Download Bernoulli Distribution

Unit-II-Distribution Funtions
Bernoulli Distribution
• An experiment consists of one trial. It can result in one of 2
outcomes: Success or Failure (or a characteristic being Present
or Absent).
• Probability of Success is p (0<p<1)
• Y = 1 if Success (Characteristic Present), 0 if not
p
p( y)  
1  p
y 1
y0
1
E (Y )   yp( y ) 0(1  p )  1 p  p
 
y 0
E Y 2  0 2 (1  p )  12 p  p
 
 V (Y )  E Y 2  E (Y )  p  p 2  p (1  p )
 
p (1  p )
2
Binomial Experiment
• Experiment consists of a series of n identical trials
• Each trial can end in one of 2 outcomes: Success or
Failure
• Trials are independent (outcome of one has no
bearing on outcomes of others)
• Probability of Success, p, is constant for all trials
• Random Variable X, is the number of Successes in the
n trials is said to follow Binomial Distribution with
parameters n and p
• Y can take on the values x=0,1,…,n
• Notation: X~Bin(n,p)
Binomial Distribution
Consider outcomes of an experiment with 3 Trials:
SSS  y  3 P( SSS )  P(Y  3)  p (3)  p 3
SSF , SFS , FSS  y  2 P ( SSF  SFS  FSS )  P (Y  2)  p (2)  3 p 2 (1  p )
SFF , FSF , FFS  y  1 P( SFF  FSF  FFS )  P (Y  1)  p (1)  3 p (1  p ) 2
FFF  y  0 P ( FFF )  P (Y  0)  p (0)  (1  p)3
In General:
n
n!
1) # of ways of arranging y S s (and (n  y ) F s ) in a sequence of n positions    
 y  y !(n  y )!
2) Probability of each arrangement of y S s (and (n  y ) F s )  p y (1  p ) n  y
n
3)  P(Y  y )  p ( y )    p y (1  p) n  y y  0,1,..., n
 y
EXCEL Functions:
p ( y ) is obtained by function:  BINOM.DIST(y, n, p, 0)
F ( y ) is obtained by function:  BINOM.DIST(y, n, p,1)
n
Binomial Expansion: (a  b)     a i b n i
i 0  i 
n
n
n
n
  p ( y )     p y (1  p ) n  y   p  (1  p )   1  "Legitimate" Probability Distribution
y 0
y 0  y 
n
n
Binomial Distribution (n=10,p=0.10)
0.5
0.45
0.4
0.35
p(y)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
y
6
7
8
9
10
Binomial Distribution (n=10, p=0.50)
0.5
0.45
0.4
0.35
p(y)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
y
6
7
8
9
10
Binomial Distribution(n=10,p=0.8)
0.35
0.3
0.25
p(y)
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
y
6
7
8
9
10
Binomial Distribution – Expected Value
f ( y) 
n!
p y q n y
y!(n  y )!
y  0,1,..., n q  1  p
n



n!
n!
y n y 
E (Y )   y 
p q    y
p y q n y 
y  0  y!( n  y )!
 y 1  y!(n  y )!

(Summand  0 when y  0)
n
n


yn!
n!
y n y 
y n y 
 E (Y )   
p q   
p q 
y 1  y ( y  1)! ( n  y )!
 y 1  ( y  1)!(n  y )!

Let y *  y  1  y  y *  1 Note : y  1,..., n  y *  0,..., n  1
n
n 1
n(n  1)!
(n  1)!
y*1 n  ( y*1)
y* ( n 1)  y*
 E (Y )   *
p
q

np
p
q


*
*
*
y * 0 y ! n  ( y  1) !
y * 0 y ! ( n  1)  y !
n 1



 np ( p  q ) n 1  np p  (1  p ) 
n 1
 np (1)  np

Geometric Distribution
• Used to model the number of Bernoulli trials needed until the
first Success occurs (P(S)=p)
– First Success on Trial 1  S, y = 1  p(1)=p
– First Success on Trial 2  FS, y = 2  p(2)=(1-p)p
– First Success on Trial k  F…FS, y = k  p(k)=(1-p)k-1 p
p ( y )  (1  p ) y 1 p


y 1
y 1
 p( y)   (1  p)
y  1,2,...
y 1

p  p  (1  p )
y 1
y 1
Setting y *  y  1 and noting that y  1,2,...  y *  0,1,...

 p
1
  p ( y )  p  (1  p )  p 
 1

y 1
y * 0
1  (1  p )  p


y*
Poisson Distribution
• Distribution often used to model the number of
incidences of some characteristic in time or space:
– Arrivals of customers in a queue
– Numbers of flaws in a roll of fabric
– Number of typos per page of text.
• Distribution obtained as follows:
–
–
–
–
–
–
Break down the “area” into many small “pieces” (n pieces)
Each “piece” can have only 0 or 1 occurrences (p=P(1))
Let l=np ≡ Average number of occurrences over “area”
Y ≡ # occurrences in “area” is sum of 0s & 1s over “pieces”
Y ~ Bin(n,p) with p = l/n
Take limit of Binomial Distribution as n  with p = l/n
Poisson Distribution - Derivation
n!
n!
l  l
p( y) 
p y (1  p ) n  y 
  1  
y!(n  y )!
y!(n  y )!  n   n 
Taking limit as n   :
y
n!
l   l 
lim p ( y )  lim
  1  
n 
n  y!( n  y )! n
   n
y
ly
n y
n y
ly
n(n  1)...( n  y  1)( n  y )!  l   n  l 
 lim
1   

y! n
n y (n  y )!
 n  n 
n
n(n  1)...( n  y  1)  l 
ly
 n  n  1   n  y  1  l 
 lim
1    lim 


