anonymous
  • anonymous
Statistics: Let X have the density function f(x) = cx(1-x) for 0<=x<=1 and f(x)=0 otherwise. Find c. Do I integrate the function between the values 0 and 1?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TuringTest
  • TuringTest
why not?\[\int_{0}^{1}cx(1-x)dx\]
TuringTest
  • TuringTest
...=1 I guess, right?
anonymous
  • anonymous
How did you get that? I brought the c outside what's being integrated and then multiplied out the function. So now I have \[c \int\limits_{0}^{1}(x^2/2)-(x^3/3)dx\] Then subbing in 1 and 0 I ended up with c*1/6. Have I done that much correctly?

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TuringTest
  • TuringTest
you have already integrated it, but you left the integral sign o-0 ?
anonymous
  • anonymous
Sorry. That's a typo :D
TuringTest
  • TuringTest
\[\int_{0}^{1}cx(1-x)dx=1\]\[c\int x-x^2dx=1\]\[c(\frac12x^2-\frac13x^3)|_{0}^{1}=1\]
anonymous
  • anonymous
right you should get c(1/2 - 1/3) =1, then solve for c
TuringTest
  • TuringTest
\[c(\frac12-\frac13)=c\frac16=1\implies c=6\]
anonymous
  • anonymous
Oh right, sorry Turing, I thought you were getting 1 as the answer for the integration. :D
anonymous
  • anonymous
So that's a rule that applies to all functions in pdf, is it? You just let them equal to 1?
TuringTest
  • TuringTest
for a probability density functions \[\int_{-\infty}^{\infty}f(x)dx=1\]since the function is zero everywhere except the interval [0,1], we only needed to integrate along those bounds
TuringTest
  • TuringTest
if you have never seen that formula before, try to think about the logic behind it.
anonymous
  • anonymous
I'm guessing it's equal to one because of the rule in statistics that all probabilities must be between 0 and 1?

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