I've done an experiment with several essays in humans, and I have the mean and variance of the results of all essays per human (each data point is not independent from each other because of the nature of the metric I'm using).  I'm creating a model that predicts human behavior, and I want to know how well it is performing. So I thought about computing the mean and variance of all means and compute how many standard deviations away is my model from the mean of means. Does this make sense? The data passed z-test from MATLAB. If it doesn't, what other kind of measure of dispersion could I use?  Mean for each human for four different categories of essays:  [[0.5450  0.6605  1.0846  1.2752]; [0.5704  0.7386  0.9530  1.0476]; [0.5307  0.7269  0.7603  0.8317]; [0.5673  0.6713  1.0126  1.1333]; [0.4641  0.5162  0.7868  0.8933]; [0.4872  0.5299  0.8521  0.9655]; [0.4610  0.5477  0.6610  0.7872]; [0.5682  0.6332  0.9416  1.3915]; [0.4832  0.4663  0.7870  0.9705]; [0.4965  0.4920  0.8236  0.8065]; [0.4576  0.5171  0.7066  0.7857]; [0.4797  0.5849  0.9257  0.9133]; [0.5316  0.7297  1.1286  1.0829]; [0.4475  0.5560  0.8381  0.3333]; [0.4863  0.5645  0.8010  1.0333]; [0.5326  0.6206  1.0022  0.9667]; [0.5063  0.6971  1.0398  1.0000]; [0.4620  0.4976  0.8451  1.5333]; [0.4559  0.5456  0.6881  1.5952]; [0.4449  0.4666  1.0014  0.6667]; [0.5060  0.5841  0.9442  NaN   ]; [0.5210  0.6592  0.9338  0.9167]; [0.5355  0.5837  0.7764  0.8095]; [0.5158  0.5478  0.9157  0.8333]]  Standard deviation for each human for four different categories of essays:  [[0.3231  0.5578  0.8506  1.4848]; [0.3297  0.5541  0.5780  1.3469]; [0.2867  0.5482  0.5948  0.7591]; [0.3661  0.5505  0.5888  0.7310]; [0.2216  0.3942  0.7027  0.7470]; [0.2497  0.4051  0.6257  0.9586]; [0.2481  0.4105  0.4684  0.4857]; [0.3348  0.4401  0.5719  1.8024]; [0.2916  0.3877  0.5794  0.5054]; [0.3020  0.3879  0.5681  0.6255]; [0.2452  0.4009  0.5164  0.5190]; [0.2612  0.4558  0.7113  1.1915]; [0.2393  0.4104  0.6467  0.9595]; [0.2475  0.3992  0.8032  0.5774]; [0.2684  0.4371  0.6703  0.7394]; [0.2363  0.4004  0.5972  1.0435]; [0.2091  0.3677  0.7398  0     ]; [0.2793  0.3872  0.3672  0.6600]; [0.2348  0.4044  0.5832  0.5724]; [0.2297  0.3957  0.5663  0.8819]; [0.2312  0.3826  0.4504  NaN   ]; [0.2219  0.4414  0.7050  1.1902]; [0.2791  0.4194  0.5304  0.7902]; [0.2901  0.4270  0.5227  0.3333]]  Thanks in advance!
It's my understanding that a z-score can only be calculated and accurately used for data sets that are normally distributed.  Since "perfect" normal distribution almost never occurs in real-world data (where "perfect" normal distribution is defined as 1. The mean, median, and mode all equal the same number, 2. the distribution is perfectly symmetrical between all standard deviations on both sides of the mean, and 3. the distribution is asymptotic), how "close" can the distribution be to perfectly normal for the z-score to still be a valid statistical measure?  If the answer is it has to be a perfectly normal distribution, then my question is why is the z-score so important since it would seldom be a valid measure on real-world data?
Content Type field on NewForm.aspx I have a custom list and I am using content types to add items to the list. I am using OOTB forms for new and edit items. One field that I see on the EditForm.aspx is the one that I hope I get on the NewForm.aspx also and it is the Content Type field. It is shown in the following screenshot. Do you know how to add that field on the NewForm.aspx page also?    Thanks. If matrix $A$ is unitary and $B^2=A$, is $B$ necessarily unitary? Apply SP3 on SQL Server 2012 AlwaysOn AAGs What is the best possible way to check if a string can be represented as a number in Python?   The function I currently have right now is:  def is_number(s):     try:         float(s)         return True     except ValueError:         return False   Which, not only is ugly and slow, seems clunky. However I haven't found a better method because calling float in the main function is even worse. Book with travellers, alien artefacts and age reversal Looking for the title of a book I read about 5 years ago though I believe the book was a lot older.  