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A continuous probability distribution of a random variable whose logarithm is normally distributed is known as a log-normal (or lognormal) distribution in probability theory. Y = ln (X) has a normal distribution if the random variable X is log-normally distributed.

Let a rv \({\rm{X}}\)indicate the \({\rm{S}}{{\rm{O}}_{\rm{2}}}\)concentration above a specific forest. Then it is assumed that \({\rm{X}}\) has the following Lognormal distribution parameters:

\(\begin{array}{l}{\rm{\mu = 1}}{\rm{.9}}\\{\rm{\sigma = 0}}{\rm{.9}}\end{array}\)

\(\begin{array}{c}{{\rm{\mu }}_{\rm{x}}}{\rm{ = exp}}\left( {{\rm{\mu + }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp}}\left( {{\rm{1}}{\rm{.9 + }}\frac{{{{{\rm{(0}}{\rm{.9)}}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp(2}}{\rm{.305)}}\\{{\rm{\mu }}_{\rm{x}}}{\rm{ = 10}}{\rm{.02}}\end{array}\)

The Variance for lognormal distribution is given as:

\(\begin{array}{c}{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{2(1}}{\rm{.9) + 0}}{\rm{.}}{{\rm{9}}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.}}{{\rm{9}}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{4}}{\rm{.61}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.81}}}}{\rm{ - 1}}} \right)\\{\rm{V(X) = 125}}{\rm{.39}}\end{array}\)

The standard deviation for a lognormal distribution is then calculated as follows:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{125}}{\rm{.39}}} \\{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 11}}{\rm{.20}}\end{array}\)

\(\begin{array}{c}{\rm{P(X}} \le 10){\rm{ = P(ln(X)}} \le \ln (10))\\{\rm{ = P(ln(X)}} \le 2.30)\\{\rm{ = P}}\left( {\frac{{\ln (X) - 1.9}}{{0.9}} \le \frac{{2.3 - 1.9}}{{0.9}}} \right)\\{\rm{ = P(Z}} \le 0.44)\\ = \phi (0.44)\end{array}\)

The cdf for the standard normal distribution rv is \(\phi {\rm{(Z)}}\). Appendix A-3 is used.

\({\rm{P(X}} \le 10) = 0.67\)

\({\rm{P(}}5 \le {\rm{X}} \le 10)\)is the probability that the concentration is between \(5\) and \(10\). Then we'll be able to write

\(\begin{array}{c}P(5 \le X \le 10){\rm{ = P(ln(5}}) \le {\rm{ln(X)}} \le \ln (10))\\{\rm{ = P(1}}{\rm{.61}} \le \ln ({\rm{X}}) \le 2.30)\\ = P\left( {\frac{{1.61 - 1.9}}{{0.9}} \le \frac{{\ln ({\rm{X}}) - 1.9}}{{0.9}} \le \frac{{2.3 - 1.9}}{{0.9}}} \right)\\{\rm{ = P(}} - 0.32 \le {\rm{Z}} \le 0.44)\\ = \phi (0.44) - \phi ( - 0.32)\end{array}\)

Here \(\phi ({\rm{Z}})\) is the cdf for standard normal distribution rv.Using Appendix A-3

\( = 0.67 - 0.3745\)

\({\rm{P(5}} \le {\rm{X}} \le 10) = 0.2955\)

Proposition : Let \({\rm{X}}\) be a continuous \({\rm{rv}}\) with \({\rm{pdff(x)}}\)and \({\rm{cdfF(x)}}\). Then for any number a,

\({\rm{P(X}} \le {\rm{a) = F(a)}}\)

And with \({\rm{a > b}}\) for any two numbers a and b.

\({\rm{P(b}} \le {\rm{X}} \le {\rm{a) = F(a) - F(b)}}\)