A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 5 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?
anonymous
 5 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?

This Question is Closed

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0for a population\[Var=\frac{\sum (xX)^2}{n}\] hmmm

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats all I can come up with at the moment :/

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what does it mean: the mean of x^2 = 9?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey try the options for the ques are80,20,144,112

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i know that much :) i just dont know what the sentence itself means

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we add up all the x^2s and divide by how many there are we get 9 perhaps?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(xX)^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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey how did u get this?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i expanded the (xX)^2 and split the fraction

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now, I believe the sum of x^2/n is the mean of x^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[Var=\frac{\sum x^2}{n}\frac{\sum 2xX}{n}+\frac{\sum X^2}{n}\] \[Var=9\frac{\sum 2x(2)}{n}+\frac{\sum (2)^2}{n}\] \[Var=9\frac{\sum 4x}{n}+4\] \[Var=13\frac{\sum 4x}{n}\] is what im guessing so far

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now, \[\sum_{1}^{n} 4x=4\sum_{1}^{n}x=4(x_1+x_2+x_3+...+x_n)=4xn\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats wrong; \[4\sum_{1}^{n} x=\frac{4n(n+1)}{2}=2n(n+1)\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0divided by n = 2n+2 var = 13  2n + 2 var = 15  2n but that doesnt seem to be applicable as an answer tho

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0well, class is starting so I gots to go; good luck :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I 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 (xu)^2 can be expanded to: x^2 2xu + u^2\[\frac{\sum (x\mu)^2}{n}=\frac{\sum (x^22x\mu+\mu^2)}{n}=\frac{\sum(x^2)}{n}\frac{\sum(2x\mu)}{n}+\frac{\sum(\mu^2)}{n}\] 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 \[94\frac{\sum(x)}{n}+4\] relpace known values \[94*2+4=5\] \[var(x)=5\] 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 :) \[94\frac{\sum(x)}{n}+4\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know why the last part is tacked on the end .... but its just spurious.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.