What is the formula for SD in quality Control for the Levey Jenning's chart?
149140
Don't know Levey Jennings, but in general the variance is the sum of the squared deviations from the mean divided by the number of samples (for large data sets) or the number of samples minus 1 (for smaller data sets).  The standard deviation (SD) is then the square root of the variance.\n\nSo, if "m" is the mean and each sample is x1..n, the variance is:\n\n[(m-x1)^2 + (m-x2)^2 + (m-x3)^2 ... (m-xn)^2]/n\n\nThen, take the square root of the variance to get the standard deviation.
458039
In the experiment of testing for the number of errors in a chapter...\n\nLet the event an error appearing in a chapter be a success..\n\nLet "x" represent the number of success.\nThe number of errors in a chapter can be either 0,1,2,...\n==> "x" can carry the values 0,1,2,3,...\n==> "x" is a discrete random variable with range = {0,1,2,..} [Countably infinite]\n\nThe probability distribution of "x" is a poisson distribution. \n[Since there are a large number of trials all of which are identical and independent with only two possible outcomes in each trial, either a success (there being an error) or a failure (there not being an error), with the probability of success in all the trials remaining the same all throught ]\n\nThe probability function of a poisson distribution\n==> P(X = x) = {e^(-m) * m^(x)}/x!\n \nParameter of the distribution\n==> Mean of the distribution\n==> Mean number of errors per chapter =0.8 \n==> m = 0.8\n\nTherefore P(X = x) = {e^(-0.8) * (0.8)^(x)}/x!\n\nProbability that there are less than 2 errors in a chapter \n==> P(x < 2) \n= P(x=0) + P(x=1)\n= {e^(-0.8) * (0.8)^(0)}/0! + {e^(-0.8) * (0.8)^(1)}/1!\n= {e^(-0.8) * 1)/1 + {e^(-0.8) * (0.8)}/1\n= e^(-0.8) + {e^(-0.8) * (0.8)}\n= e^(-0.8) [ 1 + 0.8]\n= e^(-0.8) [1.8]\n\n\nhttp://www.futureaccountant.com/probability/
180959
http://www.worldwidewords.org/qa/qa-thr1.htm\n\nhttp://www.bartleby.com/59/4/threesheetst.html
450656
The statement is true.  The formula for the slope of the regression line is\n\nr(SDy)/SDx\n\nSince SDy and SDx are always positive, then the only thing that determines the sign of the slope is r, the correlation coefficient.  So if r is positive, then the slope is positive; if r is negative, then the slope is negative.
953641
I can only say so much because those equations are too long for two in the morning. Not that it matters; I just won't be retyping them. But you certainly need a bunch of random things for this. First, you need a variable -- a char, probably, since you don't care about any input beyond that. I don't know what the compare thing is going to output, but that'll come.\n\nI suggest an array for the old and new charts.\n\nGeez, what's the syntax? int oldchart[11], or something? I'm getting too used to Java. Anyway. Since you're going in increments of 5 from 0 to 50, you'll have a 0, 5, 10, ... 45, 50 -- that is, 11 different speeds, right? So you'll need an array for the old chart and an array for the new chart, both with a size of 11. Do you get where I'm going?\n\nCalculating the chart takes a bit more thinking. We'll ignore the actual process of getting input for the moment and just deal with filling it in. I'll suggest a method, and explain afterwards:\n\nint speed;\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   oldchart[i] = <calculate wind chill>;\n}\n\nThe same thing could be applies to the newchart array. The index, i, will go from 0 to 10, matching the indexes of the array, right? Since you're going by increments of 5, the speed for slot i will be i*5. That takes you from 0 to 50.\n\nSo after you pull that off for the old and new, you've got charts saved in the variables. Printing them is kind of similar to what happened above -- you use a loop to output what's in the array, using i*5 for the speed.\n\nNow, granted, you could combine this and the above step so you wouldn't need the arrays at all--you'd just recalculate and print every time. But I find it easier to organize my thoughts by going a way that deals with one issue at a time... It depends on what you prefer at this point.\n\nfor(int i = 0; i < 11; i++)\n{\n   speed = i*5;\n   cout << "Speed: "<< speed << " Wind Chill: " << oldarray[i] << endl;\n}\n\nLet's see...