Download Problem Statement The paper “The Statistics of Phytotoxic Air

Darren Tully
Engr323
Problem 4-74
Page 1
of 3
04/09/00
BHW #11
Problem Statement
The paper “The Statistics of Phytotoxic Air Pollutants” (J. Royal Stst. Soc.,
1989, pp. 183-198) suggests the lognormal distribution as a model for SO2
concentration above a certain forest. Suppose the parameter values are µ = 1.9
and σ = 0.9.
a. What are the mean value and standard deviation of the
concentration?
b. What is the probability that concentration is at most 10?
Between 5 and 10?
Solution
The problem statement has two important phrases, which will be guides to solving the
problem.
1. Suggesting the model for the SO2 concentration to be lognormal means
that the data is a continuous random variable and the random variable
can be defined as:
X = the SO2 concentration above the rainforest
2. As we learned in class, the parameters µ and σ refer to the mean of
ln(X). We know if X is lognormally distributed, then ln(X) is
normally distributed. Since µ = 1.9 doen’t have any units, then this
suggests that it is the mean of the natural log of X. Declaring the
distribution and parameters gives:
X ~ LogNormal(µ ln x = 1.9, σln x = 0.9)
Reminder
The µ given in the problem statement is µln x , not µx , and the same for σ. Now
solving the first question will be much esier.
a. What are the mean value and standard deviation of the concentration?
To start, there are a few equations needed and they can be found in the Devore text on
page 177 under the Lognormal Distribution heading. They are as follows:
Mean = µ x = E(X) = e
µ ln x +
σ ln x
2
2
= Exp[µ ln x + (σ 2ln x /2)]
and
2
2
Variance = σ2 x = V(X) = ( e 2 µ ln x + σ ln x )( e σ ln x − 1) = Exp[2µ ln x +σ2ln x]( Exp[σ 2ln x ]-1)
Next is to plug and chug:
µ x = E(X) = Exp[1.9 + (0.9)2 /2] = 10.024 ppm
2
2
σx = √(V(X)) = ( e 2*1.9+ .09 )( e . 9 −1) = 11.198 ppm
b. There are two questions to this part.
1. What is the probability that concentration is at most 10?
Darren Tully
Engr323
Problem 4-74
Page 2
of 3
04/09/00
BHW #11
This basic question is easy to solve. First state the problem in terms of symbols.
P(X ≤ 10) = ?
Next, take the natural log of both sides of the inequality;
P(ln(X) ≤ ln(10)) = P(Y ≤ 2.303)
The Y variable is used for ease of understanding and is defined as:
Y = ln(X)
The probability is then normalized and the normal table is used to generate an answer:
P((Y-µ ln x )/σln x ≤ ((2.303 – 1.9)/0.9) = P(Z ≤ 0.447) = Φ (0.447) =
0.6736 = P(X ≤ 10)
(see figure 1)
In Excel, the function is written:
=lognormdist(x, µln x , σln x )
And for this problem it is ;
=lognormdist(10, 1.9, 0.9) = .6727
Note: Excel only provides the CDF for the lognormal
0.1
f(x)
0.08
P(X<10)
0.06
0.04
0.02
0
-5.00
0.00
5.00 10.00 15.00
X= SO2 concentration
20.00
Figure 1: Lognormal Density function, Mean = 1.9, StDev = .9
2. What is the probability that the concentration is between 5 and
10?
Darren Tully
Engr323
Problem 4-74
Page 3
of 3
04/09/00
BHW #11
This problem is almost identical to the above question except for the use of the
proposition from page 100 of the Devore text. The first step is to convert the words into
a symbolic statement.
P(5 ≤ X ≤ 10) = ?
Next is to use the process shown above.
P(ln(5) ≤ ln(X) ≤ ln(10)) = P(1.609 ≤ Y ≤ 2.303) =
P(-0.323 ≤ Z ≤ 0.447) = Φ (0.447) - Φ (-0.323) =
0.6736 – 0.3745 =
0.2991 = P(5 ≤ X ≤ 10)
(see figure 2)
In Excel, the function is written:
=lognormdist(xmax, µln x , σln x ) - lognormdist(xmin , µln x , σln x )
And for this problem it is:
=lognormdist(10, 1.9, 0.9) - lognormdist(5, 1.9, 0.9) =.2993
0.1
0.08
f(x)
P(5<X<10)
0.06
0.04
0.02
-5.00
0
0.00
5.00 10.00 15.00
X = SO2 concentration
Figure 2: Lognormal Density function, Mean = 1.9, StDev = 0.9
20.00