I want to find the 25th, 50th, and 75th percentile of this data-set.  data-set:   Am not sure how one uses the cumulative frequency percentage in order to derive what the 25th,50th, and 75th percentiles are.  I am guessing for the 50th, as I know this is generally the median, or middle number but I cannot be sure this is where 50% of the data clusters. But I would guess that p_50 = 36.6% and therefore would be on line 5.  So the median would be at 5 or at 36.6% with 605 cumulative frequency  Please help
123301
Here is the data-set:   I understand that the cum per sums all the previous data up until that point. So 153 / 1655 = 19.8%, or if looking at the cum freq a total of 328 people scored 4 or less on the vocabulary test. But how does one determine if this is the 25th percentile? At what point in the range of cum per would one know it it's the 25th percentile, 50th percentile or 75th percentile?  Please help Thank you
85054
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?
30185
Here are some grades. (They aren't test scores. They aren't a sample.) I want to test whether a normal distribution derived from the mean and variance of the grades provides a good fit for the grades. What other distribution might fit is, for now, an orthogonal issue. The question is whether the normal distribution provides a valid model.  &gt; grades   [1] NA 75 70 70 80 75 85 85 75 85 75 80 85 80 75 70 70 85 90 90 NA 85 NA 75 80  [26] 80 85 80 85 85 75 NA 75 70 80 65 80 75 70 80 85 65 85 75 75 85 75 60 85 80  [51] 85 85 70 70 80 75 80 80 NA 80 75 85 80 80 55 75 60 90 80 75 80 70 85 75 80  [76] 75 80 75 75 70 85 85 80 80 75 70 NA 75 70 75 70 65 85 70 70 80 80 85 85 70 [101] 75 70 75 75 65 70 80 75 65 65 &gt; t &lt;- table(grades) &gt; t grades 55 60 65 70 75 80 85 90  1  2  6 18 27 25 22  3   The grades come from a discrete set of possible values from five to 100 in increments of five. Because of that, a Chi-squared test might be appropriate. First, the normal distribution probabilities that any of the discrete values will occur:  &gt; p &lt;- dnorm(seq(55,90,5), mean(grades, na.rm=TRUE), sd(grades, na.rm=TRUE)) &gt; p [1] 0.0004433536 0.0031786310 0.0136901824 0.0354207364 0.0550535301 [6] 0.0514034173 0.0288322176 0.0097150119   Why are the values of p so small? Why do they not add-up to one?  &gt; tf &lt;- as.data.frame(t) &gt; tf   grades Freq 1     55    1 2     60    2 3     65    6 4     70   18 5     75   27 6     80   25 7     85   22 8     90    3 &gt; chisq.test(tf$Freq, p=p, rescale.p=TRUE)      Chi-squared test for given probabilities  data:  tf$Freq X-squared = 7.0452, df = 7, p-value = 0.4242  Warning message: In chisq.test(tf$Freq, p = p, rescale.p = TRUE) :   Chi-squared approximation may be incorrect   Why does the chisq.test function produce the warning?  I need numbers, because I have a lot of these data sets and want fit results for a large sample of them. Visual inspection isn't proof. The closest relevant question I've found is  ("Testing Whether a Binomial Distribution Fits Data")
26874
I'm trying to come up with a metric for measuring non-uniformity of a distribution for an experiment I'm running.  I have a random variable that should be uniformly distributed in most cases, and I'd like to be able to identify (and possibly measure the degree of) examples of data sets where the variable is not uniformly distributed within some margin.    An example of three data series each with 10 measurements representing frequency of the occurrence of something I'm measuring might be something like this:  a: [10% 11% 10%  9%  9% 11% 10% 10% 12%  8%] b: [10% 10% 10%  8% 10% 10%  9%  9% 12%  8%] c: [ 3%  2% 60%  2%  3%  7%  6%  5%  5%  7%]   &lt;-- non-uniform d: [98% 97% 99% 98% 98% 96% 99% 96% 99% 98%]   I'd like to be able to distinguish distributions like c from those like a and b, and measure c's deviation from a uniform distribution.  Equivalently, if there's a metric for how uniform a distribution is (std. deviation close to zero?), I can perhaps use that to distinguish ones with high variance.  However, my data may just have one or two outliers, like the c example above, and am not sure if that will be easily detectable that way.   I can hack something to do this in software, but am looking for statistical methods/approaches to justify this formally.  I took a class years ago, but stats is not my area.  This seems like something that should have a well-known approach.  Sorry if any of this is completely bone-headed.  Thanks in advance!
110112
What methods can I use to infer a distribution if I know only three percentiles?  For example, I know that in a certain data set, the fifth percentile is 8,135, the 50th percentile is 11,259, and the 95th percentile is 23,611.  I want to be able to go from any other number to its percentile.  It's not  my data, and those are all the statistics I have.  It's clear that the distribution isn't normal.  The only other information I have is that this data represents government per-capita funding for different school districts.  I know enough about statistics to know that this problem has no definite solution, but not enough to know how to go about finding good guesses.  Would a lognormal distribution be appropriate?  What tools can I use to perform the regression (or do I need to do it myself)?
