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asked in Probability by Loyal (8.7k points)
edited by | 102 views
0
0.075
0
0.1 ??

2 Answers

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Is it 1?
answered by (373 points)
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Given $v$ is uniformly distributed, Therefore P.D.F of $v$ is-

$$ \ F(v) = \left\{ \begin{array}{ll} \frac{1}{10-0} & 0\leq v\leq10 \\ \\ 0 & otherwise \end{array} \right. $$

Also $W(v) $= 0.003$v^{2}$,

$ \ E[w(v)] = \int_{0}^{10} W(v)F(v)dv \\ \\ E[w(v)]  = \int_{0}^{10} 0.003v^{2} * \frac{1}{10} dv$


solving this we get $E[w(v)] = 0.1$

 

Reference: Google Books

answered by (261 points)
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