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```Chapter 7
Lesson 7.5
Random Variables and Probability
Distributions
7.5: Binomial and Geometric Distributions
Special Distributions
Two Discrete Distributions:
Binomial and Geometric
One Continuous Distribution:
Normal Distributions
Properties of a Binomial
Experiment
1. There are a fixed number of trials
2. Each trial results in one of two mutually
We use n to denote the fixed
exclusive outcomes.
(success/failure)
number
of trials.
3. Outcomes of different trials are independent
4. The probability that a trial results in success is
constant.
The binomial random variable x is defined as
x = the number of successes when a binomial
experiment is performed
Are these binomial distributions?
1) Toss a coin 10 times and count the
Yes
2) Deal 10 cards from a shuffled deck
and count the number of red cards
No, probability does not remain constant
3) The number of tickets sold to
children under 12 at a movie theater
in a one hour period
No, no fixed number
Binomial Probability Formula:
Let
n = number of independent trials in a binomial experiment
p = constant probability that any trial results in a success
Where:
n!
x
n x
P (x ) 
p (1  p )
x ! (n  x )!
n

n
!
Technology,
 n C xsuch
 as calculators and
x
software,
x ! (n  x )!will also
 statistical
perform this calculation.
Let’s record the gender of the next 5 newborns
at Huntington Memorial Hospital and see how
many girls we get.
Is this a binomial experiment?
What is the probability of
Yes, if the births“success”?
were not multiple births
(twins, etc).
Define the random variable of interest.
x = the number of females born out of the next
5 births
What are the possible values of x?
x
0
1
2
3
4
5
Newborns Continued . . .
What is the probability that exactly 2 girls will
be born out of the next 5 births?
P (x  2) 5 C 2  0.5  0.5  .3125
2
3
What is the probability that less than 2 girls will
be born out of the next 5 births?
P (x  2)  p (0)  p (1)
5 C 0 .5 .5 5 C 1 .5 .5
0
 .1875
5
1
4
Newborns Continued . . .
Let’s construct the discrete probability
distribution table for this binomial random
variable:
x
0
1
2
3
p(x)
.03125
.15625
.3125
.3125
4
5
.15625 .03125
is the
multiplying
WhatNotice
is the that
meanthis
number
ofsame
girls as
born
in the
next five births?
n×p
Since
this is a +discrete
mx = 0(.03125)
+ 1(.15625)
2(.3125) +
distribution,
could use:
3(.3125)
+ 4(.15625)we
+ 5(.03125)
=2.5
mx   xp
Formulas for mean and standard deviation
of a binomial distribution
mx  np
x  np 1  p 
Newborns Continued . . .
How many girls would you expect in the next five
births at a particular hospital?
mx  np  5(.5)  2.5
What is the standard deviation of the number
of girls born in the next five births?
x  np (1  p )  5(.5)(.5)
 1.118
The Binomial Model (cont.)
Binomial Probability Model
n = number of trials
p = probability of success
q = 1 – p = probability of failure
X = # of successes in n trials
x
n-x
P(X = x) = nCx p q
m = np
s = npq
Independence



One of the important requirements is that the trials be
independent.
When we don’t have an infinite population and we are
sampling without replacement, the trials are not
independent.
But, there is a rule that allows us to pretend we have
independent trials:
 The 10% condition: If the trials are not independent, it
is still okay to proceed as long as the sample is smaller
than 10% of the population.
Slide 17- 13
Newborns Revisited . . .
Suppose we were not interested in the
number of females born out of the next
five births, but
which birth would result in the first
female being born?
How is this question different from a
binomial distribution?
Properties of Geometric
Distributions:
• There are two possible outcomes: a success or
failure
So what are the
• Each trial is independent of the others
possible values of x
• The probability of success is constant for all
trials.
To infinity
How far will this go?
A geometric random variable x is defined as
x = the number of trials UNTIL the FIRST success
is observed ( including the success).
x
1
2
3
4
. . .
Probability Formula for the
Geometric Distribution
Let
p = constant probability that any trial results in a success
x 1
p (x )  (1  p )
Where
x = 1, 2, 3, …
p
Suppose that 40% of students who drive to
campus at your school or university carry jumper
have jumper cables, so you decide to stop
students as they are headed to the parking lot
and ask them whether they have a pair of jumper
cables.
Let x = the number of students stopped before
finding one with a pair of jumper cables
Is this a geometric distribution?
Yes
Jumper Cables Continued . . .
Let x = the number of students stopped before
finding one with a pair of jumper cables
p = .4
What is the probability that third student
stopped will be the first student to have jumper
cables?
P(x = 3) = (.6)2(.4) = .144
What is the probability that at most three
student are stopped before finding one with
jumper cables? P(x < 3) = P(1) + P(2) + P(3) =
(.6)0(.4) + (.6)1(.4) + (.6)2(.4) = .784
The Geometric Model (cont.)
Geometric probability model for Bernoulli trials:
p = probability of success
q = 1 – p = probability of failure
X = # of trials until the first success occurs
x-1
P(X = x) = q p
1
m=
p