91影视

The unknown number is represented by a variable, which is an alphabet. It represents the worth of something. A variable is a quantity that can be altered to solve a mathematical issue.

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < - 2}}}\\{\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4x - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)}&{{\rm{ - 2}} \le {\rm{x < 2}}}\\{\rm{1}}&{{\rm{2}} \le {\rm{x}}}\end{array}} \right.\)

Then, using the cdf as a reference point, we can write:

\(\begin{array}{c}{\rm{P(X < 0) = F(0)}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4(0) - }}\frac{{{{{\rm{(0)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)\\{\rm{P(X < 0) = 0}}{\rm{.5}}\end{array}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as parameters\({\rm{(x)}}\). After that, for any number a,

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

Therefore, the value is \({\rm{0}}{\rm{.5}}\).

\(\begin{aligned} P( - 1 < X < 1) &= F(1) - F( - 1) \\ &= \left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4(1) - }}\frac{{{{\rm{1}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4( - 1) - }}\frac{{{{{\rm{( - 1)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\ &= \left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {\frac{{{\rm{11}}}}{{\rm{3}}}} \right)} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{ - }}\frac{{{\rm{11}}}}{{\rm{3}}}} \right)} \right)\\ &= \left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{{\rm{11}}}}{{{\rm{32}}}}} \right){\rm{ - }}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{11}}}}{{{\rm{32}}}}} \right)\\ &= \frac{{{\rm{11}}}}{{{\rm{16}}}}\\ P( - 1 < X < 1) &= 0 {\rm{.6875}}\end{aligned}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as a proposition\({\rm{(x)}}\). Then, using\({\rm{a < b}}\), for any two numbers a and b,

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

Therefore, the value is \({\rm{0}}{\rm{.6875}}\).

\(\begin{array}{c}{\rm{P(X > 0}}{\rm{.5) = 1 - F(0}}{\rm{.5)}}\\{\rm{ = 1 - }}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4(0}}{\rm{.5) - }}\frac{{{{{\rm{(0}}{\rm{.5)}}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\{\rm{ = 1 - }}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}{\rm{(1}}{\rm{.9583)}}} \right)\\{\rm{ = 1 - (0}}{\rm{.5 + 0}}{\rm{.1836)}}\\{\rm{P(X > 0}}{\rm{.5) = 0}}{\rm{.3164}}\end{array}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf \({\rm{F}}\)as parameters\({\rm{(x)}}\). After that, for any number a,

\({\rm{P(X > a) = 1 - F(a)}}\)

Therefore, the value is \({\rm{0}}{\rm{.3164}}\).

Because\({\rm{F(x)}}\)is constant between\({\rm{x < - 2}}\)and\({\rm{x}} \ge {\rm{2}}\),\({\rm{f(x) = 0}}\)for these intervals. In the case of\({\rm{ - 2}} \le {\rm{x < 2}}\),\({\rm{f(x)}}\)can be represented as follows:

\(\begin{array}{c}{\rm{f(x) = }}\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4x - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)} \right)\\{\rm{ = }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4 - }}\frac{{{\rm{3 \times }}{{\rm{x}}^{\rm{2}}}}}{{\rm{3}}}} \right)\\{\rm{f(x) = 0}}{\rm{.09375}}\left( {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} \right)\end{array}\)

Finally,\({\rm{f(x)}}\)as can be written as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < - 2}}}\\{{\rm{0}}{\rm{.09375}}\left( {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} \right)}&{{\rm{ - 2}} \le {\rm{x < 2}}}\\{\rm{1}}&{{\rm{2}} \le {\rm{x}}}\end{array}} \right.\)

We can ensure that\({\rm{f(x)}}\)supplied in Exercise\({\rm{3}}\)is the same as\({{\rm{F}}^{\rm{'}}}{\rm{(x)}}\)by comparing

Proposition:

If X is a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F(x)}}\), then the derivative\({{\rm{F}}^{\rm{'}}}{\rm{(x)}}\)occurs at every x.

\({\rm{f(x) = }}{{\rm{F}}^{\rm{'}}}{\rm{(x)}}\)

Therefore, \({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < - 2}}}\\{{\rm{0}}{\rm{.09375}}\left( {{\rm{4 - }}{{\rm{x}}^{\rm{2}}}} \right)}&{{\rm{ - 2}} \le {\rm{x < 2}}}\\{\rm{1}}&{{\rm{2}} \le {\rm{x}}}\end{array}} \right.\).

\({\rm{F(\tilde \mu ) = 0}}{\rm{.5}}\)

We prepare the provided cdf as follows:

\(\frac{{\rm{1}}}{{\rm{2}}}{\rm{ + }}\frac{{\rm{3}}}{{{\rm{32}}}}\left( {{\rm{4\tilde \mu - }}\frac{{{{{\rm{\tilde \mu }}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ = 0}}{\rm{.5}}\)

We obtain the following when we multiply both sides by\({\rm{32}}\):

\(\begin{array}{c}{\rm{16 + 3}}\left( {{\rm{4\tilde \mu - }}\frac{{{{{\rm{\tilde \mu }}}^{\rm{3}}}}}{{\rm{3}}}} \right){\rm{ = 16}}\\{\rm{16 + 12\tilde \mu - }}{{{\rm{\tilde \mu }}}^{\rm{3}}}{\rm{ = 16}}\\{{{\rm{\tilde \mu }}}^{\rm{3}}}{\rm{ = 12\tilde \mu }}\end{array}\)

\({\rm{\tilde \mu }}\)can take three different forms in this equation:\({\rm{0, - }}\sqrt {{\rm{12}}} {\rm{,sqrt12}}\). However, because it must fall between\({\rm{( - 2,2)}}\)and\({\rm{( - 2,2)}}\),

\({\rm{\tilde \mu = 0}}\)

Allow p to be a number between\({\rm{0}}\)and\({\rm{1}}\). The\({{\rm{(100p)}}^{{\rm{th}}}}\)percentile of a continuous\({\rm{rv}}\)X's distribution, abbreviated by\(\left( {{{\rm{\eta }}_{\rm{p}}}} \right)\), is defined as

\(\begin{array}{c}{\rm{p = F}}\left( {{{\rm{\eta }}_{\rm{p}}}} \right)\\{\rm{ = }}\int_{{\rm{ - }}\infty }^{{{\rm{\eta }}_{\rm{p}}}} {\rm{f}} {\rm{(y)dy}}\end{array}\)

The value of p for median is\({\rm{0}}{\rm{.5}}\).

Therefore, \({\rm{\tilde \mu = 0}}\).