Download ENGR 323 Beautiful Homework #4 1/5 Coleman Problem 3-11 3

ENGR 323
Coleman
Beautiful Homework #4
Problem 3-11
1/5
3-11. The sample space of a random experiment is {a, b, c, d, e, f }, and each outcome
is equally likely. A random variable is defined as follows:
outcome
x
a
0
b
0
c
1.5
d
1.5
e
2
f
3
Determine the probability mass function of X.
This question asks us to determine the probability mass function of X. In order to
do this, we must first determine the range of the random variable X. A definition of
random variable is given on page 100 of our text. In our problem the possible values of
X are x=0, x=1.5, x=2, or x=3.
A probability mass function is the function f(x) = P(X=x) from the set of possible
values of the discrete random variable X to the interval [0, 1]. In this function, P(X=x) is
the probability that the discrete random variable X satisfies the event x. This definition,
along with the properties of a probability mass function, can be found in our text on page
105.
Choosing the first value of x to equal 0, the probability mass function looks like
this:
f(0) = P(X=0)
The right side of this equation is read as the probability of finding a value x=0 in
the discrete sample space. The probability of finding any one outcome is 1/6 since there
are six different outcomes.
Since x=0 twice in our discrete sample space,
f(0) = P(X=0) = 1/6 + 1/6 = 1/3
The probabilities of finding the other values of the outcomes are found using the
same and the calculations are shown below.
f(1.5) = P(X=1.5) = 1/6 +1/6 = 1/3
f(2) = P(X=2) = 1/6
f(3) = P(X=3) = 1/6
Figure 1 shows the probability mass function for this problem.
f(x)
2/5
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.5
1
1.5
2
x
Figure 1 Probability mass function for problem 3-11
2.5
3
ENGR 323
Coleman
Beautiful Homework #4
Problem 3-12
3/5
3-12. Continuation of Exercise 3-11. Determine the following probabilities.
a.
b.
c.
d.
e.
P(X=1.5)
P(0.5<X<2.7)
P(X>3)
P(0≤X<2)
P(X=0 or X=2)
a. P(X=1.5) = 1/3
This probability was found in problem 11.
b. P(0.5<X<2.7)
The event {0.5<X<2.7} is the union of the two events {X=1.5} and {X=2}. Since
theses two events are mutually exclusive:
P(0.5<X<2.7) = P(X=1.5) + P(X=2)
= 1/3 + 1/6 = 1/2
(See Equation 2-2 on page 72 of our text to see why you can do this)
c. P(X>3)
The event {X>3} does not occur because no outcome is greater than 3. Because of
this,
P(X>3) = 0
d. P(0≤X<2)
The event {0≤X<2} is the union of the two events {X=0} and {X=1.5}. Since these
two events are mutually exclusive:
P(0≤X<2) = P(X=0) + P(X=1.5)
= 1/3 + 1/3 = 2/3
e. P(X=0 or X=2)
The event { X=0 or X=2} is the union of the two events {X=0} and {X=2}. Since
these two events are mutually exclusive:
P(X=0 or X=2) = P(X=0) + P(X=2)
= 1/3 + 1/6 = 1/2
ENGR 323
Coleman
Beautiful Homework #4
Problem 3-25
4/5
3-25. Continuation of Exercise 3-11. Determine the cumulative distribution function of
the random variable in Exercise 3-11.
This question asks us to determine the cumulative distribution function. The
definition of a cumulative distribution function can be found on page 109 of our text and
it reads,
The cumulative distribution function of a random discrete variable X, denoted as F(x),
is
F(x) = P(X≤x) = ∑ f(xI )
For a discrete random variable X, F(x) satisfies the following properties.
(1) F(x) = P(X≤x) = ∑ f(xI )
(2) 0 ≤F(x) ≤1
(3) If x ≤y, then F(x) ≤F(y)
Equation (3-2)
Example 3-12 in our book shows the proper format for a cumulative distribution
function and is very helpful for this problem. In the example you are given a cumulative
distribution function and asked to determine a probability mass function. In problem 25
we have the opposite, we already know the probability mass function and are asked to
determine the cumulative distribution function.
F(0) = f(0) = 1/3
F(1.5) = f(0) + f(1.5) = 1/3 + 1/3 = 2/3
F(2) = f(0) + f(1.5) + f(2) = 1/3 +1/3 + 1/6 = 5/6
F(3) = f(0) + f(1.5) + f(2) + f(3) = 1
{ 0,
{ 1/3,
F(x) = { 2/3,
{ 5/6,
{ 1,
x<0
0 ≤x < 1.5
1.5 ≤x < 2
2 ≤x < 3
3 ≤x
Figure 2 shows the cumulative distribution function for this problem.
5/5
1.2
1
F(x)
0.8
0.6
0.4
0.2
0
-1
0
1
2
3
x
Figure 2 Cumulative distribution function for problem 3-25
4