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 that \({\rm{X}}\)follows a lognormal distribution,

\(\begin{array}{*{20}{c}}{{\rm{\mu = 10281}}}\\{{\rm{CV = 0}}{\rm{.40}}}\end{array}\)

\({\rm{\sigma = \mu CV = 10281(0}}{\rm{.40) = 4112}}{\rm{.4}}\)

The following equations have been used to calculate the mean and variance of \({\rm{ln(X)}}\) (with \({\rm{X}}\)being a lognormal distribution):

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

The variance is the square of the standard deviation \({\rm{4112}}{\rm{.4}}\)and the mean is \({\rm{10281}}\):

\(\begin{array}{*{20}{c}}{{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}{\rm{ = 10281}}}\\{{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = 4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}}\end{array}\)

Take each side of the equation's natural logarithm:

\(\begin{array}{*{20}{c}}{{\rm{ln}}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}{\rm{ = ln10281}}}\\{{\rm{ln}}\left( {{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}}\end{array}\)

Make use of the logarithm's property \({\rm{ln}}{{\rm{e}}^{\rm{b}}}{\rm{ = b}}\):

\(\begin{array}{*{20}{c}}{{\rm{\mu + }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}{\rm{ = ln10281}}}\\{{\rm{ln}}\left( {{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}}\end{array}\)

Solve the first equation for \({\rm{\mu }}\)and substitute this expression for \({\rm{\mu }}\)in the second equation:

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{\rm{ln}}\left( {{{\rm{e}}^{{\rm{2ln10281 - }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}}\end{array}\)

Reduce the second equation to:

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{\rm{ln}}\left( {{{\rm{e}}^{{\rm{2ln10281}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}}\end{array}{\rm{ }}\)

A product's logarithm is the sum of its logarithms:

\({\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}\)

\({\rm{ln}}{{\rm{e}}^{{\rm{2ln10281}}}}{\rm{ + ln}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}\)

Make use of the logarithm's property \({\rm{ln}}{{\rm{e}}^{\rm{b}}}{\rm{ = b}}\):

\({\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}\)

\({\rm{2ln10281 + ln}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}\)

\({\rm{2ln10281}}\)is subtracted from each side of the equation:

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{\rm{ln}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}\end{array}\)

Use \({{\rm{e}}^{{\rm{lnb}}}}{\rm{ = b}}\) to take the exponential of either side of the second equation.

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1 = }}{{\rm{e}}^{{\rm{ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}}}\end{array}\)

To the second equation, add 1 to each side:

\({\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}\)

\({{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}}{\rm{ + 1}}\)

Take each side of the second equation's natural logarithm:

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = ln}}\left( {{{\rm{e}}^{{\rm{ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}}{\rm{ + 1}}} \right)}\end{array}\)

To solve the second equation, take the square root of each side:

\(\begin{array}{*{20}{c}}{{\rm{\mu = ln10281 - }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}}\\{{\rm{\sigma = }}\sqrt {{\rm{ln}}\left( {{{\rm{e}}^{{\rm{ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}}{\rm{ + 1}}} \right)} {\rm{\gg 0}}{\rm{.385253}}}\end{array}\)

Substitute \({\rm{0}}{\rm{.385253}}\)for \({\rm{\sigma }}\)in the first equation and evaluate:

\({\rm{\mu = ln10281 - }}\frac{{{\rm{0}}{\rm{.38525}}{{\rm{3}}^{\rm{2}}}}}{{\rm{2}}}{\rm{\gg 9}}{\rm{.1638429}}\)

\({\rm{\sigma = }}\sqrt {{\rm{ln}}\left( {{{\rm{e}}^{{\rm{ln4112}}{\rm{.}}{{\rm{4}}^{\rm{2}}}{\rm{ - 2ln10281}}}}{\rm{ + 1}}} \right)} {\rm{\gg 0}}{\rm{.385253}}\)

The standard deviation is\({\rm{0}}{\rm{.385253}}{\rm{.}}\) while the mean of \({\rm{ln(X) is 9}}{\rm{.1638429 }}\)

(b) The logarithm of the value reduced by the mean divided by the standard deviation is the \({\rm{z - }}\)score:

\({\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln15000 - 9}}{\rm{.1638429}}}}{{{\rm{0}}{\rm{.385253}}}}{\rm{\gg 1}}{\rm{.17}}\)

Using the normal probability table in the appendix, calculate the corresponding probability:

\({\rm{P(X\pounds15000) = P(Z < 1}}{\rm{.17) = 0}}{\rm{.8790 = 87}}{\rm{.90\% }}\)

(c) The\({\rm{z - }}\)score is calculated by dividing the logarithm of the value reduced by the mean by the standard deviation:

\({\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln10281 - 9}}{\rm{.1638429}}}}{{{\rm{0}}{\rm{.385253}}}}{\rm{\gg 0}}{\rm{.19}}\)

Using the normal probability table in the appendix, calculate the corresponding probability:

\({\rm{P(X > \mu ) = 1 - P(X < 10281) = 1 - P(Z < 0}}{\rm{.19) = 1 - 0}}{\rm{.5753 = 0}}{\rm{.4247 = 42}}{\rm{.47\% }}\)

Because the lognormal distribution is not symmetric around its mean and consequently skewed, the probability is not equal to \({\rm{0}}{\rm{.5}}\).

(d) The \({\rm{95th}}\) percentile is the value below which \({\rm{95\% }}\)of all other values fall.

The logarithm of the value reduced by the mean, divided by the standard deviation, is the \({\rm{z - }}\)score:

\({\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln17000 - 9}}{\rm{.1638429}}}}{{{\rm{0}}{\rm{.385253}}}}{\rm{\gg 1}}{\rm{.50}}\)

Using the normal probability table in the appendix, calculate the corresponding probability:

\({\rm{P(X\pounds17000) = P(Z < 1}}{\rm{.50) = 0}}{\rm{.9332 = 93}}{\rm{.32\% }}\)

Because the probability \({\rm{P(X\pounds17000)}}\) is not equal to \({\rm{0}}{\rm{.95}}\),\({\rm{17000 }}\)is not the \({\rm{95th}}\)percentile.