Essentially the main male character was on a task and was travelling with a group of people. The mode of transport if I recall was a boat and they were on a remote planet travelling via a river. One of the group was a father to a young female infant. There's a reference to alien artefact/temple that essentially reverse aged a female scientist (the baby) a day at a time and the father is trying to undo it. Also a line I remember is the father used to say see you later alligator and the daughter will reply that awhile crocodile until a day where she forgot.  Another fellow traveller was a pilgrim and his male disciple. There are several other male and female travellers all had different reasons to travel to that location.  The aliens was located within the temple (time locked?) Were invisible to humans. Killed any that entered. Got the impression that they were quadpeds with organic blades/razors that cut people.  I believe the book may be part of a series. The male author though I don't recall the name, has written other works that have received awards. The book or series may have received one too. Which is probably why I decided to read it in the first place.  Question never asked anywhere else. I am from Australia, book was a paper copy bought locally. Prove that the power set of an $n$-element set contains $2^n$ elements Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.      Proof. The proof is by induction on the numbers of elements of $X$.      For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.      Now the induction step. Suppose $|X|=n$; by the induction hypotesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$      From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.  The construction of a new set $S$ is the union of the old set with cardinality $n$ and an new element $\{a\}$, therefore the set that do not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element...  Can someone please elucidate? Are there any practical implementation of a homomorphic hashing or signature scheme? A  is a function $H : A \to B$ between two sets with some algebraic structure $(A, *)$ and $(B, \star)$ such that   $H$ is collision resistant, i.e. it is hard to find $x \neq y$ such that $H(x) = H(y)$ and $H$ is a homomorphism, i.e. $H(x * y) = H(x) \star H(y)$.   Are there any practical realizations of such a homomorphic hash function, or even a homomorphic signature scheme (i.e., where we can "add" valid signatures to get a signature of the "sum" of two messages)?  Even better, are there even any libraries implementing this? The selected task "{0}" no longer exists. To see the current tasks, click Refresh When I start my task scheduler on Windows 7 I get a dialog with the following error:     The selected task "{0}" no longer exists.  To see the current tasks,   click Refresh     Also there's a UUID, that is displayed as the name and when I try to delete it I cannot, it just hangs around.  Is there a fix for this? To prove that $f_n(x) = \frac{nx}{1+n^2x^2}$ does not uniformly converge to $f(x) = 0$ on $[0,1]$ My Approach. To prove the given statement it is sufficient to show that $$\exists \epsilon &gt; 0, \exists x \in [0,1], \forall N \in \mathbb{N}, \exists n &gt; N $$ $$ \Rightarrow | f_n(x) - f(x)| &gt; \epsilon$$    Let $\epsilon = \frac{1}{10}, x = \frac{1}{n}, n =N+1$. Hence $| f_n(x) - f(x)| = \frac{1}{2} &gt; \epsilon = \frac{1}{10}$. Hence, we have shown for every $N \in \mathbb{N}$, how to choose $n$ such that $| f_n(x) - f(x)| &gt; \epsilon$. Thus, $f_n(x)$ does not converge uniformly on $[0,1]$.    