Probability Density Function (1 Viewer)

davidgoes4wce · May 2, 2016

$$ The probability density function of the continuous random variable Z is kz for $ 1<z\leq{3}$

$a) Find the value of k$

$b) Find E(Z)$

$c) Find E(6Z+5)$

$$ d) Find the exact value of $ E(\frac{1}{1+Z^{3}})$

Spent about 15 minutes on part (b)......can't seem to get it..

Answers are: a) 1/4 b) 13/6 c) 18 d) 1/8 ln (5)

Drongoski · May 2, 2016

(b)

$E(Z) = \int ^3 _1 z\times \frac {z}{4} dz = \left[ \frac {1}{4} \times \frac {1}{3} z^3 \right]^3 _1 = \frac {13}{6}$

c) E(6Z + 5) = 6E(Z) + 5 = 6 x (13/6) + 5 = 18

(d)

$E(\frac {1}{1+Z^2}) = \int _1 ^3 \frac {z}{4}\times \frac {1}{1 + z^2}dz = \left [ \frac {1}{8}ln (1+z^2) \right ]_1 ^3 = \frac {1}{8} ln 5$

ps
(z^2 not z^3)

InteGrand · May 2, 2016

davidgoes4wce said:
$$ The probability density function of the continuous random variable Z is kz for $ 1<z\leq{3}$

$a) Find the value of k$

$b) Find E(Z)$

$c) Find E(6Z+5)$

$$ d) Find the exact value of $ E(\frac{1}{1+Z^{3}})$

Spent about 15 minutes on part (b)......can't seem to get it..

Answers are: a) 1/4 b) 13/6 c) 18 d) 1/8 ln (5)

For a), the pdf needs to integrate to 1.

So we have

$\begin{align*}\int_1 ^3 kz \text{ d}z &= 1 \\ \Longleftrightarrow k \left[\frac{z^2}{2}\right]_1 ^3 &= 1 \\ \Longleftrightarrow k \left(\frac{3^2}{2}-\frac{1^2}{2}\right) &= 1 \\ \Longleftrightarrow 4k &= 1 \\ \Longleftrightarrow k &= \frac{1}{4}. \end{align*}$

davidgoes4wce · May 2, 2016

For any of you guys that are keen , this is a Cambridge University Press published question. I could get part a). Was a bit confused about the

$z \times \frac{z}{4}$

Im assuming it has something to do with the formula :

$E(g(X))= \int g(x) f(x) dx$

Drongoski · May 2, 2016

davidgoes4wce said:
For any of you guys that are keen , this is a Cambridge University Press published question. I could get part a). Was a bit confused about the

$z \times \frac{z}{4}$

Im assuming it has something to do with the formula :

$E(g(X))= \int g(x) f(x) dx$

That's roughly correct,. In your question:

g(x) = x and f(x) = x/4

Probability Density Function (1 Viewer)

davidgoes4wce

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Drongoski

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InteGrand

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davidgoes4wce

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Drongoski

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