...
1  
y
y! n
(n  l )
y! n n  l  n  l   n  l  n 
 n
n
 n 
 n  y 1
Note : lim 
 ...  lim 

  1 for all fixed y
n  n  l
n 


 nl 
ly
 l
 lim p ( y )  lim 1  
n 
y! n n 
n
n
 a
From Calculus, we get : lim 1    e a
n 
 n
ly
e l l y
 lim p ( y )  e l 
y  0,1,2,...
n 
y!
y!

Series expansion of exponentia l function : e x  
x 0


e l
l
 e l   e l e l  1  " Legitimate " Probabilit y Distributi on
y!
y 0
y  0 y!

  p( y)  
y 0
l
xi
i!
y
EXCEL Functions :
p ( y ) :  POISSON( y, l ,0)
F ( y ) :  POISSON( y, l ,1)
y
n
y

Poisson Distribution - Expectations
el ly
f ( y) 
y!
y  0,1,2,...

 e l l y    e l l y   e l l y
l y 1
l
l l
E (Y )   y 

y


l
e

l
e
e l

  
 
y!  y 1  y!  y 1 ( y  1)!
y 0 
y 1 ( y  1)!

 e l l y  
 e l l y   e l l y
E Y (Y  1)    y ( y  1) 

   y ( y  1) 

y 0
 y!  y  2
 y!  y  2 ( y  2)!

ly 2

 l2 e l 
y 2
 
( y  2)!
 l2 e l e l  l2
 E Y 2  E Y (Y  1)   E (Y )  l2  l
 
 V (Y )  E Y 2  E (Y )  l2  l  [l ]2  l
  l
2
Probability Density Functions…
Unlike a discrete random variable which we
studied in Chapter 7, a continuous random
variable is one that can assume an uncountable
number of values.
 We cannot list the possible values because
there is an infinite number of them.
 Because there is an infinite number of values,
the probability of each individual value is
virtually 0.
8.12
Point Probabilities are Zero
Because there is an infinite number of values, the
probability of each individual value is virtually 0.
Thus, we can determine the probability of a range
of values only.
E.g. with a discrete random variable like tossing a die, it is meaningful
to talk about P(X=5), say.
In a continuous setting (e.g. with time as a random variable), the
probability the random variable of interest, say task length, takes
exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
It is meaningful to talk about P(X ≤ 5).
8.13
Probability Density Function…
A function f(x) is called a probability density function
(over the range a ≤ x ≤ b if it meets the following
requirements:
1) f(x) ≥ 0 for all x between a and b, and
f(x)
area=1
a
b
x
2) The total area under the curve between a and b is 1.0
8.14
8.15
Uniform Distribution…
Consider the uniform probability distribution
(sometimes called the rectangular probability
distribution).
f(x)
It is described by the function:
a
b
area = width x height = (b – a) x
x
=1
8.16
Example:
The amount of gasoline sold daily at a service station
is uniformly distributed with a minimum of 2,000
gallons and a maximum of 5,000 gallons.
f(x)
2,000
5,000
x
What is the probability that the service station will
sell at least 4,000 gallons?
Algebraically: what is P(X ≥ 4,000) ?
P(X ≥ 4,000) = (5,000 – 4,000) x (1/3000) = .3333
8.17
The Normal Distribution…
The normal distribution is the most important of all
probability distributions. The probability density function
of a normal random variable is given by:
It looks like this:
Bell shaped,
Symmetrical around the mean
…
8.18
The Normal Distribution…
Important things to note:
The normal distribution is fully defined by two parameters:
its standard deviation and mean
The normal distribution is bell shaped and
symmetrical about the mean
Unlike the range of the uniform distribution (a ≤ x ≤ b)
Normal distributions range from minus infinity to plus infinity
8.19
Standard Normal Distribution…
A normal distribution whose mean is zero and standard
deviation is one is called the standard normal distribution.
0
1
1
As we shall see shortly, any normal distribution can be
converted to a standard normal distribution with simple
algebra. This makes calculations much easier.
8.20
Normal Distribution…
The normal distribution is described by two
parameters:
its mean and its standard deviation .
Increasing the mean shifts the curve to the
right…
8.21
Normal Distribution…
The normal distribution is described by two
parameters:
its mean and its standard deviation .
Increasing the standard deviation “flattens” the
curve…
8.22
Calculating Normal Probabilities…
Example: The time required to build a computer is
normally distributed with a mean of 50 minutes and a
standard deviation of 10 minutes:
0
What is the probability that a computer is assembled in a
time between 45 and 60 minutes?
Algebraically speaking, what is P(45 < X < 60) ?
8.23
Calculating Normal Probabilities…
P(45 < X < 60) ?
…mean of 50 minutes and a
standard deviation of 10 minutes…
0
8.24
Calculating Normal Probabilities…
P(–.5 < Z < 1) looks like this:
The probability is the area
under the curve…
We will add up the
two sections:
P(–.5 < Z < 0) and
P(0 < Z < 1)
0
–.5 … 1
8.25
Calculating Normal Probabilities…
This table gives probabilities P(0 < Z < z)
First column = integer + first decimal
Top row = second decimal place
P(0 < Z < 0.5)
How to use Table
3… [other forms of
normal tables exist]
P(0 < Z < 1)
P(–.5 < Z < 1) = .1915 + .3414 = .5328
The probability time is between
45 and 60 minutes = .5328
8.26
Using the Normal Table
What is P(Z > 1.6) ?
P(0 < Z < 1.6) = .4452
z
0
1.6
P(Z > 1.6) = .5 – P(0 < Z < 1.6)
= .5 – .4452
= .0548
8.27
Using the Normal Table
P(0 < Z < 2.23)
What is P(Z < -2.23) ?
P(Z < -2.23)
P(Z > 2.23)
z
-2.23
0
2.23
P(Z < -2.23) = P(Z > 2.23)
= .5 – P(0 < Z < 2.23)
= .0129
8.28
Using the Normal Table
What
is P(Z < 1.52) ?
P(Z < 0) = .5
P(0 < Z < 1.52)
z
0
1.52
P(Z < 1.52) = .5 + P(0 < Z < 1.52)
= .5 + .4357
= .9357
8.29
Using the Normal Table
What is P(0.9 < Z < 1.9) ? P(0 < Z < 0.9)
P(0.9 < Z < 1.9)
z
0
0.9
1.9
P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9)
=.4713 – .3159
= .1554
8.30
Example
The return on investment is normally distributed with a
mean of 10% and a standard deviation of 5%. What is the
probability of losing money?
We want to determine P(X < 0). Thus,
 X   0  10 
P(X  0)  P


