Math Is Fun Forum

basketboy

Member

Registered: 2011-10-11

Posts: 14

solve

f(x,y)=[6-x-y/8   0<x<2,  2<y<4
           0     

find P(1<y<3]x=1

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: solve

hi basketboy,

What about your earlier problem?

I assumed that x and y could only take integers values.  I'll modify my answer if the variables are continuous.

I need some feedback from you!

Have you done it?

What answers did you get?

Bob

Last edited by Bob (2011-10-12 00:08:44)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

basketboy

Member

Registered: 2011-10-11

Posts: 14

Re: solve

thanks bob new here the question was x+y/30 not x+y/3....so i used diff figures

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: solve

hi basketboy,

new here

Welcome to the forum. 

So are you saying that you have done the first question completely?

Post your answers.  30 instead of 3, won't change much so I can still check them.

Now to the new question.

f(x,y)=[6-x-y/8   0<x<2,  2<y<4
           0

This is very unclear.  Do you want (6-x-y)/8  ??  And what's that 0 doing ?  Perhaps it is (6-x-y)/80  ??

Last time you said x is in the set {0,1,2,3} so I assumed that only x = 0, x=1, x=2, x=3 are possible.

This time you've said 0 < x < 2 so I'm assuming that all values between these limits are possible.

That will require a modified method.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

basketboy

Member

Registered: 2011-10-11

Posts: 14

Re: solve

yeah this the right question

f(x,y)=[6-x-y/8   0<x<2,  2<y<4

find

P{1<x<y<3}x+1

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: solve

f(x,y)=[6-x-y/8   0<x<2,  2<y<4

f(x,y)= (6-x-y)/8   0<x<2,  2<y<4

YES/NO

Continuous variables?

YES/NO

P{1<x<y<3}x+1

I can show P{1<x<y<3} but what am I supposed to do with x + 1 ??

Bob

Last edited by Bob (2011-10-12 00:38:08)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

basketboy

Member

Registered: 2011-10-11

Posts: 14

Re: solve

NO

f(x,y)=[6-x-y/8 ,  0<x<2,  2<y<4
                          otherwise
find

P{1<x<y<3}x=1

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: solve

The question is not making sense to me.

You've got an open bracket and no close bracket.

If it's zero otherwise then it is continuous.  See my graph below.

You would have to ignore the x + 1 unnecessary bit and then work out the area of the region that fits these boundaries:

to the right of the line x=1

below the line y = 3

to the left of the line y=x ( so that x < y )

There is no such region ?  So the answer is zero.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

basketboy

Member

Registered: 2011-10-11

Posts: 14

Re: solve

There no close bracket...yes x=1 in the question i believe you right from the graph

basketboy

Member

Registered: 2011-10-11

Posts: 14

Re: solve

can we try 1 more example

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: solve

hi basketboy,

Yes we can!  Your third question makes sense and I've started you off on that post.

But also, it's given me the clues I needed to make sene of Q1 and Q2.

Your original post for Q1 had an error in it.  You spotted it should be /30.  That makes all the difference as it makes the function into a sensible probability function.  The fractions in the revised table now add up to 1.  I have changed my original post to reflect this.  It is important that you look at the edit and revise your answers.

And I can make sense of Q2 as well.  As you have typed it it doesn't make a probability density function, but it does with a little change or two.

I think it should be:

when 0<x<2 and 2<y<4

and 0 (zero) otherwise.

Check that this comes to 1, over this range.

So we have the definition.

But what is the question?

1<y<3 is ok as it gives us a range for the integral.  But x = 1 doesn't work because the 'area' of an infinitesimally thin line is zero.

But maybe I should continue anyway?

f is zero for y between 1 and 2 so this can be re-written as

So that's it unless you find the range(s) are wrong.

Bob

Last edited by Bob (2011-10-12 05:53:02)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob