# Tag Info

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In your problem, there are five independent experiments, each of which is the sum of two die rolls. This is different from ten dice rolls. For example, you would expect a mean of 7 from your experiment, and 3.5 from the single dice rolls. In excel, create two columns of five rows of random die rolls (=INT(RAND()*6)+1 in cells A1..B5), and then add the ...

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There is a very simple explanation for this: it allows for the calculation of analytical solutions for many interesting problems. As others have pointed out before, $x^2$ is differentiable, whereas $|x|$ is not. Hence, in problems where quadratic terms are present, one can differentiate them to find optimal solutions analytically. On the other hand, with ...

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You say the variance is $\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2}{n-1}$. What if I told you the variane is $\dfrac{\sum_{i=1}^{n}(x_i-\bar x)^2} n$? You can find both in textbooks. The fact is, dividing by $n-1$ rather than $n$ is properly done (if at all) ONLY when one is estimating the population variance by using a finite sample $x_1,\ldots,x_n$ that is ...

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If you need to find standard deviations of a lot of datasets of this kind, you should consider using a statistical calculator that is programmed to find SDs, or using statistical software. R is excellent software and free from www.r-project.org. Just follow the path to install R on your Windows, Mac, or UNIX computer. The package is extensive and some of ...

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Let's work it out for the case of splicing together two data sets of size $n_x,n_y$. This generalizes easily enough. You have the averages $\overline{x},\overline{y}$ and the standard deviations $s_x,s_y$. The overall average is easy to compute: as you noticed it is just $\frac{n_x \overline{x}+n_y \overline{y}}{n_x+n_y}$. For the standard deviations, we ...

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Your calculations are correct. The extreme $z$ score you get indicates that if the percentage of households was really $33$%, then this sample is highly unrepresentative. The conclusion you would therefore draw is that the actual percentage must in reality be lower than $33$%, assuming that the sample is random and representative of the population as a whole....

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General Solution To compute mean, variance, and standard deviation you only need to keep track of three sums $s_0, s_1, s_2$ defined as follows for a set of values $X$: $$(s_0, s_1, s_2) = \sum_{x \in X} (1, x, x^2)$$ In English, $s_0$ is the number of values, $s_1$ is the sum of the values, and $s_2$ is the sum of the square of each value. Given these ...

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This answer is merely a summary of Stephen Gorard (2005), "Revisiting a 90-Year-Old Debate: The Advantages of the Mean Deviation" (PDF), which argues that we should abandon the standard deviation (SD) in favor of the mean deviation (MD). The apparent superiority of SD [standard deviation] is not as clearly settled as is usually portrayed in texts This ...

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eventually I think I have managed to solve that. The key issue is to calculate the density function $f(x,y)$ of the 2 variables. which because the variables are independent we have: $$f(x,y) = \dfrac{e^{-x^2/2}}{\sqrt{2\pi}} * \dfrac{e^{-y^2/2}}{\sqrt{2\pi}} = \dfrac{e^{-(x^2+y^2)/2}}{2\pi}=g(x^2+y^2)$$

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Special about normal distribution is: if $Y$ has normal distribution and $Z:=aY+b$ (where $a,b$ are constants and $a\neq0$) then also $Z$ has normal distribution. If we look at non-degenerate cases (as we mostly do) then the standard deviation $\sigma$ is positive. If $X$ denotes a random variable then based on it we can define a new one: $$U:=\frac{X-\mu}... 0 Begin by breaking the observation down into Noise + True value Let Y be the true value and Z be the noise, assume that the noise is normally distributed Y \sim N(\mu,\sigma^2) Z \sim N(\nu,\tau^2) X=Y+Z The sum of two normal variables is normally distributed and has a mean equal to the sum of their means and a variance equal to the sum of ... 1 It is best to use a boxplot to find outliers. The problem with using the sample mean \bar X and the sample SD S is that an outlier seriously affects the values of \bar X and S. By contrast, the boxplot uses the median and the interquartile range to detect outliers. These measures of location and dispersion, respectively, are not much affected by ... 0 Absolutely. It is a known fact that for a sufficiently long list , (denoting mean by \mu and standard deviation by \sigma) the range [\mu-3\sigma,\mu+3\sigma] encompasses about (more than) 99.73\% of the data points, so if the new value is out of this range then it is 99.7\%sure to be out of the list You can somewhat use the concept of p-value ... 0 Yes. You can use your Standard Deviation to tell you this. Think about what Standard Deviation is telling you. 5 Maximising the standard deviation or variance for a given mean is equivalent to maximising the sum of squares of the values for a given sum of the values. Meanwhile if b \ge a and \delta \gt 0,$$(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 ...

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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 ...

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

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