Question

Find E(X) , Var(X) , SD(X) if a random variable X is given by its density
function f(x), such that
f(x) = 0, if x<=0
f(x) = 3x^2, if 0 < x <= 1
f(x) = 0, if x > 1

Answer 1

variance, standard deviation, and what is "E(x)"?

Answer 2

i think it is the math expectation

Answer 3

do we have a density function given or is this general?

Answer 4

I am not sure the question is worded the same as i received it

Answer 5

the density function is f(x)

Answer 6

f(x) is a generic term meaning "any given function" so its asking for a generalization

Answer 7

yes. I will assume that it is general. I do not know where to start

Answer 8

0, if x<=0
f(x) = 3x^2, if 0 < x <= 1
0, if x > 1

this is saying that the f(x) is defined for a given domain; if the random variable is less than or equal to 0; then the probability = 0. If the random variable is greater than 1; then the probability = 0; otherwise, it is defined as 3x^2

Answer 9

ok i get that part

Answer 10

i believe this means that our distribution curve is parabolic and defined between 0 and 1; but i havent come across this b4 to be certain

Answer 11

would there be 2 point that I would need for E(x)?

Answer 12

i cant really say, i still dont have a clear understanding of what E(x) stands for :)

Answer 13

expected value ..... ok

Answer 14

expected value = mean of x*P(x)

Answer 15

we got any choices to guide is with?

Answer 16

do we know if our random variables are discrete or continuous?

Answer 17

according to my notes it is the mathematical expectation of a continuous random variable

Answer 18

if i read this right:

E(x) = \(\int_{0}^1\ 3x^2\ dx\)

Answer 19

well, I forgot an "x" in that :)

Answer 20

3x^3 integrates to (3/4)x^4; at x=1

E(x) = 3/4 if my assumption is right

Answer 21

and since E(x) is also the mean, we can see about determining the variance and sd with it

Answer 22

the variance is equal to:

\[\sum_{x=0}^{1}\ [(x-\bar x)^2P(x)]\]

Answer 23

(0-.75)^2 * 3(0)^2 + (1-.75)^2 *3(1)^2 = .1875 ; but got no clue if thats right ....

Answer 24

and of course the sd is just the square root of variance

Answer 25

some kind of way we are supposed to have an additional E(X^2) equation

Answer 26

then I think subtact them. so would the answer to E(x^2) be 3/5?

Answer 27

that resembles the "short cut" sd in my book:

\[\sqrt{\frac{n(\sum(x^2))-(\sum(x))^2}{n(n-1)}}\]

Answer 28

perhaps:

\[var(x)=\sum[(x^2)*3x^2] - E(x)\]??

Answer 29

E(x)^2 that is

Answer 30

ok now im confused lol

Answer 31

i think i understand the main part of the equation though

Answer 32

for continuous v random variable that wouold amount to:

\(\int3x^4dx\) from 0 to 1 which equals \(\frac{3}{5}x^5\) from 0 to 1 which does = 3/5

Answer 33

3/5 - (3/4)^2 = variance ?

Answer 34

\[\frac{3}{5}-\frac{9}{16}=\frac{48-45}{80} = \frac{3}{80}\]

Answer 35

I think thats correct it seems right. maybe i did it right. thank you

Answer 36

:) good luck with it all