I have done a boxplot to represent my claim frequency. However make 2 and 4 have only 1 whisker. How do I interpret that?
60563
I have a data set that has customerID and daily visits. This dataset records the number of times customers visited the site in a day.   customerID   TotalVisits  1            6  2            17  3            20  4            25   Just to explore the data further, I tried plotting the TotalVisits values in a box plot. The plot doesn't seem to have a lower whisker. q1=1 and there are no values between q1 and  q1-1.5(IQR). The plot has q1, q2 and q3. The minimum value and q1 are the same.   Is this why the boxplot doesn't have the lower whisker? Could you please help me with the interpretation? Is it correct to draw a boxplot without any whiskers?  [I am new to the site and I am not able to post the image.]
73535
So, I have been presented with this question: A sample of 100 people were asked how many days they drove their car during the last week (inc. the weekend). The resulting frequency of response is shown below: Days, frequency 0, 1 1, 5 2, 3 3, 15 4, 20 5, 25 6, 31 7, 0 I need to use a beta-binomial model in order to estimate the average number of 'car driven' days a week for a uniform prior. There are a total of 7 Bernoulli trials, each with success probability p. I'm incredibly confused at what the outcome will be. I know that for a uniform prior that alpha = 1, and beta = 1, but I don't know how to apply this all to get the average? Thank you.
120160
Figure 1: X-axis represents the time (0-23 hours) and Y-axis represents the electricity consumption in kWh. I am working with time-series power consumption data. Figure 1 shows the shapes of one user on Wednesday, Thursday and Friday. All the shapes look different that indicates the stochastic behavior of that user. I want to measure this variability in load shapes. My ultimate goal is to quantify the difference in consumption patterns across days. What kind of statistical techniques I can use for the same? Is there any library/package available in python/R? Please help me in this regard.
32729
We know that in the first innings the predicted score and in the second innings the predicted winning percentage will be calculated in the WASP.   But there is only one formula for calculating both score and percentage.  V(b,w) =r(b,w) +p(b,w) V(b+1,w+1) +(1-p(b,w)))V(b+1,w)  How is this one formula enough to get two results?  This question is not related to the . I’m not asking about logical explanation here; I just want to know, what is the change that they do in the formula to get the percentage for the second innings?  Providing an example is a great thing for this type of question.
56145
So, for fun, I am taking some of the data of calls from the call center I work at and trying to do some hypothesis testing on them, specifically the number of calls received in a week, and using a Poisson distribution to fit it. Due to the subject matter of my job, there are two types of weeks, let's call one of them on-weeks where I hypothesize there are more calls, and off-weeks where I hypothesize there are fewer.  I have a theory that the $\lambda$ from on-weeks (let's call it $\lambda_1$) is larger than that of the one from off-weeks (let's call it $\lambda_2$)  So the hypothesis I want to test is $H_0: \lambda_1 &gt; \lambda_2, H_1: \lambda_1 \leq \lambda_2 $  I know how to test for one parameter (say  $H_0: \lambda_1 &gt; 1, H_1: \lambda_1 \leq 1 $) but not so sure how to go about doing 2 given a data set. Let's say I take two week's worth of data from each one $X_1 = 2$ and $X_2 = 3$ for the off-week and $Y_1 = 2$ and $Y_2=6$ for the on-week. Can someone help walk me though this simpler version such that I can apply it to a larger data set? Any help is appreciated, thank you.
26565
4 cards are drawn from a pack without replacement. What is the probability of getting all 4 from different suits?  Here's how I tried to solve:  For the first draw, we have 52 cards, and we have to pick one suit. So, probability for this is $\frac{13}{52}$.  For the second draw, only 51 cards are left. The second suit has to be selected, so there are 13 cards from that suit. The probability is $\frac{13}{51}$.  Similarly, the third and fourth draw have probabilities $\frac{13}{50}$ and $\frac{13}{49}$ respectively.  Since the draws are independent, the total probability becomes $$\frac{13}{52} \times \frac{13}{51} \times \frac{13}{50} \times \frac{13}{49}$$  But my book says the answer is $\frac{{13\choose 1} \times {13 \choose 1} \times {13\choose1} \times {13\choose1}}{52 \choose 4}$.  My answer differs by a factor of $4!$. What did I do wrong?
72003
You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 3 balls without replacement. The balls, save for the color, are indistinguishable from each other. What are the odds that your sample contains at least a ball from each color?  I thought about finding the chance that the sample will not contain at least one color and subtracting that from one:  $$P(\text{sample is missing at least one color})=P(\text{sample is missing one color})+P(\text{sample is missing two colors}) $$  Essentially the second probability statement on the right hand side is the probability that all the balls in the sample are the same color. The case where the sample does not contain three colors doesn't make sense, so I didn't include it. $$=\binom{3}{1} \left[\left(\frac{12}{18}\right) \cdot \left(\frac{11}{17}\right) \cdot\left(\frac{10}{16}\right)\right]+\binom{3}{2}\left[\left(\frac{6}{18}\right) \cdot \left(\frac{5}{17}\right) \cdot\left(\frac{4}{16}\right)\right] \approx 0.882353$$  When all of this is done, we simply subtract the probability from one to get $0.117647$.  What am I doing wrong?
