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Engineering 323
Morelli
Beautiful Homework Set 7
Problem 4.11
1 of 7
4.11 The cdf of checkout duration X as described in Exercise 1 is
ì0
x2
F( x) = í
4
1
î
x<0
0≤x≤2
2≤x
4.1 Let X denote the amount of time for which a book on 2-hour reserve at a
college library is checked out by a randomly selected student and suppose X
has density function
ì.5 x
f ( x) = í
0
0≤x≤2
otherwise
Before we begin, let’s explore some useful definitions, propositions, and their equations
that will help us solve these problems. We will be mainly using the highlighted
definitions and equations, but the others will help to understand the concepts.
DEFINITIONS
“continuous random variable”- An rv X is said to be continuous if its set of possible
values are an entire interval of numbers—that is, if for some A < B, any number x
between A and B is possible. (p.138)
“probability density function”- Let X be a continuous rv. Then a probability
distribution or probability density function (pdf) of X is a function f(x) such that for
any two numbers a and b with a ≤ b,
P(a ≤ X ≤ b) =
b
a
f ( x)dx
That is, the probability that X takes on a value in the interval [a, b] is the area under the
graph of the density function, as illustrated in Figure 4.2, of Devore (p140). The graph of
f(x) is often called the density curve.(p.140)
Engineering 323
Morelli
2 of 7
Beautiful Homework Set 7
Problem 4.11
“uniform distribution”- A continuous rv X is said to have uniform distribution
(p.141) on the interval [A, B] if the pdf of X is
ì 1
f ( x; A, B) = í B − A
î 0
A≤ x≤ B
otherwise
Lets look at the graph to gain a greater understanding of this concept.
P( A ≤ X ≤ B)
f(x)
1/B-A
A
0
B
x
Figure 1. General pdf for uniform distribution
“cumulative distribution function”- The cumulative distribution function F(x) for a
continuous rv X is defined for every number x by
F ( x) = P ( X ≤ x) =
x
−∞
f ( y )dy
For each x, F(x) is the area under the density curve to the left of x. This is illustrated in
Figure 4.5, of Devore (p.145), where F(x) increases smoothly as x increases.
“(100p)th percentile”- Let p be a number between 0 and 1. The (100p)th percentile
(p.148) of the distribution of a continuous rv X, denoted by η(p), is defined by
p = F (η ( p)) =
η ( p)
−∞
f ( y )dy
Engineering 323
Morelli
Beautiful Homework Set 7
Problem 4.11
3 of 7
“median”- The median of a continuous distribution, denoted by µ (but with a squigly
over it), is the 50th percentile, so the median satisfies .5 = F(µ). That is, half of the area
under the density curve is to the left of the median and half is to the right.(p.149)
“expected value”- The expected or mean value (p.150) of a continuous rv X with pdf
f(x) is
µ x = E( X ) =
∞
x ⋅ f ( x)dx
−∞
“variance and standard deviation”- The variance (p.151) of a continuous rv X with
pdf f(x) and mean value µ is
σ x2 = V ( X ) =
∞
−∞
( x − µ ) 2 ⋅ f ( x)dx = E[( X − µ ) 2 ]
The standard deviation (SD) (p.151) of X is
σ x = V (X )
PROPOSITIONS
If X is a continuous rv, then for any number c, P(X = c) = 0. Furthermore, for any two
numbers a and b with a<b, (p.141)
P ( a ≤ X ≤ b) = P ( a < X ≤ b ) = P ( a ≤ X < b) = P ( a < X < b )
Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any two numbers a, b with
a<b, (p.146)
P (a ≤ X ≤ b) = F (b) − F (a )
If X is a continuous rv with pdf f(x) and cdf F(x), then at every x at which the derivative
F′(x) exists, F′(x) = f(x).(p.148)
If X is a continuous rv with pdf f(x) and h(X) is any function of X, (p.150) then
E[h( X )] = µ h ( x ) =
∞
−∞
h( x) ⋅ f ( x)dx
For variance (p.151) the following proposition states
V ( X ) = E ( X 2 ) − [ E ( X )] 2
Engineering 323
Morelli
4 of 7
Beautiful Homework Set 7
Problem 4.11
Problem 4.11
Use the pdf and cdf given below to compute the following:
x<0
ì0
x2
F ( x) = í
4
1
î
0≤ x≤2
ì.5 x
f ( x) = í
0
0≤ x≤2
otherwise
2≤ x
First, lets define our random variable:
X = the amount of time (hours) for which a book an 2-hour reserve is checked out by a
randomly selected student
Now lets graph our cdf:
1.2
F(x)
1
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
x
Figure 2. CDF of problem 4.11
a. P(X≤1)
Engineering 323
Morelli
5 of 7
Beautiful Homework Set 7
Problem 4.11
12
P( X ≤ 1) =
f ( y )dy = F (1) = = .25
−∞
4
So, we can see that 1 falls in the interval 0≤x≤2 and we used the equation x2/4
accordingly.
1
1.2
f(x)
1
P ( X ≤ 1)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
x
Figure 3. PDF for problem 4.11
b. P(.5 ≤ X ≤ 1)
P(.5 ≤ X ≤ 1) =
1
.5
12 .5 2 .25
f (u )du = F (1) − F (.5) = −
=
= .1875
4
4
4
1.2
f(x)
1
P (.5 ≤ X ≤ 1)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
x
Figure 4. PDF for problem 4.11
Engineering 323
Morelli
6 of 7
Beautiful Homework Set 7
Problem 4.11
c. P(X > .5)
P( X > .5) =
∞
.5
f (u )du = 1 − P( X ≤ .5) = 1 −
.5
f (u )du = 1 − F (.5) = 1 −
−∞
.25
= .9375
4
1.2
f(x)
1
P( X > .5)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
µ~
2.5
x
Figure 5. PDF of problem 4.11
d. The median checkout duration [solve .5=F(median)]
µ2
P( X ≤ µ~ ) = .5 = F ( µ~ ) =
4
µ~ 2 = .5(4)
µ~ = 1.414
Notice we used the definition of the median of a continuous distribution and substituted µ
for x. See Figure 5. for graphical interpretation.
e. F′(x) to obtain the density function f(x)
ì0
2x
F ′( x) = í
4
0
î
x<0
ìx
0 ≤ x ≤ 2 = f ( x) = í 2
î0
2≤ x
0≤x≤2
otherwise
Engineering 323
Morelli
Beautiful Homework Set 7
Problem 4.11
7 of 7
We took the derivatives of each of the cdf intervals corresponding functions to find the
pdf.