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Binomial Distribution
Definition 4.1. A random variable is a variable
that assumes numerical values associated with
the random outcomes of an experiment, where
one and only one numerical value is assigned
to each sample point. Random variables that
can assume a countable or finite number of
value are called discrete. Random variables
that can assume values corresponding to all of
points contained in one or more intervals are
called continuous.
Binomial Distribution
Example 4.1. Identify the following variables as
discrete or continuous.
(a) The reaction time difference to the stimulus
before and after training.
continuous
(b) The number of violent crimes committed per
month in your community.
discrete
(c ) The number of commercial aircraft nearmisses per month.
discrete
Binomial Distribution
(d) The number of winners each week in a
state lottery.
discrete
(e) The number of free throws made per
game by a basketball team. discrete
(f) The distance traveled by a school bus
each day.
continuous
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To completely describe a discrete random
variable one must specify the possible
values that the random variable can
assume and the probability associated
with each value.
Binomial Distribution
The probability distribution of a discrete
random variable must satisfy the following
two rules.
(1) p(x)  0 for all x
(2) p(x) = 1where the summation of p(x)
is over all possible values of x.
Binomial Distribution
Example 4.2. In each case determine
whether the given values can serve as the
probabilities for a random variable that
can take on the values 1, 2, 3, 4.
(a) p(1) = .2, p(2) = .8, p(3) = .2, p(4) = -.2
(b) P(1) = .25, p(2) = .17, p(3) = .39, p(4)
= .19
Binomial Distribution
Sometimes it helps to table the discrete
probability distribution(s):
X
P1(X)
P2(X)
1
.2
.25
2
.8
.17
3
.2
.39
4
-.2
.19
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Solution. In both (a) and (b),
p(1) + p(2) + p(3) + p(4) = 1. However, in
(a) p(4) = -.2. Recall that one of the
rules is that probabilities are nonnegative. So only the set in (b) can serve
as the probabilities of a random variable
with possible values 1, 2, 3, 4.
Binomial Distribution
Definition 4.2. The mean, or expected
value, of a discrete random variable x is 
= E(x) = x p(x). The variance of a discrete
random variable is 2 = E[(x - )2] =
 [(x - )2 ]p(x). The standard deviation of a
discrete random variable is equal to the
square root of the variance, i.e.,  = sqrt(2).
Binomial Distribution
Example 4.3. Suppose that the
probabilities are .4, .3, .2, and .1 that 1, 2,
3, or 4 new anti-inflammatory drugs
respectively will be approved by the FDA
in the year 2003.
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(a) Find the mean of this distribution.
Solution.  =  x p(x) = (1)(.4) + (2)(.3) +
(3)(.2) + (4)(.1) = 2
(b) Find the variance of this distribution.
Solution. 2 = (x - )2p(x) = (1 – 2)2(.4) +
(2 – 2)2(.3) + (3 –2)2(.2) + (4 – 2)2(.1) = 1
Binomial Distribution
Let x be a discrete random variable with
probability distribution p(x), mean , and
standard deviation . Then, depending
on the shape of p(x), the following
probability statements can be made:
Binomial Distribution
Chebyshev’s
Rule
P( -  < x <  + )  0
P( - 2 < x <  +2)  ¾
P( - 3 < x <  + 3)  8/9
Empirical
Rule
 .68
 .95
 1.00
Chebyshev’s rule applies to any probability distribution.
The empirical rule applies to distributions that are
mound-shaped and symmetric.
Binomial Distribution
Recall that we had that if events A and B
are independent, p(AB) = p(A)p(B).
One can also show that if we have four
events – A, B, C, D – then if each pair of
the events is independent then
p(AB CD) = p(A) p(B) p(C) p(D).
Binomial Distribution
Example 4.4. Consider a population of
voters that is .4 Democrats and .6
Republicans. Suppose we choose a voter
at random from the population, put the
voter back in the population, and repeat
the process a total of four times.
Binomial Distribution
Each time we take a voter there are 2
possible outcomes – D and R. We take a
voter 4 times. Therefore there are 2 x 2 x
2 x 2 = 16 outcomes.
Binomial Distribution
DDDD
DDDR
DDRR
DDRD
DRDD
DRRD
DRDR
DRRR
RRRR
RRRD
RRDD
RRDR
RDRR
RDDR
RDRD
RDDD
Binomial Distribution
Since we put a voter back after polling her
or him, P(D) = .4 and P(R) =.6 on each of
the four draws. So the four draws are
independent, i.e., the probability of a D
(or R) on a given draw does not depend
on what the outcomes of the previous
draws were. So we get the probability of the
intersection of any set of four results on the
four draws by multiplying.
