anonymous
  • anonymous
Continuous random variable X has density given by f(x)=c⋅(x+1)*(x−3) if x∈[−1,3] and 0 otherwise. 1) Find DF(2;X) . distribution function 2) Find E[X] .
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Zarkon
  • Zarkon
what have you tried?
anonymous
  • anonymous
i try to use the integral by putting this together as x^2-2x-3
IrishBoy123
  • IrishBoy123
\(f(x)=c(x+1)(x−3)\) \(F(x) = \int\limits_{-1}^{3} \; dx \quad c(x^2-2x-3)\)

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More answers

anonymous
  • anonymous
\[\frac{ x^3 }{ 3 }- x^2 -3x\]
IrishBoy123
  • IrishBoy123
for the next bit, you might try \(G(x) = \int\limits_{-1}^{3} \; x \; dx \quad c(x^2-2x-3) = \int\limits_{-1}^{3} \; dx \quad c(x^3-2x^2-3x)\)
anonymous
  • anonymous
ok
Zarkon
  • Zarkon
you need to find the value of \(c\) such that \[\int\limits_{-1}^{3} c(x^2-2x-3)dx=1\]
anonymous
  • anonymous
is |dw:1446854671991:dw|
anonymous
  • anonymous
\[(3-9-9 )- (-\frac{ 1 }{ 3 }-1+3)\]
Zarkon
  • Zarkon
\[\frac{3^3}{3}=9\]
anonymous
  • anonymous
\[(-9)-(-\frac{ 5 }{ 3 })\]
anonymous
  • anonymous
32/2
Zarkon
  • Zarkon
\[(-9)-(\frac{ 5 }{ 3 })=-\frac{32}{3}\]
anonymous
  • anonymous
\[c=-\frac{ 3 }{ 32 }\]
Zarkon
  • Zarkon
yes
Zarkon
  • Zarkon
if you want the distribution function (CDF) then start by computing \[\int\limits_{-1}^{x} -\frac{3}{32}(t^2-2t-3)dt\]
anonymous
  • anonymous
isnt the integral ? \[\int\limits_{-2}^{2}\]
anonymous
  • anonymous
maybe I'm wrong ?
Zarkon
  • Zarkon
what is this notation "DF(2;X)"
anonymous
  • anonymous
yes
Zarkon
  • Zarkon
I don't think i asked a yes or no question What does that notation mean?
anonymous
  • anonymous
let be think about it
anonymous
  • anonymous
i guess the notion mean that (2) - prob (x<=2) and 1-prob (x>2)

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