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Hi need help for this maths question

Hi need help for this maths question-example-1

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a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have


\displaystyle \int_(-\infty)^\infty f(y) \, dy = \frac14 \int_0^\infty e^(-\frac y4) \, dy = -\left(\lim_(y\to\infty)e^(-\frac y4) - e^0\right) = \boxed{1}

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,


\displaystyle \int_(-\infty)^4 f(y) \, dy = \frac14 \int_0^4 e^(-\frac y4) \, dy = -\left(e^(-1) - e^0\right) = (e - 1)/(e) \approx \boxed{0.632}

c) The mean is given by the integral,


\displaystyle \int_(-\infty)^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^(-\frac y4) \, dy

Integrate by parts, with


u = y \implies du = dy


dv = e^(-\frac y4) \, dy \implies v = -4 e^(-\frac y4)

Then


\displaystyle \int_(-\infty)^\infty y f(y) \, dy = \frac14 \left(\left(\lim_(y\to\infty)\left(-4y e^(-\frac y4)\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^(-\frac y4) \, dy\right)


\displaystyle \cdots = \int_0^\infty e^(-\frac y4) \, dy


\displaystyle \cdots = -4 \left(\lim_(y\to\infty) e^(-\frac y4) - e^0\right) = \boxed{4}

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