5 
 
 P( Z   2)
 .5  P(0  Z  2)
 .5  .4772
 .0228
8.31
Finding Values of ZA…
Often we’re asked to find some value of Z for a
given probability, i.e. given an area (A) under the
curve, what is the corresponding value of z (zA)
on the horizontal axis that gives us this area?
That is:
P(Z > zA) = A
8.32
Finding Values of Z…
What value of z corresponds to an area under
the curve of 2.5%? That is, what is z.025 ?
Area = .50
Area = .025
Area = .50–.025 = .4750
If you do a “reverse look-up” on Table 3 for .4750,
you will get the corresponding zA = 1.96
Since P(z > 1.96) = .025, we say: z.025 = 1.96
8.33
Finding Values of Z…
Other Z values are
Z.05 = 1.645
Z.01 = 2.33
Will show you shortly how to use the “t-tables”
with infinite degrees of freedom to find a bunch
of these standard values for Zα
8.34
Using the values of Z
Because z.025 = 1.96 and - z.025= -1.96, it follows that we
can state
P(-1.96 < Z < 1.96) = .95
The old Empirical Rule stated about 95% within + 2σ
P(-2 < Z < 2) = .95
From now on we will use the 1.96 number for this
statement unless you are just talking in general terms
about how much of a population in with + 2σ
Similarly
P(-1.645 < Z < 1.645) = .90
8.35
Problems: Standard Normal “Z”
If the random variable Z has a standard normal
distribution, calculate the following probabilities.
P(Z > 1.7) =
P(Z < 1.7) =
P(Z > -1.7) =
P(Z < -1.7) =
P(-1.7 < Z < 1.7)
8.36
Problems: Normal Distribution
If the random variable X has a normal distribution
with
mean 40 and std. dev. 5, calculate the following
probabilities.
P(X > 43) =
P(X < 38) =
P(X = 40) =
P(X > 23) =
8.37
Problem: Normal
The time (Y) it takes your professor to drive home each
night is normally distributed with mean 15 minutes and
standard deviation 2 minutes. Find the following
probabilities. Draw a picture of the normal distribution and
show (shade) the area that represents the probability you are
calculating.
P(Y > 25) =
P( 11 < Y < 19) =
P (Y < 18) =
8.38
Problem: Target the Mean
The manufacturing process used to make “heart pills” is known
to have a standard deviation of 0.1 mg. of active ingredient.
Doctors tell us that a patient who takes a pill with over 6 mg. of
active ingredient may experience kidney problems. Since you
want to protect against this (and most likely lawyers), you are
asked to determine the “target” for the mean amount of active
ingredient in each pill such that the probability of a pill
containing over 6 mg. is 0.0035 (0.35%). You may assume that
the amount of active ingredient in a pill is normally distributed.
*Solve for the target value for the mean.
8.39