Environmental Engineering Reference
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the concentrations are measured in mg/l. (a) Estimate
the mean, standard deviation, and skewness of the mea-
sured concentration data; (b) What is the probability of
the concentration exceeding 30 mg/l?; and (c) How
would the estimated exceedance probability in Part (b)
be affected if the measured data were assumed to have
a normal distribution?
c
−
µ
σ
30 20 4
6 28
−
.
x
z
=
=
=
1 53
.
.
x
1
2
[
B
=
1 0.196854(1.53) 0.115194(1.53)
+
+
2
]
=
−
+
0.000344(1.53)
3
+
0.01
9527(1.53)
4
0 063
.
which indicates that the the probability of the
sample concentrations exceeding 30 mg/l is 0.063
or 6.3%. This result is not dramatically different
from that obtained by representing the samples as
being drawn from a log-normal distribution, but the
difference in these exceedance estimates would
increase as the skewness of the data increases.
Solution
From the given data:
μ
y
= 2.97 and
σ
y
= 0.301.
(a) using Equations (10.31-10.34) yields
σ
2
2
=
0 301
2
.
=
Many water-quality variables exhibit a marked skew-
ness, largely because they cannot be negative. Whereas
the normal distribution allows the random variable to
range without limit from negative infinity to positive
infinity, the log-normal distribution has a lower limit of
zero and an unlimited upper bound, which is more con-
sistent with water-quality data. The log-normal distribu-
tion has been found to describe the occurrence of
bacteria (coliforms) in public water distribution systems
(e.g., Christian and Pipes, 1983) and bacteria in streams
(e.g., Chin, 2009). A limitation of the log-normal distri-
bution is that once the mean and variance are specified,
the value of the skewness is fixed, as evidenced by Equa-
tions (10.33) and (10.34). Consequently, if the skewness
of the observed data does not match that of the log-
normal distribution (based on the mean and variance),
then an alternative population distribution should be
sought.
y
µ
=
exp
µ
+
exp
2 97
.
+
20 4 mg/L
.
x
y
2
σ
=
µ
2
[exp(
σ
2
)
−
1
]
=
20 4
.
2
[exp( .
0 301
2
)
−
1
]
x
x
y
=
6 28
. mg/L
σ
µ
6 28
20 4
.
x
C
v
=
=
=
0 308
.
.
x
g
x
=
3
C C
+
3
=
3 0 308
( .
)
+
( .
0 308
)
3
=
0 953
.
v
v
Hence, the mean, standard deviation, and skewness
of the concentration measurements are 20.4 mg/l,
6.28 mg/l, and 0.953, respectively.
(b) For a concentration of
c
= 30 mg/l, the exceedance
probability
is determined from the following
calculations,
(
)
=
ln
c
=
ln
30
3 40
.
10.3.3 Uniform Distribution
The
uniform distribution
describes the behavior of a
random variable in which all possible outcomes are
equally likely within the range [
a
,
b
]. For a continuous
random variable,
x
, the uniform probability density
function,
f
(
x
), is given by
ln
c
−
µ
3 40 2 97
0 301
.
−
.
y
z
=
=
=
1 43
.
σ
.
y
1
2
2
3
B
=
1 0 196854
+
.
z
+
0 115194
.
z
+
0 000344
.
z
−
4
4
+
0 019527
.
z
1
(10.39)
f x
( )
=
,
a
≤ ≤
x b
1
2
b a
−
[
=
1
+
0 196854 1 43
.
( .
)
+
0 115194 1 43
.
( .
)
2
]
=
−
where the parameters
a
and
b
define the range of the
random variable. The shape of the uniform distribution
is illustrated in Figure 10.3. The mean,
μ
x
, and variance,
σ
2
, of a uniformly distributed random variable are
given by
+
0 000344 1 43
.
( .
)
3
+
0 019527
.
( .
1 43
)
4
0 075
.
Hence, the probability of the sample concentration
exceeding 30 mg/l is 0.075 or 7.5%.
(c) If the concentration is assumed to be drawn from
a normal population with
μ
x
= 20.4 mg/l and
σ
x
= 6.28 mg/l, then
1
2
1
12
2
)
2
(10.40)
µ
=
(
a b
+
),
σ
=
(
b a
−
x
x
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