Thread: cumulative distribution function

Question attached.

It cannot be done unless you complete that table.
You have given no values for .

Sorry for confusion:

0 = 1/27
1 = 6/27
2 = 12/27
3 = 8/27

So for an answer maybe something along the lines...

F(x) = 0 if x<0
F(x) = 1/27 if 0<=x<1
F(x) = 6/27 if 1<=x<2
F(x) = 12/27 if 2<=x<3
F(x) = 8/27 if 3<=x

It will look like this:

Now you fill in the blanks.

F(x) = 0 if x<0
F(x) = 1/27 if 0<=x<1
F(x) = 6/27 if 1<=x<2
F(x) = 12/27 if 2<=x<3
F(x) = 1 if 3<=x

Okay I see I did a mistake... it should be 1, not 8/27 for the 3<=x

No you missed the accumulation.

Ok I get it now. ACCUMULATION.... aka adding.

So we get 0+1/27+6/27+12/27=19/27.

Is my 2<=x<3 incorrect though?

Originally Posted by DINOCALC09

Is my 2<=x<3 incorrect though?

It is wrong also. Add the 'jump'.
When you finish must be continuous on the right.

What do you mean by jump?

I know what continuous is kinda, but not sure how it relates to inequalities.

Originally Posted by DINOCALC09

What do you mean by jump?

It should go from to .

That is a jump of .

yes, i see that. however, don't see what's wrong with the inequality necessarily. Should 2 not be inclusive?

Frankly this is tiresome. It is


I am done with the thread.