22014
I was doing my homework and the following problem came up! We have the CDF of an exponential distribution that is shifted $L$ units where $L&gt;0$ and $x&gt;=L$. The CDF is:  $$1-e^{-\lambda(x-L)}$$  The question says that we should assume that the following data are lifetimes of electric motors, in hours, which are:  $$\begin{align*} 153.52,103.23,31.75,28.91,37.91,7.11,99.21,31.77,11.01,217.40 \end{align*}$$  Please note that the $mean$ of these numbers is: $72.182$  Now the question has two parts which I will go through one by one:  Part1: Evaluate the log likelihood for the data when $\lambda=0.02$ and $L=3.555$. Now the way I approached the problem was to take the derivative of the CDF with respect to $\lambda$ to get the PDF which is:  $$(x-L)e^{-\lambda(x-L)}$$  Then since we have $n$ observations where $n=10$, we have the following joint pdf, due to independence:  $$(x_i-L)^ne^{-\lambda(x_i-L)n}$$ which can be rewritten as the following log likelihood:  $$n\ln(x_i-L)-\lambda\sum_{i=1}^n(x_i-L)$$ Is this the correct approach? Because it would take quite a while and be pretty cumbersome to evaluate $n\ln(x_i-L)$ for every observation?  Part2: The question also asks for the ML Estimate of $L$. So assuming the log likelihood is correct, we can take the derivative with respect to $L$ and get:  $\frac{n}{x_i-L}+\lambda=0$ and solve for $L$? Is this correct? Because I am not quite sure on how I should proceed?  Thanks so much for your help! I greatly appreciate it :)
23908
I have a problem with the normality of some data I have:  I've done a Kolmogorov test which says it isn't normal with p=.0000,  I don't understand: the skewness of my distribution =-.497, and the kurtosis =-0,024  Here is the plot of my distribution which looks very much normal ...  (I have three scores, and each one of this scores isn't normal with a significant p-value for the Kolmogorov test ... I really don't understand )
12896
I am using r and have ~400 observations divided into 3 groups.  Group A contains 199 observations, while B contains 173 and C contains 24. I have generated a boxplot of their corresponding measures and would like to compare the 3rd quartile between them to see if they are significantly different. I've included the boxplot here.  I'm not sure what test I should use to compare the 3rd quartile. Does anyone know?
114943
Let $X_1, X_2, \ldots, X_{16}$ be  independent with $N(3,4)$ distributions and $\bar{X}$ denote the sample mean. Find:   $P(-8 &lt; 2 \bar{X} &lt; -4)$. A number $K$ such that $P(-K &lt; 2 \bar{X} &lt; K) = 0.99$. $P(-10 &lt; 3 X_1+X_7 &lt; -8)$. A number $K$ such that $P(3 X_2+X_5 &lt; K) = 0.3$.   If someone could give me some help or hints on where to start with (a) or any of the problems that would be really helpful and I'd really appreciate it.
30822
I've received a results from a Mann-Whitney rank test that I don't understand.  The median of the 2 populations is identical (6.9). The uppper and lower quantiles of each population are:   6.64 &amp; 7.2 6.60 &amp; 7.1   The p-value resulting from the test comparing these populations is 0.007. How can these populations be significantly different? Is it due to the spread about the median? A boxplot comparing the 2 shows that the second one has far more outliers than the first. Thanks for any suggestions.
97350
I was wondering, given two normal distributions with $\sigma_1,\ \mu_1$ and $\sigma_2, \ \mu_2$   how can I calculate the percentage of overlapping regions of two distributions? I suppose this problem has a specific name, are you aware of any particular name describing this problem?  Are you aware of any implementation of this (e.g., Java code)?
122346
I am trying to find the correlation between a dichotomous and a continuous variable.  From my ground work on this I found that I have to use independent t-test and the precondition for it is that the distribution of the variable has to be normal.  I performed Kolmogorov-Smirnov test for testing the normality and found that the continuous variable is non-normal and is skewed (for about 4,000 data points).   I did the Kolmogorov-Smirnov test for the entire range of variables. Should I split them into groups and do the test? I.e., say if I have risk level (0 = Not risky, 1 = Risky) and cholesterol levels, then should I:    Divide them into two groups, like  Risk level =0 (Cholestrol level) -&gt; Apply KS Risk level =1 (Cholestrol level) -&gt; Apply KS  Take them together and apply the test? (I performed it on the whole dataset only.)   After that, what test should I do if it is still non-normal?    EDIT: The above scenario was just a description I tried to provide for my problem. I have a dataset which contains more than 1000 variables and about 4000 samples. They are either continuous or categorical in nature. My task is to predict a dichotomous variable based on these variables (maybe come up with a logistic regression model). So I thought the initial investigation would involve finding the correlation between dichotomous and a continuous variable.  I was trying to see how the distribution of the variables are and hence tried to go to t-test. Here I found the normality as an issue. The  Kolmogorov-Smirnov test gave a significance value of 0.00 in most of these variables.  Should I assume normality here? The skewness and kurtosis of these variables also show that the data is skewed (>0) in almost all cases.  As per the note given below I will investigate the point-biserial correlation further. But about the distribution of variables I am still unsure.
