Probability Distributions

[khalil]

So its just an apprx value, its not exactly 0? (does that make sense?)

[m@tty]

yer i sketched it, so the larger the q value the closer the graph is to 0.
so does that mean we can conclusively make it 0...it can never reach 0
im confused

Yes but only because e is to the power of negative q

i.e. as q approaches approaches 0
as can be read off the graph

[khalil]

yer i sketched it, so the larger the q value the closer the graph is to 0.
so does that mean we can conclusively make it 0...it can never reach 0
im confused

Yes but only because e is to the power of negative q

i.e. as q approaches approaches 0
as can be read off the graph

Oh yerr, thanks M@tty
TT your welcome to elaborate

[/0]

You don't need to know about the rigorous definition of a limit in high school, you're expected to use the graph to extrapolate to infinity.

[TrueTears]

yer i sketched it, so the larger the q value the closer the graph is to 0.
so does that mean we can conclusively make it 0...it can never reach 0
im confused

Yes but only because e is to the power of negative q

i.e. as q approaches approaches 0
as can be read off the graph

Oh yerr, thanks M@tty
TT your welcome to elaborate

Like /0 said the rigious definition of limit (symbolically )
is not required.

As far as methods 3/4 go, taking the limit is taking the value of what the expression "approaches". In this case is approaching 0. Therefore the limit of to infinity is 0.

[chuckjefster90]

got another question

chpt 17C

q 13

f(x)= hybrid function, x for 0<x<1 and 2-x for 1<x<2 and 0 elsewhere

a) what is the expected value of x
b) find median value of X
c) find mode of x

thanks

[Damo17]

got another question

chpt 17C

q 13

f(x)= hybrid function, x for 0<x<1 and 2-x for 1<x<2 and 0 elsewhere

a) what is the expected value of x
b) find median value of X
c) find mode of x

thanks

a)


b)






c) the mode of x is the max of over the interval [0,1] and the max of over the interval [1,2] such that for all other x.
So for over the interval [0,1] the max is at 1 and the same for over the interval [1,2]

So the mode of x is 1.

[chuckjefster90]

q16

f(x)= { 0.3 for -1<x<0
          0.2 + 1.2x for 0<x<1
          0  elsewhere

q16b why is the median value the positive one and not the negative one

and C) how do u find the mode

[chuckjefster90]

anyone?

[m@tty]

Your question is not very clear.
short answers.

b) because the point with half area(0.5) either side is greater than 0. This can be figured out using the same method as damo17, letting the integral between the the lowest point of the function, in this example -1, and an unknown(m) equal to 0.5

The rest are details....

as falls outside domain which is

c) The mode is the value of x with highest probability, when reading off a graph the highest point.

I am not really sure what the max of the function is though, as it never really equals 1.