Solve this problem that involves a probability density function

[chwala]

Homework Statement

see attached

Relevant Equations

understanding of continous and discrete distribution

I am refreshing on this; ..after a long time...

Note that i do not have the solution to this problem.

I will start with part (a).

$f(u)= 3u-\dfrac{3u^2}{2k}$ with limits $0≤u≤k$

it follows that,

$3k - \dfrac{3k}{2}=1$

$\dfrac{3k}{2}=1$

$k=\dfrac {2}{3}$

For part (b),

$E(T)=\int_0^{\frac{2}{3}} u⋅(3-\dfrac{9}{2}u )du=\left[\dfrac{3}{2}×\dfrac{4}{9}-\dfrac{3}{2}×\dfrac{8}{27}\right]=\dfrac{6-4}{9}= \dfrac{2}{9}$Ok let me know if that's correct...

 

[Office_Shredder]

Those both look OK to me. Do you know how to do the last two parts?

One minor  issue

$f(u)= 3u-\dfrac{3u^2}{2k}$ with limits $0≤u≤k$

This is bad notation, since $f$ it's already defined to be the pdf, not the cdf.

 

[chwala]

Those both look OK to me. Do you know how to do the last two parts?

One minor  issue

This is bad notation, since $f$ it's already defined to be the pdf, not the cdf.

I will work on the last parts too...refreshing on this...should be fine ...Will post once done.

 

[chwala]

Those both look OK to me. Do you know how to do the last two parts?

One minor  issue

This is bad notation, since $f$ it's already defined to be the pdf, not the cdf.

I thought pdf is generally defined to be $f_{u}$ being the derivative of the continuous distribution $f(u)$ ...different books have different notations/language that I pretty find to be time waster...my focus is on the concept...

Which is the right notation? Thanks @Office_Shredder

 

[haider]

Which is the right notation? Thanks @Office_Shredder

In my stats class we used $f_{X}(x)$ for the PDF and $F_{X}(x)$ for the CDF for a given continuous random variable $X$. I think that this is fairly standard, because most books and lecture notes I found online used it.

 

[chwala]

Those both look OK to me. Do you know how to do the last two parts?

One minor  issue

This is bad notation, since $f$ it's already defined to be the pdf, not the cdf.

$E(T^2)=\int_0^{\frac{2}{3}} u^2⋅(3-\dfrac{9}{2}u )du$

$=\int_0^{\frac{2}{3}} (3u^2-\dfrac{9u^3}{2} )du=\left[u^3-\dfrac{9u^4}{8} \right]$

$=\left[\dfrac{8}{27}-\dfrac{9}{8}×\dfrac{16}{81}\right]=\dfrac{8-6}{27}= \dfrac{2}{27}$

$⇒Var (T) = \dfrac{24-4}{81}=\dfrac{20}{81}$

...this part is easy and straightforward. Bingo!

 

[Office_Shredder]

Where does the 24 in the variance come from? Shouldn't it be a 6?

 

[chwala]

Where does the 24 in the variance come from? Shouldn't it be a 6?

True...let me amend that.

 

[chwala]

$E(T^2)=\int_0^{\frac{2}{3}} u^2⋅(3-\dfrac{9}{2}u )du$

$=\int_0^{\frac{2}{3}} (3u^2-\dfrac{9u^3}{2} )du=\left[u^3-\dfrac{9u^4}{8} \right]$

$=\left[\dfrac{8}{27}-\dfrac{9}{8}×\dfrac{16}{81}\right]=\dfrac{8-6}{27}= \dfrac{2}{27}$

$⇒Var (T) = \dfrac{6-4}{81}=\dfrac{2}{81}$

...this part is easy and straightforward. Bingo!