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5

Maximising the standard deviation or variance for a given mean is equivalent to maximising the sum of squares of the values for a given sum of the values. Meanwhile if $b \ge a$ and $\delta \gt 0$, $$(a-\delta)^2 +(b+\delta)^2 $$ $$= a^2-2a\delta+\delta^2+b^2+2b\delta + \delta^2 $$ $$= a^2+b^2 +2(b-a)\delta+2\delta^2 $$ $$\gt a^2+b^2$$ so the sum of ...


3

First, you might as well assume that the mean is zero, so that $a < 0 < b$; that just simplifies things. (Otherwise: let $a' = a - \mu; b' = b - \mu$ and you can convert your problem for $[a, b], \mu$ to mine for $[a, b], 0$ by subtracting $\mu$ from all the sample values, etc.) The SD is $$ s = \sum x_i^2 $$ and you want to maximize this subject to ...


3

Let $y_i = x_i - \bar{x}$ then by Cauchy-Schwarz we have the following: $\sum_{i=1}^n |y_i| \frac{1}{n} \leq \|y\|_2 \sqrt{\sum_{i=1}^n \frac{1}{n^2}} = \sqrt{\frac{1}{n}\sum_{i=1}^n |x_i - \bar{x}|^2}$


2

If $X_i$s are independent and you have already their standard deviations, say $\sigma_i$, for $X_i$, which seem to be the same, then the variance of each $X_i$, for $i=1,\cdots, n$ is $\sigma^2_i$, and so the variance of $X$ is equal to $\sum_{i=1}^n\sigma^2_i=n\times \sigma^2_i$. As a result, the standard deviation of $X$ is $\sqrt{n\times \sigma^2_i}=\...


1

It is best to use a boxplot to find outliers. The problem with using the sample mean $\bar X$ and the sample SD $S$ is that an outlier seriously affects the values of $\bar X$ and $S$. By contrast, the boxplot uses the median and the interquartile range to detect outliers. These measures of location and dispersion, respectively, are not much affected by ...


1

Obviously if $d$ is the greatest distance from any of the values to the mean, then every $(x_i - \bar{x})^2 \le d^2$; plug into the formula for SD to find that SD $\le d$. Moreover, it can only be $d$ when all distances are equal to $d$, no less; and since the average must still be $d$, this means half the values are at one extreme and half at the other. And ...


1

You should interpret the statement Experience suggests that the standard deviation is more stable than the mean as the assertion that even when the machine is not working properly, the standard deviation of a package weight remains unchanged. Judging from the context, it's a statement about the population $\sigma$, not the sample standard deviation. So the ...



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