92617
I am using excel and I am trying to find both the average age and median age.  I have two columns. 1 for the category and the other for the number of people in each category.  Under 5       6,360 5-9           6,360 10-14        10,986 15-17         5,204 18-24         7,886 25-34         9,463 35-44        17,349 45-54        18,926 55-64        13,406 65-74         6,309 75 and over   5,520       total 107,769
105925
We report 95th percentile and 50th percentile of the time series data for every week. The data is time-series data in the following format date. sensor_id. sensor_location. hardware_version. sensor_value 2021-01-01. S1. L1. V1. 10. 2021-01-01. S2. L2. V2. 20. 2021-01-01. S3. L1. V1. 12. 2021-02-01. S1. L1. V3. 30. 2021-02-01. S2. L2. V3. 40. 2021-02-01. S3. L1. V2. 32. The problem we are facing is to understand what caused the change in percentile week over week. For example some weeks the p95 is very high and we would like to know what sensor or which location or which version caused the big jump in p95. Could you please help what's the best way to solve this problem?
76998
I'm learning probability, specifically transformations of random variables, and need help to understand the solution to the following exercise:      Consider the continuous random variable $X$ with probability density function $$f(x) = \begin{cases} \frac{1}{3}x^2 \quad  -1 \leq x \leq 2, \\ 0 \quad \quad \text{elsewhere}. \end{cases}$$ Find the cumulative distribution function of the random variable $Y = X^2$.   The author gives the following solution:   For $0 \leq y \leq 1: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-\sqrt y \leq X \leq \sqrt y) = \int_{-\sqrt y}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{2}{9}y\sqrt y.$  For $1 \leq y \leq 4: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-1 \leq X \leq \sqrt y) = \int_{-1}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{1}{9} + \frac{1}{9}y\sqrt y.$  For $y &gt; 4: F_{Y}(y) = 1.$    Previous to this exercise, I've managed to follow the solutions of two similar (obviously simpler) problems for a strictly increasing and strictly decreasing function of $X$, respectively. However in this problem, I don't understand the computations being done, specifically:    How does the three intervals $0 \leq y \leq 1$, $1 \leq y \leq 4$ and $y &gt; 4$ are determined? In the two previous problems I've encountered, we only considered one interval which was identical to the interval where $f(x)$ was non-zero. In the case where $0 \leq y \leq 1$, why does $P(X^2 \leq y) = P(-\sqrt y \leq X \leq \sqrt y)$ and not $P(X \leq \sqrt y)$? I have put question marks above the equalities that I don't understand.   I think I have not understand the theory well enough. I'm looking for an answer that will make me understand the solution to this problem and possibly make the theory clearer.
8875
I understand that when sampling from a finite population and our sample size is more than 5% of the population, we need to make a correction on the sample's mean and standard error using this formula:  $\hspace{10mm} FPC=\sqrt{\frac{N-n}{N-1}}$  Where $N$ is the population size and $n$ is the sample size.  I have 3 questions about this formula:   Why is the threshold set at 5%? How was the formula derived? Are there other online resources that comprehensively explain this formula besides  paper?
97405
I recently read the advice that you should generally use median not mean to eliminate outliers. Example: The following article  has 16 reviews at the moment: review = c(5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 3, 2, 1, 1) summary(review)  ## &quot;ordinary&quot; summary  Min. 1st Qu.  Median    Mean 3rd Qu.    Max.  1.000   3.750   5.000   4.062   5.000   5.000   Because they use Mean the article gets 4 stars but if they would use Median it would get 5 stars. Isn't the median a 'fairer' judge?  An experiment shows that median's error is always bigger than mean. Is median worse? library(foreach)  # the overall population of bookjudgments n &lt;- 5 p &lt;- 0.5 expected.value &lt;- n*p peoplesbelieve &lt;- rbinom(10^6,n, p)  # 16 ratings made for 100 books ratings &lt;- foreach(i=1:100, .combine=rbind) %do% sample(peoplesbelieve,16) stat    &lt;- foreach(i=1:100, .combine=rbind) %do% c(mean=mean(ratings[i,]),                     median=median(ratings[i,]))  # which mean square error is bigger? Mean's or Median's? meansqrterror.mean   &lt;- mean((stat[,&quot;mean&quot;]-expected.value)^2) meansqrterror.median &lt;- mean((stat[,&quot;median&quot;]-expected.value)^2)  res &lt;- paste(&quot;mean MSE&quot;,meansqrterror.mean) res &lt;- paste(res, &quot;| median MSE&quot;, meansqrterror.median) print(res)