Environmental Engineering Reference
In-Depth Information
Hence
z
= 0.6667,
B
= 0.2524, and Equation (10.22)
yields the probability that the concentration is less
than 12 mg/l as
B z
B z
,
,
≤
0
F z
( )
≈
0
(10.22)
1
−
≥
F
( .
0 6667
)
= −
1
B
= −
1 0 2524
.
=
0 7476
.
where
The probability that the concentration exceeds
12 mg/l is therefore equal to 1 − 0.7476 = 0.2524, or
approximately 25%.
(b) let
x
95
be the concentration that is exceeded 5% of
the time and
z
95
be the corresponding standard
normal deviate, then
F z
95
1
2
2
3
B
=
1 0 196854
+
.
z
+
0 115194
.
z
+
0 000344
.
z
−
4
4
+
0 019527
.
z
(10.23)
and the error in
F
(
z
) using Equation (10.22) is less than
0.00025. The cumulative distribution function (CDF)
F
(
z
) can be readily and conveniently obtained from
most spreadsheet and statistical programs and many
online utilities. For normally distributed variables,
approximately 68% of the values lie within 1 standard
deviation of the mean and approximately 95% lie within
two standard deviations of the mean.
(
)
= .
0 95
= −
(
)
= .
B
1
F z
0 05
95
95
using Equation (10.23) requires that
1
2
2
3
B
=
1 0 196854
+
.
z
+
0 115194
.
z
+
0 000344
.
z
95
95
95
95
−
4
4
+
0 019527
.
z
9
5
EXAMPLE 10.1
1
2
2
3
0 05
.
=
1 0 196854
+
.
z
+
0 115194
.
z
+
0 000344
.
z
95
95
95
Water-quality samples in a lake show that the concen-
trations of a pollutant fluctuate randomly and can be
approximated by a normal distribution with a mean of
10 mg/l and a standard deviation of 3 mg/l. (a) Esti-
mate the probability that the concentration of the pol-
lutant will exceed 12 mg/l; and (b) estimate the
concentration of the pollutant that is likely to be
exceeded only 5% of the time.
−
4
z
95
4
+
0 019527
.
which can be solved numerically (such as with a
solver utility in a spreadsheet) to yield
z
95
= 1.643,
and hence
x
− µ
σ
95
x
z
=
95
x
Solution
x
−
10
95
1 643
.
=
3
From the given data:
μ
x
= 10 mg/l and
σ
x
= 3 mg/l.
The population distribution from which the samples
are drawn is assumed to be a normal distribution.
which yields
x
95
= 14.9 mg/l. Hence, a concentra-
tion of 14.9 mg/l will be exceeded only 5% of the
time. Higher concentrations will obviously be
exceeded less regularly than 5% of the time.
(a) For a concentration of
x
= 12 mg/l, the correspond-
ing standard normal deviate,
z
, and the correspond-
ing value of
B
are given by
Conditions under which any random variable can be
expected to follow a normal distribution are specified
by the
central limit theorem
, which states that:
x
µ
σ
−
12 10
3
−
x
z
=
=
=
0 6667
.
x
1
2
2
3
B
=
1 0 196854
+
.
z
+
0 115194
.
z
+
0 000344
.
z
If
S
n
is the sum of
n
independently and identically dis-
tributed random variables
X
i
, each having a mean
μ
and
variance
σ
2
, then in the limit as
n
approaches infinity,
the distribution of
S
n
approaches a normal distribution
with mean
nμ
and variance
nσ
2
.
−
4
4
+
0 019527
.
z
1
2
[
2
=
1
+
0 196854 0 6667
.
( .
)
+
0 115194 0 6667
.
( .
)
]
−
4
+
0 000344 0 6667
.
( .
)
3
+
0
.019527 0 6667
( .
)
4
Although it can seldom be assured that water-quality
variables are the sum of
=
0 2524
.
independent
identically
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