32788
In Pillow Problems and a Tangled Tale Lewis Carroll presents the following problem:     A bag contains a counter, known to be either white or black. A white counter is put in, the bag is shaken, and a counter is drawn out, which proves to be white. What is now the chance of drawing a white counter?   One can easily reason that in the end, the bag must be in one of three states:   Black remains, new white taken New white remains, original white taken Original white remains, new white taken   In two of those states, a white counter remains to be drawn. Therefore, the chance is 2/3.  How can one arrive to this answer using Bayes' theorem?
5288
When writing (a blog post, script, etc..) what is the proper way to indicate two or more instances of a single letter? For instance, in Monty Python's Bookshop Sketch:     C: I wonder if you might have a copy of "Rarnaby Budge"?      P: No, as I say, we're right out of Edmund Wells!      C: No, not Edmund Wells - Charles Dikkens.      P: (pause - eagerly) Charles Dickens??      C: Yes.      P: (excitedly) You mean "Barnaby Rudge"!      C: No, "Rarnaby Budge" by Charles Dikkens.  That's Dikkens with two Ks, the      well-known Dutch author.      P: (slight pause) No, well we don't have "Rarnaby Budge" by Charles Dikkens      with two Ks, the well-known Dutch author, and perhaps to save time I      should add that we don't have "Karnaby Fudge" by Darles Chickens, or      "Farmer of Sludge" by Marles Pickens, or even "Stickwick Stapers" by Farles      Wickens with four M's and a silent Q!!!!!  Why don't you try W. H. Smith's?      C: Ah did, They sent me here.   I had always believed that plural never uses an apostrophe before the 's' (it's only used for possession), but I have rarely seen in written material the format "four Ms". (On a side note, whoever wrote  also used "two Ks".)  On a side note, and perhaps this should be a separate question, if a Compact Disc is a CD, then two Compact Discs would be two CDs right? (I see "CD's" written everywhere)
113832
I am having some trouble in understanding odds, and I would like just a basic explanation for how to interpret them.  I have found various posts related to odds but most of them are more complex than what I am trying to understand. Here is an example of how I am interpreting odds: if the odds of an event happening are 3 to 1, then the event will happen 3 times for every 1 time that it doesn't happen. I do not know if this interpretation would be correct. So, any guidance and more examples on interpreting odds would be greatly appreciated.
105191
Context: I need to solve one variation of the . Problem: Two players ($A$ and $B$) throw a coin until one of the victory sequences appears. For player $A$ the victory sequence is $HTT$, for player $B$ the victory sequence is $TTH$. What is the probability that $A$ wins and what is the expected number of throws given that $A$ won? My attempt: Intuitively, it seems that player $B$ is more likely to win. Basically, I reliase that we can use the Markov-Chains, but my idea is only draw a binary tree and consider the sequences on the n-th step. My another idea was to calculate the expected number of throws to get, say, $HH$ that would be equal to $r_{HH}$: $$r_{HH} =  \frac{1}{4}E(number|HH) + \frac{1}{4}E(number|HH)+ \frac{1}{4}E(number|TT)+ \frac{1}{4}E(number|TH) = \frac{1}{4}(9 + \frac{5}{2} r_{HH})$$  And get $r_{HH}$. My issues: I don't really understand how to join these ideas to get a proper solution.
52482
If I have a $w \times h$ matrix where each value is an integer  $0 \lt n \lt 20$,  how can I count the number of distinct configurations, where $2$ configurations are "distinct" if there is no way to reshuffle the rows and columns that would produce the same matrix?  Can this be counted with the stars and bars method?    For example, these are equal (we swapped a row, then a column):  0 0 0    2 0 4 0 2 4    0 0 0   but these are distinct (no way to swap rows or columns to produce the other):  0 0 0    2 0 0 0 2 4    0 4 0   It seems like there ought to be a way to count the rows or columns as "bins" and the values as balls.  I realize that in this case there are $18$ different colored balls, but even if the only values possible were $1$ and $0$, (ball or no ball) I can't see how to represent it as stars and bars.
