anonymous
  • 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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
pls try
amistre64
  • amistre64
for a population\[Var=\frac{\sum (x-X)^2}{n}\] hmmm
anonymous
  • anonymous
ya......

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amistre64
  • amistre64
thats all I can come up with at the moment :/
amistre64
  • amistre64
what does it mean: the mean of x^2 = 9?
anonymous
  • anonymous
hey try the options for the ques are-80,20,144,112
anonymous
  • anonymous
x square
amistre64
  • amistre64
i know that much :) i just dont know what the sentence itself means
anonymous
  • anonymous
oh,pls try
amistre64
  • amistre64
if we add up all the x^2s and divide by how many there are we get 9 perhaps?
amistre64
  • 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
anonymous
  • anonymous
hey how did u get this?
amistre64
  • amistre64
i expanded the (x-X)^2 and split the fraction
amistre64
  • amistre64
now, I believe the sum of x^2/n is the mean of x^2
amistre64
  • 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
amistre64
  • amistre64
now, \[\sum_{1}^{n} 4x=4\sum_{1}^{n}x=4(x_1+x_2+x_3+...+x_n)=4xn\]
amistre64
  • amistre64
thats wrong; \[4\sum_{1}^{n} x=\frac{4n(n+1)}{2}=2n(n+1)\]
amistre64
  • amistre64
divided by n = 2n+2 var = 13 - 2n + 2 var = 15 - 2n but that doesnt seem to be applicable as an answer tho
amistre64
  • amistre64
well, class is starting so I gots to go; good luck :)
anonymous
  • anonymous
ok..try afterward
amistre64
  • 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 :) \[9-4\frac{\sum(x)}{n}+4\]
amistre64
  • amistre64
i dont know why the last part is tacked on the end .... but its just spurious.
anonymous
  • anonymous
ya thats what i did

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