UnkleRhaukus
  • UnkleRhaukus
find \[\langle \theta \rangle \] for \(0≤\theta≤\pi/2\)
Probability
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
\[\langle x\rangle=\int\limits_0^{\pi/2}\theta\cdot\rho(\theta)\cdot\text d\theta \]
hartnn
  • hartnn
whats \(\rho(\theta)\) ?
UnkleRhaukus
  • UnkleRhaukus
probability density

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UnkleRhaukus
  • UnkleRhaukus
\[1=\int\limits_0^{\pi/2}\rho(\theta)\cdot\text d\theta\]
UnkleRhaukus
  • UnkleRhaukus
assuming a unifirm probability distribution \[\rho(\theta)=A\] \[\frac 1A=\int\limits_0^{\pi/2}\text d\theta\]
hartnn
  • hartnn
A=2/\(\pi\) then
UnkleRhaukus
  • UnkleRhaukus
\[\frac 1A=\theta|_0^{\pi/2}=\frac\pi2\] \[A=\frac 2\pi\]
UnkleRhaukus
  • UnkleRhaukus
so, \[\langle x\rangle=\frac 2\pi\int\limits_0^{\pi/2}\theta\cdot\text d\theta\]
hartnn
  • hartnn
pi/4 ?
UnkleRhaukus
  • UnkleRhaukus
\[=\frac 2\pi \frac{\theta^2}{2} |_0^{\pi/2}\] \[=\frac 2\pi \frac{\left(\pi/2\right)^2}{2}\] \[=\frac\pi 4\]
hartnn
  • hartnn
nice, pdf was assumed or given? assuming it as uniform simplified it a lot.....
UnkleRhaukus
  • UnkleRhaukus
i probably should have stated uniform probability distribution in the question
UnkleRhaukus
  • UnkleRhaukus
so for \(0≤\theta≤2\pi\)\[\langle \theta \rangle =\pi\] the expected angle in a circle is 180°
UnkleRhaukus
  • UnkleRhaukus
makes sense
hartnn
  • hartnn
uniform pdf will always give you mid-point as expectation....
hartnn
  • hartnn
right ?
UnkleRhaukus
  • UnkleRhaukus
right

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