91Ó°ÊÓ

Probability simply refers to the likelihood of something occurring. We may talk about the probabilities of particular outcomes—how likely they are—when we're unclear about the result of an event. Statistics is the study of occurrences guided by probability.

Given:

\(\begin{array}{l}{\rm{F(x;\alpha ,\beta ) = }}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (x - a)/\beta }}}}}}\\{\rm{\alpha = 150}}\\{\rm{\beta = 90}}\end{array}\)

(a) Evaluate \({\rm{F}}\)at \(150\) and \({\rm{300}}\) :

\(\begin{array}{l}{\rm{F(150;150,90) = }}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (150 - 150)/90}}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 1}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{e}}}\\{\rm{F(300;150,90) = }}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (300 - 150/90}}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - 1/}}{{\rm{e}}^{{\rm{5/3}}}}}}\end{array}\)

The cumulative distribution function \({\rm{F}}\)at \({\rm{X = x}}\) describes the probability to the left of \({\rm{X = x}}\):

\(\begin{aligned}P(X£150)&= F(150,150,90) \\&= \frac{1}{e}\gg 0.3679 \\&= 36.79\% \\P(X£300) &= F(300,150,90)\\&= {e^{ - 1/{e^{5/3}}}}\gg 0.8279 \\&= 82.79\% \\P(150£X£300) &= F(300,150,90) - F(150,150,90) \\&= {e^{ - 1/{e^{5/3}}}} - \frac{1}{e}\gg 0.4600 \\&= 46.00\% \\\end{aligned} \)

(b) We are looking for the value\({\rm{x}}\)for which \({\rm{90\% }}\) of the data values are below it and thus for which the cumulative distribution function is equal to \({\rm{90\% }}\) or\({\rm{0}}{\rm{.90}}\):

\({\rm{F(x;150,90) = 0}}{\rm{.90}}\)

Replace \({\rm{F}}\)by its known expression:

\({{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (x - 150)/90}}}}}}{\rm{ = 0}}{\rm{.90}}\)

Take the natural logarithm of each side:

\( - {e^{ - (x - 150)/90}} = {\rm{l}}n{e^{ - {e^{ - (x - 150)/90}}}} = \ln 0.90\)

Multiply each side by\({\rm{ - 1}}\):

\({{\rm{e}}^{{\rm{ - (x - 150)/90}}}}{\rm{ = - ln0}}{\rm{.90}}\)

Take the natural logarithm of each side:

\({\rm{ - }}\frac{{{\rm{x - 150}}}}{{{\rm{90}}}}{\rm{ = ln}}{{\rm{e}}^{{\rm{ - (x - 150)/90}}}}{\rm{ = ln( - ln0}}{\rm{.90)}}\)

Multiply each side by\({\rm{ - 90}}\):

\({\rm{x - 150 = - 90ln( - ln0}}{\rm{.90)}}\)

Add 50 to each side:

\({\rm{x = 150 - 90ln( - ln0}}{\rm{.90)\gg 352}}{\rm{.5331}}\)

(c) The density function of \({\rm{X}}\) is the derivative of the cumulative distribution function \({\rm{F}}\):

\(\begin{aligned}f(x) &= \frac{d}{{dx}}F(x;\alpha ,\beta ) \\&= \frac{d}{{dx}}{e^{ - {e^{ - (x - \alpha )/\beta }}}} \\ &= {e^{ - {e^{ - (x - \alpha )/\beta }}}}\frac{d}{{dx}}\left( { - {e^{ - (x - \alpha )/\beta }}} \right) \\&= - {e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}}\frac{d}{{dx}}\left( { - \frac{{x - \alpha }}{\beta }} \right) \\&= \frac{1}{\beta }{e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} \\\end{aligned} \)

(d) The mode is the value where the density function of \({\rm{X}}\)is highest and thus is the maximum a root of the derivative of the density function.

\(\begin{aligned}{f^¢}(x) &= \frac{d}{{dx}}f(x) = \frac{d}{{dx}}\frac{1}{\beta }{e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} \\&= \frac{1}{\beta }{e^{ - (x - \alpha )/\beta }}\frac{1}{\beta }{e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} - \frac{1}{\beta }\frac{1}{\beta }{e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} \\ &= \frac{1}{{{\beta ^2}}}{e^{ - 2(x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} - \frac{1}{{{\beta ^2}}}{e^{ - (x - \alpha )/\beta }}{e^{ - {e^{ - (x - \alpha )/\beta }}}} \\\end{aligned} \)

The maximum is a root of the derivative: \(\frac{{\rm{1}}}{{{{\rm{\beta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - 2(x - \alpha )/\beta }}}}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}}}{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{\beta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}}}{\rm{ = 0}}\)

Divide each side of the equation by\(\frac{{\rm{1}}}{{{{\rm{\beta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}{{\rm{e}}^{{\rm{ - }}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}}}\): \({{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}{\rm{ - 1 = 0}}\)

Add \({\rm{1}}\) to each side: \({{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}{\rm{ = 1}}\)

Take the natural logarithm of each side:

\({\rm{ - }}\frac{{{\rm{x - \alpha }}}}{{\rm{\beta }}}{\rm{ = ln}}{{\rm{e}}^{{\rm{ - (x - \alpha )/\beta }}}}{\rm{ = ln1 = 0}}\)

Multiply each side by \({\rm{ - \beta }}\) : \({\rm{x - \alpha = 0}}\)

Add \({\rm{\alpha }}\) to each side: \({\rm{x = \alpha }}\)

Thus, the maximum occurs at\({\rm{x = \alpha }}\)and thus the mode is at \({\rm{x = \alpha }}\).

(e) Given:

\({\rm{E(X)\gg 0}}{\rm{.5772\beta + \alpha }}\)

Replace \({\rm{\alpha }}\) by \({\rm{150}}\) and \({\rm{\beta = 90}}\) :

\({\rm{E(X)\gg 0}}{\rm{.5772(90) + 150 = 201}}{\rm{.948}}\)

The median is \({\rm{x = 150 - 90ln( - ln0}}{\rm{.50)\gg 182}}{\rm{.986}}\) (which can be found by replacing \({\rm{0}}{\rm{.9}}\) by \({\rm{0}}{\rm{.5}}\) in part (b)) and the mode is\({\rm{x = \alpha = 150}}\).

We can see that the mean is higher than both the mean and the median, indicating a right-skewed distribution (or positively skewed).