Transcript Chapter 4 - Pegasus @ UCF

Chapter 4 Discrete Random Variables
4.1
4.2
4.3
4.4
4.5
4.6
Discrete and Continuous
Probability Distribution
Expectation and Variance
Binominal
Poisson
Hypergeometric
Homework: 3,5,7,9,11,13,15,21,23,27,31,
43,52,53,61,63,68,69,77
1
Last chapter we discussed several useful
concepts of dealing with probability problems.
However, it is very difficult to write down sample
spaces of some random experiments. In these
cases the concept random variable is very useful.
A random variable is a variable that assumes
values associated with the random outcomes of a
random experiment, where one and only one
numerical values is assigned to each sample point.
It is random because we can not predict the
outcome of a random experiment. It is a variable
because there are more than one possible sample
points in a random experiment.
2
<Example 4.1>: (Basic)
One hundred fair coins are tossed and the up
faces are observed.
(a) Is it convenient to write down the sample space
for this random experiment?
(b) Do we need to write down the sample space if we
are interested in counting the number of heads in
this random experiment?
<Solutions>:
(a) No, it takes vary long time to write the sample space of
this random experiment.
(b) No, we can use the concept of random variable to solve
our problem.
3
Section 4.1: Discrete and Continuous Random
Variable
Some random variable can assume values on
countable many numbers (such as integers) and
some random variable can assume values on one
or more intervals. For example, the distance
between you home and UCF is between 0 and 100
miles that is an interval, i.e. the distance between
your home and UCF is a continuous random
variable. But the number of head in coin tossing
experiment is a countable number, i.e. the number
of heads in a coin tossing experiment is a discrete
random variable.
4
<Example 4.2> (Basic)
List five discrete random variables and five
continuous random variables.
5
Sec 4.2: Probability Distributions for Discrete
Random Variables
This chapter will focus on the discussion of discrete
random variable. A complete description of a discrete
random variable requires that we specify all the possible
values the random variable can assume and the probability
associated with each value. Usually, we can use the
following four steps to complete a probability table.
Step 1: Find out the variable of interest.
Step 2: List all the sample points in the sample space.
Step 3: List all the possible values of this random variable.
Step 4: Assign the probabilities to all the possible values.
6
<Example 4.3>: (Basic)
A company has five applicants for two
positions: three from UCF and two from UF.
Suppose that the five applicants are equally
qualified and no preference is given for choosing
either school. Let x be the number of UCF
graduates chosen to fill the two positions.
(a) What is the random variable of interest?
(b) Write down the sample space.
(c) Write the probability table.
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The probability distribution of a discrete random
variable is a graph, a table, or a formula that specifies the
probability associated with each possible value the random
variable can assume. The probability distribution should
not include values that have zero probabilities. The rules
to assign probability discussed in Section 3.2 should be
followed as well. Thus, the probability of any value of a
random variable is between 0 and 1 and the sum of the
probabilities of all possible values of a random variable is
equal to one.
8
<Example 4.4>: (Basic)
A random variable has the following
probability table:
x
0
1
2
3
4
p(x)
0.1 0.2 0.3 ?
0.15
(a) Find P(x=3).
(b) Is x a continuous random variable?
9
<Example 4.5>: (Advance)
A lady claims that she can taste the difference
between PEPSI and COKE.
Therefore, we
conduct an experiment to confirm her claim. Four
cups of cola that some are COKE colas and some
are PEPSI colas are displayed in front of her.
After tasting these colas, she needs to identify the
contents in each cups. We are interested in the
correct decisions made by her in this experiment.
(a) What is the random variable of interest?
(b) Write down the sample space.
(c) Write the probability table.
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Sec 4.3: Expectation and Variance of a Discrete
Random Variable
We discussed how to obtain the sample mean, the
sample variance, and the sample standard deviation in
