Download 4. Probability Distributions

Slide 3.1
Probability distributions
• We extend the probability analysis by considering
random variables (usually the outcome of a
probability experiment)
• These (usually) have an associated probability
distribution
• Once we work out the relevant distribution,
solving the problem is usually straightforward
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.2
Random variables
• Most statistics (e.g. the sample mean) are
random variables
• A random variable is a numeric event whose
value is determined by a chance process or an
experiment
• The event should not be under the control of the
observer; i.e., the value of the random variable is
unknown before the experiment is carried out.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.3
Random variables
• Example 1: Suppose we roll two dice and take the sum of
the numbers showing up.
• This sum is clearly a random variable because its value is
determined by chance
• Such an experiment produces 11 possible values (what
are they?)
• Example 2: Consider a random experiment in which a
coin is tossed three times. Let X be the number of heads.
Let H represent the outcome of a head and T the outcome
of a tail.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.4
Random variables
• The sample space for such an experiment will be:
TTT, TTH, THT, THH, HTT, HTH, HHT, HHH.
• Thus the possible values of X (number of heads)
are x = 0,1,2,3.
• Many random variables have well-known
probability distributions associated with them.
• To understand random variables, we need to
know about probability distributions.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.5
Discrete & continuous random variables
• A discrete random variable is a variable that can
assume only certain clearly separated values
resulting from a count of some item of interest
(countably finite or infinite).

Example: Let X be the number of heads when a coin is tossed
3 times. Here the values for X are x = 0,1,2,3.
• A continuous random variable is a variable that can
assume one of an infinitely large number of values.


Assumes any value within a designated range of values.
Example: Height of a student in this class.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.6
Probability Distribution for Discrete RVs
• Each value of a random variable has an associated
probability
• The probability distribution of a random variable
lists all the possible values of the random variable
and their corresponding probabilities
• If P (X) is the probability that X is the value of the
random variable, then
 P( X )  1
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.7
Discrete probability distributions
• Suppose we toss a coin three times and want
to observe the number of heads.
• The possible values are 0, 1, 2, 3.
• What is the probability distribution of the
number of heads?
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.8
Discrete probability distributions
No. of heads
Frequency
Probability
X
0
1
P (X)
1/8 = .125
1
3
3/8 = .375
2
3
3/8 = .375
3
1
1/8 = .125
8/8 = 1.00
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.9
Expected value of discrete random variable
• The mean of a random value is called its expected
value
• For discrete random variables, it is the weighted
mean of all possible values of the random variable,
with weights being the probabilities
 X  E( X )   XP( X )
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.10
Expected value
• For the three tosses of a coin, we have
E( X )   XP( X )  0(.125)  1(.375)  2(.375)  3(.125)  1.5
• The expected value is not the value we “expect”
on any single toss
• Rather it is the long run average value (when the
experiment is done a large number of times)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.11
Variance for a random variable
• The variance is given by
 X2   [ X  E ( X )]2 P( X )
• Or for computational ease, we use
   X P( X )  [ XP( X )]
2
X
2
2
• Take square root to obtain standard deviation
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Variance
Slide 3.12
X
P(X) X P(X) X-E(X) [X-E(X)]2
[X-E(X)]2P(X)
X2
X2P(X)
0
.125
0
-1.5
2.25
.2813
0
0
1
.375
.375
-.5
.25
.0938
1
.375
2
.375
.75
.5
.25
.0938
4
1.5
3
.125
.375
1.5
2.25
.2813
9
1.125
1.5
.7502
3.0
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.13
Variance
   X P( X )  [ XP( X )]
2
X
2
2
• = 3.0 – (1.5)2 = .75
• Standard deviation = .87
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
6-14
Slide 3.14
Practice
• The Managing Director
of Perfect Painters, a
painting firm in Accra,
has studied his records
for the past 20 weeks
and reports the
following number of
houses painted per
week. Compute the
mean and variance of
the number of houses
painted per week.
# of Houses
Painted
10
Weeks
11
6
12
7
13
2
5
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
6-15
Slide 3.15
Solution
• Probability Distribution:
Number of houses
painted, X
10
Probability, P(X)
11
.30
12
.35
13
.10
Total
1
.25
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
6-16
Slide 3.16
Solution cont’d
• Mean number of houses painted per week:
  E ( x )  [ xP( x )]
 (10)(.25)  (11)(.30)  (12)(.35)  (13)(.10)
 11.3
• Variance of the number of houses painted
per week:
 2  [( x   ) 2 P( x )]
.4225.0270.1715.2890
.91
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.17
Some standard probability distributions
• Binomial distribution (discrete)
• Normal distribution (continuous)
• Poisson distribution (discrete)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.18
When do they arise?
• Binomial - when the underlying probability
experiment has only two possible outcomes
(e.g. tossing a coin)
• Normal - when many small independent factors
influence a variable (e.g. IQ, influenced by genes,
diet, etc.)
• Poisson - for rare events, when the probability of
occurrence is low
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.19
The Binomial distribution
• Applied when?



