Question

If the probability density of a random variable is given by: f(x)= K(x+1) for -1 12 years ago

Answer 1

K(x+1) is just some function that you have to prolly solve for K to define it as a probability

Answer 2

you integrate f(x) as it is defined, and integrate it along the stated interval .. K is just a constant. Solve for K so that the integration is equal to 1

Answer 3

I keep getting the wrong answer when I solve it, my key says it should be 1/2 but I am getting 1 as my answer

Answer 4

@amistre64 ^^^^^

Answer 5

int K(x+1) from -1 to 1

int Kx + K from -1 to 1

Kx^2/2 + Kx from -1 to 1 = 1

K/2 + K - K/2 + K = 1

K(1/2 + 1 - 1/2 + 1) = 1

K(2) = 1

K = 1/2

Answer 6

\[\int_{1}^{-1}K(x+1)~dx=1\]
solve for the constant K

Answer 7

This is how I set it up \[\int\limits_{-1}^{1}(\frac{ x^2 }{ 2 }-1)\]

Answer 8

why would you setup up something different then what the information in the question provides?

Answer 9

idk that how our teacher taught us to solve it. he told us to set it up so that k is on the outside of the \[\int\limits_{-1}^{1}\]

Answer 10

then how does K on the outside change (x+1) into (x^2/2 - 1) ?

Answer 11

\[\int_{1}^{-1}K(x+1)~dx=1\]
\[K\int_{1}^{-1}x+1~dx=1\]

Answer 12

and then you integrate the (x+1), im confused on how to integrate the +1. like if the equation was \[\int\limits_{-1}^{1}kx(x+1)\] then you would get \[\int\limits_{-1}^{1}(x^2)+x\]

Answer 13

the integrate will give you \[\int\limits_{-1}^{1}(\frac{ x^3 }{ 3 }-\frac{ x^2 }{ 2 })\] idk thats just what i have in my notes i was trying to follow and then you solve for k

Answer 14

hmm i think I see what your talking about I typed a negative sign by mistake???

Answer 15

the probability density of a random variable is given by: f(x)= K(x+1) for -1<x<1

so your notes have the modified version you are presenting? If so, work it on the information in the stated problem. Is it that you dont know how to integrate? or that you are simply integrating the wrong thing?

Answer 16

i am not sure with one of the rules of integration when dealing with a whole number. I know that if I have a variable I plus one to that and divide by that plus one. but with a whole number do I do that also or just leave the number?

Answer 17

the constant rule of integration eh ..

consider a derivative that produces a constant:

nx, for some constant n derives to: n

if n=1 thats just: 1x derives to 1
if n=4 thats just: 4x derives to 4

etc

Answer 18

you could just do a clean up .. let u = (x+1) , then du = dx

Answer 19

\[\int Ku~du\]
\[K\int u~du\]
\[K(\frac12 u^2)~:~[a,b]\]

Answer 20

im getting the thought that you are taking a stats class, and are rusty or have not taken a calc 2 class maybe?

Answer 21

yea I haven't taken calc in 3 years so i am rusty and in a stats class..lol

Answer 22

:) the +1 just integrates up to x is all if we want to go that route.

Answer 23

Thank you soooooo much!!!!!

Answer 24

good luck :)