chapter 2. Now, we introduce the formulas of getting the
population mean, the population variance, and the
population standard deviation of a discrete random
variable. Suppose X is a discrete random variable with
probability distribution p(x). The expectation of X is the
population mean of X. Let m, s ,2 and s be the population
mean, the population variance, and the population standard
deviation of X, respectively. Then
m = E(x) = S xp(x),
2
2
2
2
s = E[(x-m) ] = S (x-m) p(x), and s = s
11
<Example 4.6>: (Basic)
Consider the probability table of random
variable x below.
x
p(x)
1
0.1
3
0.2
5
0.3
6
0.3
10
0.1
12
(a) Find the expectation of this random variable.
<Solution>:
x
p(x) xp(x)
1
0.1 0.1
3
0.2 0.6
5
0.3 1.5
6
0.3 1.8
10
0.1 1.0
mean = m = Sxp(x) = 0.1 + 0.6 + 1.5 + 1.8 + 1.0 =
5
13
(b) Find the standard deviation of this random
variable.
<Solution>:
x-m (x-m)2
(x-m)2p(x)
-4
16
1.6
-2
4
0.8
0
0
0
1
1
0.3
5
25
2.5
the variance = s 2 =S ( x  m ) 2
p(x) =
1.6+0.8+0+0.3+2.5=5.2 and the standard deviation
= s = 2.28.
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(c) What is the probability that x falls within the
interval (m2s, m+2s)?
(d) Does the result satisfy the Chebyshev’s Rule?
(e) Does the result satisfy the Empirical Rule?
Explain.
<Solutions>:
(c) m2s = 5  2*2.28 = 0.44
m+2s = 5 + 2*2.28 = 9.56
Thus, the probability that x falls within the interval (m2s,
m+2s) is 0.9 (0.9=0.1+0.2+0.3+0.3).
(d) Yes.
(e) No, because the random variable x does not have a
mound-shape distribution.
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<Example 4.7>: (Basic)
You need to pay one dollar to buy an instant lottery
ticket. In this instant lottery game, you have 10%
chance to win a one dollar bill and 5% chance to
win a five dollar bill. You are interested in the
money which you can win in a single play.
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(a) Write down the probability table.
<Solution>:
x p(x)
1 0.1
5 0.05
0 0.85
Note: The lottery official does (not?) want you
to know that you have 85% chance to “win”
nothing.
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(b) Find the mean.
<Solution>:
x p(x) xp(x)
1 0.1 0.1
5 0.05 0.25
0 0.85 0
mean = m = Sxp(x) = 0.1+0.25+0 = 0.35
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(c) Find the standard deviation of the game.
<Solution>:
x-m
(x-m)2
(x-m)2p(x)
0.65
0.4225
0.04225
4.65
21.6225
1.081125
-0.35
0.1225
0.104125
the variance = s**2 = S (xm)**2 p(x) =
0.04225+1.081125+0.104125 = 1.2275 and the
standard deviation = s = 1.10793.
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(d) Do you believe the lottery officials claim ``the
more you play the more you win’’?
<Solution>:
Clearly, I don’t believe it because lottery
revenue is another form of taxes for the lottery
players.
20
<Example 4.8>: (Basic)
A study selected a sample of fifth grade pupils
and recorded how many years of school they
eventually completed. Let X be the highest year
of school that a randomly selected fifth grader
completes. (Students who go on to college are
included in the outcome of x=12.) The probability
is as follows:
x 4 5 6
7
8
9 10 11 12
p(x) .01 .007 .007 ? .032 .068 .070 .041 .752
(a) Find P(X = 7).
<Solution>: P(X=7)=1-(0.01+0.007+0.013+0.032+
0.068+0.070+0.041+0.752) = 0.007.
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(b) Find the mean and standard deviation.
<Solutions>:
x
p(x)
x * p(x)
4
5
6
7
8
9
10
11
12
0.01
0.007
0.007
0.013
0.032
0.068
0.070
0.041
0.752
0.04
0.035
0.042
0.091
0.256
0.612
0.700
0.451
9.024
(x-m)2
52.577001
39.075001
27.573001
18.071001
10.569001
5.067001
1.565001
0.063001
0.561001
11.251
(x-m)2p(x)
0.52577001
0.273525007
0.193011007
0.234923013
0.338208032
0.344556068
0.10955007
0.002583041
0.421872752
2.44400
Thus, m=11.251, s2 = 2.44400, and s = 1.563.
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(c) Find P(x >= 9).
(d) Can you apply the Empirical rule to find the
probability of X falls into the interval (m2s,
m+2s)?
<Solutions>:
(c) P(X 9) = .068+.070+.041+.762 = 0.931
(d) m2s = 11.251  2* 1.563 8.124
m+2s = 11.251 + 2 * 1.563 14.378
Thus, the probability that x falls within the interval
(m2s, m+2s) is 0. 934. Although the probability is close
to 0.95, we can only apply the chebyshev’s Rule because
the empirical distribution of this random variable is not
mound-shape.
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Sec 4.4: The Binomial Random Variable
The responses of many experiments have only
two alternatives such as "Yes or No”, "True or