Only two mutually exclusive outcomes are possible
in each trial (success or failure)
The outcomes in the series of trials are independent
The probability of success, denoted P, in each trial
remains constant from trial to trial
• The objective of using the BD is to determine the
probability values for various possible number of
successes (X), given the number of trails (n) and
the known (and constant) probability of success (P)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.20
The Binomial distribution
• Consider five tosses of a coin. We can draw a tree
diagram for this experiment, but we won’t.
• Recall from last lecture, we can write the probability of
1 Head in 2 tosses as the probability of a head and a
tail (in that order) times the number of possible
orderings (# of times that event occurs).

P (1 Head) = ½  ½  2C1 = ¼  2 = ½
• We can apply same technique to calculate the
probability for the number of heads in five tosses of a
coin.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.21
The Binomial distribution
• P (X Heads in five tosses of a coin)

P(X = 0) = (½)0  (½)5  5C0 = /32  1 = /32

P(X = 1) = (½)1  (½)4  5C1 = /32  5 = /32

P(X = 2) = (½)2  (½)3  5C2 = /32  10 =
10
/32

P(X = 3) = (½)3  (½)2  5C3 = /32  10 =
10
/32

P(X = 4) = (½)4  (½)1  5C4 = /32  5 = /32

P(X = 5) = (½)5  (½)0  5C5 = /32  1 = /32
1
1
1
1
1
1
1
5
5
1
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Probability distribution of 5
tosses of a coin
Slide 3.22
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
Number of heads
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Binomial distribution with
different parameters
Slide 3.23
1
• Eight tosses of an unfair coin (P = /6 )
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
8
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.24
The Binomial ‘family’
• Like other distributions, the Binomial is a family of
distributions, members being distinguished by
their different parameters.
• The parameters of the Binomial are:


P - the probability of ‘success’
n - the number of trials
• Notation: X ~ B(n, P)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.25
Formula for Calculating Binomial Probabilities
• For the Binomial distribution, X ~ B(n, P), we can
calculate the probability of X successes as
P(X) = nCx * Px * (1-P)(n-x)
n!
P( X ) 
P x (1  P) n  x
x!(n  x)!
• E.g. X ~ B(5, ½ ) means P(X) = 5Cx(½)x (1- ½ )(5-x)
and from this we can work out P(X) for any value
of X.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.26
Example
• Probability of obtaining two heads in three tosses
of a coin: X=2 and n=3
• X ~ B(3, ½ )
• P(X=2) = 3C2 (.5)2(.5) = 3×.25×.5 = .375
• Note that, nCX is the combination formula, which
is equal to
n!
x ! n  x !
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.27
Using complement rule
• When the BD is used, it is typically because
we wish to determine the probability or “X or
more” successes [i.e., P(X ≥ xi)] or “ X or
fewer successes [i.e., P (X ≤ xi)]
• If the individual probabilities to be summed is
large, it is easier to use the complement rule
• Example: P (X ≥ xi) = 1 - P(X < xi)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.28
Example
• Suppose the probability is .05 that a randomly selected
student of the University of Ghana owns a car. What is the
probability of observing two or more student car-owners in
a random sample of 20 students?
• P(X≥2) = P(X=2) + P(X=3) + …..+ P(X=20)
• P(X≥2) = 1 – P(X<2) = 1- P(X=0, 1)
• = 1 – [P(X=0) +P(X=1)]
• Now


P(X=0) = 20C0 (.05)0(.95)20 = .3585
P(X=1) = 20C1 (.05)(.95)19 = .3774
• So P(X≥2) = 1 – (.3585 + .3774) = .2641
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.29
Mean and variance of the Binomial