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Outcome
P
DDDD .4x.4x.4x.4
DDDR .4x.4x.4x.6
DDRR .4x.4x.6x.6
DDRD .4x.4x.6x.4
DRDD .4x.6x.4x.4
DRRD .4x.6x.6x.4
DRDR .4x.6x.4x.6
DRRR .4x.6x.6x.6
Outcome P
RRRR .6x.6x.6x.6
RRRD .6x.6x.6x.4
RRDD .6x.6x.4x.4
RRDR .6x.6x.4x.4
RDRR .6x.4x.6x.6
RDDR .6x.4x.4x.6
RDRD .6x.4x.6x.4
RDDD .6x.4x.4x.4
Binomial Distribution
Outcome
P
DDDD (.4)4x(.6)0
DDDR (.4)3x(.6)1
DDRR (.4)2x(.6)2
DDRD (.4)3x(.6)1
DRDD (.4)3x(.6)1
DRRD (.4)2x(.6)2
DRDR (.4)2x(.6)2
DRRR (.4)1x(.6)3
Outcome P
RRRR (.4)0x(.6)4
RRRD
(.4)1x(.6)3
RRDD (.4)2x(.6)2
RRDR (.4)1x(.6)3
RDRR (.4)1x(.6)3
RDDR (.4)2x(.6)2
RDRD (.4)2x(.6)2
RDDD (.4)3x(.6)1
Binomial Distribution
Now the random variable we are
interested in is X = the number of
Democrats we get in four draws from the
voter pool. X has the following set of
possible values: {0, 1, 2, 3, 4}. What are
the probabilities of each value?
Binomial Distribution
X
0
1
2
3
4
P(X)
1 . (.4)0 . (.6)4
4 . (.4)1 . (.6)3
6 . (.4)2 . (.6)2
4 . (.4)3 . (.6)1
1 . (.4)4 . (.6)0
Binomial Distribution
4
4
But 1 = 4!/0!4! = ( ) = ( )
0
4
4 = 4!/1!3! = ( ) =
1
6 = 4!/2!2! =
4
4
4
( ) , and
3
( ).
2
So we can write the probability distribution
for x as:
Binomial Distribution
X
0
P(X)
4
( ).
0
1
(.4)0 . (.6)4
4
( ). (.4)1 . (.6)3
1
2
4
( ). (.4)2 . (.6)2
2
3
4
4
( ) . (.4)3 . (.6)1
3
4
( ) . (.4)4 . (.6)0
4
Binomial Distribution
So we can say that in choosing four
voters one at a time with replacement the
probability that the number of Democrats
is x, that is, p(X = x) =
4
( ) . (.4)x . (.6)4 - x for x = 0, 1, 2, 3, 4.
x
Binomial Distribution
Now we chose 4 voters but suppose
instead that we were interested in n.
Also, we said that p(D) and p(R) in a
single draw were .4 and .6 respectively.
Suppose instead that they were p and
(1- p) = q respectively where p is any number
such that 0 < p < 1.
Binomial Distribution
Then we can say that in choosing n
voters one at a time with replacement,
where P(D) = p and P(R) = 1 – p = q, the
probability that the number of Democrats
is x, that is, P(X = x) =
n
( ) . px qn - x for x = 0, 1, 2, . . ., n.
x
Binomial Distribution
We call this the binomial probability
distribution of x successes in n trials or
b(n, p).
Binomial Distribution
Notice there were four key assumptions in
developing this distribution:
1. There is a fixed number, n, of
identical repetitions or trials. In our case
a trial was drawing a voter.
Binomial Distribution
2. There are only two possible outcomes
on each trial. We will denote one
outcome by S (for Success) and the other
by F (for failure).
Binomial Distribution
3. The probability of S, p, is the same for each
trial, as is the probability of F, 1 – p = q. In our
case we made this true by saying that we
replace the voter we drew on one trial before
the next one. If we are polling voter populations
with very large numbers of Democrats and
Republicans this assumption will be
approximately true even without replacement.
Binomial Distribution
4. The trials are all independent. Again,
this was true in our case by virtue of
replacement but it will generally be
approximately true when polling voter
populations with very large numbers of
Democrats and Republicans.
Binomial Distribution
The binomial random variable x is the
number of S’s in n trials. We call the
probability distribution of x b(n, p) to
indicate that it depends on n and p.
