Update So I've done some of my own work on transformational methods, and the best I can get is what I call an s transform as detailed in this workbook; however, various attempts at trying to mean adjust on the Summary page show that sometimes the values get fudged/overlap each other at .5...    Update: On further thought, the Monte Carlo method has seemed to fail me.  I was able to derive the integral of the dataset, but it's mean is not .5..  This is in the realm of data normalizing.  Here's a work in progress; I did borrow some ideas from integrals.    Update: People marking questions as duplicates ruins any new chance of conversation on a re-clarified topic, therefore, I deleted the old question.  It was suggested that this is the possible solution:    but I disagree.  I am not using random sampling.  I have the entirety of the data set.  This is not an me observing variables, but rather have a dataset and would like to transform it using said methods and then mean adjust the final output to .5  Update I now agree I can use the monte carlo simulation concept.  I just have to supply the entirety of my data set.  ...[deleted]  This was also suggested as a solution Scaling a Series of Numbers   but rank (and I have a method called Empircal Cumulative Distribution Function) ignores the size of the distance between elements. This is maybe where it's "not possible".  Maybe what I'm expecting isn't possible and can only be achieved with such a ranking function which destroys any individual distance between values.  Original Post:  I've figured out how to do a minmax transformation to achieve normalized data from 0 to 100%  Basically  (x-min)/(max-min)   However, it's mean can be anywhere, and not near 50% like I want.  I've kind of carried my transformation further, and decided to do a minmax transformation around average like so  if x &lt;= average value = (x-min)/(average-min) else (x-average)/(max-average)   I do a follow transform around median  if x x&lt;= median (x-min)/(median-min) else (x-average)/(max-average)   What this does is achieves a nice transformed distribution around .5 mean.  Not quite .5 though.  I can apply this concept to an entire matrix of values, and achieve quite pleasant normalized data that I can start to use on an nth Dimensional Problem.  However, one thing that bothers me, and causes issues down the line when combining these nth Dimensions together for an average of values...  My mean is never quite .5  Is there a way I can transform the data proportionally and achieve a .5 mean by still keeping the values that are above 50% above 50% and those below below?  I had an idea to do this using ratio's  I took the weighted ratio's of  1, 2, 3 NumOfElements = 3 sum = 6  1/6 = .166 2/6 = .33 3/6 = .5  Sum of ratio's = 1  Average of Sum = .333  If I wanted a .5 average, I would multiple each % by the NumOfElements 1/6*3 = .5 2/6*3 = 1 3/6*3 = 1.5 average = 1 desired average = .5 Divide by 2 (universal rule) achieve's .5 mean.   However... this doesn't respect my upper boundaries.  Applying such logic to my distribution merely factored up all the values, which probably could have been derived easier  by measuring the difference between the sum of the data elements compared to: .5 * number of elements, and factor up accordingly.  However, I don't want to factor "up" from 100%, and I have the problem of having normalized two sides of my data using different minmax anchors...  So, I figured.  If I could do one half in this fashion (&lt;50% values) and the other half (>50%) using some method, maybe I can still achieve a .5 mean normalized mean.  Old Solution to the problem I would recursively do a  mean adjust from the difference of the matrix's mean from .5  ex... x= x + difference  That would give me a .5 mean.  However, it has the added affect of moving my min and max outside of 0 and 100% respectively, I would then just do another minmax around 50%.
Suppose I have 30 numbers that vary between 0 and 1.0 and which sum to 1.0.  The mean is obviously 0.033.  A client wants these scaled to lie between 0 and 1.0 but to have a mean of 0.5.  By the way, I'll probably have to do this for any set of numbers (positive, negative, outside the range 0 - 1) so that they lie in the range 0 - 1 and have a mean 0.5.  Any suggestions?
