Transcript Chapter 6 Lesson 6.3 and 6.4

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
number of heads
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
At a particular hospital, let’s record the gender
of the next 5 newborns at this hospital.
Is this a binomial experiment?
Yes, if the
births
were
not multiple
What
is the
probability
of births
(twins, etc).
“success”?
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) =.3125
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) = 0.1875
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 for Bernoulli trials:
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
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
s = npq
Homework
• Pg.344: #7.43 (π=.9 means p=.9)
44, 46, 51, 52
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
cables. Your car has a dead battery and you don’t
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
Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
s=
q
p2
Homework
• Pg.344: #7.56-58