91影视

The calculation of an integral is known as integration. Integrals are used in mathematics to calculate a variety of useful quantities such as areas, volumes, displacement, and so on. When we talk about integrals, we usually mean definite integrals.

(a) Determine the \({\rm{f(x;\theta ,\tau )}}\) equation with \({\rm{\tau = 80}}\) and \({\rm{\theta = 4}}\) :

\(\begin{gathered}f(x;4,80) &= \left\{ {\begin{array}{*{20}{c}}{\frac{4}{{80}}{{\left( {1 - \frac{x}{{80}}} \right)}^{4 - 1}}}&{0 \leqslant x < 80} \\ 0&{{\text{ }}otherwise{\text{ }}} \end{array}} \right. \\ &= \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{20}}{{\left( {1 - \frac{x}{{80}}} \right)}^3}}&{0 \leqslant x < 80} \\ 0&{{\text{ }}otherwise{\text{ }}} \end{array}} \right. \\ \end{gathered} \)

Determine the \({\rm{f(x;\theta ,\tau )}}\) equation with \({\rm{\tau = 80}}\) and \({\rm{\theta = 1}}\) :

\(\begin{gathered}f(x;1,80) &= \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{80}}{{\left( {1 - \frac{x}{{80}}} \right)}^{1 - 1}}}&{0 \leqslant x < 80} \\ 0&{{\text{ }}otherwise{\text{ }}} \end{array}} \right. \\ &= \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{80}}{{\left( {1 - \frac{x}{{80}}} \right)}^0}}&{0 \leqslant x < 80} \\ 0&{{\text{ }}otherwise{\text{ }}}\end{array}} \right. \\ &= \left\{ {\begin{array}{*{20}{c}}{\frac{1}{{80}}}&{0 \leqslant x < 80} \\ 0&{{\text{ }}otherwise{\text{ }}} \end{array}} \right. \\ \end{gathered} \)

\({\rm{\theta = 0}}{\rm{.5}}\)

Determine the \({\rm{f(x;\theta ,\tau )}}\) equation with \({\rm{\tau = 80}}\) and \({\rm{\theta = 0}}{\rm{.5}}\) :

\(\begin{aligned}{\rm{f(x;0}}.5,80) &= \left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{0}}{\rm{.5}}}}{{{\rm{80}}}}{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{{\rm{80}}}}} \right)}^{{\rm{0}}{\rm{.5 - 1}}}}}&{{\rm{0}} \le {\rm{x < 80}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\\&= \left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{1}}}{{{\rm{160}}}}{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{{\rm{80}}}}} \right)}^{{\rm{ - 0}}{\rm{.5}}}}}&{{\rm{0}} \le {\rm{x < 80}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\end{aligned}\)

Therefore,

Because the graph is a horizontal line, we can see that the distribution is uniform if \({\rm{\theta = 1}}\).

If \({\rm{\theta > 1}}\), the distribution is skewed to the right (or positively skewed), because the peak is to the left for \({\rm{\theta = 4}}\).

Because the peak in the distribution is to the right for \({\rm{\theta = 0}}{\rm{.5}}\), the distribution is skewed to the left (or negatively skewed) \({\rm{\theta < 1}}\).

(b)The integral of the pdf\({\rm{f}}\)over all values up to\({\rm{x}}\)is the cumulative distribution function of\({\rm{X}}\)at\({\rm{x}}\):

\(\begin{array}{c}{\rm{F(x;\theta ,\tau ) = }}\int_{\rm{0}}^{\rm{x}} {\rm{f}} {\rm{(x;\theta ,\tau )dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{x}} {\frac{{\rm{\theta }}}{{\rm{\tau }}}} {\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{{\rm{\theta - 1}}}}{\rm{dx}}\end{array}\)

Let\({\rm{u = 1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}{\rm{ }}\)and\({\rm{du = - }}\frac{{\rm{1}}}{{\rm{\tau }}}{\rm{dx:}}\)

\(\begin{array}{l}{\rm{ = }}\int_{{\rm{x = 0}}}^{{\rm{x = x}}} {\rm{ - }} {\rm{\theta }}{{\rm{u}}^{{\rm{\theta - 1}}}}{\rm{du}}\\{\rm{ = - \theta }}\int_{{\rm{x = 0}}}^{{\rm{x = x}}} {{{\rm{u}}^{{\rm{\theta - 1}}}}} {\rm{du}}\\{\rm{ = - }}\left. {{\rm{\theta }}\left( {\frac{{{{\rm{u}}^{\rm{\theta }}}}}{{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\\{\rm{ = - }}\left. {\left( {{{\rm{u}}^{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\\{\rm{ = - }}\left. {\left( {{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)}^{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\\{\rm{ = - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}{\rm{ + (1)}}\\{\rm{ = 1 - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}\end{array}\)

Therefore,

When \({\rm{x}}\) is smaller than \({\rm{0}}\), the cumulative distribution function \({\rm{F}}\) is zero, and when \({\rm{x}}\) is larger than \({\rm{\tau }}\), the cumulative distribution function \({\rm{F}}\) is one:

(c)Part (b) revealed the cumulative distribution function:

\({\rm{F(x;\theta ,\tau ) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 0}}}\\{{\rm{1 - }}{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)}^{\rm{\theta }}}}&{{\rm{0}} \le {\rm{x < \tau }}}\\{\rm{1}}&{{\rm{x}} \ge {\rm{\tau }}}\end{array}} \right.\)

The median is defined as the value x at which the cumulative distributive function equals\(\frac{{\rm{1}}}{{\rm{2}}}\).