Mean = E(X) = n  P
Variance = σ2 = n  P  (1-P)
• On average, you would expect 10 Heads from
(n = ) 20 tosses of a fair (P = ½) coin (10 = 20 
½)
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
6-21
Slide 3.30
Practice
• The Ministry of Employment reports that 20% of the
labour force in Ghana is unemployed. From a sample
of 14 members of the labour force, calculate the
following probabilities using the formula for the
binomial probability distribution:
 three are unemployed

at least three are unemployed.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.31
Solution
• Three are unemployed:
• P(x=3)=.250
• At least three are unemployed:



P(x ≥ 3) = 1 - P(x<3)
= 1 – [P(x=0) +P(x=1) + P(x=2)]
= 1- (.044 +.154 +.250) =.552
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.32
The Poisson distribution
• It is a sampling process in which events occur over time or
space.
• The Poisson distribution is used to describe a number of
processes or events such as




The distribution of telephone calls going through a switch board
The demand of patients for service at a health facility
The arrival of vehicles at a tollbooth
The number of accidents occurring at a road intersection, etc.
• The Poisson is the limiting case of the Binomial when the
probability of success is very small, i.e. BD becomes more
and more skewed to the right as probability of success
becomes smaller and smaller (rare events).
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.33
The Poisson distribution
• Characteristics defining a Poisson random
variable are



The experiment consists of counting the number of
times a particular event occurs during a given time
interval
The probability that the event occurs in one time
interval is independent of the probability of the event
occurring in another time interval
The mean number of events in each unit of time is
proportional to the length of the time interval.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.34
The Poisson distribution
• The probability that the Poisson random variable will
assume the value X is given by
x 
e
P  x 
x!
• For X = 0, 1, 2,3 …….
• λ is the mean of the distribution (mean number of
events occurring in a given unit of time)
• e is approximately 2.7183 and is the base of the
natural logarithms.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.35
Example
• Suppose an average of 2 calls per minute are
received at a switchboard during a designated time
interval, then the probability that exactly 3 calls are
received in a randomly sampled minute is:
3 2
2 e 8(.1353)
P  x  3 

 .1804
3!
6
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.36
Poisson
• As with the BD, the PD typically involves determining the
probability of “X or more” or “X or fewer” number of events.
• To calculate, we sum the appropriate probability values
• The use of the complement rule may also come in handy
• For example, we may want to calculate the probability of
receiving 3 or more calls in a three-minute interval
• That is P(X≥3) = P(X=3) + P(X=4) + ……………
• Using the complement rule, we have
• P(X≥3) = 1 - P(X<3) = 1 – [P(x=0) + P(x=1) + P(x=2)]
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.37
Poisson
• From our proposition 3, we know the mean number of
occurrences is proportional to the length of the time
interval
• So if we expect a mean of 2 calls per minute, in three
minutes we must expect 6 calls
• For an interval of 30 seconds, the mean number of calls is
one
• The probability of 5 calls in a three-minute interval implies
λ = 6, so P(X=5 / λ=6) = .1606
• The probability of no calls in an interval of 30 seconds
implies that λ = 1, so P(X=0 / λ=1) = .3679
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.38
Expected value and variance
• The expected value and variance for a
Poisson random variable are both equal to the
mean number of events for the time interval of
interest
• E (X) = λ
• Var (X) = λ
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.39
Poisson approximation of Binomial
• When the probability of occurrence (success)
is very small (P<.05)
• And the number of trials is large (n>20)
• So that λ = nP and we apply the Poisson
formula
• Some say use the Poisson in place of the
Binomial when nP < 5
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.40
Example
• A manufacturer claims a failure rate of 0.2% for
its hard disk drives. In an assignment of 500
drives, what is the probability that, none are
faulty, one is faulty, etc?
• On average, 1 drive (0.2% of 500) should be
faulty, so λ = nP = 1.
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006
Slide 3.41
Example (continued)
• The probability of no faulty drives is
10 e1
P  x  0 
 0.368
0!
• The probability of one faulty drive is
11 e1
P  x  1 
 0.368
1!
• and
12 e1
P  x  2 
 0.184
2!
Barrow, Statistics for Economics, Accounting and Business Studies, 4th edition © Pearson Education Limited 2006