11167
I'm studying a course on probability and statistics and at some point this symbol comes up without introduction. It looks like the number one, but slightly bigger and with a double vertical line.  First time it comes up is when discussing stochastic/random variables that are neither continuous nor discreet in an example:   And somewhat later in a proof:    Anyone got an idea what this symbol represents?  EDIT: Thanks for the fast answers, I was still editing the question for better (larger) images :)
94791
Another problem from quora.  What can be said about $v =\prod_{s=2}^{\infty} \zeta(s) $?  Wolfy says that $v \approx 2.294856591673313794183 $.  The Inverse Symbolic Calculator () finds nothing.  Here are the close values:  $2294856391585950 = (0064) sum((7/6*n^3-3*n^2+65/6*n-3)/(Fibo(n)+1),n=1..inf)\\ 2294856397653578 = (0001) GAM(5/6)*BesI(1,2)^{GAM(7/12)}\\ 2294856469349552 = (0006) 1/4357571\\ 2294856473388934 = (0324) 1/10*(10+14^{1/2}*10^{1/4})^{1/2}*10^{3/4}\\ 2294856488490537 = (0001) BesJ(0,1)^{GAM(1/3)/GAM(5/12)}\\ 2294856524215021 = (0314) sin(Pi*8/45)-sin(Pi*14/51)\\ 2294856591673313 \text{ would be here}\\ 2294856740350177 = (0011) sum((2*n^3-5*n^2+22*n-14)/(n!+2),n=1..inf)\\ 2294856991319600 = (0324) 23^{1/2}/(12^{3/4}-3^{2/3})^{1/2}\\ 2294857017919943 = (0001) Feig2*Li4(1/2)/GAM(5/6)\\ 2294857089879260 = (0248) F(11/26;27/50;1) \\ 2294857493303450 = (0404) Psi(1/21)+Psi(19/21)+Psi(13/14)\\ $  I tried using Euler's zeta product and partition sum, but this didn't seem to help. Here is what resulted:  $\zeta(s) =\dfrac1{\prod_p (1-p^{-s})} $.  $\begin{array}\\ v &amp;=\prod_{s=2}^{\infty} \zeta(s)\\ &amp;=\prod_{s=2}^{\infty} \dfrac1{\prod_p (1-p^{-s})}\\ &amp;=\prod_{s=2}^{\infty} \prod_p\dfrac1{ (1-p^{-s})}\\ &amp;= \prod_p\prod_{s=2}^{\infty}\dfrac1{ (1-p^{-s})}\\ \end{array} $  Euler's product identity states that $\prod_{n=1}^{\infty}\dfrac1{1-zx^n} =\sum_{n=1}^{\infty} \dfrac{x^nz^n}{\prod_{k=1}^n(1-x^k)} $. Putting $x=z=\dfrac1{p} $, $\prod_{n=2}^{\infty}\dfrac1{1-p^{-n}} =\sum_{n=1}^{\infty} \dfrac{p^{-2n}}{\prod_{k=1}^n(1-p^{-k})} $.  That's as far as I can go.
48830
I have $k_1$ red balls and $k_2$ blue balls in a box. Let $p_r$ be probability that randomly select $r$ balls in the box without replacement, $\sum_{r=1}^{m} p_r=1$. A trial is succeeds if the trial can draw at least one red ball in the trial. Find expected number of successful trail in $N$ times.   This is my solution  Let $X$ be event that draw $r$ ball in the box which has at least one red ball.  $$P(X)=p(r) \times\sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$  Finally, we got the expected value as $$E(X)= N \times \sum_{r=1}^{m} p(r) \sum_{i=0}^{k_1}\frac{\binom{k_1}{i}\binom{k_2}{r-i}}{\binom{k_1+k_2}{n}}$$  Is it right?
42162
Mr. Jones has three shirts: Red, Green and White. Each day he picks randomly a red shirt with probability of $\frac{1}{2}$, green with probability of $\frac{1}{3}$ and white with probability of $\frac{1}{6}$.     What is the probability that he wears all three shirts after 6 days?   My try:   $$\binom{6}{1}\binom{5}{1}\binom{4}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}\cdot[\binom{3}{2}(\frac{1}{2})^2\cdot[\frac{1}{3}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{3})^2\cdot[\frac{1}{2}+\frac{1}{6}]+[\binom{3}{2}(\frac{1}{6})^2\cdot[\frac{1}{2}+\frac{1}{3}]+\binom{3}{1}\binom{2}{1}\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}]$$  First, we pick possessions for red, green and white. Then, we multiply it by the other combinations possible of the other three possessions left, and summing it up (I didn't count all possibilities). The problem is the answer is bigger than 1, so it must be wrong. Also, I guess there is more elegant way.  Thanks in advance.
81055
I'm analyzing a residual plot  of the residuals vs the fitted values.  I'm not quite sure how to interpret this plot since there looks like there is a pattern and the average is not actually zero. There are diagonal straight parallel lines.    I'm working with several ordinal IV's.  Any ideas?