\(\begin{array}{c}{\rm{1 - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\\{\rm{ - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}{\rm{ = - }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(Subtract 1 from each side)}}\\{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{(Multiply each side by - 1)}}\\{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}{\rm{ = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{{\rm{1/\theta }}}}{\rm{(Take }}\frac{{\rm{1}}}{{\rm{\theta }}}{\rm{th power of each side)}}\\{\rm{ - }}\frac{{\rm{x}}}{{\rm{\tau }}}{\rm{ = }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{{\rm{1/\theta }}}}{\rm{ - 1(Subtract 1 from each side)}}\\\frac{{\rm{x}}}{{\rm{\tau }}}{\rm{ = 1 - }}{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)^{{\rm{1/\theta }}}}{\rm{(Multiply each side by - 1)}}\\{\rm{x = \tau }}\left( {{\rm{1 - }}{{\left( {\frac{{\rm{1}}}{{\rm{2}}}} \right)}^{{\rm{1/\theta }}}}} \right){\rm{(Multiply each side by \tau )}}\\{\rm{x = \tau - }}\frac{{\rm{\tau }}}{{{{\rm{2}}^{{\rm{1/\theta }}}}}}\end{array}\)

Therefore, the median is \({\rm{\tau - }}\frac{{\rm{\tau }}}{{{{\rm{2}}^{{\rm{1/\theta }}}}}}\).

(d)The integral of the pdf\({\rm{f}}\)over all values up to\({\rm{x}}\)is the cumulative distribution function of\({\rm{X}}\)at\({\rm{x}}\):

\(\begin{array}{c}{\rm{F(x;\theta ,\tau ) = }}\int_{\rm{0}}^{\rm{x}} {\rm{f}} {\rm{(x;\theta ,\tau )dx}}\\{\rm{ = }}\int_{\rm{0}}^{\rm{x}} {\frac{{\rm{\theta }}}{{\rm{\tau }}}} {\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{{\rm{\theta - 1}}}}{\rm{dx}}\end{array}\)

Let \({\rm{u = 1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}{\rm{ }}\)and \({\rm{du = - }}\frac{{\rm{1}}}{{\rm{\tau }}}{\rm{dx:}}\)

\(\begin{array}{l}{\rm{ = }}\int_{{\rm{x = 0}}}^{{\rm{x = x}}} {\rm{ - }} {\rm{\theta }}{{\rm{u}}^{{\rm{\theta - 1}}}}{\rm{du}}\\{\rm{ = - \theta }}\int_{{\rm{x = 0}}}^{{\rm{x = x}}} {{{\rm{u}}^{{\rm{\theta - 1}}}}} {\rm{du}}\\{\rm{ = - }}\left. {{\rm{\theta }}\left( {\frac{{{{\rm{u}}^{\rm{\theta }}}}}{{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\\{\rm{ = - }}\left. {\left( {{{\rm{u}}^{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\end{array}\)

\(\begin{array}{l}{\rm{ = - }}\left. {\left( {{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)}^{\rm{\theta }}}} \right)} \right|_{{\rm{x = 0}}}^{{\rm{x = x}}}\\{\rm{ = - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}{\rm{ + (1)}}\\{\rm{ = 1 - }}{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)^{\rm{\theta }}}\end{array}\)

When\({\rm{x}}\)is smaller than\({\rm{0}}\), the cumulative distribution function\({\rm{F}}\)is zero, and when\({\rm{x}}\)is larger than\({\rm{\tau }}\), the cumulative distribution function\({\rm{F}}\)is one:

\({\rm{F(x;\theta ,\tau ) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 0}}}\\{{\rm{1 - }}{{\left( {{\rm{1 - }}\frac{{\rm{x}}}{{\rm{\tau }}}} \right)}^{\rm{\theta }}}}&{{\rm{0}} \le {\rm{x < \tau }}}\\{\rm{1}}&{{\rm{x}} \ge {\rm{\tau }}}\end{array}} \right.\)

The difference of the cumulative distribution function at either endpoint is then used to calculate the probability\({\rm{P(50}} \le {\rm{X}} \le {\rm{70)}}\):

\(\begin{aligned}{\rm{P(50}} \le {\rm{X}} \le 70) &= F(70,4,80) - F(50,4,80) \\ &= 1 - {\left( {{\rm{1 - }}\frac{{{\rm{70}}}}{{{\rm{80}}}}} \right)^{\rm{4}}}{\rm{ - }}\left( {{\rm{1 - }}{{\left( {{\rm{1 - }}\frac{{{\rm{50}}}}{{{\rm{80}}}}} \right)}^{\rm{4}}}} \right)\\ &= \frac{{{\rm{4095}}}}{{{\rm{4096}}}}{\rm{ - }}\frac{{{\rm{4015}}}}{{{\rm{4096}}}}\\ &= \frac{{{\rm{80}}}}{{{\rm{4096}}}}\\ &= \frac{{\rm{5}}}{{{\rm{256}}}}\\ \approx {\rm{0}}{\rm{.195}}\\{\rm{ = 1}}{\rm{.95\% }}\end{aligned}\)

Therefore, the probability is \({\rm{1}}{\rm{.95\% }}\).