I heard median of any number of data. Suppose we have a data      [13,23,23,23,54] The median of this data is is 23. How we calculate the weighted median of same data (as mentioned above).  **  (this is i get when i Google it. But things are not clear).
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I was given some data that represents the number of lines in a document as well as the line count per hour (which is the lines in the document divided by the number of hours that the document was worked on).  Considering the following data:  lines LCPH 57    133 62    135 33    137 76    139 78    141   Typically, I would express the median LCPH as 137 as it is the 3rd of 5 figures.  However, the median is currently being reported as 139 because it is not until the fourth document that more than half the lines in all documents were seen.  This figure is currently being referred to as the Weighted Median LCPH instead of Median LCPH.  Is there a different term for this sort of figure?
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Here, have a look:  You can see exactly where the training data ends. Training data goes from $-1$ to $1$.  I used Keras and a 1-100-100-2 dense network with tanh activation. I calculate the result from two values, p and q as p / q. This way I can achive any size of number using only smaller than 1 values.  Please note I am still a beginner in this field, so go easy on me.
101933
I've looked through a lot of literature available online, including this forum without any luck and hoping someone can help a statistical issue I currently face:  I have 5 lists of of ranked data, each containing 10 items ranked from position 1 (best) to position 10 (worst). For sake of context, the 10 items in each lists are the same, but in different ranked orders as the technique used to decide their rank is different.  Example data:              List 1      List 2      List 3     ... etc Item 1     Ranked 1    Ranked 2    Ranked 1      Item 2     Ranked 3    Ranked 1    Ranked 2 Item 3     Ranked 2    Ranked 3    Ranked 3 ... etc   I am looking for a way to interpret and analyse the above data so that I get a final result showing the overall rank of each item based on each test and its position, e.g.  Result Rank 1 = Item 1 Rank 2 = Item 3 Rank 3 = Item 4 ... etc   So far I have attempted to interpret this information from performing Pearson's Correlation, Spearman's Correlation, Kendall Tau's B, and Friedman tests. I have found however, that these results have generally paired my lists (i.e. compared list 1 to list 2, then list 1 to list 3 .. etc), or have produced results such as Chi-Square, P-Values etc about the overall data.   Does anyone know how I can interpret this data in a statistically sound method (at a post graduate / PhD applicable level) so that I can understand the overall ranks signalling the importance of each item in the list across the 5 tests please? Or, if there is another type of technique or statistical test I can look into I would appreciate any hints or guidance.   (It maybe also worth noting, I have also performed the simpler mathematical techniques such as sums, averaging, minimum - maximum tests etc, but do not feel these are statistically important enough at this level).  Any help or advice would be greatly appreciated, thank you for your time.
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I have a data set with a sample size over three million numeric values. Close to 20% are either 0 or 1, with the maximum being nearly 18500. So the data is clearly quite heavily positively skewed.  I am trying to categorize some of this data by putting it into bins of equal width, so I decided to try and find the optimal number of bins. Using the Freedman-Diaconis rule it gave me a value of 126044.0262335108, this is clearly a ridiculously large number of bins for the data.   Breaking the set into the Inter-decile range also proved fruitless giving me [0, 1, 1, 2, 3, 5, 8, 17, 47]  Reading elsewhere the square root of the sample size was suggested, this gave 1732.05081 which is more reasonable. However the method is quite crude.  I also looking into Doane's formula given . But reading up on this method it seems to have been based on an incorrect .  How should I deal with this level of skew in the data?  What is the best way to categorize this data?
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If so, what? If not, why not?  For a sample on the line, the median minimizes the total absolute deviation. It would seem natural to extend the definition to R2, etc., but I've never seen it. But then, I've been out in left field for a long time.
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I'm a mathematician self-studying statistics and struggling especially with the language.   In the book I'm using, there is the following problem:  A random variable $X$ is given as $\text{Pareto}(\alpha,60)$-distributed with $\alpha&gt;0$. (Of course, you could take any distribution depending on one parameter for the sake of this question.) Then a sample of five values $14$, $21$, $6$, $32$, $2$ is given.   First part: "Using the method of maximum likelihood, find an estimate $\hat{\alpha}$ of $\alpha$ based on [the sample]." This was no problem. The answer is $\hat{\alpha}\approx 4.6931$.  But then: "Give an estimate for the standard error of $\hat{\alpha}$."  What is meant by this? Since $\hat{\alpha}$ is just a fixed real number, I don't see in what way it could have a standard error. Am I to determine the standard deviation of $\text{Pareto}(\hat{\alpha},60)$?  If you think the question is not clear, this information would help me as well.
