91影视

The average of a group of numbers is simply defined as the mean. In statistics, the mean is also one of the indicators of central tendency.

\({\rm{f(x) = }}\frac{{\rm{k}}}{{{{\rm{x}}^{\rm{4}}}}}\)

If the integral of over all possible values of\({\rm{x}}\)equals one, the function\({\rm{f}}\)is a valid pdf.

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

The integral must be equal to one, as follows:

\(\frac{{\rm{k}}}{{\rm{3}}}{\rm{ = 1}}\)

Multiply each side by three to get the following result:

\({\rm{k = 3}}\)

Therefore, \({\rm{k = 3}}\).

\(\begin{array}{c}{\rm{F(x) = }}\int_{{\rm{ - }}\infty }^{\rm{x}} {\rm{f}} {\rm{(x)dx}}\\{\rm{ = }}\int_{\rm{1}}^{\rm{x}} {\frac{{\rm{3}}}{{{{\rm{x}}^{\rm{4}}}}}} {\rm{dx}}\\{\rm{ = }}\left. {\left( {\frac{{\rm{3}}}{{{\rm{ - 3}}{{\rm{x}}^{\rm{3}}}}}} \right)} \right|_{\rm{1}}^{\rm{x}}\\{\rm{ = }}\left. {\left( {{\rm{ - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{3}}}}}} \right)} \right|_{\rm{1}}^{\rm{x}}\\{\rm{ = - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{3}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{1}}^{\rm{3}}}}}\\{\rm{ = 1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{3}}}}}\end{array}\)

So, because cumulative distribution function is zero whenever the pdf is zero, we get:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{x}}^{\rm{3}}}}}}&{{\rm{x > 1}}}\\{\rm{0}}&{{\rm{x}} \le {\rm{1}}}\end{array}} \right.\)

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

Rule of complements:

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

Use the complement rule and the cdf from part (b) to solve this problem:

\(\begin{aligned}P(X > 2) &= 1 - P(X \le {\rm{2)}}\\ &= 1 - F(2) \\ &= 1 - \left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{3}}}}}} \right)\\ &= \frac{{\rm{1}}}{{\rm{8}}}\\ &= 0 {\rm{.125}}\\P(2 < x < 3) &= P(X < 3) - P(X < 2) \\ &= F(3) - F(2) \\ &= \left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{3}}^{\rm{3}}}}}} \right){\rm{ - }}\left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{\rm{2}}^{\rm{3}}}}}} \right)\\ &= \frac{{\rm{1}}}{{\rm{8}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{27}}}}\\ &= \frac{{{\rm{19}}}}{{{\rm{216}}}}\\ \approx {\rm{0}}{\rm{.0880}}\end{aligned}\)

Therefore, the value is \( \approx {\rm{0}}{\rm{.0880}}\).

\(\begin{aligned}\mu &= \int_{{\rm{ - }}\infty }^{{\rm{ + }}\infty } {\rm{x}} {\rm{f(x)dx}}\\ &= \int_{\rm{1}}^{{\rm{ + }}\infty } {\rm{x}} \frac{{\rm{3}}}{{{{\rm{x}}^{\rm{4}}}}}{\rm{dx}}\\ &= \int_{\rm{1}}^{{\rm{ + }}\infty } {\rm{3}} {{\rm{x}}^{{\rm{ - 3}}}}{\rm{dx}}\\ &= \left. {\left( {\frac{{{\rm{3}}{{\rm{x}}^{{\rm{ - 2}}}}}}{{{\rm{ - 2}}}}} \right)} \right|_{\rm{1}}^{{\rm{ + }}\infty }\\ &= \frac{{\rm{3}}}{{\rm{2}}}\\ &= 1{\rm{.5}}\end{aligned}\)