False", "Male or Female”, and "Failure or
Success". These types of experiments have some
characteristic in common. First, they consist of n
identical and independent trials. Second, there
are only two possible outcomes, denoted by S and
F on each trail. Third, the possibility of each
outcome remains unchanged from trial to trial, that
is, the probability of S is p and probability of F
is q=(1-p) in each trial.
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Fourth, we are interested in the random variable x
represented the number of S happened in n trails
(n is a fixed number). Therefore, it is worth to
develop a special probability model to deal with
this kind of random variables. Any random
variable that has these four characteristics is called
binomial random variable and can be dealt with by
using this special probability model.
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<Example 4.9>: (Basic) List several random
variables that have only two possible outcomes.
<Solutions>:
Gender of a student in STA 3023;
Win or Loss in a football game;
Pass or Fail in an exam;
Hit or Miss in a state lottery drawing;
True or False to answer a question;
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<Example 4.10>: (Basic)
For each of the
following situations, indicate whether a binomial
distribution is a reasonable probability model for the
random variable X.
(a) A couple decides to continue to have children until their
first girl is born; X is the total number of children the
couple has.
(b) Fifty students are taught about binomial probabilities by a
television program. After completing their study, all
students take the same examination; X is the number of
students who passed this exam.
(c) A chemist repeats a solubility test 10 times on the same
substance. Each test is conducted at a temperature 10
degrees higher than the previous test.
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Suppose that X is a binomial random variable.
The probability of success on any single trial is p
and there are n trials in this random experiment.
The probability density function of X is
 n x ( n  x)
P ( X = x) = p( x) =   p q
x = 0,1,2,, n
 x
where
p = the probability of success on any single trial
n = total number of trials
q=1-p
x = number of successes in n trials.
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Let m and s be the mean and standard
deviation of the binomial random variable X. In
stead of using the expectation summation rules to
calculate m and s, we can find m and s easily
using the formulas
m = np,
s**2 = npq = np(1-p), and
s = npq = np(1  p).
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<Example 4.11>: (Basic) To test the side effect of a
newly developed medicine, we conduct an animal
experiment. Five dogs are given this drug and
each dog has 20% chance to develop certain
symptoms. We are interested in the number of
dogs that develop this symptom.
(a) Is this a binomial random variable?
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(b) Write down the probability table of this random
variable.
<solution to part (b)>:
Probability Table:
X
P(X=x)
0
0.32768
1
0.4096
2
0.2048
3
0.0512
4
0.0064
5
0.00032
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(c) Find the mean and standard deviation of this
random variable.
<solution to part (c)>:
m = np = 5 * 0.2 = 1;
s**2 = npq = 5 * 0.2 * 0.8 = 0.8;
s = 0.894.
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<Example 4.12>: (Basic)
A firm receives a shipment of 500 hi-fi
speakers. For any randomly selected sample of 9
speakers, if 2 or more of the speakers are defective
then rejects this shipment. What is the probability
that this firm will accept the shipment if the
proportion of defective is
(a) 0.20.
(b) 0.10.
(c) 0.05.
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<Example 4.13>: (Basic) An oil exploration firm
plans to drill six holes. Due to experience, the
probability of each hole yielding oil is 0.12. Since
the holes are in quite different locations, the
outcome of drilling one hole is statistically
independent of drilling of any other holes.
(a) Give the expectation and standard deviation of
the number of holes that results in oil.
<Solution to part (a)>:
(a) n = 6 and p = 0.12;
m = np = 6 * 0.12 = 0.72;
s = np(1  p) = 6  012
.  (1  012
. ) = 0.796;
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(b) If the firm will be able to stay in business only if
two or more holes produce oil, what is the
