
. Each outcome is weighted by its probability in order to take into account outcomes that are not equally likely.
The larger the variance of
X
, the more scattered the values of
X
on average. The positive square root of the variance gives the
standard deviation
σ
of
X
.
Variance of a discrete random variable
For a discrete random variable
X
with probability distribution and mean
µ X ,
the variance
σ
2 of
X
is found by multiplying each squared deviation of
X
by its probability and then adding all the products.
A basketball player shoots three free throws. The random variable
X
is the number of baskets successfully made.
µ X
= 1.5.
The variance
σ
2 of
X
is
Value of X Probability 0 1/8 1 3/8 2 3/8
σ
2 = 1/8*(0−1.5) 2 + 3/8*(1−1.5) 2 + 3/8*(2−1.5) 2 + 1/8*(3−1.5) 2 = 2*(1/8*9/4) + 2*(3/8*1/4) = 24/32 = 3/4 = .75
3 1/8
Rules for means and variances
If
X
is a random variable and
a
and
b
are fixed numbers, then
µ a
+
bX
=
a
+
bµ X σ
2
a+bX
=
b
2
σ
2
X
If
X
and
Y
are two independent random variables, then
µ
X+Y =
µ
X +
µ
Y
σ
2
X+Y
=
σ
2
X
+
σ
2
Y σ
2
X-Y
=
σ
2
X
+
σ
2
Y
If
X
and
Y
are NOT independent but have correlation
ρ,
then
µ
X+Y =
µ
X +
µ
Y
σ
2
X+Y
=
σ
2
X
+
σ
2
Y
+ 2
ρσ X σ Y σ
2
X-Y
=
σ
2
X
+
σ
2
Y
- 2
ρσ X σ Y
Binomial setting
Binomial distributions are models for some categorical variables, typically representing the number of successes in a series of
n
trials. The observations must meet these requirements: The total number of observations
n
is fixed in advance. The outcomes of all
n
observations are statistically independent. Each observation falls into just one of 2 categories: success and failure. All
n
observations have the same probability of “success,”
p
. We record the next 50 births at a local hospital. Each newborn is either a boy or a girl; each baby is either born on a Sunday or not.
Binomial distribution
The distribution of the count
X
of successes in the binomial setting is the binomial distribution with parameters
n
and
p: B
(
n,p
). The parameter
n
is the total number of observations.
The parameter
p
is the probability of success on each observation.
The count of successes
X
can be any whole number between 0 and
n
.
A coin is flipped 10 times. Each outcome is either a head or a tail. The variable
X
is the number of heads among those 10 flips, our count of “successes.” On each flip, the probability of success, “head,” is 0.5. The number
X
of heads among 10 flips has the binomial distribution
B
(
n
= 10,
p
= 0.5).
Applications for binomial distributions
Binomial distributions describe the possible number of times that a particular event will occur in a sequence of observations. They are used when we want to know about the occurrence of an event, not its magnitude. In a clinical trial, a patient’s condition may improve or not. We study the number of patients who improved, not how much better they feel.
Is a person ambitious or not? The binomial distribution describes the number of ambitious persons, not how ambitious they are. In quality control we assess the number of defective items in a lot of goods, irrespective of the type of defect.
Binomial probabilities
The number of ways of arranging
k
successes in a series of
n
observations (with constant probability
p
of success) is the number of possible combinations (unordered sequences). This can be calculated with the
binomial coefficient
:
n k
n C k
n
!
k
!
(
n
k
)!
Where k = 0, 1, 2, ..., or n.
Binomial formulas
The binomial coefficient “
n
_choose_
k
” uses the
factorial
“
!
”. notation The factorial
n
! for any strictly positive whole number
n
is:
n
! =
n
× (
n
− 1) × (
n
− 2) × · · · × 3 × 2 × 1 For example:
5
! =
5
× 4 × 3 × 2 × 1 = 120 Note that 0! = 1.
