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.

Given: \({\rm{X}}\)has a lognormal distribution with a standard deviation of \({\rm{X}}\).

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.16}}}\\{{\rm{\sigma = 1}}{\rm{.03}}}\\{{\rm{\gamma = 1}}{\rm{.0}}}\end{array}\)

Then, with \({\rm{\mu = 2}}{\rm{.16 and \sigma = 1}}{\rm{.03}}\) we know that \({\rm{Y = X - \gamma }}\) has a (regular) lognormal distribution, as follows:

\({\rm{X = Y + \gamma }}\)

The (regular) lognormal distribution's mean and variance. The formulas then provide \({\rm{Y}}\):

\(\begin{array}{*{20}{c}}{{\rm{E(X) = }}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}}\\{{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)}\end{array}\)

Substitute \({\rm{\mu = 2}}{\rm{.16 and \sigma = 1}}{\rm{.03}}\) :

\(\begin{array}{*{20}{c}}{{\rm{E(X) = }}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.16 + 1}}{\rm{.0}}{{\rm{3}}^{\rm{2}}}{\rm{/2}}}}{\rm{\gg 14}}{\rm{.7383}}}\\{{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = }}{{\rm{e}}^{{\rm{2(2}}{\rm{.16) + 1}}{\rm{.0}}{{\rm{3}}^{\rm{2}}}}}\left( {{{\rm{e}}^{{\rm{1}}{\rm{.0}}{{\rm{3}}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{\gg 410}}{\rm{.3177}}}\end{array}\)

The mean and variance of the linear combination \({\rm{W = aX + b}}\) have the following properties:

\(\begin{array}{*{20}{c}}{{\rm{E(W) = aE(X) + b}}}\\{{\rm{V(W) = }}{{\rm{a}}^{\rm{2}}}{\rm{V(X)}}}\end{array}\)

We can then calculate the mean and variance of \({\rm{X}}\) using these properties:

\(\begin{array}{*{20}{c}}{{\rm{\mu = E(X) = E(Y + \gamma ) = E(Y) + \gamma = 14}}{\rm{.7383 + 1}}{\rm{.0 = 15}}{\rm{.7383}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = V(X) = V(Y + \gamma ) = V(Y) = 410}}{\rm{.3177}}}\end{array}\)

The square root of the variance is the standard deviation:

\({\rm{\sigma = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} {\rm{ = }}\sqrt {{\rm{410}}{\rm{.3177}}} {\rm{\gg 20}}{\rm{.2563}}\)

Given: \({\rm{X}}\)has a lognormal distribution with a standard deviation of \({\rm{X}}\).

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.16}}}\\{{\rm{\sigma = 1}}{\rm{.03}}}\\{{\rm{\gamma = 1}}{\rm{.0}}}\end{array}\)

Then, with \({\rm{\mu = 2}}{\rm{.16 and \sigma = 1}}{\rm{.03}}\) we know that \({\rm{Y = X - \gamma }}\) has a (regular) lognormal distribution, as follows:

\({\rm{X = Y + \gamma }}\)

We want to know how likely it is that \({\rm{X}}\)is greater than 2. Write the probability in \({\rm{Y}}\) terms:

\({\rm{P(X > 2) = P(Y + \gamma > 2) = P(Y > 2 - \gamma ) = P(Y > 2 - 1}}{\rm{.0) = P(Y > 1)}}\)

The logarithm of the value reduced by the mean divided by the standard deviation (the standard deviation is the square root of the variance) yields the z-score:

\({\rm{z = }}\frac{{{\rm{lny - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln1 - 2}}{\rm{.16}}}}{{{\rm{1}}{\rm{.03}}}}{\rm{\gg - 2}}{\rm{.10}}\)

Using the normal probability table in the appendix (and the complement rule), calculate the corresponding probability:

\(\begin{array}{*{20}{c}}{{\rm{P(X > 2) = P(Y > 1) = P(Z > - 2}}{\rm{.10) = 1 - P(Z < - 2}}{\rm{.10)}}}\\{{\rm{ = 1 - 0}}{\rm{.0179 = 0}}{\rm{.9821 = 98}}{\rm{.21\% }}}\end{array}\)

Given: \({\rm{X}}\)has a lognormal distribution with a standard deviation of \({\rm{X}}\).

