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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!


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. $$\sigma_p = \sqrt{s/20} = 7.08237$$


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 ...


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 $$\sigma_X=5K_2.$$


1

$P(z \geq -4.55) = 1- P(z < -4.55) \approx 0.999 $ $P(z < 6.82) \approx 1$


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

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 ...



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