probability that it can survive.
<Solution to part (b)>:
P(X  2) = P(X=2) +(X=3)+P(X=4)+P(X=5)+
P(X=6) = 0.129534197 +0.023551672+
0.00240869376+0.000131383296
+0.000002985984 @ 0.156
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Note:
(1) We can not use the Binomial probabilities Table
to obtain this probability because p = 0.12 is not in
the Table.
(2) We can obtain this probability much easier with
the concept of complement event:
P(X  2) = 1 - P(X1)
= 1 - P(X=0) - P(X=1)
= 1 - 0.4644044086 - 0.37996698
@ 0.156
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Section 4.5: The Poisson Random Variable
The random variables produced by many
random experiments can be well described by
using Poisson probability model.
Typical examples are as follows:
(1) the number of customers served per hour in a
given restaurant,
(2) the number of alcohol related traffic accidents per
month at a busy intersection,
(3) the number of diseased trees per acre in a certain
national park,
(4) the number of telephone calls received per
minute during your lunch hour.
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Poisson random variable has the following
common characteristics.
(1). The experiment consists of counting the
number of times that a certain event occurs during
a given unit of time or in a given area or volume.
(2). The probability of an event occurs in a given
unit of time is same for all time units.
(3). The number of events that occur in one unit of
time, area, or volume is independent of the
number that occur in other units.
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The probability density function of a Poisson
random variable is
lxe l
p( x) =
x = 0,1,2,.
x!
Both the mean and the variance of a Poisson
random variable equals to l, i.e. m = l and s2 = l.
39
<Example 4.14>: (Basic)
Suppose x is a Poisson random variable, use Table
III on page 804 to find the following probabilities.
(a) P(x 2) when l = 1.
<Solution to part (a)>:
part of Table III
x =0 1
2
3
4
5
6
l = 1 .368 .736 .920 .981 .996 .999 1.000
Thus, P(X2) = 0.920.
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(b) P(x  2) when m = 2.
<Solution to part (b)>
part of Table III
l=2
x =0 1 2 3
4
5
6
7
8
.135 .406 .677 .857 .956 .987 0.997 0.999 1.000
Since l = m = 2, P(x  2) = 1 - P(x  1) = 1 0.406 = 0.594.
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(c) P(x >3) where s2 = 3.
<Solution to part (c)>:
part of Table III
x=0
l=3
1
2
3
4
5
6
7
8
0.050 0.199 0.423 0.647 0.815 0.916 0.996 0.988 0.996
Since s2 = m = 3, P(x > 3) = 1 - P(x  2) = 1 0.423 = 0.577.
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Note:
For Poisson random variable, we can find P(x  a)
from the table directly if a is an integer. We need
to apply the concept of complement event to find
the probability of P(x > a) or P(x  a). We need to
know that P(x > a) = 1 - P(x  a) and P(x  a ) =
1 - P(x  a-1).
43
<Example 4.15>: (basic) We know that the mean of
a Poisson random variable is equal to 2. Find the
probabilities of x equal to 1, 2, and 3.
<Solutions>:
We can use the probability function to compute
the probability of a Poisson random variable as
well.
lxe l 21e 2
p( x = 1) =
=
@ 0.271;
x!
1!
lxe l 22 e 2
p( x = 2) =
=
@ 0.271;
x!
2!
lxe l 23 e 2
p( x = 3) =
=
@ 0180
. .
x!
3!
44
<Example 4.16>: (Intermediate)
According to the records of an airline, the
number of people who buy tickets but fail to show
up for the early morning flight between Orlando
and Washington D.C. is a Poisson random
variable. We know that the standard deviation of
this random variable is 2.
Determine the
probabilities that the number of no shows in an
early flight
(a) is equal to 5,
(b) is less than or equal to 3,
(c) is greater than or equal to 6.
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<Solutions to EX 4.16>:
l = s2 = 2 * 2 = 4
x l
5 4
l
e
4
e
(a)
p( x = 5) =
=
@ 0156
. .
x!
5!
(b)
p(x  3) = 0.433. (From Table III)
(c )
p(x6) = 1 - p(x5) = 1 - 0.785 = 0.215.
46
Section 4.6 The Hypergeometric Random
Variable
Hypergeometric random variable is another
popular discrete random variable. Suppose there
are a total of N balls: r red balls and (N-r) white
balls, in a bag. And n balls are randomly selected
from this bag without replacement. Let X denote
the number of red balls in the n balls selected.
Then the distribution of X is called a
Hypergeometric distribution,
with parameters N, r and n.
47
The probability density function of this hypogeometric distribution
is given by:
p( x) =
 r
 