Binomial Distribution
The picture of a binomial distribution
b(n, p) depends on n and p. For our
example n was 4 and p was .4. The
probabilities work out as:
Binomial Distribution
X
0
1
2
3
4
P(X)
1 . (.4)0 . (.6)4 = .130
4 . (.4)1 . (.6)3 = .346
6 . (.4)2 . (.6)2 = .346
4 . (.4)3 . (.6)1 = .154
1 . (.4)4 . (.6)0 = .026
Binomial Distribution
0.35
X
0
1
2
3
4
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
b(4, .4)
P(X)
.130
.346
.346
.154
.026
Binomial Distribution
Since calculations with the formula can
become tedious, tables of cumulative
binomial probabilities for n = 5-10, 15, 20,
and 25 have been constructed and are
available as Table II on pp. 798-801 of the
book.
Binomial Distribution
In the cumulative tables you first find the
value of n in which you are interested.
Then you find the value of p, in the top
row of the n table you chose. Then the
vertical k values give the cumulative
binomial probability from x = 0 up to k.
Binomial Distribution
Example 4.6. Looking at n = 6, p = .7, and
k = 3 we find the entry .256. What this
means is that for the b(n, p) = b(6, .7)
distribution, p(0) + p(1) + p(2) = p(3) =
.256.
Binomial Distribution
One can also use the cumulative table to
find the probability of a single value of x.
Binomial Distribution
Example 4.7. In Example 4.5 we used the
binomial formula to calculate p(4) for
b(n, p) = (6, .3) and found it to be  .06.
But in this case, p(4) = [p(0) + p(1) + p(2)
+ p(3) + p(4)] – [p(0) + p(1) + p(2) + p(3)]
=, from the cumulative table, .989 - .930 =
.059  .06.
Binomial Distribution
To translate word problems into questions
about the binomial distribution and then to
use the cumulative binomial table to
answer the questions is an art which
requires practice.
Binomial Distribution
Example 4.8. Experience has shown that 30%
of the rocket launchings at a NASA base have
to be delayed due to weather conditions. Use
Table II to determine the probabilities that
among ten rocket launchings at that base
(a) At most three will have to be delayed due to
weather conditions
(b) At least six will have to be delayed due to
weather conditions.
Binomial Distribution
Solution. p = .3 n = 10
(a) “At most three” means 0, 1, 2, or 3.
From the Table II with n = 10 and p = .3
we find that p(0) + p(1) + p(2) + p(3) =
.650
Binomial Distribution
(b) “At least six” means 6, 7, 8, 9, or 10.
From the table with n = 10 and p = .3 we
seek p(6) + p(7) + p(8) + p(9) + p(10) = 1
– [p(0) + p(1) + p(2) + p(3) + p(4) = p(5)]
= 1 - .953 = .047.
Binomial Distribution
In the case of a b(n, p) distribution a
theorem gives us some special results:
 =  x p(x) = np
2 = (x - )2p(x) = npq
 = the square root of npq
Binomial Distribution
Example 4.9. If 80% of certain videocasette
recorders will function successfully through the
90-day warranty period, find the mean and
standard deviation of the number of these
videocasette recorders, among 10 randomly
selected, which will function successfully
through the 90-day warranty period, using:
Binomial Distribution
(a) Table II, the formula that defines ,
and the formula that defines 2.
Solution. p = .8 and n = 10
 =  x p(x) = (0)p(0) + (1)p(1) + (2)p(2)
+ (3)p(3) + (4)p(4) + (5)p(5) + (6)(p(6) +
(7)p(7) + (8)p(8) + (9)p(9) + (10)p(10) =, by
Table II, (0)(0) + (1)(0) + (2)(0) + (3)(.001) +
(4)(.005) + (5)(.027) + (6)(.088) + (7)(.201) +
(8)(.302) + (9)(.269) + (10)(.107) = 8
Binomial Distribution
2 = (x - )2p(x) =, by Table II,
(-8)2(0) + (-7)2(0)+ (-6)2(0) + (-5)2(.001) +
(-4)2(.005) + (-3)2(.027) + (-2)2(.088) +
(-1)2(.201) + (0)2(.302) + (1)2(.269) +
(2)2(.107) = 1.598 so  = the square root
of 1.598 = 1.264
Binomial Distribution
(b) The special formulas for the mean and the
standard deviation of the binomial distribution.
Solution. p = .8 and n = 10
 = np = 10 . .8 = 8
2 = npq = 10 . .8 x .2 = 1.6
 = the square root of npq = the square
root of 1.6 = 1.265