B Why is V(1/5X) equal to 1/25*V(X)?

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The discussion centers on the calculation of variance, specifically why V(1/5X) equals 1/25*V(X). The confusion arises from the property of variance that states V(aX) = a^2V(X), which means that when scaling a random variable by a factor, the variance is scaled by the square of that factor. In this case, since 1/5 is the scaling factor, the variance becomes V(1/5X) = (1/5)^2*V(X) = 1/25*V(X). The definition of variance is also highlighted, emphasizing its dependence on the squared values of the random variable. Understanding this property clarifies the calculation of variance in this context.
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I don't understand how V(1/5X) can be turned into 1/25*V(X).
Shouldn't I just extract the 1/5 so:
V(1/5X) = 1/5*V(X) ?

V stands for variance.
In my book, when calculating the variance of X = (x_1 + x_2 + x_3 + x_4 + x_5)/5
in an example it says:

V(X) = V(1/5(X_1 + X_2 + X_3 + X_4 + X_5)) = 1/25*V(X_1) + 1/25*V(X_2) + 1/25*V(X_3) + 1/25*V(X_4) + 1/25*V(X_5) = 1/5Ф

I don't understand how V(1/5X) can be turned into 1/25*V(X), shouldn't I just extract the 1/5 so:
V(1/5X) = 1/5*V(X) ?
 
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According to the properties of the variance it is ##V(aX)=a^2V(X)##. This follows from the definition of variance and the properties of the mean E(X).
It is ##V(X)=E(X^2)-E(X)^2##
and $$V(aX)=E((aX)^2)-E(aX)^2=E(a^2X^2)-a^2E(X)^2=a^2E(X^2)-a^2E(X)^2=a^2(E(X^2)-E(X)^2)=a^2V(X)$$
 
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Summary:: I don't understand how V(1/5X) can be turned into 1/25*V(X).
Shouldn't I just extract the 1/5 so:
V(1/5X) = 1/5*V(X) ?

V stands for variance.

In my book, when calculating the variance of X = (x_1 + x_2 + x_3 + x_4 + x_5)/5
in an example it says:

V(X) = V(1/5(X_1 + X_2 + X_3 + X_4 + X_5)) = 1/25*V(X_1) + 1/25*V(X_2) + 1/25*V(X_3) + 1/25*V(X_4) + 1/25*V(X_5) = 1/5Ф

I don't understand how V(1/5X) can be turned into 1/25*V(X), shouldn't I just extract the 1/5 so:
V(1/5X) = 1/5*V(X) ?
What's definition of variance?
 
Variance is based on values of ##X^2## and ##(1/5*X)^2 = 1/25*X^2##.
 

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