Transcript Binomial Distribution

Math 1107
Introduction to Statistics
Lecture 9
The Binomial Distribution
Math 1107 Binomial Distribution
If we are interested in determining the
probability of events with binary outcomes
(only 2 possibilities), we use the binomial
distribution.
Examples of outcomes:
– Yes/no
– Success/failure
– Respond/No Respond
Math 1107 Binomial Distribution
A binomial probability distribution results from a
procedure that meets all the following requirements:
1. The procedure has a fixed number of trials.
2. The trials must be independent. (The outcome of any
individual trial doesn’t affect the probabilities in the other trials.)
3. Each trial must have all outcomes classified into two categories.
4. The probabilities must remain constant for each trial.
Math 1107 Binomial Distribution
P(x) =
n!
•
(n – x )!x!
px •
n-x
q
for x = 0, 1, 2, . . ., n
where
n = number of trials
x = number of successes among n trials
p = probability of success in any one trial
(Note that x and p must indicate the same event)
q = probability of failure in any one trial (q = 1 – p)
Math 1107 Binomial Distribution
Lets go through a few examples…
Example 1:
You are presented with a multiple choice test in
nuclear physics. The test has 4 questions, with 5
answers for each question. Given that you know
nothing about nuclear physics, what is the probability
that you would pass the test and answer 3 out of 4
answers correctly?
Math 1107 Binomial Distribution
First step is to verify that the binomial distribution
makes sense to apply to this question. Does it? Why or
why not?
1. The number of “trials” is fixed (4);
2. The 4 trials are independent of each other;
3. Each of the 4 trials have only 2 possible outcomes;
4. The probability of a correct answer is the same for
each trial.
Math 1107 Binomial Distribution
The second step is to identify the values of n,x,p and q.
In this example:
n= 4
X= 3
p= .2
q= 1-p or .8
Math 1107 Binomial Distribution
Now, we plug the numbers into the formula:
P(x) =
n!
•
(n – x )!x!
px
•
qn-x
for x = 0, 1, 2, . . ., n
And we get:
P(x) =
4!
•
(4 – 3 )!3!
.23 •
4-3
.8
Math 1107 Binomial Distribution
Our result is:
P(3) = (4!/1!3!)*.008*.8 = .0256
Therefore, you have a 2.56% chance of passing the
test if you are guessing at random. You had better
study!
Math 1107 Binomial Distribution
Example 2:
You are presented with a multiple choice test in
nuclear physics. The test has 4 questions, with 5
answers for each question. Given that you know
nothing about nuclear physics, what is the probability
that you would fail the test and score 0, 1, 2 out of a
possible 4?
Math 1107 Binomial Distribution
In this example:
n= 4
x= 0, 1 and 2
p= .2
q= .8
So…
[4!/((4-0)!*0!)]*.20*.84= .4096
[4!/((4-1)!*1!)]*.21*.83= .4096
[4!/((4-2)!*2!)]*.22*.82= .1536
.4096+.4096+.1536 = .9728
You had better study!
Math 1107 Binomial Distribution
Fun EXCEL Exercise
Math 1107 Binomial Distribution
Example 3:
Across all airlines, the on time arrival percentage is
74%. If you commute to NY for work, and fly 10
times a month, what is the chance that 8 of those 10
flights will be on time?
Math 1107 Binomial Distribution
P(x) =
n!
•
(n – x )!x!
px •
n-x
q
for x = 0, 1, 2, . . ., n
In this example:
So…
[10!/((10-8)!*8!)]*.748*.262
n= 10
x= 8
p= .74
q= .26
= .2735