Transcript X - Energy E-Learning Courses

Statistical Applications
 n x
p( x)    p (1  p)n x
 x
.40
.30
.20
.10
0
4
1
2
3
Binominal and Poisson’s Probability
distributions
x 
f ( x) 
E(x) =  = xf(x)
 e
x!
Learning Objectives
After the session the students should be able to:




Evaluate discrete probability distributions
from realistic data
Use the binominal distribution to evaluate
simple probabilities
Evaluate probabilities using the Poisson’s
distribution
Answer examination type question
pertaining to these distributions
Recap- Types of data
• Discrete (A variable controlled by a fixed set of values)
• Continuous data (A variable measured on a continuous scale )
• These data may be collected (ungrouped) and then grouped
together in particular form so that can be easily inspected
• But how would we collect data?
Recap: Sampling Techniques
Simple random sampling
Stratified sampling
Cluster sampling
Quota sampling
Systematic sampling
Mechanical sampling
Convenience sampling
Random Variables
A random variable is a numerical description of the
outcome of an experiment.
A discrete random variable may assume either a
finite number of values or an infinite sequence of
values.
A continuous random variable may assume any
numerical value in an interval or collection of
intervals.
Examples
Question
Family
size
Random Variable x
x = Number of dependents
reported on tax return
Type
Discrete
Distance from x = Distance in miles from
home to store
home to the store site
Continuous
Own dog
or cat
Discrete
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)
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 equation.
Discrete Probability
Distributions cont…
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
Discrete Uniform Probability
Distribution
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
Relative frequency

Say a shop uses past knowledge to produce a tabular
representation of the probability distribution for TV sales:
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
Mean Value
 The or mean or
expected value, of
a random variable
 is a measure of its
central location.
E(x) =  = xf(x)
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
Mean cont…
• Graphical Representation of Probability Distribution
Probability
.50
.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
Variance & Standard
deviation

The variance summarizes the variability in the
values of a random variable.
Var(x) =  2 = (x - )2f(x)
NOTE: The standard deviation, , is defined as the positive
square root of the variance.
Binomial Distribution
•
Four Properties of a Binomial Experiment
1. The experiment consists of a sequence of n
identical trials.
2. 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.
stationary
assumption
4. The trials are independent.
Binomial Probability
Function
•
•
Of interest is the number of
success occurring in n trials
Let x be the number of
successes
n!
f (x) 
p x (1  p)( n  x )
x !(n  x )!
Jacob Bernoulli
where:
f(x) = the probability of x successes in n trials
n = the number of trials
p = the probability of success on any one trial
Binomial Probability
Function cont…
• Evaluation of probabilities using the distribution
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
px (1  p)( n  x )
Probability of a particular
sequence of trial outcomes
with x successes in n trials
Binomial probability function
alternative notation
• Evaluation of probabilities using the distribution
function:
P X  r   Cr p (1  p)
n
No of
combinations
Number of experimental
outcomes providing exactly
x successes in n trials
r
nr
Notice the pattern
of numbers
Probability of a particular
sequence of trial outcomes
with x successes in n trials
Mean and Variance
 It is useful to note that for a binominal
distribution the following are valid:
 Expected Value
E(x) =  = np
 Variance
Var(x) =  2 = np(1  p)
 Standard Deviation
  np(1  p)
Example #1 :

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.
Solution:


Using the Binomial Probability Function
Choosing 3 hourly employees at random,
what is the probability that 1 of them will leave the
company this year?
Let: p = 0.1, n = 3, x = 1
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
3!
f (1) 
(0.1)1 (0.9)2  3(.1)(.81)  .243
1!(3  1)!
Exercise #1
 A milling machine is know to produce 9%
defective components, if a random
sample of 5 components are taken,
evaluate the probability of no more than
2 components being defective
Exercise #1: Solution
•
•
•
•
•
Find the required parameters, namely:
p=0.09
n=5
P X  3 | X  Bi( 0.09,5)  p
X<3
Here you will need to use a little intelligence, i.e.:


p  P( X  0 X  Bi(0.09,5) ) 
P( X  1 X  Bi(0.09,5) )  P( X  2 X  Bi(0.09,5) )
Now put the numbers into:
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
Exercise #1: Solution
cont…
n!
f ( x) 
p x (1  p ) (n  x )
x !( n  x )!
 Using this standard formula gives
5!
0
5
0.09 0.81
P( X  0) 
0!(5)!
Notice the
pattern:
Top and bottom
equal the top
4+1=5
5!
5!
2
3
1
4
0.09 0.81
0.09 0.81 P( X  2) 
P( X  1) 
2!(3)!
1!(4)!
P( X  0)  P( X  1)  P( X  2)  0.349  0.1937  0.043
 0.586
Further example
 A machine produces on average 1 defective parts out of 8. 5 samples
are collected from this machine. Find the probability that 2 of them
are defective.
P X  3 | X  Bi(1/ 8,5)  p
Solution:
P( X  2)  C (0.125) (1  0.125)
5
2
2
3
 5! 

