91影视

The number of values below 'x' divided by the total number of values yields the percentile.

(a)The integral of the pdf of all values up to x is the cumulative distribution function.

\(\begin{aligned}F(x) &= \int_{{\rm{ - }}\infty }^{\rm{x}} {\rm{f}} {\rm{(x)dx}}\\&= \int_{\rm{1}}^{\rm{x}} {\rm{2}} \left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}}}} \right){\rm{dx}}\\&= \left. {\left( {{\rm{2x - }}\frac{{{\rm{2}}{{\rm{x}}^{{\rm{ - 1}}}}}}{{{\rm{ - 1}}}}} \right)} \right|_{\rm{1}}^{\rm{x}}\\&= \left. {\left( {{\rm{2x + }}\frac{{\rm{2}}}{{\rm{x}}}} \right)} \right|_{\rm{1}}^{\rm{x}}\\&= 2x + \frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}\end{aligned}\)

Since the cumulative distribution function is zero for all values smaller than the pdf's first nonzero value and one for all values larger than the pdf's last nonzero value, we get:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 1}}}\\{{\rm{2x + }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}}&{{\rm{1}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{1}}&{{\rm{x > 2}}}\end{array}} \right.\)

Therefore, the function is \({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 1}}}\\{{\rm{2x + }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}}&{{\rm{1}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{1}}&{{\rm{x > 2}}}\end{array}} \right.\).

Since the cumulative distribution function is zero for all values smaller than the pdf's first nonzero value and one for all values larger than the pdf's last nonzero value, we get:

\[{\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 1}}}\\{{\rm{2x + }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}}&{{\rm{1}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{1}}&{{\rm{x > 2}}}\end{array}} \right.\]

Therefore, the function is \[{\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{\rm{0}}&{{\rm{x < 1}}}\\{{\rm{2x + }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}}&{{\rm{1}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{1}}&{{\rm{x > 2}}}\end{array}} \right.\].

(b) The \({\rm{(100p)th}}\) percentile is defined as the value of \({\rm{x}}\) for which the probability of values smaller than \({\rm{x}}\) equals \({\rm{p}}\).

\(\begin{array}{c}{\rm{F(x) = P(Y}} \le {\rm{x)}}\\{\rm{ = p}}\end{array}\)

Fill in the following formula for\({\rm{F(x)}}\):

\(\begin{array}{c}{\rm{p = F(x)}}\\{\rm{ = 2x + }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ - 4}}\end{array}\)

\({\rm{p}}\)and\(\frac{{\rm{2}}}{{\rm{x}}}\)are subtracted from each side of the equation as follows:

\({\rm{ - }}\frac{{\rm{2}}}{{\rm{x}}}{\rm{ = 2x - 4 - p}}\)

Each side of the equation should be multiplied by x:

\({\rm{ - 2 = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ - (4 + p)x}}\)

To each side, add two more:

\({\rm{0 = 2}}{{\rm{x}}^{\rm{2}}}{\rm{ - (4 + p)x + 2}}\)

Using the quadratic formula, find the roots:

\(\begin{array}{c}{\rm{x = }}\frac{{{\rm{ - b \pm }}\sqrt {{{\rm{b}}^{\rm{2}}}{\rm{ - 4ac}}} }}{{{\rm{2a}}}}\\{\rm{ = }}\frac{{{\rm{(4 + p) \pm }}\sqrt {{{{\rm{(4 + p)}}}^{\rm{2}}}{\rm{ - 4(2)(2)}}} }}{{{\rm{2(2)}}}}\\{\rm{ = }}\frac{{{\rm{(4 + p) \pm }}\sqrt {{{\rm{p}}^{\rm{2}}}{\rm{ + 8p}}} }}{{\rm{4}}}\end{array}\)

Because only the positive root produces an\({\rm{x - }}\)value between\({\rm{1}}\)and\({\rm{2}}\)(with\({\rm{0 < p < 1}}\)):

\({\rm{x = }}\frac{{{\rm{(4 + p) + }}\sqrt {{{\rm{p}}^{\rm{2}}}{\rm{ + 8p}}} }}{{\rm{4}}}\)

with \({\rm{p = 0}}{\rm{.5}}\), \({\rm{\tilde \mu }}\) is the percentile:

\(\begin{aligned}\tilde \mu &= x \\ &= \frac{{{\rm{(4 + 0}}{\rm{.5) + }}\sqrt {{\rm{0}}{\rm{.}}{{\rm{5}}^{\rm{2}}}{\rm{ + 8(0}}{\rm{.5)}}} }}{{\rm{4}}}\\&= \frac{{{\rm{9 + }}\sqrt {{\rm{17}}} }}{{\rm{8}}}\\ \approx {\rm{1}}{\rm{.6404}}\end{aligned}\)

Therefore, the percentile is \(\frac{{{\rm{(4 + p) + }}\sqrt {{{\rm{p}}^{\rm{2}}}{\rm{ + 8p}}} }}{{\rm{4}}}\) and \( \approx {\rm{1}}{\rm{.6404}}\).