Is there any logical flaw or wrong steps in my proof. Any help would be appreciated. Icon missing when application is launched I've created an application launcher in /usr/share/applications that launches rdesktop. The file specifies an icon name and I've added the icons to:   /usr/share/icons/HighContrast/48x48/apps /usr/share/icons/HighContrast/22x22/apps /usr/share/icons/HighContrast/256x256/apps /usr/share/icons/HighContrast/32x32/apps /usr/share/icons/HighContrast/scalable/apps-extra /usr/share/icons/HighContrast/24x24/apps /usr/share/icons/HighContrast/16x16/apps /usr/share/icons/hicolor/scalable/apps /usr/share/icons/hicolor/128x128/apps /usr/share/icons/hicolor/16x16/apps   I then drug the launcher to the taskbar and the launcher with the correct logo is visible in the taskbar. However, when I start the task, a new icon is created in the taskbar, a square with a question mark in the center.  What icons do I need and where, so that the taskbar displays the correct icon for the application when it is launched?  The .desktop file contents:   [Desktop Entry] Version=1.0 Terminal=false Icon=mswindows Type=Application Categories=Network; Exec=rdesktop win7shared1 -g 1680x1000  Name=win7shared1 GenericName=rdesktop Comment=Open windows desktop  Keywords=Windows StartupNotify=false What color would you see if you place 2 mirrors in opposit when one is a one way mirror Two players, $A$ and $B$, alternately and independently flip a coin and the first player to get a head wins. Assume player $A$ flips first.  If the coin is fair, what is the probability that $A$ wins?   So $A$ only flips on odd tosses. So the probability of winning would be $$ P =\frac{1}{2}+\left(\frac{1}{2} \right)^{2} \frac{1}{2} + \cdots+ \left(\frac{1}{2} \right)^{2n} \frac{1}{2}$$  Is that right? It seems that if $A$ only flips on odd tosses, this shouldn't matter. Either $A$ can win on his first toss, his second toss, ...., or his $n^{th}$ toss. So the third flip of the coin is actually $A$'s second toss. So shouldn't it be $$P = \frac{1}{2} + \left(\frac{1}{2} \right)^{2} + \left(\frac{1}{2} \right)^{3} + \cdots$$ Why do I have to enter my password to log in? Using the Stack Exchange login, why do I have to enter my password again and again to log in? Wasn’t the whole purpose of OpenID to do away with this?  This doesn’t even only happen when visiting a new Stack Exchange site – it also happens for sites where I already have an account but which I don’t visit regularly. Here’s what I mean:    Of course I don’t remember my password (it’s a long random string). And so I have to go to my password manager and manually copy the password from openid.stackexchange.com’s account into the login form. – And incidentally, this is what my password manager looks like now:    (And, by the way, those are the only entries in my keychain so far, but by far not the only SE sites where I have an account.)  … I feel that this defeats the purpose of site-wide login and OpenID. In the olden times when MyOpenID was still working properly, I’d simply enter my OpenID on the login form, and the software would take care of the rest – namely, redirect to the OpenID service, (automatically) verify that I was authorised to use it, and redirect back to the original website, having logged me in. It was literally one button click for me. It’s substantially more work now, and I’d like to understand why this regression is necessary.  (Note: of course once a site’s password has landed in my password manager I (probably?) won’t have to enter it again. But still: wasn’t site-wide login and OpenID supposed to solve this already?) Is there a dollar amount that, when adding Massachusetts Sales Tax, precisely equals $200? how to make clean without any modification? How to EFI install Ubuntu? On trying to install Ubuntu 12.10 Desktop amd64 to a desktop pc using a ga a75 ud4h motherboard (which is using an AMD 3870k APU with a new 1TB HDD) following the advice in the  and  have now got to grub option screen for EFI install. Booting live dvd in EFI mode gives &quot;Secure Boot is not enabled&quot; just before grub kernel option list shows.  