## anonymous 4 years ago 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?

1. anonymous

pls try

2. amistre64

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

3. anonymous

ya......

4. amistre64

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

5. amistre64

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

6. anonymous

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

7. anonymous

x square

8. amistre64

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

9. anonymous

oh,pls try

10. amistre64

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

11. 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

12. anonymous

hey how did u get this?

13. amistre64

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

14. amistre64

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

15. 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

16. amistre64

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

17. amistre64

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

18. amistre64

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

19. amistre64

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

20. anonymous

ok..try afterward

21. 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 :) $9-4\frac{\sum(x)}{n}+4$

22. amistre64

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

23. anonymous

ya thats what i did