Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #22
Fall 2000
Page 1
The weekly demand for propane gas (in 1000s of gallons) from a particular facility is a
random variable with a probability density function as follows.
2[1 – (1/x2 )]
1≤x≤2
f(x) =
0
Otherwise
A ) Compute the cumulative distribution function of X.
B ) Obtain the expression for the (100p) th percentile. What is the value of µ?
C ) Compute E(X) and V(X).
D ) If 1.5 thousand gallons is in stock at the beginning of the week an no new supply is
due in during the week, how much of the 1.5 thousand gallons is expected to be left at the
end of the week?
CUMULATIVE DISTRIBUTION FUNCION
F(x) for a continuous random variable X is defined for every number X as follows
(Devore, p145).
∞
F(x) = P(X ≤ x) = ∫ f(y) dy
-∞
Figure 1 is a graphical representation of the of the CDF
Figure 1. Graphical representation of the PDF and CDF (Devore, p145).
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #22
Fall 2000
Page 2
For a continuous random variable X with pdf f(x) and cdf F(x). The following applies for
for any two numbers a and b with a < b (Devore, p 146).
P(a ≤ X ≤ b) = F(b)- F(a)
If X is a continuous random variable with pdf f(x) and cdf F(x), then at every x at which
the derivative F`(x) exists, F`(x) = f(x) (Devore, p148).
Let P be a number between 0 and 1. The (100p)th percentile of the distribution of a
continuous random variable X, denoted by η(p), is defined as follows (Devore, p148).
η(p)
P = F(η(p)) = ∫ f(y) dy
-∞
The median of a continuous random variable, denoted by α, is the 50th percentile, so α
satisfies .5 = F (α). This means that half of the area under the density curve is to the left
of α and half is to the right of α (Devore, p149).
The expected value or mean of a continuous random variable X with pdf f(x) is defined
as follows (Devore, p150).
∞
µ = E(X) = ∫ x • f(x) dx
-∞
If the value of X happens to be some function of X the following applies (Devore, p150)
∞
E(h(x)) = ∫ h(x) • f(x) dx
-∞
The variance of a continuous random variable X (denoted by σ) with pdf f(x) and mean
value µ is defined as follows (Devore, p 151).
∞
σ = V(X) = ∫ (x - µ) • f(x) dx = E [(X - µ) ]
-∞
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #22
Fall 2000
Page 3
The standard deviation of X is the square root of σ, as shown below.
σ = [V(X)]1/2
Figure 2 is a graphical representation of the PDF.
2
f(x)
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
x= weekly demand for propane gas (1000's
gallons)
Figure 2. PDF for problem 22
A ) Compute the CDF of X
∞
x
F(x) = P(X ≤ x) = ∫ f(y) dy
x
F(x) = ∫ 2[1- (1/x )]
2
-∞
1
1
F(x) = 2x + (2/x) - 4
Below is a mathematical representation of the CDF
F(x) =
0
2x + (2/x) – 4
0
F(x) = [2x + (2/x)]
x<1
1≤x≤2
x>2
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #22
Fall 2000
Page 4
Figure 3 is a graphical representation of the CDF
1.2
1
F(x)
0.8
0.6
0.4
0.2
0
-0.2 0
0.5
1
1.5
2
2.5
x= weekly demand for propane gas (in
1000's of gallons)
Figure 3. CDF for problem 22
B ) Obtain the expression for the (100p)th percentile. What is the value of the median?
η(p)
p = F(η(p)) = ∫ f(y) dy
η
p = ∫ [2 – (2/x^2)] dx
-∞
η
p = [2x + (2/x)]
1
1
2η2 -η(4+p) +2
p = 2η + (2/η) – 4
η= .25[4+p+(p2 +8p)1/2 ]
The median is the 50 percentile, so plug in p=.5
Median = η(.5) = .25[4 + .5 + (.52 + 8(.5))1/2 = 1.64 thousand gallons
C ) Compute E(X) and V(X)
∞
µ = E(X) = ∫ x • f(x) dx
-∞
2
µ = ∫ x [2 – (2/x )] dx
2
µ = x - 2Ln (x)
2
2
1
µ = 4 – 2Ln(2) – 1 + 2Ln(1) = 1.614 thousand gallons
1
Scott N Gillespie
Engineering 323
Beautiful Homework #8
Chapter 4 Problem #22
Fall 2000
Page 5
∞
σ2 = V(X) = ∫ (x - µ) • f(x) dx
(or V(X) = E (x2 ) – [E(x)]2 )
-∞
∞
E(X^2) = ∫ x2 [2- (2/x2 )] dx
-∞
2
∫ [2x2 – (2)] dx
2
[(2/3)x3 –2x]
1
1
E(X^2) = 5.33 - 4 – (2/3) + 2 = 2.66 thousand gallons
V(X) = 2.66 – (1.614)2 = .055 (thousands of gallons)2
D ) If 1.5 thousand gallons is in stock at the beginning of the week an no new supply is
due in during the week, how much of the 1.5 thousand gallons is expected to be left at the
end of the week?
The weekly expected value for propane is 1.614. the propane place has 1.5 thousand
gallons in stock so the remaining amount is as follows.
1.5 – E(X) = 1.5 – 1.614 = -.114 thousand gallons
This value is negative so they can expect to have no propane left by the end of the week