Calculations for binomial probabilities
The binomial coefficient counts the number of ways in which k successes can be arranged among n observations. The
binomial probability P(X = k)
is this count multiplied by the probability of any specific arrangement of the
k
successes:
P
(
X
k
)
n k p k
( 1
p
)
n
k
The probability that a binomial random variable takes any range of values is the sum of each probability for getting exactly that many successes in
n
observations.
P
(
X
≤ 2) =
P
(
X
= 0) +
P
(
X
= 1) +
P
(
X
= 2)
X
0 1 2 …
k
…
n Total P(X) n C
0
p
0
q
n =
q n n C
1
p
1
q n
-1
n C
2
p
2
q n
-2 …
n C x p k q n-k
…
n C n p n q
0 =
p n 1
Finding binomial probabilities: tables
You can also look up the probabilities for some values of n and p in Table C in the back of the book. The entries in the table are the probabilities P(
X
=
k
) of individual outcomes.
The values of
p
that appear in Table C are all 0.5 or smaller. When the probability of a success is greater than 0.5, restate the problem in terms of the number of failures.
Color blindness
The frequency of color blindness (dyschromatopsia) in the Caucasian American male population is estimated to be about 8%. We take a random sample of size 25 from this population. What is the probability that exactly five individuals in the sample are color blind?
Use Excel’s “=BINOMDIST(number_
s
,trials,probability_
s
,cumulative)”
P
(
x
= 5) = BINOMDIST(5, 25, 0.08, 0) = 0.03285
P
(
x
= 5) = (
n
! /
k
!(
n
k
)!)
p k
(1
p
)
n-k
= (25! / 5!(20)!) 0.08
5 0.92
5
P
(
x
= 5) = (21*22*23*24*24*25 / 1*2*3*4*5) 0.08
5 0.92
20
P
(
x
= 5) = 53,130 * 0.0000033 * 0.1887 = 0.03285
Binomial mean and standard deviation
The center and spread of the binomial distribution for a count
X
are defined by the mean m and standard deviation s
:
m
np
s
npq
np
( 1
p
)
Effect of changing p when n is fixed.
a)
n
= 10,
p
= 0.25
b)
n
= 10,
p
= 0.5
c)
n
= 10,
p
= 0.75
For small samples, binomial distributions are skewed when
p
is different from 0.5.
0.3
0.25
0.2
0.15
0.1
0.05
0 a) 0.3
0.25
0.2
0.15
0.1
0.05
0 0 1 2 3 4 5 6 7 Number of successes 8 9 10 b) 0.3
0.25
0.2
0.15
0.1
0.05
0 0 1 2 3 4 5 6 7 Number of successes 8 9 10 c) 0 1 2 3 4 5 6 7 Number of successes 8 9 10
The Poisson setting
A count
X
setting: of successes has a Poisson distribution in the Poisson The number of successes that occur in any unit of measure is independent of the number of successes that occur in any non overlapping unit of measure.
The probability that a success will occur in a unit of measure is the same for all units of equal size and is proportional to the size of the unit.
The probability that 2 or more successes will occur in a unit approaches 0 as the size of the unit becomes smaller.
Poisson distribution
The distribution of the count the Poisson distribution with
X
of successes in the Poisson setting is mean
μ
. The parameter
μ
is the mean number of successes per unit of measure. The possible values of
X
are the whole numbers 0, 1, 2, 3, ….If
k
any whole number 0 or greater, then is P(
X
=
k
) = (
e μ μ k
)/
k
! where
e
2 .
72
P
(
X
k
)
e
m m
k k
!
The standard deviation of the distribution is the square root of
μ
.
Poisson distribution: example
Number accidents
X
happening in a certain town in 6 –month period is known to be a Poisson distribution with mean
μ=2
. What is the probability that no accident in past 6 months in the town?
P
(
X
0 )
e
2 2 0 0 !
e
2 0 .
135 What is the probability that only 1 accident in past 6 months in the town?
P
(
X
1 )
e
2 2 1 0 !
0 .
271 What is the probability that more than 1 accident in past 6 months in the town?
P
(
X
1 ) 1
P
(
X
0 )
P
(
X
1 ) 1 0 .
135 0 .
271 0 .
594 Poisson probabilities can be found in statistical tables