91Ó°ÊÓ

The Weibull Distribution is a continuous probability distribution that can be used to analyse life statistics, model failure times, and determine product reliability.

Given: \({\rm{X}}\)has a Weibull distribution with

\(\begin{array}{l}{\rm{\alpha = 1}}{\rm{.817}}\\{\rm{\beta = 0}}{\rm{.863}}\end{array}\)

(a) \({\rm{P(X}} \ge 0.5)\)

Cumulative distribution function of a (regular) Weibull distribution:

\({\rm{F(x;\alpha ,\beta )}} = \left\{ {\begin{array}{*{20}{c}}0&{x < 0}\\{{\rm{1 - }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}&{x \ge 0}\end{array}} \right.\)

Evaluate at \({\rm{x = 0}}{\rm{.5}}\):

\(\begin{array}{c}{\rm{F(0}}{\rm{.5;\alpha ,\beta ) = F(0}}{\rm{.5;1}}{\rm{.817,0}}{\rm{.863)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - (0}}{\rm{.5/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}} \approx 0.309911\end{array}\)

Complement rule:

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

Use the complement rule to determine the probability:

\(\begin{array}{c}{\rm{P(X}} \ge 0.5) = {\rm{1 - P(X}} \le 0.5)\\{\rm{ = 1 - F(0}}{\rm{.5;\alpha ,\beta )}}\\{\rm{ = 1 - 0}}{\rm{.309911}}\\{\rm{ = 0}}{\rm{.690089}}\\{\rm{ = 69}}{\rm{.0089\% }}\end{array}\)

Therefore,The probability is \({\rm{69}}{\rm{.0089\% }}\)

(b) \({\rm{P(X > \mu + \sigma )}}\)

\(\begin{array}{l}{\rm{E(Y) = \beta \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)\\{\rm{V(Y) = }}{{\rm{\beta }}^{\rm{2}}}\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\rm{\alpha }}}} \right){\rm{ - }}{{\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)} \right)}^{\rm{2}}}} \right)\end{array}\)

The mean and variance of a (regular) Weibull distribution can be calculated using the following formula:

Evaluate for \({\rm{\alpha = 1}}{\rm{.817}}\)and \({\rm{\beta = 0}}{\rm{.863}}\) :

\(\begin{array}{l}{\rm{E(Y) = \beta \Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)\\{\rm{ = 0}}{\rm{.863\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{1}}{\rm{.817}}}}} \right) \approx 0.767116\\{\rm{V(Y) = }}{{\rm{\beta }}^{\rm{2}}}\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{\rm{\alpha }}}} \right){\rm{ - }}{{\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{\alpha }}}} \right)} \right)}^{\rm{2}}}} \right)\\ = 0.86{{\rm{3}}^{\rm{2}}}\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{2}}}{{{\rm{1}}{\rm{.817}}}}} \right){\rm{ - }}{{\left( {{\rm{\Gamma }}\left( {{\rm{1 + }}\frac{{\rm{1}}}{{{\rm{1}}{\rm{.817}}}}} \right)} \right)}^{\rm{2}}}} \right) \approx 0.191194\end{array}\)

Rewrite the given probability:

\(\begin{array}{c}{\rm{P(X > \mu + \sigma ) = P(X > 0}}{\rm{.767116 + 0}}{\rm{.191194)}}\\{\rm{ = P(X > 0}}{\rm{.958310)}}\end{array}\)

A (regular) Weibull distribution's cumulative distribution function is:

\({\rm{F(x;\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{c}}0&{x < 0}\\{{\rm{1 - }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}&{x \ge 0}\end{array}} \right.\)

Evaluate at \({\rm{x = 0}}{\rm{.5}}\):

\(\begin{array}{c}{\rm{F(0}}{\rm{.958310;\alpha ,\beta ) = F(0}}{\rm{.958310;1}}{\rm{.817,0}}{\rm{.863)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - (0}}{\rm{.958310/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}} \approx 0.701703\end{array}\)

Complement rule:

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

To calculate the likelihood, use the complement rule:

\(\begin{array}{c}{\rm{P(X > \mu + \sigma ) = P(X > 0}}{\rm{.767116 + 0}}{\rm{.191194)}}\\{\rm{ = P(X > 0}}{\rm{.958310)}}\\ = 1 - P(X \le 0.958310)\\{\rm{ = 1 - F(0}}{\rm{.958310;\alpha ,\beta )}}\\{\rm{ = 1 - 0}}{\rm{.701703}}\\{\rm{ = 0}}{\rm{.298297}}\\{\rm{ = 29}}{\rm{.8297\% }}\end{array}\)

Therefore,The probability is \({\rm{29}}{\rm{.8297\% }}\)

(c) The median is defined as the value of \({\rm{x}}\) for which \({\rm{50\% }}\) of the data values are less than \({\rm{x}}\).

