What are the mean and standard deviation of the probability density function given by p(x)k=sin⁡x for x∈[0,π], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Explanation: The recipe is: Step 1 Determine the probability distribution function pdf, p(x) such that: P(x)=∫x1x2p(x)dx=1 over [x1,x2] 1=∫0πksin⁡(x)dx=2k;k=12 so your pdf is p(x)=12sin⁡x;[0,π] Step 2 Determine the mean μ and standard deviation σ a) Mean: μ=12∫0πxsin⁡xdx=sin⁡x−xcos⁡x2=π2 b) var: σ2=12∫0π(x−μ)2sin⁡xdx σ2=4(2x−π)sin⁡x−((2x−π)2−8)cos⁡x2=π2−84 σ=12π2−8

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pozicijombx

Answered question

2022-01-27

What are the mean and standard deviation of the probability density function given by

\frac{p (x)}{k} = \sin x

for

x \in [0, π]

, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Answer & Explanation

Micheal Hensley

Beginner2022-01-28Added 10 answers

Explanation:
The recipe is:
Step 1
Determine the probability distribution function pdf, p(x) such that:
$P (x) = \int_{x_{1}}^{x_{2}} p (x) d x = 1$ over $[x_{1}, x_{2}]$
$1 = \int_{0}^{π} k \sin (x) d x = 2 k; k = \frac{1}{2}$
so your pdf is $p (x) = \frac{1}{2} \sin x; [0, π]$
Step 2
Determine the mean $μ$ and standard deviation $σ$
a) Mean: $μ = \frac{1}{2} \int_{0}^{π} x \sin x d x = \frac{\sin x - x \cos x}{2} = \frac{π}{2}$
b) var: $σ^{2} = \frac{1}{2} \int_{0}^{π} {(x - μ)}^{2} \sin x d x$
$σ^{2} = \frac{4 (2 x - π) \sin x - ({(2 x - π)}^{2} - 8) \cos x}{2} = \frac{π^{2} - 8}{4}$
$σ = \frac{1}{2} \sqrt{π^{2} - 8}$

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