(0.015625)(0.669921875)  0.10475

 2!3! 
Notice here
the new
nomenclature
C 25
Poisson’s
distribution
 This distributions is named
after the famous French
mathematician who
formulated it:
Siméon Denis Poisson
A Poisson distributed random variable is often
useful in estimating the number of occurrences
over a specified interval of time or space
It is a discrete random variable that may assume
an infinite sequence of values (x = 0, 1, 2, . . . ).
Poisson’s random
variables
 They can be time dependent or not!
Examples of a Poisson distributed random variable:
the number of knotholes in 14 linear feet of
pine board
the number of vehicles arriving at a
toll booth in one hour
Poisson distribution
function
 Just as with the binominal distribution this allows
the calculation of probabilities!
f ( x) 
 x e 
x!
where:
f(x) = probability of x occurrences in an interval
 = mean number of occurrences in an interval
e = 2.71828
Poisson’s cumulative
distribution function
 By definition this is given by:
PX  r X  Po( )
0
1
2
3







PX  r X  Po( )  e     ...
 0! 1! 2! 3! 
Remembering this pattern helps in the evaluation of the
required probabilities since each term in the series are
respectively: P(X=0), P(X=1), P(X=2), P(X=3)
Example #2:
 Patients arrive at the
Casualty department of a
hospital at the average
rate of 6 per hour on
weekend evenings. What
is the probability of 4
arrivals in 30 minutes on a
weekend evening?
Example #2: Solution
o Simply use the Poisson’s
distribution function:
 = 6/hour = 3/half-hour, x = 4
34 (2.71828)3
f (4) 
 .1680
4!
MERCY
Poisson’s distribution
cont…

Poisson Distribution of Arrivals
0.20
NB: The Poisson’s
distribution has
the very special
property of the
mean and
variance being
equal!
0.15
=2
Poisson Probabilities
Probability
0.25
0.10
0.05
0.00
0
1
2
3
4
5
6
7
8
9
Number of Arrivals in 30 Minutes
10
Also when n>50, i.e.
large and np<5, i.e.
small. Then this
distribution
approximates the
Binominal.
Exercise #2:
 A serviceman is “beeped” each time there is a call for service.
The number of beeps per hour is Poisson distributed with a
mean of 2 per hour. Find the probability that he gets beeped 3
times in the next 2 hours.
Solution: The units of interval need to be uniform.
So, the mean beep rate will be 4 per 2 hour intervals.
Application of the Poisson’s probability function,
renders:
e 4 43 (0.0183)(64)
P( X  3) 
3!

6
 0.195
Further Example
 A garage workshop has an expensive machine tool which
is used on average 1.6 times per 8-hour day for a four
hour period. How many days in 60 day work period is the
tool required no more than twice.
p  1  P( X  0)  P( X  1)  P( X  2)
0
1
2

1.6 1.6 
1.6 1.6
 1 e 



1!
2! 
 0!
 1  0.202  0.323  0.258  0.217
Hence the required no. of days
0.217  60  13
Examination type
questions
1.
A machine is know to produce 10% defective
components, if a random sample of 12 components
are taken, evaluate the probability of:
a)
b)
2.
No components being defective [2]
more than 3 components being defective [3]
If jobs arrive at a machine at random average intervals
of 10/hr, estimate the probability of the machine
remaining idle for a 1.5 hour period [4].
a)
State the standard deviation of this distribution [1].
Further examination
type questions
Over a long period of time it is known that 5%
of the total production are below standard. If 6
are chosen at random, evaluate the probability
that at least 2 are defective [5].
4. A machine is known to produce 2% defective
components. In a packet of 100 what is the
probability of obtaining over 2 defective
components [5].
3.
Solutions:
1.
Here a simple application of the Binominal is required
thus:
a)
12!
0.100 0.9012  0.2824
0!(12)!
b)0.2824 12 C1 (0.1)1 (0.9)11 12 C2 (0.1) 2 (0.9)10 12 C3 (0.1)3 (0.9)9
 0.2842  0.3765  0.2301  0.0852
2.
Here let X be a Poisson RV, thus:
pX  0 X  Po(15)
150 e 15

0
0!
Solutions:
3.
This is the binominal model:
1  P( X  0)  P( X  1)  1 6 C0 (0.05)0 (0.95)6 6 C1 (0.05)1 (0.95)5
1  (0.95)6  6(0.05)1 (0.95)5  1  0.7351  0.2321
4.
We could use the binominal here but it is also a
Poisson’s approximation with np=100×0.02, thus:
2


2
2
1  pX  2 X  Po(2)  1  e 1  2  
2! 

 1  5e  2  0.3233
Alternative solution
4:
4.
We could use the binominal here also, with n=100
and p=0.02:
1  P( X  2 X  Bi( 0.02,100)) 
1 100 C0 (0.02) 0 (0.98)100 100 C1 (0.02)1 (0.98)99 100 C2 (0.02) 2 (0.98)98
 (0.98)100  100(0.02)1 (0.98)99 100 C2 (0.02) 2 (0.98)98 
 1  0.1326  0.2706  0.2734
 0.3234
Note: It’s worth noticing the Poisson’s
approximation if it turns up, “less calculations”!
Summary
Have we met out learning objectives? Specifically are
you able to:




Evaluate discrete probability distributions from
realistic data
Use the binominal distribution to evaluate simple
probabilities
Evaluate probabilities using the Poisson’s
distribution
Answer examination type question pertaining to
these distributions