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,

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

The formulas for calculating the mean and variance of \({\rm{X}}\)are as follows:

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

Substitute \({\rm{2}}{\rm{.05 for \mu and 0}}{\rm{.06 for }}{{\rm{\sigma }}^{\rm{2}}}\):

\(\begin{array}{*{20}{c}}{{\rm{E(X) = }}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.05 + 0}}{\rm{.06/2}}}}{\rm{\gg 8}}{\rm{.00447}}}\\{{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = }}{{\rm{e}}^{{\rm{2(2}}{\rm{.05) + 0}}{\rm{.06}}}}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.06}}}}{\rm{ - 1}}} \right){\rm{\gg 3}}{\rm{.96196}}}\end{array}\)

The square root of the variance is the standard deviation:

\({\rm{SD(X) = }}\sqrt {{\rm{V(X)}}} {\rm{ = }}\sqrt {{\rm{3}}{\rm{.96196}}} {\rm{\gg 1}}{\rm{.99047}}\)

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

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

Rule of complements:

\({\rm{P(notA) = 1 - P(A)}}\)

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:

\({\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln12 - 2}}{\rm{.05}}}}{{\sqrt {{\rm{0}}{\rm{.06}}} }}{\rm{\gg 1}}{\rm{.78}}\)

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

\({\rm{P(X > 12) = P(Z > 1}}{\rm{.78) = 1 - P(Z < 1}}{\rm{.78) = 1 - 0}}{\rm{.9625 = 0}}{\rm{.0375 = 3}}{\rm{.75\% }}\)

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

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

The formulas for calculating the mean and variance of \({\rm{X}}\)are as follows:

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

Substitute \({\rm{2}}{\rm{.05 for \mu and 0}}{\rm{.06 for }}{{\rm{\sigma }}^{\rm{2}}}\):

\(\begin{array}{*{20}{c}}{{\rm{E(X) = }}{{\rm{e}}^{{\rm{\mu + }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/2}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.05 + 0}}{\rm{.06/2}}}}{\rm{\gg 8}}{\rm{.00447}}}\\{{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right){\rm{ = }}{{\rm{e}}^{{\rm{2(2}}{\rm{.05) + 0}}{\rm{.06}}}}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.06}}}}{\rm{ - 1}}} \right){\rm{\gg 3}}{\rm{.96196}}}\end{array}\)

The square root of the variance is the standard deviation:

\({\rm{SD(X) = }}\sqrt {{\rm{V(X)}}} {\rm{ = }}\sqrt {{\rm{3}}{\rm{.96196}}} {\rm{\gg 1}}{\rm{.99047}}\)

The following is one standard deviation above/below the mean:

\(\begin{array}{*{20}{c}}{{\rm{E(X) - SD(X) = 8}}{\rm{.00447 - 1}}{\rm{.99047 = 6}}{\rm{.014}}}\\{{\rm{E(X) + SD(X) = 8}}{\rm{.00447 + 1}}{\rm{.99047 = 9}}{\rm{.99494}}}\end{array}\)

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:

\(\begin{array}{*{20}{c}}{{\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln9}}{\rm{.99494 - 2}}{\rm{.05}}}}{{\sqrt {{\rm{0}}{\rm{.06}}} }}{\rm{\gg 1}}{\rm{.03}}}\\{{\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln6}}{\rm{.014 - 2}}{\rm{.05}}}}{{\sqrt {{\rm{0}}{\rm{.06}}} }}{\rm{\gg - 1}}{\rm{.04}}}\end{array}\)

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

\(\begin{array}{*{20}{c}}{{\rm{P(\mu - \sigma < X < \mu + \sigma ) = P(6}}{\rm{.014 < X < 9}}{\rm{.99494) = P( - 1}}{\rm{.04 < Z < 1}}{\rm{.03)}}}\\{{\rm{ = P(Z < 1}}{\rm{.03) - P(Z < - 1}}{\rm{.04) = 0}}{\rm{.8485 - 0}}{\rm{.1492 = 0}}{\rm{.6993 = 69}}{\rm{.93\% }}}\end{array}\)

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

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

The median has the property that the \({\rm{50\% }}\)of the values are less than it.

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

Determine the z-score that corresponds to a probability of \({\rm{0}}{\rm{.50}}\) in the normal probability table in the appendix:

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

The appropriate number is the mean multiplied by the standard deviation and the z-score:

\({\rm{lnx = \mu + z\sigma = 2}}{\rm{.05 + 0}}\sqrt {{\rm{0}}{\rm{.06}}} {\rm{ = 2}}{\rm{.05}}\)

Take each side of the previous equation's exponential:

\({\rm{x = }}{{\rm{e}}^{{\rm{lnx}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.05}}}}{\rm{\gg 7}}{\rm{.7679}}\)

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

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

The \({\rm{99\% }}\)of the values are below \({{\rm{P}}_{{\rm{99}}}}\), which makes the 99th percentile \({{\rm{P}}_{{\rm{99}}}}\)

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

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

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

The appropriate number is the mean multiplied by the standard deviation and the z-score:

\({\rm{lnx = \mu + z\sigma = 2}}{\rm{.05 + 2}}{\rm{.33}}\sqrt {{\rm{0}}{\rm{.06}}} {\rm{\gg 2}}{\rm{.62073}}\)

Take each side of the previous equation's exponential:

\({\rm{x = }}{{\rm{e}}^{{\rm{lnx}}}}{\rm{ = }}{{\rm{e}}^{{\rm{2}}{\rm{.62073}}}}{\rm{\gg 13}}{\rm{.7458}}\)

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

\(\begin{array}{*{20}{c}}{{\rm{\mu = 2}}{\rm{.05}}}\\{{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 0}}{\rm{.06}}}\end{array}\)

Rule of complements:

\({\rm{P(notA) = 1 - P(A)}}\)

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:

\({\rm{z = }}\frac{{{\rm{lnx - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{ln8 - 2}}{\rm{.05}}}}{{\sqrt {{\rm{0}}{\rm{.06}}} }}{\rm{\gg 0}}{\rm{.12}}\)

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

\({\rm{P(X > 8) = P(Z > 0}}{\rm{.12) = 1 - P(Z < 0}}{\rm{.12) = 1 - 0}}{\rm{.5478 = 0}}{\rm{.4522 = 45}}{\rm{.22\% }}\)

We can infer that these \({\rm{10 }}\) objects are independent because they were chosen at random. Then we should expect \({\rm{45}}{\rm{.22\% }}\)of the \({\rm{10 }}\) products to take longer than \({\rm{8}}\) months:

\({\rm{45}}{\rm{.22\% \times 10 = 0}}{\rm{.4522 \times 10 = 4}}{\rm{.522}}\)

As a result, we predict \({\rm{45}}{\rm{.22\% }}\)of the \({\rm{10 }}\)goods to take longer than 8 months.