91Ó°ÊÓ

A lognormal distribution is a discrete and continuous distribution of a random variable with a normally distributed logarithm. In other words, lognormal distribution is based on the idea that the logarithms of the raw data generated are also normally distributed, rather than the original raw data being normally distributed.

Allow a rv \({\rm{X}}\)to represent the drill lifespan. Then it is assumed that \({\rm{X}}\) has the following Lognormal distribution parameters:

\(\begin{array}{l}{\rm{\mu = 4}}{\rm{.5}}\\{\rm{\sigma = 0}}{\rm{.8}}\end{array}\)

\(\begin{array}{c}{{\rm{\mu }}_{\rm{x}}}{\rm{ = exp}}\left( {{\rm{\mu + }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp}}\left( {{\rm{4}}{\rm{.5 + }}\frac{{{{{\rm{(0}}{\rm{.8)}}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp(4}}{\rm{.82)}}\\{{\rm{\mu }}_{\rm{x}}}{\rm{ = 123}}{\rm{.96}}\end{array}\)

The lognormal distribution's variance is calculated as follows:

\(\begin{array}{l}{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{2(4}}{\rm{.5) + 0}}{\rm{.}}{{\rm{8}}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.}}{{\rm{8}}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{9}}{\rm{.64}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.64}}}}{\rm{ - 1}}} \right)\\{\rm{V(X) = 13776}}{\rm{.53}}\end{array}\)

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

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{13776}}{\rm{.53}}} \\{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 117}}{\rm{.37}}\end{array}\)

The chance that one's lifetime is at most 100 can be expressed as If \({\rm{X}}\) has a lognormal distribution, we know that \({\rm{ln(X)}}\) is normally distributed.

\(\begin{array}{l}{\rm{P(X}} \le 100) = {\rm{P(ln}}(X) \le \ln (100))\\ = P(ln(X) \le 4.605) = {\rm{P}}\left( {\frac{{\ln ({\rm{X}}) - 4.5}}{{0.8}} \le \frac{{4.605 - 4.5}}{{0.8}}} \right)\\{\rm{ = P(Z}} \le 0.13)\\ = \phi (0.13)\end{array}\)

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

\({\rm{P(X}} \le 100) = 0.5517\)

Proposition : Assume \({\rm{X}}\) is a continuous rv with pdf and.

Then, for any \({\rm{a}}\) number,

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

The probability of a lifetime of at least 200 years can be expressed as \({\rm{P(X}} \ge 200)\). If \({\rm{X}}\) has a lognormal distribution, we know that \({\rm{ln(X)}}\) is normally distributed.

\(\begin{array}{l}{\rm{P(X}} \ge 200) = {\rm{P(ln(X)}} \ge \ln (200))\\{\rm{ = P(ln(X}}) \ge 5.298)\\{\rm{ = P}}\left( {\frac{{\ln (X) - 4.5}}{{0.8}} \ge \frac{{5.298 - 4.5}}{{0.8}}} \right)\\{\rm{ = P(Z}} \ge 1)\\ = 1 - \phi (1)\end{array}\)

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

\(\begin{array}{l}{\rm{P(X}} \ge 200) = 1 - 0.8413\\{\rm{P(X}} \ge 200) = 0.1587\end{array}\)

Because \({\rm{X}}\)is continuous, the chance of \({\rm{X}}\) equaling a fixed value is nil.

\(\begin{array}{l}{\rm{P(X > 200) = P(}}X \ge 200)\\{\rm{P(X > 200) = }}0.1587\end{array}\)