Download Scott N Gillespie Beautiful Homework #8 Fall 2000 Engineering 323

Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #1
Fall 2000
Page 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 that X has a density function
as follows.
.5x
0≤x≤2
0
Otherwise
f(x) =
Calculate the following probabilities:
A) P(X ≤ 1)
B) P(.5 ≤ X ≤ 1.5)
C) P(1.5 < X)
CONTINUOUS RANDOM VARIABLES
A random variable is said to be continuous if its set of possible values is an entire
interval of numbers. That is, for some set A to B, any number x between A and B is
possible.
Definition:
Let X be a continuous random variable. Then X has a Probability Density Function
f(x) for any two number A and B (where A is less that B). This is denoted in the
function below (Devore, p 140).
b
P(a ≤ X ≤ b) = ∫ f(x) dx
a
This is the probability that the random variable X takes on a value within the interval
from a to b. Figure 1 is a representation of this probability.
Figure 1 Graphical representation of
P(a < X <b) (Devore, p140).
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #1
Fall 2000
Page 2
For a continuous random variable the probability that X is equal to a certain single value
is always 0. These functions have to be evaluated over an interval to determine a positive
probability. As shown in Figure1, there has to be an “area” underneath the graph. A thin
vertical line would represent a single value. For example, let’s let a and b both be equal
to five. This statement is valid for all X.
b
∫ f(x) dx = F(b) – F(a) = F(5) – F(5) = 0
a
The graph of the PDF is shown in Figure 2
1.2
1
f(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
X = Hours that a book is checked out
2.5
Figure 2 Probability Density Function
for problem 1.
Expected Value [E(X)]
∝
∝
∝
E(X) = ∫ x f(x) dx = ∫ x .5x dx = ∫ .5 x
2
-∝
-∝
2
2
dx = (.5/3)x
= 4/3 Hours
0
0
Variance [V(X)]
V(X) = E(X2 ) - [E(X)]2
∝
∝
2
E(X2 ) = ∫ x2 f(x) dx = ∫ x2 .5x dx = ∫ .5x3 dx = (.5/4) x4
-∝
-∝
0
V(X) = 2 – (4/3)2 = 2/9 (Hours)2
2
= 2 Hours
0
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #1
Fall 2000
Page 3
Cumulative Distribution Function
X
X
F(x) = P (X ≤ x) = ∫ f(y) dy = ∫ .5y dy = .25y2
-∞
0
X
0
Mathematical Representation:
F(x) =
0
.25 x2
1
x<0
0≤x≤2
x>2
This probability is shown in Figure 3
1.2
1
F(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
2.5
x = hours that a book is checked out
Figure 3 Cumulative Distribution Function
for problem 1.
A ) P(X ≤ 1)
What is the probability that the continuous random variable X is less than or equal to 1?
Let X equal the amount of time that a book on 2 hour reserve is checked out.
B
1
P(a ≤ X ≤ b) = ∫ f(x) dx
P(0 ≤ X ≤ 1) = ∫ .5x dx
A
0
P(0 ≤ X ≤ 1) = .25x
1
2
= .25 Hours
0
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #1
Fall 2000
Page 4
This probability is show in Figure 2
1.2
1
f(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
X = Hours that a book is checked out
2.5
Figure 2 Probability that (0<x<.5).
B ) P(.5 ≤ X ≤ 1.5)
What is the probability that the continuous random variable X is greater than or equal to
.5 and less than or equal to 1.5?
Let X equal the amount of time that a book on 2 hour reserve is checked out.
B
P(a ≤ X ≤ b) = ∫ f(x) dx
1.5
P(.5 ≤ X ≤ 1.5) = ∫ .5x dx
A
.5
P(.5 ≤ X ≤ 1.5) = .25x2
1
= .5 Hours
0
This probability is show in Figure 3
1.2
1
f(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
X = Hours that a book is checked out
Figure 3 Probability that (.5<x<1.5).
2.5
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #1
Fall 2000
Page 5
C ) P(X > 1.5)
What is the probability that the continuous random variable X is greater that 1.5?
Let X equal the amount of time that a book on 2 hour reserve is checked out.
B
2
P(a ≤ X ≤ b) = ∫ f(x) dx
P(1.5 ≤ X ≤ 2) = ∫ .5x dx
A
1.5
P(1.5 ≤ X ≤ 2) =
1
.25x2
= .4375 Hours
0
This probability is shown in Figure 4
1.2
1
f(x)
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
X = Hours that a book is checked out
Figure 4 Probability that (1.5<x<2).
2.5