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Nonpoint source loads are chemical masses that travel to the main stem of a river and its tributaries in flows that are distributed over relatively long stream reaches, in contrast to those that enter at well-defined and regulated points. An article suggests that for a certain time period and location, X = nonpoint source load of total dissolved solids could be modeled with a lognormal distribution having mean value 10,827 kg/day/km and a coefficient of variation CV = 0.40 CV = σX μX .What are the mean value and standard deviation of ln(x)?

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rodolforodriguezr

Answer:

Mean 9.2898

Standard deviation 0.006

Step-by-step explanation:

Let m and s the mean and standard deviation of the non-logarithmized  

distribution and [tex]\bf \mu[/tex], [tex]\bf \sigma[/tex]  the mean and standard deviation of the logarithmized one.

m = 10,827

Since the CV=0.4, then

s/m=0.4 and  

s = 10,827*0.4 = 4,330.8

The mean of the logarithmized distribution is

[tex]\bf \mu=ln\left( \frac{m}{\sqrt{1+s/m^2}}\right)=ln\left( \frac{10827}{\sqrt{1+4330.8/10827^2}}\right)=9.2898[/tex]

and the standard deviation is

[tex]\bf \sigma=\sqrt{ln\left(1+\frac{s}{m^2} \right)}=\sqrt{ln(1+\frac{4330.8}{10827^2})}=0.006[/tex]

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