Mathematics
OpenStudy (anonymous):

random variable x s properties(i)the mean of X is 2. and (2) the mean of x^2 is 9.what is variance of 4x?

OpenStudy (anonymous):

pls try

OpenStudy (amistre64):

for a population$Var=\frac{\sum (x-X)^2}{n}$ hmmm

OpenStudy (anonymous):

ya......

OpenStudy (amistre64):

thats all I can come up with at the moment :/

OpenStudy (amistre64):

what does it mean: the mean of x^2 = 9?

OpenStudy (anonymous):

hey try the options for the ques are-80,20,144,112

OpenStudy (anonymous):

x square

OpenStudy (amistre64):

i know that much :) i just dont know what the sentence itself means

OpenStudy (anonymous):

oh,pls try

OpenStudy (amistre64):

if we add up all the x^2s and divide by how many there are we get 9 perhaps?

OpenStudy (amistre64):

(x-X)^2 = x^2 -2xX + X^2 $Var=\frac{\sum x^2}{n}-\frac{\sum 2xX}{n}+\frac{\sum X^2}{n}$ might be useful if its correct

OpenStudy (anonymous):

hey how did u get this?

OpenStudy (amistre64):

i expanded the (x-X)^2 and split the fraction

OpenStudy (amistre64):

now, I believe the sum of x^2/n is the mean of x^2

OpenStudy (amistre64):

$Var=\frac{\sum x^2}{n}-\frac{\sum 2xX}{n}+\frac{\sum X^2}{n}$ $Var=9-\frac{\sum 2x(2)}{n}+\frac{\sum (2)^2}{n}$ $Var=9-\frac{\sum 4x}{n}+4$ $Var=13-\frac{\sum 4x}{n}$ is what im guessing so far

OpenStudy (amistre64):

now, $\sum_{1}^{n} 4x=4\sum_{1}^{n}x=4(x_1+x_2+x_3+...+x_n)=4xn$

OpenStudy (amistre64):

thats wrong; $4\sum_{1}^{n} x=\frac{4n(n+1)}{2}=2n(n+1)$

OpenStudy (amistre64):

divided by n = 2n+2 var = 13 - 2n + 2 var = 15 - 2n but that doesnt seem to be applicable as an answer tho

OpenStudy (amistre64):

well, class is starting so I gots to go; good luck :)

OpenStudy (anonymous):

ok..try afterward

OpenStudy (amistre64):

I want to say its 20. to recap: a mean is an average, an average is adding up all the values and dividing by how many there are. $avg(x_1,x_2,x_3,...,x_n)=\frac{x_1+x_2+x_3+...+x_n}{n}=\frac{\sum x}{n}=\mu$ given that:$\mu=2;\ and\ \frac{\sum(x^2)}{n}=9$what is variance of 4x? From algebra we know that (x-u)^2 can be expanded to: x^2 -2xu + u^2$\frac{\sum (x-\mu)^2}{n}=\frac{\sum (x^2-2x\mu+\mu^2)}{n}=\frac{\sum(x^2)}{n}-\frac{\sum(2x\mu)}{n}+\frac{\sum(\mu^2)}{n}$ replace this with known values: $9-\frac{\sum(2x*2)}{n}+\frac{\sum(2^2)}{n}$ constants can be factored out and the average of a constant is itself $9-4\frac{\sum(x)}{n}+4$ relpace known values $9-4*2+4=5$ $var(x)=5$ we want to know the variance of 4 times x which I would say is 4*5 = 20 but i reserve the right to be complety wrong :) $9-4\frac{\sum(x)}{n}+4$

OpenStudy (amistre64):

i dont know why the last part is tacked on the end .... but its just spurious.

OpenStudy (anonymous):

ya thats what i did