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The second part of the question is asking the following: if the standard deviation of a single bottle is $10$ ml, then what is the standard deviation of the average of $25$ bottles? Intuitively, the average of $25$ bottles will tend to have less variability than the variability of a single bottle, because the individual variabilities of the bottles will ...

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Hint: You can transform the random variable. Z | Pr(Y = y) 12 | 0.4 9 | 0.5 6 | 0.1 Now you can calculate $Var(Z)$ directly.

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Since $(-1)^3=-1$, $0^3=0$, and $1^3=1$, $Y^3$ has the same distribution as $Y$. Therefore $$\mathrm{Var}(9-3Y^3) = 3^2\mathrm{Var}(Y^3) = 9\mathrm{Var}(Y).$$ Since $$\mathrm{Var}(Y) = \mathbb E[Y^2] - \mathbb E[Y]^2 = \frac12 - \left(\frac3{10}\right)^2 = \frac{41}{100},$$ it follows that $$\mathrm{Var}(Z) = 9\cdot\frac{41}{100} = \frac{369}{100}.$$

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No, there isn't. Counterexample: The list (10, 10, 10, ..., 10) has the standard deviation 0. The list (-50, -10, -50, -10, ..., -50) has a standard deviation of approximately 19.9555. The list (10, 50, 10, 50, ..., 10) has a standard deviation of approximately 19.9555. The list (-10, -10, -10, ..., -10) has the standard deviation 0. More generally, ...

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A little hard to understand what you mean by the words "mean anything by itself"... Anyway, relative to some real life problem one shall note that the units of measurement of the standard deviation are the same as of the random variable itself. So in some very crude meaning it can be understood as the range of the random outcomes that you'll get on average ...

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Let $f(x,Y)$ denote the points a player x will score against a team Y. Let's attempt to do a basic prediction under the following assumption: All teams have the same lineup throughout every match of the season $f(x,Y) = R(x,Y) + [1+PC(Y, p(x))]*PS(X)$ $R(x,Y)$ - Random factor $PC(Y, p(x))$ - Percentage boost/fall that team Y concedes to a player ...

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