\(\begin{array}{c}{\rm{P(X}} \le x) = 50\% \\ = 0.5\end{array}\)

A (regular) Weibull distribution's cumulative distribution function is:

\({\rm{F(x;\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{c}}0&{x < 0}\\{1 - {e^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}&{x \ge 0}\end{array}} \right.\)

The cumulative distribution function evaluates at \({\rm{x}}\), therefore the probability to the left of \({\rm{x}}\) is the same:

\({\rm{F(x;\alpha ,\beta ) = 0}}{\rm{.5}}\)

\({\rm{\alpha }}\)is \(1.817\)and \({\rm{\beta }}\) is \(0.863\):

\({\rm{F(x;1}}{\rm{.817,0}}{\rm{.863) = 0}}{\rm{.5}}\)

Use the following formula to calculate the cumulative distribution function:

\({\rm{1 - }}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = 0}}{\rm{.5}}\)

SOLVING THE EQUATION:

Subtract 1 from each side of the equation:

\({\rm{ - }}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = - 0}}{\rm{.5}}\)

Multiply each side by \( - 1\):

\({{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = 0}}{\rm{.5}}\)

Take each side's natural logarithm:

\({\rm{ln}}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = ln0}}{\rm{.5}}\)

Use the power property of logarithm \(\left( {{\rm{ln}}{{\rm{a}}^{\rm{b}}}{\rm{ = blna}}} \right)\):

\({\rm{ - }}{\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{lne = ln0}}{\rm{.5}}\)

\({\rm{e}}\)'s natural logarithm is 1:

\({\rm{ - }}{\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{ = ln0}}{\rm{.5}}\)

Multiply each side by \( - 1\) :

\({\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{ = - ln0}}{\rm{.5}}\)

Take each side's \(1.817\)th root:

\(\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}{\rm{ = ( - ln0}}{\rm{.5}}{{\rm{)}}^{{\rm{1/1}}{\rm{.817}}}}\)

Multiply each side by \(0.863\):

\({\rm{x = 0}}{\rm{.863( - ln0}}{\rm{.5}}{{\rm{)}}^{{\rm{1/1}}{\rm{.817}}}} \approx 0.705355\)

Thus , the median is \(0.705355\).

(c) The \({\rm{100p}}\)th percentile is the value \({\rm{x}}\) for which \({\rm{100p\% }}\) of all data value lie below \({\rm{x}}\).

\(\begin{array}{c}{\rm{P(X}} \le x){\rm{ = 100p\% }}\\{\rm{ = }}\frac{{{\rm{100p}}}}{{{\rm{100}}}}{\rm{ = p}}\end{array}\)

A (regular) Weibull distribution's cumulative distribution function is:

\({\rm{F(x;\alpha ,\beta ) = }}\left\{ {\begin{array}{*{20}{c}}0&{x < 0}\\{1{\rm{ - }}{{\rm{e}}^{{\rm{ - (x/\beta }}{{\rm{)}}^{\rm{\alpha }}}}}}&{x \ge 0}\end{array}} \right.\)

The cumulative distribution function evaluates at \({\rm{x}}\), therefore the probability to the left of \({\rm{x}}\)is the same:

\({\rm{F(x;\alpha ,\beta ) = 0}}{\rm{.5}}\)

\({\rm{\alpha }}\)is \(1.817\)and \({\rm{\beta }}\) is \(0.863\) :

\({\rm{F(x;1}}{\rm{.817,0}}{\rm{.863) = p}}\)

Use the following formula to calculate the cumulative distribution function:

\({\rm{1 - }}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = p}}\)

SOLVING THE EQUATION:

Subtract 1 from each side of the equation:

\({\rm{ - }}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = p - 1}}\)

Multiply each side by $-1$ :

\({{\mathop{\rm e}\nolimits} ^{ - {{(x/0.863)}^{1.817}}}} = 1 - p\)

Take each side's natural logarithm:

\({\rm{ln}}{{\rm{e}}^{{\rm{ - (x/0}}{\rm{.863}}{{\rm{)}}^{{\rm{1}}{\rm{.817}}}}}}{\rm{ = ln(1 - p)}}\)

Use the logarithm's power property \(\left( {{\rm{ln}}{{\rm{a}}^{\rm{b}}}{\rm{ = blna}}} \right)\):

\({\rm{ - }}{\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{lne = ln(1 - p)}}\)

\({\rm{e}}\)'s natural logarithm is l:

\({\rm{ - }}{\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{ = ln(1 - p)}}\)

Multiply each side by \( - 1\):

\({\left( {\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}} \right)^{{\rm{1}}{\rm{.817}}}}{\rm{ = - ln(1 - p)}}\)

Take each side's \(1.817\)th root:

\(\frac{{\rm{x}}}{{{\rm{0}}{\rm{.863}}}}{\rm{ = ( - ln(1 - p)}}{{\rm{)}}^{{\rm{1/1}}{\rm{.817}}}}\)

Multiply each side by \(0.863\):

\({\rm{x = 0}}{\rm{.863( - ln(1 - p)}}{{\rm{)}}^{{\rm{1/1}}{\rm{.817}}}}\)

Thus ,\({\rm{100p}}\)the percentile is \({\rm{x = 0}}{\rm{.863( - ln(1 - p)}}{{\rm{)}}^{{\rm{1/1}}{\rm{.817}}}}\)