Download Binomial.s03

Bernoulli Distribution
A Bernoulli distribution arises from a random
experiment which can give rise to just two possible
outcomes. These outcomes are usually labeled as
either “success” or “failure.” If p denotes the
probability of a success and the probability of a
failure is (1 - p ), the the Bernoulli probability
function is
P(0)  (1  p) and P(1)  p
Mean and Variance of a
Bernoulli Random Variable
The mean is:
 X  E ( X )   xP( x)  (0)(1  p)  (1)  p
X
And the variance is:
 X2  E[( X   X ) 2 ]   ( x   X ) 2 P( x)
X
 (0  p) (1  p )  (1  p ) p  p (1  p )
2
2
Sequences of x Successes in n
Trials
The number of sequences with x successes in n
independent trials is:
n!
C 
x!(n  x)!
n
x
Where n! = n x (x – 1) x (n – 2) x . . . x 1 and 0! = 1.
These C xn sequencesare mutually exclusive,
since no two of them can occur at the same time.
Binomial Distribution
Suppose that a random experiment can result in two possible
mutually exclusive and collectively exhaustive outcomes, “success”
and “failure,” and that  is the probability of a success resulting in a
single trial. If n independent trials are carried out, the distribution
of the resulting number of successes “x” is called the binomial
distribution. Its probability distribution function for the binomial
random variable X = x is:
P(x successes in n independent trials)=
n!
x
( n x )
P( x) 
p (1  p)
x!(n  x)!
for x = 0, 1, 2 . . . , n
Mean and Variance of a Binomial
Probability Distribution
Let X be the number of successes in n independent
trials, each with probability of success . The x follows
a binomial distribution with mean,
X  E( X )  np
and variance,
  E[( X   ) ]  np(1  p)
2
X
2
Binomial Probabilities
- An Example –
(Example 5.7)
An insurance broker, Shirley Ferguson, has five contracts,
and she believes that for each contract, the probability of
making a sale is 0.40.
What is the probability that she makes at most one sale?
P(at most one sale) = P(X  1) = P(X = 0) + P(X = 1)
= 0.078 + 0.259 = 0.337
5!
P(no sales)  P(0) 
(0.4) 0 (0.6) 5  0.078
0!5!
5!
P(1 sale)  P(1) 
(0.4)1 (0.6) 4  0.259
1!4!
Binomial Probabilities, n = 100, p =0.40
(Figure 5.10)
Sample size
100
Probability of success
0.4
Mean
40
Variance
24
Standard deviation
4.898979
Binomial Probabilities Table
X
36
37
38
39
40
41
42
43
P(X)
0.059141
0.068199
0.075378
0.079888
0.081219
0.079238
0.074207
0.066729
P(<=X)
0.238611
0.30681
0.382188
0.462075
0.543294
0.622533
0.69674
0.763469
P(<X)
0.179469
0.238611
0.30681
0.382188
0.462075
0.543294
0.622533
0.69674
P(>X)
0.761389
0.69319
0.617812
0.537925
0.456706
0.377467
0.30326
0.236531
P(>=X)
0.820531
0.761389
0.69319
0.617812
0.537925
0.456706
0.377467
0.30326