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Engineering 323
Morelli
Beautiful Homework Set 7
Problem 4.14
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4.14 The article “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants”
(Water Research, 1984, pp. 1169-1174) suggests the uniform distribution on the interval (7.5,
20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region.
a.
b.
c.
d.
What are the mean and variance of depth?
What is the cdf of depth?
What is the probability that observed depth is at most 10? Between 10 and 15?
What is the probability that observed depth is within 1 SD of the mean value? Within 2
SDs?
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)
“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.
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P( A ≤ X ≤ B)
f(x)
1/B-A
0
A
B
x
Figure 1. Graph of pdf of general 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
η ( p)
p = F (η ( p)) =
−∞
f ( y )dy
“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
−∞
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“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
Problem 4.14
First, lets define our random variable:
X = depth (cm) of the bioturbation layer in sediment in a certain area
The problem states that we have uniform distribution so,
X~Uniform(A = 7.5, B = 20)
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We can construct our pdf using the definition of uniform distribution given above:
ì 1
f ( x; A, B ) = í B − A
î 0
A≤ x≤ B
otherwise
So,
ì 1
f ( x) = í 20 − 7.5
î 0
7.5 ≤ x ≤ 20
otherwise
We will look at the graphs of f(x) and F(x) later in the solution.
Now lets solve the problem:
a. What are the mean and variance of depth?
µ x = E( X ) =
∞
−∞
uf (u )du =
7.5
−∞
u 0du +
20
7.5
u
1
du +
20 − 7.5
∞
20
u 0du = 13.75
From the last proposition above:
V ( X ) = E ( X 2 ) − [ E ( X )] 2
E( X 2 ) =
∞
−∞
u 2 f (u )du =
20
7.5
u2
1
du = 202.08
20 − 7.5
σ 2 = V ( X ) = 202.08 − (13.75) 2 = 202.08 − 189.06 = 13.02
σ = 3.61
See Figure 4. below for graph of mean.
b. What is the cdf of depth?
We have all ready found our pdf, now the cdf is a easy to compute:
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x < 7.5
ì
F ( x) = P( X ≤ x) =
x
0
1
x − 7.5
f (u )du = 0du +
du = í
−∞
7.5 20 − 7.5
20 − 7.5
î 1
7.5
−∞
x
7.5 ≤ x ≤ 20
x > 20
1.2
F(x)
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
x
Figure 5. CDF for problem 4.14
c. What is the probability that observed depth is at most 10? Between 10 and 15?
Now we can use the cdf to avoid having to integrate:
P( X ≤ 10) =
10
7.5
1
10 − 7.5
du = F (10) =
= .2
20 − 7.5
20 − 7.5
P (10 ≤ X ≤ 15) =
1
15 − 7.5
du = F (15) − F (10) =
− .2 = .4
10 20 − 7.5
20 − 7.5
15
d. What is the probability that observed depth is within 1SD of the mean value? Within 2SDs?
For 1 SD:
µ ± σ = (10.14,17.36)
P( µ − σ ≤ X ≤ µ + σ ) = F (17.36) − F (10.14) = .5776
For 2 SDs:
µ ± 2σ = (6.53 ≤ X ≤ 20.97) = 1
Note that this interval spans beyond our interval and therefore the probability must be 1.
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Lets look at a graph to finish up:
P(7.5 ≤ X ≤ 20)
P(10.14 ≤ X ≤ 17.36)
1/12.5
f(x)
0
5
10
15
µx
Figure 4. PDF for problem 4.14
20
25
x