Download Descriptive Indices - National University of Singapore

Equivalent ly, P( X  x, Y  y)  P( X  x) P(Y  y)
E ( X ) is also called the population mean of X and
can be interprete d as the long - run average of X ,
i.e., if we sample repeatedly and independen tly
according to the distributi on p( x) to obtain
X 1 , X 2 ,..., X n , then the sample mean
X n  ( X 1  ...  X n )  E ( X ) as n increases.
1
n
Furthermor e,
4. E ( XY )  E ( X ) E (Y ) if X and Y are independen t.
As in the case of the mean,
the sample variance S  var( X )
as the sample size increases.
2
Standard deviation : SD( X )  var( X )
Topic 7
Binomial and Poisson
Distribution
Binomial Probability Distribution
Consider a sequence of independent events with
only two possible outcomes called success (S) and
failure (F)
Example: outcome of treatment (cured/not cured)
opinion (yes/no) S=yes, F=no
gender (boy/girl) S=boy, F=girl
Let p be the probability of S
Consider n number of such independent events.
Then the total no of success out of n such events is
a random variable called Binomial r.v.
Binomial Probability Distribution
Let p = probability of having a boy
q = probability of having a girl
Assume the outcome of the first child (boy or girl)
does not affect the outcome of the second and
subsequent children i.e. events are independent ,
we can multiply the probabilities together to get
Pr(BBB) = p x p x p = Pr(3 boys)
Pr(GGG) = q x q x q = Pr(No boy)
More generally, there are
n
n!
 
 k  k!( n  k )!
ways of ordering k B’s and n-k G’s. So
Pr (k boys out of n children)
n
   Pr( B
...BG...G )


k 
k
nk
n k
nk
   p (1  p )
k 
Binomial Probability
For a 3-child family,
1 boy = BGG or GBG or GGB
Pr(1 boy) = Pr(BGG)+Pr(GBG)+Pr(GGB)
= pqq + qpq +qqp
= 3pqq
Similarly,
Pr(2 boys) = Pr(BBG)+Pr(BGB)+Pr(GBB)
= 3ppq
Example 1
Choose a group of 7 old individuals
randomly from the population of 65-74
years old in US. Suppose 12.5% of the
population in that age is diabetic. The total
no. of persons out of the 7 selected who
suffers from Diabetes has a binomial
distribution.
Let X = # diabetic
Binomial( n = 7, p = 0.125)
7
P[ X  2]   0.1252 (1  0.125)5
 2
 0.1683
Example 2
Past records indicate that 70% of patients responded
to treatment. What is the probability that 16 out of
the next 20 patients will respond to treatment?
# patients responding ~ Bin( n = 20, p = 0.7)
P(16 out of 20 responded)
 20 
  0.716 (1  0.7) 4 = 0.1304
 16 
Poisson distribution:
Consider a binomial random variable Y with n very
large and p small and np is moderate equal to . Then the
probabilities can be approximated by what is called a
Poisson random variable.
n k
e   k
  p (1  p) n  k 
k!
k 
Y ~ Poisson ( ) if
e   k
P (Y  k ) 
k!
for k  0,1,2,...
 is called the rate.
Often used to model count data such as
a) # industrial accidents in a plant per month
b) # chromosome interchanges within cells
c) # insurance claims
The number of deaths attributable to typhoid
fever follows a Poisson distribution at a rate of
4.6 deaths per year
X = # deaths in 1 year ~ Poisson(4.6)
P( X  2)  e 4.6
4.6 2
 0.1063
2!
Y = # deaths in 6 months ~ Poisson(2.3)
P(Y  2)  e 2.3
2.3 2
 0.265
2!