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

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

\(\begin{array}{l}{\rm{\alpha = 2}}{\rm{.2}}\\{\rm{\beta = 1}}{\rm{.1}}\\{\rm{\gamma = 0}}{\rm{.5P(1 < X < 2)}}\end{array}\)

Then we know that \({\rm{Y = X - }}\gamma \) has a (regular) Weibull distribution with \({\rm{\alpha = 2}}{\rm{.2}}\) and \({\rm{\beta = 1}}{\rm{.1}}\), thus:

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

Rewrite the given probability in terms of \({\rm{Y}}\):

\(\begin{aligned} P(1 < X < 2) &= P(1 < Y + \gamma < {\rm{2)}}\\ &= P(1 - \gamma < {\rm{Y}} < 2 - \gamma )\\ &= P(Y < 2 - \gamma ){\rm{ - P(Y}} \le 1 - \gamma )\\ &= P(Y \le 2 - \gamma ){\rm{ - P(Y}} \le 1 - \gamma )\\ &= P(Y \le 2 - 0.5){\rm{ - P(Y}} \le 1 - 0.5)\\ &= P(Y \le 1.5){\rm{ - P(Y}} \le 0.5)\\ &= {\rm{F(1}}{\rm{.5) - F(0}}{\rm{.5)}}\end{aligned}\)

Cumulative distribution function of a (regular) Weibull distribution:

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

Evaluate at \({\rm{x = 1}}{\rm{.5}}\) and \({\rm{x = 0}}{\rm{.5}}\):

\(\begin{array}{c}{\rm{F(1}}{\rm{.5;\alpha ,\beta ) = F(1}}{\rm{.5;2}}{\rm{.2,1}}{\rm{.1)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - (1}}{\rm{.5/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}} \approx 0.861724\\F(0.5;\alpha ,\beta ) = {\rm{F(0}}{\rm{.5;2}}{\rm{.2,1}}{\rm{.1)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - (0}}{\rm{.5/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}} \approx 0.161776\end{array}\)

Therefore,The probability then becomes:

\(\begin{aligned}P(1 < X < 2) &= F(1}}{\rm{.5) - F(0}}{\rm{.5)}}\\ &= 0 {\rm{.861724 - 0}}{\rm{.161776}}\\ &= 0{\rm{.699948}}\\ &= 69{\rm{.9948\% }}\end{aligned}\)

Given: The Weibull distribution for \({\rm{X}}\) is universal.

\(\begin{array}{l}{\rm{\alpha = 2}}{\rm{.2}}\\{\rm{\beta = 1}}{\rm{.1}}\\\gamma {\rm{ = 0}}{\rm{.5P(X > 1}}{\rm{.5)}}\end{array}\)

Then we know that \({\rm{Y = X - }}\gamma \) has a (regular) Weibull distribution with \({\rm{\alpha = 2}}{\rm{.2}}\) and \({\rm{\beta = 1}}{\rm{.1}}\), thus:

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

Complement rule:

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

Use the complement rule to rewrite the given probability in terms of \({\rm{Y}}\):

\(\begin{array}{c}{\rm{P(X}} > 1.5){\rm{ = P(Y}} + \gamma > 1.5)\\{\rm{ = P(Y > }}1.5 - \gamma )\\{\rm{ = 1 - P(Y}} \le 1.5 - \gamma )\\{\rm{ = 1 - P(Y}} \le 1.5 - 0.5)\\{\rm{ = 1 - P(Y}} \le 1)\\{\rm{ = 1 - F(1)}}\end{array}\)

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 $x=1$ :

\(\begin{array}{c}{\rm{F(1;\alpha ,\beta ) = F(1;2}}{\rm{.2,1}}{\rm{.1)}}\\{\rm{ = 1 - }}{{\rm{e}}^{{\rm{ - (1/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}} \approx 0.555516\end{array}\)

Therefore,The likelihood then becomes:

\(\begin{array}{c}{\rm{P(X > 1}}{\rm{.5) = 1 - F(1)}}\\{\rm{ = 1 - 0}}{\rm{.555516}}\\{\rm{ = 0}}{\rm{.444484}}\\{\rm{ = 44}}{\rm{.4484\% }}\end{array}\)

The number \({\rm{X}}\) represents the amount of time (in days) that passes before the individual becomes contagious. Also, \({\rm{X}}\) has the following Weibull distribution parameters:

\(\begin{array}{l}{\rm{\alpha = 2}}{\rm{.2}}\\{\rm{\beta = 1}}{\rm{.1}}\\\gamma = 0.5\end{array}\)

The cdf of given Weibull rv can be written as:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}0&{x < 0}\\{{\rm{1 - }}{{\rm{e}}^{{\rm{ - ((x - 0}}{\rm{.5)/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}}}&{x \ge 0.5}\end{array}} \right.\)

Let us denote the \({90^{{\rm{th }}}}\)percentile by \({\mathop{\rm p}\nolimits} \)

Then using the cdf, we can write \({\mathop{\rm P}\nolimits} (X \le 6)\)as:

\(\begin{array}{c}{\rm{P(X}} \le p){\rm{ = 0}}{\rm{.91 - }}{{\rm{e}}^{{\rm{ - ((p - 0}}{\rm{.5)/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}}\\{\rm{ = 0}}{\rm{.9}}{{\rm{e}}^{{\rm{ - ((p - 0}}{\rm{.5)/1}}{\rm{.1}}{{\rm{)}}^{{\rm{2}}{\rm{.2}}}}}}\\{\rm{ = 0}}{\rm{.1}}{\left( {\frac{{{\rm{p - 0}}{\rm{.5}}}}{{{\rm{1}}{\rm{.1}}}}} \right)^{{\rm{2}}{\rm{.2}}}}\\{\rm{ = - ln(0}}{\rm{.1)}}\\{\rm{p = 0}}{\rm{.5 + (1}}{\rm{.1) \times ( - ln(0}}{\rm{.1)}}{{\rm{)}}^{{\rm{1/2}}{\rm{.2}}}}\\{\rm{p = 2}}{\rm{.107}}\end{array}\)

Definition : Let p be a number between 0 and l. The \({{\rm{(100p)}}^{{\rm{th}}}}\)percentile of the distribution of a continuous rv \({\rm{X}}\), denoted by \({{\rm{\eta }}_{\rm{p}}}\), is defined by

\({\rm{p = F}}\left( {{\eta _p}} \right) = \int_{ - \infty }^{{\eta _p}} f {\rm{(y)dy}}\)

Therefore ,The \({\rm{90th}}\) percentile of the distribution is \(2.107\)

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

\(\begin{array}{l}{\rm{\alpha = 2}}{\rm{.2}}\\{\rm{\beta = 1}}{\rm{.1}}\\\gamma = 0.5\end{array}\)

Then we know that \({\rm{Y = X - }}\gamma \) has a (regular) Weibull distribution with \({\rm{\alpha = 2}}{\rm{.2}}\)and \({\rm{\beta = 1}}{\rm{.1}}\), thus:

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

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

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

Evaluate for \({\rm{\alpha = 2}}{\rm{.2}}\) and \({\rm{\beta = 1}}{\rm{.1}}\):

\(\begin{array}{l}E(Y) = \beta \Gamma \left( {1 + \frac{1}{\alpha }} \right) = 1.1\Gamma \left( {1 + \frac{1}{{2.2}}} \right) \approx 0.974187\\V(Y) = {\beta ^2}\left( {\Gamma \left( {1 + \frac{2}{\alpha }} \right) - \left( {\Gamma {{\left( {1 + \frac{1}{\alpha }} \right)}^2}} \right) = {{1.1}^2}\left( {\Gamma \left( {1 + \frac{2}{{2.2}}} \right) - {{\left( {\Gamma \left( {1 + \frac{1}{{2.2}}} \right)} \right)}^2}} \right) \approx 0.218503} \right.\end{array}\)

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

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

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

\(\begin{aligned} \mu &= E(X) \\ &= E(Y + \gamma )\\ &= E(Y) + \gamma }}\\ &= 0 {\rm{.974187 + 0}}{\rm{.5}}\\ &= 1 {\rm{.474187}}\\{{\rm{\sigma }}^{\rm{2}}} &= V(X) \\{\rm{ = V(Y + \gamma )}}\\ &= V(Y)\\ &= 0 {\rm{.218503}}\end{aligned}\)

The square root of the variance is the standard deviation:

\(\begin{array}{c}\sigma = \sqrt {{\sigma ^2}} \\ = \sqrt {0.218503} \approx 0.467443\end{array}\)

Therefore, The mean and standard deviation

\(\begin{array}{l}{\rm{\mu = 1}}{\rm{.474187}}\\{\rm{\sigma = 0}}{\rm{.467443}}\end{array}\)