If the option to &quot;Install Ubuntu&quot; or &quot;Try Ubuntu without installing&quot; is selected gives &quot;can not read cd/0&quot; and &quot;the kernel must be loaded first&quot; errors. Using DVD or USB installation media gives same results and there has never been at any point an UEFI entry in bios/firmware settings or boot device list/s. It might be helpful to say that have had an alternative non-linux OS installing EFI mode on this hardware giving a GPT installation automagically using USB installation media when the hdd space was completely unallocated space. The USB device was listed under hard drives in bios/firmware one time boot list and did not have an UEFI entry. i.e. it installed EFI although no EFI or UEFI entry was assigned to the device like what is shown in many tutorials regarding UEFI installs. The only constructive information regarding this question is found at  and  but can not work out the answer. Checking md5sum and sha256 checksums on downloaded .iso and used &quot;Check disk for defects&quot; to check burned .iso when booting DVD in BIOS mode. No errors found. &quot;Check disk for defects&quot; will not run (or any other options) when DVD booted in EFI mode. Due to trying different distributions as well as older and newer flavours of Ubuntu all giving similar results in that no options from grub option screen in EFI mode are bootable (if getting that far) as well as spending time searching Ubuntu Forums have come to conclusion the firmware on this system is hardcoded to handle Microsoft EFI files only. After installing Windows EFI mode here is a  and here is a link to a thread i started at  which may be helpful to others building a dual boot system. Installing BIOS mode and using Boot-Repair to install grub-efi is not a complete solution due to the inability of using installation media EFI mode for recovery and maintenance due to the &quot;you need to load kernel first&quot; and &quot;can not read cd/files&quot; errors as described when trying to boot installation media using EFI. Secure Boot is not an issue. Using msinfo32 from windows confirms Secure Boot is not supported on this hardware. How to edit kernel or load kernel for EFI install? = How to EFI install Ubuntu? PLEASE CONSIDER: This is not a dupe!  The linked question that is indicated to having a working answer does not have a working answer to &quot;###How to edit kernel or load kernel for EFI install? = How to EFI install Ubuntu?&quot; therefor the workaround of installing using BIOS then manually install grub-efi without renaming bootloaders which works to get Ubuntu booting using EFI or using boot-repair to do this (and many unnecessary &quot;fixes&quot; if recommended repair is not changed in advanced settings) also works.  This goes some way to confirming hardcoding of firmware is not stopping grub the boot manager seeing the vmlinuz.efi files in /casper on DVD.  I would like an answer to this so would like this question to be reopened. How does Linux know where its swap partition is? I've read that you need to put the swap partition on HDD rather than SSD.   My questions are the following:    When and how is the "checking" done by the distribution (or something else) to find its Swap partition?  Does it happen during boot?  It just checks all available disks and searches for a partition with 'swap' flag?  What happens if there are several partitions like that?  Also, how many swap partitions do I need to have if I run, for example, two different distributions on the same disk, let's say Fedora and Ubuntu? How do I make log4j clear a log at startup? What is a simple example of a limit in the real world? Asymptotes in a plot I'm using TikZ to plotting some functions for a document. For example, I want to plot 1/(x+1). So I did something like this:  \begin{tikzpicture}[&gt;=stealth,x=3mm,y=3mm] \draw (0,5) node[above]{\fimg $y=f(x+1)=\dfrac{1}{x+1}$}; \clip (-5,-5) rectangle (5,5); \ejes[1]{-4}{4}{-4}{4} \draw[samples=100] plot(\x,{1/(\x+1)}); \end{tikzpicture}   where \ejes[style]{xmin}{xmax}{ymin}{ymax} is command, that I have defined before to draw the axes (join with \fimg). My expected output for the code below was just the function, but I get this:    So, TikZ automagically put the vertical asymptote at x=-1. That isn't a bad thing, but now I want to have some control over it. For example I want dotted lines for vertical asymptotes. How can I do that?