(c) The sum (integral) of each possibility \({\rm{x}}\) with its probability \({\rm{P(x)}}\) is the expected value (or mean) \({\rm{\mu }}\):

\(\begin{aligned}E(X) &= \int_{{\rm{ - }}\infty }^{{\rm{ + }}\infty } {\rm{x}} {\rm{f(x)dx}}\\&= \int_{\rm{1}}^{\rm{2}} {\rm{x}} {\rm{2}}\left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}}}} \right){\rm{dx}}\\&= \left. {\left( {\frac{{{\rm{2}}{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - 2 ln x}}} \right)} \right|_{\rm{1}}^{\rm{2}}\\&= \left. {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 2 ln x}}} \right)} \right|_{\rm{1}}^{\rm{2}}\\& = 3 - ln 4\end{aligned}\)

The expected value of the squared variation from the mean is the variance:

\(\begin{array}{c}{\rm{V(X) = }}\int_{{\rm{ - }}\infty }^{{\rm{ + }}\infty } {{\rm{(x - \mu )f(x)}}} {\rm{dx}}\\{\rm{ = }}\int_{\rm{1}}^{\rm{2}} {{\rm{(x - (}}} {\rm{3 - ln4)}}{{\rm{)}}^{\rm{2}}}{\rm{2}}\left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}}}} \right){\rm{dx}}\\{\rm{ = - }}\frac{{{\rm{19}}}}{{\rm{3}}}{\rm{ - l}}{{\rm{n}}^{\rm{2}}}{\rm{4 + ln4096}}\\ \approx {\rm{0}}{\rm{.0626}}\end{array}\)

Therefore, the values are \({\rm{3 - ln 4}}\) and \( \approx {\rm{0}}{\rm{.0626}}\).

(d) If the demand exceeds \({\rm{1}}{\rm{.5}}\) thousand gallons, there is no more available: \({\rm{h(x) = 0}}\) when \({\rm{x}} \ge {\rm{1}}{\rm{.5}}\).

The probability of attaining a value of at most\({\rm{x}}\)is given by the cumulative distribution function\({\rm{F}}\)at\({\rm{x}}\):

\(\begin{array}{c}{\rm{F(X) = P(X}} \le {\rm{x)}}\\{\rm{ = P(X < x)}}\end{array}\)

Rule of complements:

\({\rm{P(not A) = 1 - P(A)}}\)

Use the complement rule and the result of part (a) to solve this problem:

\(\begin{array}{c}{\rm{P(X}} \le {\rm{1}}{\rm{.5) = F(1}}{\rm{.5)}}\\{\rm{ = }}\left( {{\rm{2(1}}{\rm{.5) + }}\frac{{\rm{2}}}{{{\rm{1}}{\rm{.5}}}}{\rm{ - 4}}} \right)\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{3}}}\\ \approx {\rm{0}}{\rm{.3333 }}\end{array}\)

\(\begin{array}{c}{\rm{P(X > 1}}{\rm{.5) = 1 - P(X}} \le {\rm{1}}{\rm{.5)}}\\{\rm{ = 1 - F(1}}{\rm{.5)}}\\{\rm{ = 1 - }}\left( {{\rm{2(1}}{\rm{.5) + }}\frac{{\rm{2}}}{{{\rm{1}}{\rm{.5}}}}{\rm{ - 4}}} \right)\\{\rm{ = }}\frac{{\rm{2}}}{{\rm{3}}}\\ \approx {\rm{0}}{\rm{.6667}}\end{array}\)

When demand is less than\({\rm{1}}{\rm{.5}}\), the amount remaining equals the difference between\({\rm{1}}{\rm{.5}}\)thousand gallons and the demand:

\({\rm{h(x) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{1}}{\rm{.5 - f(x)}}}&{{\rm{1}} \le {\rm{x < 1}}{\rm{.5}}}\\{\rm{0}}&{{\rm{x}} \ge {\rm{1}}{\rm{.5}}}\end{array}} \right.\)

The sum (integral) of each possibility\({\rm{x}}\)with its probability\({\rm{P(x)}}\)is the expected value (or mean)\({\rm{\mu }}\):

\(\begin{aligned}E(h(x)) &= E(1{\rm{.5 - f(x))}}\\&= 1{\rm{.5 - E(f(x))}}\\&= 1 -\int_{\rm{1}}^{{\rm{1}}{\rm{.5}}} {\rm{x}} {\rm{f(x)dx}}\\&= 1 - \int_{\rm{1}}^{{\rm{1}}{\rm{.5}}} {\rm{x}} {\rm{2}}\left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{2}}}}}} \right){\rm{dx}}\\&= 1 - \left. {\left( {\frac{{{\rm{2}}{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - 2lnx}}} \right)} \right|_{\rm{1}}^{{\rm{1}}{\rm{.5}}}\\ &= 1 - \left. {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 2lnx}}} \right)} \right|_{\rm{1}}^{{\rm{1}}{\rm{.5}}}\\&= 1 - (1{\rm{.5}}{{\rm{)}}^{\rm{2}}}{\rm{ + 2ln1}}{\rm{.5 + }}{{\rm{1}}^{\rm{2}}}{\rm{ - 2ln1}}\\& = -\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + 2ln1}}{\rm{.5 - 2ln1}}\\ & = -\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + ln1}}{\rm{.}}{{\rm{5}}^{\rm{2}}}{\rm{ - ln}}{{\rm{1}}^{\rm{2}}}\\ &= -\frac{{\rm{1}}}{{\rm{4}}}{\rm{ + ln}}\frac{{{\rm{1}}{\rm{.}}{{\rm{5}}^{\rm{2}}}}}{{{{\rm{1}}^{\rm{2}}}}}\\& = - \frac{{\rm{1}}}{{\rm{4}}}{\rm{ + ln}}\frac{{\rm{9}}}{{\rm{4}}}\\ \approx {\rm{0}}{\rm{.5609}}\end{aligned}\)

Therefore, the value is\( \approx {\rm{0}}{\rm{.5609}}\).