Inspired by this ....  I am the sole person using my system with 12.04. Every time I issue a sudo command; the system asks for the user password (which is good in its own way).  However I was thinking; without activating the root account; how can I execute the sudo commands which will not ask for user password to authenticate.  NOTE: I want to execute sudo command without authenticating via password; only when they are executed via terminal. I don't want to remove this extra layer of security from other functions such a while using 'Ubuntu software center' or executing a bash script by drag-drop something.sh file to the terminal. Remove automatically added keyboard inputs and prevent them from coming back (Windows 10) I have 2 languages installed on my computer, both with a single input method.  I have 2 keyboards: CES-CSQ and ENG-US.    However lately (maybe after the last Win10 update) I started to see 2 additional keyboards in my systray  - namely CES-CS and ENG-CSQ.    Is there there a way to remove those 2 input options (CES-CS, ENG-CSQ) which I do not see in the Control panel's Language options?  Solution described at  question does not apply to my problem because I do not see input options that I want remove in Control panel's Language options. doesn't has to /have to Javascript && operator as if statement? I was pondering a bit more about  question regarding being able to &quot;omit&quot; the base case in a proof by strong induction due to vacuous truth. The post states:  Strong induction proves a sequence of statements $P(0), P(1), …$ by proving the implication &quot;If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true.&quot; for every nonnegative integer $n$. There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.  However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds. Statistics and causal inference? In his 1984 paper , Paul Holland raised one of the most fundamental questions in statistics:     What can a statistical model say about   causation?   This led to his motto:     NO CAUSATION WITHOUT MANIPULATION   which emphasized the importance of restrictions around experiments that consider causation.  Andrew Gelman makes :     "To find out what happens when you change something, it is necessary to change it."...There are things you learn from perturbing a system that you'll never find out from any amount of passive observation.   His ideas are summarized in .  What considerations should be made when making a causal inference from a statistical model? ORA-00979 not a group by expression $f:I=[\pi,2\pi]\to \mathbb{R}^2$ be given by $f(t)=(\cos t,\sin t)$  which of the following are necessarily correct?  $1$. $\exists t_0\in I$ such that $f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))$  $2$. $\nexists t_0\in I$ such that $f'(t_0)=\frac{1}{\pi}(f(2\pi)-f(\pi))$  $3$.$\exists t_0\in I$ such that $\pi||f'(t_0)||\ge ||(f(2\pi)-f(\pi)||$  $4$. $f'(t)=(-\sin t,\cos t)\forall t$  $4$ is clearly true, $1$ is true by Mean Value Theorem as $f$  satisfies all the conditions of MVT? have no idea of $2$ and $3$, Thank you for help I am reading  that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:      Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in   \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+  \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on    $\mathcal M$ (a linear map from  finite dimensional space is always    continuous) which extends to a member of $\mathcal X^*$,  by the    Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let    $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.   I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.  My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case. Overfull box and broken pagination with page that contains only section headings How do I change the default file manager back to Nautilus? Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.    I have tried to using the definition but failed. Can someone help me please. The level 4 spell Polymorph requires concentration to maintain.  The subject retains their personality and alignment, but they have the mental statistics of the new form, which must be a beast and cannot speak or cast spells.  If a caster targets themselves with this spell, can they concentrate to maintain it, or does this depend on the mental statistics of the beast form they choose?  This situation certainly occurs in practice, since Wizards reaching level 10 in the School of Transmutation gain the Shapechanger feature.  This allows them to cast Polymorph, but only upon themselves and only to change into a beast of CR 1 or lower.  Beasts with CR 1 or less range in Intelligence from the Giant Eagle (Int 8) to the Frog (Int 1). What does the word "SARGable" really mean? What is the rigor behind U subsitution? I'm trying to generate a secret key to be used for HMAC SHA-256 signature processing. I've seen many sample of keys with variable length from 32 characters to 96 characters.  What is the ironclad rule for this key size? In what occasion could the word “precious” be taken disapprovingly, or sarcastically? I understand that the word “precious” in the following quote of Maureen Dowd’s article “Heart of Darkness” (NYT, March 20) literally means “2. valuable or important and not to be wasted” as defined in OALD:     Congressman Jones read an e-mail from a former boss of General   Allen’s, giving the congressman this unvarnished assessment:   “Attempting to find a true military and political answer to the   problems in Afghanistan would take decades. Would drain our nation of   precious resources, with the most precious being our sons and daughters. Simply put, the United States cannot solve the Afghan   problem, no matter how brave and determined our troops are.   But I didn’t know “precious” has another meaning as defined in OALD: 5. (disapproving, people and their behavior) very formal, exaggerated and not natural, and Readers English Dictionary: 2.(colloquial) terrible, almost worthless, until I was told by my respected forum mate a few days ago.   Being encouraged by his suggestion for me to “ask another question if I’m not sure of what he means by describing his input as “precious” is actually rather “precious” phrasing, I would like to ask:   What is an example of “precious” being used as “disapproving” or sarcastic way (as neither OALD nor CALD provides examples)? Is there an easy way to discern the instance “precious” being used as “disapproving” from the case being used in the sense of genuine “great value” as used in the above quote other than judging from the context? I'm wondering how it is correct to structure sentence and what punctuation should be used. In particular, is the next sentence correct:     I was wondering if there's any progress on the issue.    Or should I put a question mark in the end of it. Also, can it be restructured to:     I was wondering: is there any progress on the issue? "Track changes" in LaTeX I am writing a scientific paper with my collaborator. We have submitted a version to a journal and now we got a review from them, so we need to fix a few things. In Microsoft Word there is the option to "track changes", which displays the deleted text as crossed out and the new text bolded (if I remember the details correctly). Is there a way to achieve a similar result in LaTeX?  I was thinking to create a new file for the revised version and then write a script which would basically compare the two files and make the differences either bold or crossed out before sending it to latex. Is there a tool that already does it? I guess it is too difficult a question to ask about the cardinality of $S_{\mathbb{N}}$ so I would like to ask whether it is countable or not.  I tried to prove it is uncountable somewhat mimicking the Cantor's diagonal argument but failed.