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Asked Oct 1, 2019
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The mean of a continuous random variable is Y with probability density function
OO
fry) is E(Y) 0yfy(v) dy
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The mean of a continuous random variable is Y with probability density function OO fry) is E(Y) 0yfy(v) dy

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Step 3

 Now, we need to find the E(Y) of the giv...

3
(а) Нere f:(y) — (1 - у)?, 0 < уs 2.
Е(Y) — S yfrу) dy
5 ух а- у? ду
Now
2
3 с2
- 2у2 + у3) dy
2 0
3
yч)
2 2
4 Jo
-(2-54)
16
= 1
Hence, E(Y) = 1
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3 (а) Нere f:(y) — (1 - у)?, 0 < уs 2. Е(Y) — S yfrу) dy 5 ух а- у? ду Now 2 3 с2 - 2у2 + у3) dy 2 0 3 yч) 2 2 4 Jo -(2-54) 16 = 1 Hence, E(Y) = 1

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