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.16}}}\\{{\rm{\sigma = 1}}{\rm{.03}}}\\{{\rm{\gamma = 1}}{\rm{.0}}}\end{array}\)

Then, with \({\rm{\mu = 2}}{\rm{.16 and \sigma = 1}}{\rm{.03}}\) we know that \({\rm{Y = X - \gamma }}\) has a (regular) lognormal distribution, as follows:

\({\rm{X = Y + \gamma }}\)

The median is the \({\rm{X}}\)number for which we know that \({\rm{50\% }}\)of the data values are below it.

\({\rm{P(X < MEDIAN) = 50\% = 0}}{\rm{.50}}\)

Rephrase the probability in terms of \({\rm{y}}\):

\({\rm{P(X < MEDIAN) = P(Y + \gamma < MEDIAN) = P(Y < MEDIAN - \gamma ) = P(Y < MEDIAN - 1}}{\rm{.0) = 0}}{\rm{.50}}\)

compute the z-score for a probability of \({\rm{0}}{\rm{.50}}\) in the normal probability table in the appendix.

\({\rm{z = 0}}\)

The mean is then increased by the product of the z-score and the standard deviation:

\({\rm{z = lny = \mu + z\sigma = 2}}{\rm{.16 + 0(1}}{\rm{.03) = 2}}{\rm{.16}}\)

Take each side of the previous equation's exponential:

\({\rm{y = }}{{\rm{e}}^{{\rm{lny}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.16}}}}\)

Since \({\rm{y}}\) is the value for which \({\rm{P(Y < y) = 0}}{\rm{.50}}\) is calculated,

\({\rm{MEDIAN - 1}}{\rm{.0 = }}{{\rm{e}}^{{\rm{2}}{\rm{.16}}}}\)

To either side of the equation, add \({\rm{1}}\):

\({\rm{MEDIAN = 1 + }}{{\rm{e}}^{{\rm{2}}{\rm{.16}}}}{\rm{\gg 9}}{\rm{.67114}}\)

The 99th percentile is the value of \({\rm{X}}\) below which \({\rm{99\% }}\) of all data values fall.

\({\rm{P}}\left( {{\rm{X < }}{{\rm{P}}_{{\rm{99}}}}} \right){\rm{ = 99\% = 0}}{\rm{.99}}\)

Write the probability in \({\rm{Y}}\)terms:

\({\rm{P}}\left( {{\rm{X < }}{{\rm{P}}_{{\rm{99}}}}} \right){\rm{ = P}}\left( {{\rm{Y + \gamma < }}{{\rm{P}}_{{\rm{99}}}}} \right){\rm{ = P}}\left( {{\rm{Y < }}{{\rm{P}}_{{\rm{99}}}}{\rm{ - \gamma }}} \right){\rm{ = P}}\left( {{\rm{Y < }}{{\rm{P}}_{{\rm{99}}}}{\rm{ - 1}}{\rm{.0}}} \right){\rm{ = 0}}{\rm{.99}}\)

Determine the z-score for a probability of \({\rm{0}}{\rm{.99}}\) in the normal probability table in the appendix:

\({\rm{z = 2}}{\rm{.33}}\)

The mean is then multiplied by the product of the z-score and the standard deviation to get the equivalent value:

\({\rm{lny = \mu + z\sigma = 2}}{\rm{.16 + 2}}{\rm{.33(1}}{\rm{.03) = 4}}{\rm{.5599}}\)

Take each side of the previous equation's exponential:

\({\rm{y = }}{{\rm{e}}^{{\rm{lny}}}}{\rm{ = }}{{\rm{e}}^{{\rm{4}}{\rm{.5599}}}}\)

Given that \({\rm{y}}\)is the value for which \({\rm{P(Y < y) = 0}}{\rm{.99}}\), \({\rm{y}}\)must be the \({\rm{99th}}\) percentile lowered by:

\({{\rm{P}}_{{\rm{99}}}}{\rm{ - 1}}{\rm{.0 = }}{{\rm{e}}^{{\rm{4}}{\rm{.5599}}}}\)

To either side of the equation, add \({\rm{1}}\):

\({{\rm{P}}_{{\rm{99}}}}{\rm{ = 1 + }}{{\rm{e}}^{{\rm{4}}{\rm{.5599}}}}{\rm{\gg 96}}{\rm{.5739}}\)