\n\nDisplaying the menu is easy--just use cout << "your menu stuff" << endl;\nInput is a bit stranger, I think. For something like this, just use a cin and check afterwards if it's right. Um, let's see... I'll give this to you in pseudocode--you should be able to work out the actual stuff.\n\nwhile ( the user doesn't want to exit )\n//this is just a logic thing -- if the user is done, then what?\n{\n     take the user's input;\n\n     if the input is 'o' or 'O'\n     {\n          calculate and print the old chart\n     }\n\n     else if the input is 'n' or 'N'\n     {\n          calculate and print the new chart\n     }\n\n     else if the input is 'c' or 'C'\n     {\n          calculate and print both the old and new charts (?) (whatever compare does)\n     }\n\n     else if the user wants to exit\n     {\n          end the program (usu. with return 0 or something of that nature)\n     }\n\n     else the user's input was bad\n}\n\nMind that for this to go well, you'd have to mind a few issues -- as I wrote above, if the user doesn't enter what you expect them to. Also, you'd have to deal with a case where the user decided to print charts before entering data into the old or new charts. For that, you should give the temperature a default value, like 0, so the charts can still be calculated.\n\nYou can see that I wrote in a calculation for every print, which is what I was saying before with it not being necessary. You could technically use the for loop I gave for printing but just replace the reference to the array with the calculation for the wind chill.\n\n...Don't know if I covered all of it, but I should be getting to bed. ^^;
905994
deltaG = deltaH - TdeltaS\n\nWhen delta G equals zero,\n\ndeltaH = TdeltaS\n\ndeltaS = deltaH/T  = (44.0 kJ/mol)/373K = 0.11796 kJ/mol.K\n\nSo I'd pick e.
1153957
The website below seems to be what you're looking for.  The spreadsheet you can download uses the NORMDIST function to do the graph.  Check out the help feature for the function to see its equation.\n\nI assume you know how to calculate the mean and standard deviation for your data.  The AVERAGE and STDEV Excel functions are good for this.
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
908590
http://www.usps.com/news/online/0907_3.htm\n\nThere is no chart but from this info, you should be able to make a chart.
858220
Depends on what variable you're differentiating with respect to:\ndV/dr : 2/3*pi*r*h\ndV/dh:  1/3*pi*r^2
220083
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
479611
just use SPSS 13.0  ;)
1071099
Here is a link to a users manual-specifically the statistics section. You can go to the full contents.\nhttp://www.infj.ulst.ac.uk/NI-Maths/Guides/TexTI-85/Ch3/THBStats.html
400981
first, you find out the equation of Keq and delta G. I think it should be Keq = e^ [-deltaG/(RT)].\nThis equation and constant R can be found in any general chemistry textbook. And T is usu. 25+273=298K\nthen you can find out deltaG for the equation.\n\nwe also have deltaG= [- delta G for Ag(NH3)2^+1(aq) + (-32.0)] - [-26.2 - 3.8]\n\nand there comes the answer!!! ;)
262206
range = mean +- (t/sqrt(N))*st.dev.\n\nWhere t is found in a chart according to the confidence level in question, N is the number of data points, and st.dev. is the standard deviation of the data set.
913451
A box and whisker plot!? No problem! \n\nThe Box is from Quartile 1(25th percentile) to Quartile 3(75th percentile),.\n\nThe Whiskers are Q1 minus 1.5 Interquartile ranges and Q3 plus 1.5 Inter Quartile ranges.\n\nIQR (Interquartile ranges) is defined as Q3-Q1\n\nWell say that Q1 is 12 and Q3 is 25 then the IQR is 13, and 1.5IQR is 19.5\n\nTherefore, Anything below -7.5 (12-19.5) and above 44.5 would be considered an outlier. \n\nAs far as drawing the damn thing, just draw a box with the outside edges being Q1 adn Q3, and then draw the whiskers 1.5IQR long.\n\nAdditional info:\n\nThe box and whisker is only 1 criterion for outliers, + or - 3 standard deviations from the mean is another common way, which can get other results.\n\nHope that helps.
488339
Try here:\n\nhttp://edweb.sdsu.edu/courses/ET650_online/MAPPS/GDBasics.html