Вероятности и случайни величини

[babyqq7]

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Моля някой който разбира да ми помогне!!! Благодаря предварително!

[Добромир Глухаров]

$\begin{array}{ccccc}X&-5&0&5&7\\P&0,2&0,3&0,1&0,4\end{array}$

$F_Y(t)=\begin{cases}0;\ t\le-4\\0,3;\ t\in(-4;1]\\0,6;\ t\in(1;2]\\1;\ t>2\end{cases}$

$P(Y=u)=F(u+0)-F(u)$

$\begin{array}{cccc}Y&-4&1&2\\P&0,3&0,3&0,4\end{array}$

$F_X(t)=\begin{cases}0;\ t\le-5\\0,2;\ t\in(-5;0]\\0,5;\ t\in(0;5]\\0,6;\ t\in(5;7]\\1;\ t>7\end{cases}$

$P(2X-3Y+6)=P(X).P(Y)$

$\begin{array}{c|c|c|c|c|c|c}X&-5&-5&-5&0&0&0\\Y&-4&1&2&-4&1&2\\2X-3Y+6&2.(-5)-3.(-4)+6=8&2.(-5)-3.1+6=-7&2.(-5)-3.2+6=-10&18&3&0\\P(2X-3Y+6)&0,2.0,3&0,2.0,3&0,2.0,4&0,3.0,3&0,3.0,3&0,3.0,4\end{array}$

$\begin{array}{c|c|c|c|c|c|c}X&5&5&5&7&7&7\\Y&-4&1&2&-4&1&2\\2X-3Y+6&28&13&10&32&17&14\\P(2X-3Y+6)&0,1.0,3&0,1.0,3&0,1.0,4&0,4.0,3&0,4.0,3&0,4.0,1\end{array}$


$E(2X-3Y+6)=8.0,2.0,3-7.0,2.0,3-10.0,2.0,4+18.0,3.0,3+3.0,3.0,3+0+28.0,1.0,3+13.0,1.0,3+10.0,1.0,4+32.0,4.0,3+17.0,4.0,3+14.0,4.0,1=9,22$

$P(X-2Y+3)=P(X).P(Y)$

$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c}X&-5&-5&-5&0&0&0&5&5&5&7&7&7\\Y&-4&1&2&-4&1&2&-4&1&2&-4&1&2\\X-2Y+3&6&-4&-6&11&1&-1&16&6&4&18&8&6\\P(X-2Y+3)&0,2.0,3&0,2.0,3&0,2.0,4&0,3.0,3&0,3.0,3&0,3.0,4&0,1.0,3&0,1.0,3&0,1.0,4&0,4.0,3&0,4.0,3&0,4.0,1\end{array}$

$E(X-2Y+3)=6.0,2.0,3-4.0,2.0,3-6.0,2.0,4+11.0,3.0,3+1.0,3.0,3-1.0,3.0,4+16.0,1.0,3+6.0,1.0,3+4.0,1.0,4+18.0,4.0,3+8.0,4.0,3+6.0,4.0,1=4,78$

$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c}X&-5&-5&-5&0&0&0&5&5&5&7&7&7\\Y&-4&1&2&-4&1&2&-4&1&2&-4&1&2\\(X-2Y+3)^2&36&16&36&121&1&1&256&36&16&324&64&36\\P(X-2Y+3)&0,06&0,06&0,08&0,09&0,09&0,12&0,03&0,03&0,04&0,12&0,12&0,04\end{array}$

$E((X-2Y+3)^2)=36.0,06+16.0,06+36.0,08+121.0,09+1.0,09+1.0,12+256.0,03+36.0,03+16.0,04+324.0,12+64.0,12+36.0,04=74,5$

$D(X-2Y+3)=E((X-2Y+3)^2)-(E(X-2Y+3))^2=74,5-4,78^2=51,6516$

[Добромир Глухаров]

А може да се използват и формулите:

$E(2X-3Y+6)=2E(X)-3E(Y)+6$

$E(X-2Y+3)=E(X)-2E(Y)+3$

$D(X-2Y+3)=D(X)+2^2D(Y)=D(X)+4D(Y)$

$E(X)=-5.0,2+0+5.0,1+7.0,4=2,3$

$E(Y)=-4.0,3+1.0,3+2.0,4=-0,1$

$D(X)=(-5)^2.0,2+5^2.0,1+7^2.0,4=5+2,5+1,96=9,46$

$D(Y)=16.0,3+0,3+4.0,4=6,7$