BH #4 Problem 3-31
ENGR 323
HOPPER
Page 1/4
Verify that the following functions are cumulative distribution
functions, and determine the probability mass function and the requested
probabilities.
F (x) = 0
F (x) = 0.25
F (x) = 0.75
F (x) = 1
when x < -0.1
when -0.1< x < 0.3
when 0.3 < x < 0.5
when 0.5 < x
First, the Cumulative Distribution Function (CDF) was graphed from
the given information (see Figure 1). Examples of CDF graphs are on pages
110, and 111 of the text.
Cumulative Distribution Graph
F(x)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
x
Figure 1: The cumulative distribution function showing the probability of
X<x for problem 3-31.
The definition of the cumulative distribution function (see p. 109 of the
text) states that :
F(x)= P(X<x)= ∑ f (x)
x
Or, in words, that the cumulative distribution function is equal to the
probability of the random variable X being less than or equal to some value x.
This probability is also equal to the sum of the probability mass function for
all values less than or equal to the selected value x.
BH #4 Problem 3-31
ENGR 323
HOPPER
Page 2/4
What this means is that the probability mass function is the difference
between say, F(x ), and F(x ) for each successive F(x).
See Table 1 for a summary of the probability mass function
calculations.
1
2
Table 1: The calculation of the probability mass function given the cumulative
distribution function values.
x
F(x)
f(x)
-0.1
0.25
0.25 - 0 = 0.25
0.3
0.75
0.75 - 0.25 = 0.5
0.5
1
1 - 0.75 = 0.25
Figure 2 shows a graph of the probability mass function as calculated in Table
1.
Probability Mass Graph
f(x)
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.1
0
0.1
0.2
0.3
0.4
0.5
x
Figure 2: The probability mass function, graphed from the calculated values
given in table one.
BH #4 Problem 3-31
ENGR 323
HOPPER
Page 3/4
Finally, the probabilities of the given values can be solved for by
referring to Table 1, or to the CDF graph, shown again in Figure 3 for ease of
reference.
Cumulative Distribution Graph
F(x)
1
0.95
0.9
0.85
0.8
0.75
0.7
0.65
0.6
0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
x
Figure 3: CDF graph for problem 3-31 for use in calculating the
probabilities below.
a) P(X< 0.5)= 1
From the definition of the CDF, and as shown in
Figure 3 and Table 1.
b) P(X < 0.4)= 0.75
On Figure 3, follow x=0.4 up to the CDF function
line, and over to find F(0.4)=0.75.
BH #4 Problem 3-31
ENGR 323
HOPPER
Page 4/4
c) P(0.4< X < 0.6)
In lecture we learned that the probability of X greater than a, less than b,
was equal to the CDF at b minus the CDF at a. Or in symbols:
P(a<X <b)=F(b)-F(a)
This translates into the following for part c:
P(0.4< X < 0.6)= F(0.6)-F(0.4)=1 - 0.75 = 0.25
d) P(X< 0)= 0.25
This solution is shown in Figure 3. Follow x=0
up to the CDF function line, and then over to find
F(0)=0.25.
e) P(0 < X < 0.1)
This result can also be found though the application of the equation
used in part c.
P(0 <X <0.1)= F(0.1)-F(0)= 0.25 - 0.25 = 0
f) P(-0.1 < X < 0.1)
Again, the equation learned in lecture applies.
P(-0.1< X<0.1)=F(0.1)-F(-0.1)= 0.25 - 0.25 = 0