To describe random phenomena

Where Only integer values

Repeated experiment: Two outcomes:

Repetitions allowedIndependent trialsProbability remains constant

Success or failure

Consider an expt. Of n trialsLet

Xj=1 (if jth expt resulted in a success)And

Xj=0 (if jthexptresulted in a failure)thus probability:

P(X1,X2,X3…….Xn) = P1(X1).P2(X2)………Pn(Xn)And

Pj(Xj) = P(Xj) = {P, Xj=1, j=1,2….n

1-P=Q Xj=0, j=1,2….n0, otherwise

For 1 trial, this is called the bernoulli distribution

Flipping a coin. In this context, obverse ("heads") conventionally denotes success and reverse ("tails") denotes failure.

Rolling a die, where a six is "success" and everything else a "failure".

In conducting a political opinion poll, choosing a voter at random to ascertain whether that voter will vote "yes" in an upcoming referendum.

Mean and variance:•Mean:

E(Xj)= 0.q+1.p=P•Variance:

V(Xj)=[(0.q)+(12.p)]-p2=P(1-P)

Repetition of trials

Two outcomes

Independent trials

Probability remains constant

•Random variable X denoting no. of n bernoulli trials has a binomial distribution given by:

P(X) = {•Taking outcome with all successes(S) and failures(F)We get:

( ) = n!/x!(n-x)!

Where P(SSS….SS FF…….FF)=pxqn-x

( ) pX qn-X ,x=0,1,2…..n

0 ,otherwise

n x

nx

Mean and variance:

X – sum of n independnet bernoulli random variables

•Mean:

E(X)=P+P……+P=nP

•Variance:

V(X)=PQ+PQ+……+PQ=nPQ

•when n = 1, the binomial distribution is a Bernoulli distribution

P(6)=1/6

P(6’)=5/6

P(6)=1/6

P(6’)=5/6

P(6)=1/6

P(6’)=5/6

P(6)=1/6 P(6)=1/6

P(6’)=5/6

P(6)=1/6

P(6’)=5/6

P(6’)=5/6

•Assume Bernoulli trials

•Let X denote the number of trials until the first success. Then, the probability mass function of X is:

P(X=x)=(1−p)x−p

For x = 1, 2, ... In this case, we say that X follows a geometric distribution.

Mean and variance:•The mean of a geometric random variable X is:

E(X)=1/p

•The variance of a geometric random variable X is:V(X)=1−p/p2

Assume Bernoulli trials

•Let X denote the number of trials until the rth success.

P(X=x)=(x−1r−1)(1−p)x−rpr

for x = r, r + 1, r + 2, ... In this case, we say that X follows a negative binomial distribution.

Note two things:(1) There are (theoretically) an infinite number of negative binomial distributions. (2) A geometric distribution is a special case of a negative binomial distribution with r = 1.

Mean and variance:The mean of a negative binomial random variable X is:E(X)=r/pThe variance of a negative binomial random variable X is:V(x)=r(1−p)/p2

If X is a Poisson random variable

P(x)=e−λλx/x!

for x = 0, 1, 2, ... and λ > 0

Examples

•Let X equal the number of typos on a printed page.

•Let X equal the number of cars passing through the intersection of Allen Street and College Avenue in one minute.

•Let X equal the number of customers at an ATM in 10-minute intervals.

•Let X equal the number of students arriving during office hours.

Mean and variance:

The mean of a Poisson random variable X is λ.

The variance of a Poisson random variable X is λ.

To describe random phenomena

Where Any value

A continuous random variable X has a uniform distribution, denoted U(a, b), if its probability density function is:

P(x)=1/b−a

for two constants a and b, such that a < x < b.

The mean of a continuous uniform random variable defined over the support a < x < b is:E(X)=a+b/2The variance of a continuous uniform random variable defined over the support a < x < b is:V (X)=(b−a)2/12

Describes time between events in a poisson’s process.

Memoryless

Mean:

Variance:

To sample exponentional,uniform,triangular dist.

Form empirical dist.

Straight forward