 x
nr
m=
N
 N  r


 n  x
 N
 
 n
x = max(0, n  N _ r ), , min( r , n),
2
s =
48
r ( N  r ) n( N  n)
N 2 ( N  1)
.
<Example 4.17>: (Basic) Given that x is a
hypergeometric random variable, compute p(x), m, and
s**2 for each of the following cases:
(a) N=5, n=3, r=3, x=1.
<Solutions>:
 3  5  3
  

 1  3  1 3  1
P( X = 1) = p(1) =
=
= 0.3.
10
 5
 
 3
3 3
m=
= 18
.
5
s2 =
r ( N  r ) n( N  n)
2
N ( N  1)
=
3  (5  3)  3  (5  3)
2
5  (5  1)
49
= 0.36.
(b) N=9, n=5, r=3, x=3.
<Solution to part (b)>:
P( X = 3) = p(3) =
 3
 
 3
 9  3


 5  3 1  15
=
= 0119
. .
126
 9
 
 5
5 3
m=
= 167
.
9
s2 =
r ( N  r ) n( N  n)
2
N ( N  1)
=
3  (9  3)  5  (9  5)
2
9  (9  1)
50
@ 0.556.
Collection of Definitions:
Random Variable
A random variable is a rule that assigns one and only one
numerical value to each sample point in a random
experiment.
Discrete Random Variable
A discrete random variable is a random variable that can
assume only countable number of values.
Continuous Random Variable
A continuous random variable is a random variable that
can assume values in one or more intervals.
51
Probability Distribution
The probability distribution of a discrete random variable
is a way, such as a graph, a table, or a formula, that
specifies the probability associated with each possible
value the random can assume.
Expectation of a Discrete Random Variable
The expectation of a discrete random variable is
the population mean of this random variable. We
can use the following formula to compute the
expectation of a discrete random variable
m = E(x) = S xp(x).
52
Variance of Discrete Random Variable
The variance of a discrete random variable is the
population variance of the random variable, given
by formula
2
2
2
s = E[(x-m) ] = S (x-m) p(x).
Standard Deviation of Discrete Random Variable
The standard deviation of a discrete random
variable is equal to the square root of the variance
of this random variable, i.e. s = s 2 .
53
Binomial Distribution
The probability density function of a binomial random
variable is
 n x ( n  x)
P ( X = x) = p( x) =   p q
.
 x
Here
p = the probability of success on any single trial ;
n = total number of trials;
x = number of successes in n trials;
q = 1 - p;
The mean of a binomial random variable is
np, i.e. m = np;
The variance of a binomial random variable is
npq, i.e. s**2 = npq = np(1-p).
54
Poisson Random Variable
The probability density function of a Poisson
random variable is
lx e  x
p( x) =
x!
x = 0,1,2,.
Both the mean and the variance of a Poisson
random variable equals to l, i.e.
m = l and s**2 = l.
55
Hypergeometric Random Variable
The probability density function of a Hypergeometric
random variable is
p( x) =
 r
 
 x
 N  r


 n  x
 N
 
 n
x = max(0, n  N _ r ), , min( r , n),
where
N = total number of balls in the bag;
r = the number of red balls in the bag;
n = the number of balls drawn without replacement;
x = the number of red balls in the n balls selected.
56
The mean of a Hypergeometric random variable is
nr
m=
N
and the variance of a Hypergeometric random
2
variable is s =
r ( N  r ) n( N  n)
N 2 ( N  1)
57
.