91影视

the proportion of the total number of conceivable outcomes to the number of options in an exhaustive collection of equally likely outcomes that cause a given occurrence.

(a) For the distribution of \({\rm{arv}}\)\({\rm{X}}\), the median \({\rm{(\tilde \mu )}}\)is as follows:

\({\rm{P(X\pounds\tilde \mu ) = 0}}{\rm{.5}}\)

We already know that \({\rm{X}}\)has a lognormal distribution. That example, because \({\rm{ln(X)}}\)has a normal distribution, the \({\rm{ cdf of f(z)}}\) can be written as:

\(\begin{array}{*{20}{c}}{{\rm{P(X\pounds\tilde \mu ) = P(ln(X)\poundsln(\tilde \mu ))}}}\\{{\rm{ = P}}\left( {\frac{{{\rm{ln(X) - \mu }}}}{{\rm{\sigma }}}{\rm{\pounds}}\frac{{{\rm{ln(\tilde \mu ) - \mu }}}}{{\rm{\sigma }}}} \right)}\\{{\rm{P(X\pounds\tilde \mu ) = P}}\left( {{\rm{Z\pounds}}\frac{{{\rm{ln(\tilde \mu ) - \mu }}}}{{\rm{\sigma }}}} \right)}\end{array}\)

Given that this probability equals \({\rm{0}}{\rm{.5}}\),

\({\rm{P}}\left( {{\rm{Z\pounds}}\frac{{{\rm{ln(\tilde \mu ) - \mu }}}}{{\rm{\sigma }}}} \right){\rm{ = 0}}{\rm{.5}}\)

Now, according to appendix\({\rm{ - A3,}}\) the z-value for the probability of \({\rm{0}}{\rm{.5}}\)is \({\rm{0}}\)

\(\begin{array}{*{20}{c}}{\frac{{{\rm{ln(\tilde \mu ) - \mu }}}}{{\rm{\sigma }}}{\rm{ = 0}}}\\{{\rm{ln(\tilde \mu ) = \mu }}}\\{{\rm{\tilde \mu = }}{{\rm{e}}^{\rm{\mu }}}}\end{array}\)

We have \({\rm{\mu = 9}}{\rm{.164}}\)for the load distribution

\({\rm{\tilde \mu = }}{{\rm{e}}^{{\rm{9}}{\rm{.164}}}}{\rm{ = 9547}}{\rm{.17}}\)

(b) It is assumed that \({{\rm{z}}_{\rm{\alpha }}}\)is the notation for the standard normal distribution's\({\rm{100(1 - \alpha )}}\)percentile. The lognormal distribution's \({\rm{100(1 - \alpha )}}\)percentile is denoted as \({{\rm{x}}_{\rm{\alpha }}}\). Then

\(\begin{array}{*{20}{c}}{{\rm{P}}\left( {{\rm{Z\pounds}}{{\rm{z}}_{\rm{\alpha }}}} \right)}&{{\rm{ = 1 - \alpha }}}\\{{\rm{P}}\left( {\frac{{{\rm{ln(X) - \mu }}}}{{\rm{\sigma }}}{\rm{\pounds}}{{\rm{z}}_{\rm{\alpha }}}} \right)}&{{\rm{ = 1 - \alpha }}}\\{{\rm{P}}\left( {{\rm{ln(X)\pounds\mu + \sigma }}\left( {{{\rm{z}}_{\rm{\alpha }}}} \right)} \right)}&{{\rm{ = 1 - \alpha }}}\\{{\rm{P}}\left( {{\rm{X\pounds}}{{\rm{e}}^{{\rm{\mu + \sigma }}\left( {{{\rm{z}}_{\rm{\alpha }}}} \right)}}} \right)}&{{\rm{ = 1 - \alpha }}}\end{array}\)

Hence

\({{\rm{x}}_{\rm{\alpha }}}{\rm{ = }}{{\rm{e}}^{{\rm{\mu + \sigma }}\left( {{{\rm{z}}_{\rm{\alpha }}}} \right)}}\)

The \({\rm{99th}}\)percentile is the value that load will surpass just \({\rm{1\% }}\) of the time. Let's call it \({{\rm{x}}_{{\rm{0}}{\rm{.01}}}}\), which is the same as\({\rm{\alpha = 0}}{\rm{.01}}\) in the previous statement. To begin, we must calculate \({{\rm{z}}_{{\rm{0}}{\rm{.01}}}}\) in such a way that

\({\rm{P}}\left( {{\rm{Z\pounds}}{{\rm{z}}_{{\rm{0}}{\rm{.01}}}}} \right){\rm{ = f}}\left( {{{\rm{z}}_{{\rm{0}}{\rm{.01}}}}} \right){\rm{ = 0}}{\rm{.99}}\)

We can write the following:

\({{\rm{z}}_{{\rm{0}}{\rm{.01}}}}{\rm{ = 2}}{\rm{.33}}\)

\({\rm{\mu = 9}}{\rm{.164;\sigma = 0}}{\rm{.385}}\),

We get by substituting all the values.

\(\begin{array}{*{20}{c}}{{{\rm{x}}_{{\rm{0}}{\rm{.01}}}}{\rm{ = }}{{\rm{e}}^{{\rm{9}}{\rm{.164 + 0}}{\rm{.385(2}}{\rm{.33)}}}}}\\{{\rm{ = }}{{\rm{e}}^{{\rm{10}}{\rm{.061}}}}}\\{{{\rm{x}}_{{\rm{0}}{\rm{.01}}}}{\rm{ = 23413}}{\rm{.08}}}\end{array}\)