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Another way of saying the above is that you need to keep a count, an incremental sum of values and an incremental sum of squares. Let $N$ be the count of values seen so far, $S = \sum_{1,N} x_i$ and $Q = \sum_{1,N} x_i^2$ (where both $S$ and $Q$ are maintained incrementally). Then any stage, the mean is $\frac{S}{N}$ and the variance is $\frac{Q}{N} - ... 0 No,$\sigma$is not the average$\ell^2$distance from$\mu$. The$\ell^2$distance from$x$to$\mu$is$\sqrt{(x-\mu)^2} = |x-\mu|$and the average of those is $$\sum_{x\in X} p(x)|x-\mu|.$$ Rather$\sigma$is the$\ell^2$distance from the tuple of$x$values to the tuple in which every component is$\mu$. A reason for the use of the mean squared ... 1 Fitting a model such $$y= \sum_{k=1}^{m} a_k\,x^{b_k}$$ using$n$data points$(x_i,y_i)$,$n \gt 2m$, is difficult because the model is nonlinear with respect to its parameters (the$b_k$'s) and nonlinear regression require reasonable starting guesses. As you noticed, the problem is simple if you assign specific values to the$b_k$'s since the problem ... 1 Note that the standard form of the Gaussian is $$pe^{-\dfrac{(x-q)^2}{2r^2}}$$ In your equation, we have$-\frac{1}{2r^2}=-b^2$, thus$r^2=\frac{1}{2b^2}$, thus$r=\sigma=\pm\sqrt{\frac{1}{2b^2}}. However, the standard deviation is always positive. Therefore the results are: $$\sigma=\sqrt{\frac{1}{2b^2}}=\frac{\sqrt{2}}{2b}$$ $$\mathrm{FWHM} = 2 ... 0 Effects of sunshine, rain and humidity on the sales of various products correspond to some parameters of your statistical model, which you, then, each estimate by some function f_i of the data in the sample. [of course, to be a good guess of the real value of parameter, the function f_i has to be carefully chosen — that's what statistical theory is ... 1 Step 1 Use "n-1". See http://stats.stackexchange.com/questions/3931/intuitive-explanation-for-dividing-in-n-1-when-calculating-sd . If you use n you have a biased estimator of the s.d. Since n> n-1 the bias is negative (towards too small s.d.s). I.e., use \sigma_m \approx \sigma/\sqrt{n-1}. Step 2 (Added by edit on 20150715, based on ... 1 The expectation is$$E[X]=K_1+K_2(t-10).$$The variance is$$\sigma_X^2=E[(X-E[X])^2]=E[\{K_1+K_2(t-10)-(K_1+K_2(t-T_b)\}^2]==K_2^2E[(10-T_b)^2]=25K_2^2.$$So, the standard deviation$$\sigma_X=5K_2.0 Note that for x_i \not\in I we have |x_i -x| \ge 3 \sigma. Hence, \begin{align*} \sigma^2 &= \frac 1n \sum_{i=1}^n (x_i- x)^2\\ &\ge \frac 1n \sum_{x_i \not\in I} (x_i - x)^2\\ &\ge \frac 1n k \cdot 9\sigma^2\\ \iff n &\ge 9k \end{align*} The percentage of items in I, is by definition, \frac {n-k}n \cdot ... 3 Test your sum of square s = 60.84 + \dots 51.84$. I get it as$s = 1003.20$. Then depending of the standard deviation type you get for the sample std. dev. $$\sigma_s = \sqrt{s/19} = 7.26636$$ or the population std. dev. $$\sigma_p = \sqrt{s/20} = 7.08237$$ 4 There are two ways to calculate variance. First one is biased estimator and other is unbiased estimator of variance. In e-gadgets, probably estimator is unbiased estimator and that's what you are getting inconsistent result. Try dividing variance by$n-1$, the number of data points instead of$n\$ and then take square root. Good Luck!