42215
I have MEAN, MEDIAN, NSUB (number of subjects) and SD data coming from different studies and different populations. These are the only available information in the studies. My data set contains the following columns:  STUDY   NSUB   MEAN   MEDIAN   SD 1       10     1.5    1.7      0.4 2       20     1.5    1.55     0.3 3       7      2.6    3        0.5 4       15     3      3.01     0.1    And so on.  I want to judge whether the mean and median are significantly different at significance level (alpha=0.05). This is so I can make a decision whether the data from each study were come from a normal or log-normal distribution.   I thought of doing a two sided t-test. My hypothesis would be:  NULL: H0: MEAN=MEDIAN  Alternative: HA: MEAN doesn't equal MEDIAN  The standard error of the mean (SEM) can be calculated as SD/NSUB where NSUB is the number of subjects. The degree of freedom would be NSUB-1. The scenario I am presenting here is as if I am comparing two means that have the same standard deviation (SD).  Is there a way where I can apply this automatically in R for the set of data that I have and have the result added to the data as t.test.result being TRUE/FALSE (significantly different/insignificantly different). Something close to this:  STUDY   NSUB   MEAN   MEDIAN   SD     t.test.result 1       10     1.5    1.7      0.4    RESULT1? 2       20     1.5    1.55     0.3    RESULT2? 3       7      2.6    3        0.5    RESULT3? 4       15     3      3.01     0.1    RESULT4?   Also, is there a better way to test whether each study population was come from a normal or log-normal distribution?
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I'm working through a statistics textbook while learning R and I've run into a stumbling block on the following example:    After looking at ?quantile I attempted to recreate this in R with the following:  &gt; nuclear &lt;- c(7, 20, 16, 6, 58, 9, 20, 50, 23, 33, 8, 10, 15, 16, 104) &gt; quantile(nuclear)    0%   25%   50%   75%  100%    6.0   9.5  16.0  28.0 104.0    Given that the text and R have different results, I'm gathering that R is utilizing the median in the calculation of the first and third quartiles.  Question:  Should I include the median in calculating the first and third quartiles?  More specifically, does the textbook or R have this correct? If the textbook has this correct, is there a way to properly achieve this in R?  Thanks in advance.
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I'm just being curious, i have seen on some places, statistical results where they give them in form of "10 of 100.000" or "5 of every 10".  I don't have any statistical knowledge over my personal experience/reading around. But basic maths tell me that such values could be simplified as any fraction we used to do:  10 of 100.000 -> 1 of 10.000 // 5 of every 10 -> 1 of every 2.  Is there any reason not to do it?  Edit:  The numbers come from, if i remember correctly:   10 of 100.000 people per year  with some bad reaction to medication 5 of every 10 people who actively smokes on ¿Spain?
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Suppose we have a set $S$ of real numbers. Show that   $$\sum_{s\in S}|s-x| $$   is minimal if $x$ is equal to the median.    This is a sample exam question of one of the exams that I need to take and I don't know how to proceed.
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Provided a sample size "N" that I plan on using to forecast data. What are some of the ways to subdivide the data so that I use some of it to establish a model, and the remainder data to validate the model?  I know there is no black and white answer to this, but it would be interesting to know  some "rules of thumb" or usually used ratios. I know back at university, one of our professors used to say model on 60% and validate on 40%.
22199
Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$  of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.  Why do the medians solve $$ \min_{c \in \mathbb{R}} E(|X-c|) \, ? $$
101154
Suppose I have a discrete set of (finite) data (values are only positive integers) (which always remains discrete whenever the observation/survey is taken), divided into some identical categories . Now, I want to calculate the mean values of some random variable for each category, hence getting a sequence of mean values. Considering mean as a random variable (let's call it $M$), it's taking decimal values also (obviously possible). But I am not sure if $M$ is a discrete or a continuous random variable.   My intuition suggest me that it should be discrete. Since, my original data from which I calculated the sets of mean is discrete so the mean can take finitely many values which I suppose is not the case with a continuous random variable where the mean can take infinitely many values within an interval. So is my intuition right or wrong?  Is it true that a discrete random variable takes finitely many values and continuous R.V. can take infinitely many possible values?
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I was solving a question which technically reduces to the following     Given $N$ items, it is equiprobable for each item to be good or bad, what is the probability that number of good items in the set are greater than or equal to $K$.   This can be reduced to $\dfrac{x}{2^n}$ where $\displaystyle x = \sum_{p = k}^{n} \binom{n}{p}$. Is there a more simplified form which is easier to calculate for large values of $N, K$?  Note: It may be safe to assume that we do not require extremely high precision while calculating(first 5-10 digits only).  Thanks!
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I have a large set of data points of the form (mean, stdev). I wish to reduce this to a single (better) mean, and a (hopefully) smaller standard deviation.  Clearly I could simply compute $\frac{\sum data_{mean}}{N}$, however this does not take in to account the fact that some of the data points are significantly more accurate than others.  To put it simply, I wish to preform a weighted average of these data points, but do not know what the weighting function should be in terms of the standard deviation.
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My knowledge of statistics are small so I need a little bit of help.   I have some data showing the relationship between age and walking speed. The numbers of walking speed are presented only as the 15th and the 50th percentile.  For example:  Child(0-12 years): 1.07m/s (15th percentile)  1.33m/s (50th percentile).   I need to find how many percent is between 1.0m/s - 1.2m/s and 1.2m/s - 1.4 m/s etc.  That is the likelihood of observing a child walking between 1.0 and 1.2 m/s etc.   Is that possible? Please help me out.
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What is the difference?For what kind of purpose do we use the weighted average method?Why don't we instead always use the normal average method?and also,how to get the weighted average of some numbers?Also,do we have first to get the mean(normal average) of numbers in order to get their weighted average?  Thank you so much.