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

\(\begin{aligned}{{\rm{\sigma }}^{\rm{2}}} &= \int_{{\rm{ - }}\infty }^{{\rm{ + }}\infty } {{{{\rm{(x - \mu )}}}^{\rm{2}}}} {\rm{f(x)dx}}\\ &= \int_{\rm{1}}^{{\rm{ + }}\infty } {{{{\rm{(x - 1}}{\rm{.5)}}}^{\rm{2}}}} \frac{{\rm{3}}}{{{{\rm{x}}^{\rm{4}}}}}{\rm{dx}}\\ &= \int_{\rm{1}}^{{\rm{ + }}\infty } {\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - 3x + 2}}{\rm{.25}}} \right)} {\rm{3}}{{\rm{x}}^{{\rm{ - 4}}}}{\rm{dx }}\\ &= \int_{\rm{1}}^{{\rm{ + }}\infty } {\left( {{\rm{3}}{{\rm{x}}^{{\rm{ - 2}}}}{\rm{ - 9}}{{\rm{x}}^{{\rm{ - 3}}}}{\rm{ + 6}}{\rm{.75}}{{\rm{x}}^{{\rm{ - 4}}}}} \right)} {\rm{dx}}\\ &= \left. {\left( {\frac{{{\rm{3}}{{\rm{x}}^{{\rm{ - 1}}}}}}{{{\rm{ - 1}}}}{\rm{ + }}\frac{{{\rm{ - 9}}{{\rm{x}}^{{\rm{ - 2}}}}}}{{{\rm{ - 2}}}}{\rm{ + }}\frac{{{\rm{6}}{\rm{.75}}{{\rm{x}}^{{\rm{ - 3}}}}}}{{{\rm{ - 3}}}}} \right)} \right|_{\rm{1}}^{{\rm{ + }}\infty }\\ &= 3 - \frac{{\rm{9}}}{{\rm{2}}}{\rm{ + }}\frac{{{\rm{6}}{\rm{.75}}}}{{\rm{3}}}\\ &= \frac{{\rm{3}}}{{\rm{4}}}\\ &= 0 {\rm{.75}}\end{aligned}\)

The square root of the variance is the standard deviation:

\(\begin{array}{c}{\rm{\sigma = }}\sqrt {{{\rm{\sigma }}^{\rm{2}}}} \\{\rm{ = }}\sqrt {\frac{{\rm{3}}}{{\rm{4}}}} \\{\rm{ = }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\\ \approx {\rm{0}}{\rm{.8660}}\end{array}\)

Therefore, the value is \( \approx {\rm{0}}{\rm{.8660}}\).

\(\begin{aligned} mu - \sigma &= \frac{{\rm{3}}}{{\rm{2}}}{\rm{ - }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\\ &= \frac{{{\rm{3 - }}\sqrt {\rm{3}} }}{{\rm{2}}}\\ mu + \sigma &= \frac{{\rm{3}}}{{\rm{2}}}{\rm{ + }}\frac{{\sqrt {\rm{3}} }}{{\rm{2}}}\\ &= \frac{{{\rm{3 + }}\sqrt {\rm{3}} }}{{\rm{2}}}\end{aligned}\)

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

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

Using the cdf from section (b) and the knowledge that\(\frac{{{\rm{3 - }}\sqrt {\rm{3}} }}{{\rm{2}}}\)is smaller than one:

\(\begin{aligned}{\rm{P}}\left( {\frac{{{\rm{3 - }}\sqrt {\rm{3}} }}{{\rm{2}}}{\rm{ < x < }}\frac{{{\rm{3 + }}\sqrt {\rm{3}} }}{{\rm{2}}}} \right) &= P \left( {{\rm{X < }}\frac{{{\rm{3 + }}\sqrt {\rm{3}} }}{{\rm{2}}}} \right){\rm{ - P}}\left( {{\rm{X < }}\frac{{{\rm{3 - }}\sqrt {\rm{3}} }}{{\rm{2}}}} \right)\\ &= F\left( {\frac{{{\rm{3 + }}\sqrt {\rm{3}} }}{{\rm{2}}}} \right){\rm{ - F}}\left( {\frac{{{\rm{3 - }}\sqrt {\rm{3}} }}{{\rm{2}}}} \right)\\&= \left( {{\rm{1 - }}\frac{{\rm{1}}}{{{{{\rm{((3 + }}\sqrt {\rm{3}} {\rm{)/2)}}}^{\rm{3}}}}}} \right){\rm{ - 0}}\\ &= \frac{{{\rm{10}}\sqrt {\rm{3}} }}{{\rm{9}}}{\rm{ - 1 - 0}}\\ &= \frac{{{\rm{10}}\sqrt {\rm{3}} }}{{\rm{9}}}{\rm{ - 1}}\\ \approx {\rm{0}}{\rm{.9245}}\end{aligned}\)

Therefore, the value is \( \approx {\rm{0}}{\rm{.9245}}\).