Mathematics, 12.07.2019 16:10 Shadowhawk681
Let x be a random variable with mean x = 25 and x = 6 and let y be a random variable with mean y = 30 and y = 4. it is known that x and y are independent random variables. suppose the random variables x and y are added together to create new random variable w (i. e., w = x + y). what is the standard deviation of w?
Answers: 3
has mean ![\mu_X=E[X]=25](/tpl/images/0081/5238/ca223.png)
and standard deviation ![\sigma_x=\sqrt{\mathrm{Var}[X]}=6](/tpl/images/0081/5238/e46f6.png)
, and 
has means ![\mu_Y=E[Y]=30](/tpl/images/0081/5238/84bfd.png)
and standard deviation ![\sigma_Y=\sqrt{\mathrm{Var}[Y]}=4](/tpl/images/0081/5238/12ea2.png)
.
, then 
has mean![E[W]=E[X+Y]=E[X]+E[Y]=55](/tpl/images/0081/5238/9af55.png)
![\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2](/tpl/images/0081/5238/b613e.png)
![\mathrm{Var}[X]=36](/tpl/images/0081/5238/0b42a.png)
and ![\mathrm{Var}[Y]=16](/tpl/images/0081/5238/8902b.png)
, we have![\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661](/tpl/images/0081/5238/b0f2c.png)
![\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916](/tpl/images/0081/5238/e8315.png)
![E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2]](/tpl/images/0081/5238/ded03.png)
![E[XY]=E[X]E[Y]](/tpl/images/0081/5238/a04e7.png)
, and![E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077](/tpl/images/0081/5238/8143b.png)
![\mathrm{Var}[W]=3077-55^2=52](/tpl/images/0081/5238/30e7f.png)
![\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}](/tpl/images/0081/5238/b88fd.png)
![\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y]](/tpl/images/0081/5238/a372f.png)
![\mathrm{Var}[W]=36+16=52](/tpl/images/0081/5238/3161f.png)
, and so the standard deviation is again 
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Let x be a random variable with mean x = 25 and x = 6 and let y be a random variable with mean y...
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