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author | bnewbold <bnewbold@robocracy.org> | 2010-01-24 08:07:57 +0000 |
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committer | User <bnewbold@daemon.robocracy.org> | 2010-01-24 08:07:57 +0000 |
commit | 88763d7db3f803b9e5b6351e01c186a98e50bbf2 (patch) | |
tree | 14ca02b562ebf3c153403ac5f0e779bcc1e92df5 | |
parent | 4cfee64407fce9a4687565b952e4cad5336ab8c0 (diff) | |
download | knowledge-88763d7db3f803b9e5b6351e01c186a98e50bbf2.tar.gz knowledge-88763d7db3f803b9e5b6351e01c186a98e50bbf2.zip |
tex math fixes
-rw-r--r-- | math/statistics.page | 21 |
1 files changed, 10 insertions, 11 deletions
diff --git a/math/statistics.page b/math/statistics.page index 446450f..b3b5b3d 100644 --- a/math/statistics.page +++ b/math/statistics.page @@ -4,7 +4,7 @@ Statistics Basic Measures ------------------------- The sample distribution has finite size and is what has been measured; the -parent distribution is inifinite and smooth and is the limit case of the +parent distribution is infinite and smooth and is the limit case of the sample distribution. The mean, or average, is (of course): @@ -25,7 +25,7 @@ probability of getting 'yes' for a single attempt. $$P(x;n,p) = \frac{n!}{x! (n-x)!} p^x (1-p)^{n-x}$$ -The mean of this distribution is $\mu = np$, and $\sigma$ = \sqrt{np (1-p)}. +The mean of this distribution is $\mu = np$, and $\sigma = \sqrt{np (1-p)}$. Poisson Distribution ------------------------ @@ -40,7 +40,7 @@ The classic! Also called a normal distribution. $$P(x;\mu,\sigma) = \frac{1}{2\pi \sigma} e^{-\left(\frac{(x-\mu)^2}{2\sigma^2}\right)}$$ -The mean is $\mu$ and the deviation is $\sigma=\sqrt(\mu)$. +The mean is $\mu$ and the deviation is $\sigma=\sqrt{\mu}$. Lorentzian Distribution --------------------------- @@ -60,21 +60,20 @@ the error on the mean should get smaller. More elaborately, if the errors are different for each individual measurement, the mean will be: $$\bar{x}= - \frac{ \sum_{i=1}^{N} x_i / \simga_{i}^2}{\sum_{i=1}^{N} 1/\simga_{i}^2} - \pm \sqrt{ \frac{1}{\sum_{i=1}^{N} 1/\simga_{i}^2}}$$ + \frac{ \sum_{i=1}^{N} x_i / \sigma_{i}^2}{\sum_{i=1}^{N} 1/\sigma_{i}^2} + \pm \sqrt{ \frac{1}{\sum_{i=1}^{N} 1/\sigma_{i}^2}}$$ -$\Chi^2$ Distribution +$\chi^2$ Distribution ------------------------ -$\Chi^2$ is often writen "chi-squared" and is a metric for how well a fit +$\chi^2$ is often written "chi-squared" and is a metric for how well a fit curve matches uncertain data. -$$\Chi^2 = \sum_{i=1}^{N}\left(\frac{x_i-\mu_i}{\sigma{i}}\right)^2$$ +$$\chi^2 = \sum_{i=1}^{N}\left(\frac{x_i-\mu_i}{\sigma{i}}\right)^2$$ The number of degrees of freedom of the system is the number of measurements $N$ minus the number of variable parameters in a curve fit $N_c$: $\nu = N-N_c$. -The reduced $\Chi^2$ value is $\Chi^{2}_r = \Chi^2 /\nu$. You want $\Chi^{2}_r$ +The reduced $\chi^2$ value is $\chi^{2}_r = \chi^2 /\nu$. You want $\chi^{2}_r$ to be around (but not exactly!) 1; if it is significantly larger there are probably too many degrees of freedom, while if significantly smaller the fit is -bad. - +bad.
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