Transcript Binomial

The Binomial Distribution
Permutations:
How many different pairs of two items are possible from these four letters: L, M. N, P.
L,M
L,N
L,P
M,L
M,N
M,P
N,L
N,M
N,P
P,L
P,M
P,N
Pr N 
N!
( N  r )!
4!
P 
(4  2)!
4
2
4 X 3(2!)
P 
(2)!
4
2
P24  12
P = # of Permutations
N = # of items
r = how many taken at a time
Combinations
L,M
L,N
L,P
M,N
M,P
N,P
(order is not important)
CrN 
CrN 
N!
r !( N  r )!
C = # of Combinations
N = # of items
r = how many taken at a time
4!
2!(4  2)!
4 X 3 X 2 X 1 24
C 

6
2 X 1X 2 X 1 4
N
r
If we flip a single coin, there are two possible outcomes: head or tail (H or T).
P(one head) = .5
P(no head) = .5
Together the probabilities of these possible outcomes sum to 1.0.
If we flip two coins there are four possible outcomes.
H,H
H,T
T,H
T,T
P(two heads) = .25
P(one head) = .50
P(no heads) = .25
Total = 1.0
Can be calculated by…
P = P(H) and Q = P(T)
P2 
( P  Q) 2  ( P  Q)( P  Q)
Or
P  2 PQ  Q
2
2PQ 
Q2 
Probability of 2 heads
Probability of 1head and 1 tail
Probability of 2 tails
2
* Think about the frequency distribution.
Flip 3 Coins: 8 possible outcomes
H,H,H
H,H,T
H,T,H
T,H,H
T,T,H
T,H,T
H,T,T
T,T,T
P(3 heads) = 1/8 or .13
P(2 heads) = 3/8 or .37
P(1 head) = 3/8 or .37
P(0 heads) = 1/8 or .37
Total = 1.0
Can be calculated by…..
( P  Q) 3
or
( P  Q)( P  Q)( P  Q)
P 3  3P 2 Q  3PQ2  Q3
Binomial Distribution
A binomial distribution is produced when each of a number of Independent trials
results in one of two Mutually Exclusive outcomes.
A single flip of a coin is a Bernoulli trial.
They reflect a discrete variable.
P( X )  CrN P X Q( N  X )

N!
p X Q( N  X )
X !( N  X )!
P(X) = Probability of X number of an outcome
N = Number of trials
P = Probability of success on any given trial
Q = (1 – P)
CXN  The number of combinations of N
things taken X at a time
Example:
Guess if I am thinking of the colour black or white. Because I randomly
choose one of the two colours, the probability that I am thinking of black is .5
on any given trial. There will be 10 trials. What is the probability of guessing
8 of them correctly?
10!
P(8) 
(.58 )(.52 )
8!(2!)

10 * 9(8!)
(.0039062)(.25)
8!(2!)
= 45(.009765)
= .0439425
This is the probability of * OR MORE correct, but of exactly 8 out of 10 correct.
Binomial Distribution & Testing a Hypothesis
Let us repeat the previous example, but now the question is what is the probability
of guessing 8 or more of the trials correctly.
We know that P(8) = .44
That is less than .05, But that is P(8) against 10, including P(9) and P(10).
We need to test the probability of 8 or more correct.
P(8) = .044
P(9) = .010
P(10) = .001
Total P = .055
The probability of guessing 8 or more correctly out of 10 turns out to
have a probability greater than .05, thus we fail to reject the null hypothesis.
Which was what?
Binomial Distribution
We can calculate the probability of 0 to 10 correct.
# Correct Probability
0
1
2
3
4
5
6
7
8
9
10
.001
.010
.044
.117
.205
.246
.205
.117
.044
.010
.001
0.25
0.2
0.15
P
0.1
0.05
0
0
1
2
3
4
5
6
7
8
9 10
Mathematical Sampling Distribution
As N gets large, all binomial distributions approach normality,
regardless of P.
Thus,
Mean = NP
For example, 10(.5) = 5
That is, if you examine the data on the previous slide, you
will see that the mean of the distribution is 5 (the number
of correct trials).
Variance = NPQ
Standard Deviation =
For example, 10(.5)(.05) = 2.25
NPQ
= 1.58