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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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Hint $\overline{x}_1 = 249, \overline{x}_2 = 252, \overline{x}_3 = 253, \overline{x}_4 = 248,\overline{x}_5 = 245,\overline{x}_6 = 253 \Rightarrow m = \dfrac{\overline{x}_1+\overline{x}_2+\cdots +\overline{x}_6}{6} = 250 \Rightarrow s_{\overline{x}} = \sqrt{\dfrac{\displaystyle \sum_{k=1}^6 \left(\overline{x}_k-250\right)^2}{6-1}}$


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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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You need to use the $t$-distribution. So $t_{\alpha/2} = 2.947$ ($df = 16-1 = 15$, $\alpha = 0.01$, $2 \text{ tails})$, and $n = 16, s = 2.9$. So $E = t_{\alpha/2}\cdot \dfrac{s}{\sqrt{n}} = 2.947\cdot \dfrac{2.9}{\sqrt{16}} = 2.1366 \approx 2.14.$


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To do a Confidence Interval you'd need this formula $$\mu\pm t_{\alpha\over2}\cdot{\sigma\over\sqrt n}$$ Where$$\mu=\text{mean}=7$$$$t_{\alpha\over2}=\text{t-Score of Confidence Interval}=2.947$$$$\sigma=\text{Standard Deviation}=2.9$$ $$\text{and}$$$$n=\text{sample size}=16$$ Furthermore $${\sigma\over\sqrt n}=\text{Margin of Error}$$ So in your case the ...


2

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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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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Your formula for $MSD_k$ does not appear correct. Standard deviation is the square root of variance, and variance is a moment of the distribution, and so it involves expected values. It appears more likely to be (introducing also the average part of the moving average) $$[MSD_k]^2 \equiv E\left(MA_k\right) ^2- \left[E\left(MA_k\right)\right]^2$$ $$=\frac ...


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Here's what I would try. Find estimates of $E[X^2]$ for each group and then an overall estimate using a weighting scheme by group count. Do the same for the mean. Then estimate the overall variance with $$\hat \sigma^2=E[X^2]-(E[X])^2.$$


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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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If you know what the distribution of the values is, then, given the appropriate statistics, you can essentially do inverse interpolation in the cumulative distribution to find the counts in given ranges. If the data is normally distributed, the statistics you mention are enough to do what you want.


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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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$\displaystyle \sum_{i=1}^N \left(\frac{x_i-\mu}{\sigma}\right)^2$ has a $\chi_N^2$ distribution (chi-squared with $N$ degrees of freedom) as as the sum of $N$ independent standard normal random variables. So if $\mu=0$ and $\displaystyle s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2$ then $N s_2^2 / \sigma^2$ also has a $\chi_N^2$ distribution.


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Indeed, due to the Central Limit Theorem, the sample average $X_n$ follows approximately (as $n$ grows bigger, at least $n>30$ for a good approximation, which applies here since $n=81>>30$) the normal distribution with mean $$μ_n=μ=8.2$$ and standard deviation $$σ_n=\frac{σ}{\sqrt{n}}=\frac{1.5}{\sqrt{81}}=\frac{1.5}{9}=\frac{1}{6}$$ in symbols ...



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