standard-deviation

问题 In Excel I have two columns. One has a player's name and the other has points scored. Like this: Player Points Foo 10 Bar 11 Foo 23 Test 9 Joe 1 Foo 2 What I'm trying to do is get the standard deviation of the points for a combined list of players. For example, if I had this list: Foo Bar I would want the standard devation of 10, 11, 23, and 2, since those are the values that match those two players. I have tried this formula: =STDEV(IF(OR(A:A="Foo",A:A="Bar"),B:B,"")) but I get a different

问题 I am having trouble understanding the error estimates between the variables when performing linear regression. Referring to How to find error on slope and intercept using numpy.polyfit say I am fitting a straight line with numpy.polyfit (code below). As mentioned in the question in the link, the square root of the diagonals of the covariance matrix are the estimated standard-deviation for each of the fitted coefficients, and so np.sqrt(V[0][0])) is the standard deviation of the slope. My

问题 How can I convert z-score to probability using ruby? Example: z_score = 0 probability should be 0.5 z_score = 1.76 probability should be 0.039204 回答1: According to this https://stackoverflow.com/a/16197404/1062711 post, here is the function that give you the p proba from the z score def getPercent(z) return 0 if z < -6.5 return 1 if z > 6.5 factk = 1 sum = 0 term = 1 k = 0 loopStop = Math.exp(-23) while term.abs > loopStop do term = 0.3989422804 * ((-1)**k) * (z**k) / (2*k+1) / (2**k) * (z**

问题 I am looking at Accelerate to compute mean and standard deviation of arrays in Swift. I can do the mean. How do I do the standard deviation? let rr: [Double] = [ 18.0, 21.0, 41.0, 42.0, 48.0, 50.0, 55.0, 90.0 ] var mn: Double = 0.0 vDSP_meanvD(rr, 1, &mn, vDSP_Length(rr.count)) print(mn) // prints correct mean as 45.6250 // Standard Deviation should be 22.3155 回答1: You can compute the standard deviation from the mean value and the mean square value (compare https://en.wikipedia.org/wiki

问题 I want to calculate the pooled (actually weighted) standard deviation for all the unique sites in my data frame. The values for these sites are values for single species forest stands and I want to pool the mean and the sd so that I can compare broadleaved stands with conifer stands. This is the data frame (df) with values for the broadleaved stands: keybl n mean sd Vest02DenmDesp 3 58.16 6.16 Vest02DenmDesp 5 54.45 7.85 Vest02DenmDesp 3 51.34 1.71 Vest02DenmDesp 3 59.57 5.11 Vest02DenmDesp 5

问题 I am trying to use the StDev function and am getting blank results. I am using it as... SELECT StDev(fldMean) FROM myTable Where fldMean contains a value of 2.3 and should evaluate to 0 but instead I am simply getting an empty result. I can't seem to understand how expressions are to be used in the function, Microsoft's manual really didn't help. 回答1: SELECT StDev(fldMean) FROM myTable will return Null if [myTable] has only one row because the Standard Deviation cannot be computed from a

问题 I am trying to get the standard deviation of a user input string. I have as follows, but it returns the wrong value for SD. The calculation should go as follows: Sum values/number values = mean Square (sum each value-mean) Sum squares/number values. Assistance appreciated (and explanation if possible): function sum() { var val = document.getElementById('userInput').value; var temp = val.split(" "); var total = 0; var v; var mean = total / temp.length; var total1 = 0; var v1; var temp23; var

问题 I want to write a function that gets a time series and a standard deviation as parameters and returns an adjusted time series which looks like a forecast. With this function I want to test a system for stability, which gets a forecasted time series list for weather as input parameter. My approach for such a function, which is described below: vector<tuple<datetime, double>> get_adjusted_timeseries(vector<tuple<datetime, double>>& timeseries_original, const double stddev, const double dist_mid

问题 I have a shell script: dir=$1 cd $dir grep -P -o '(?<=<rating>).*' * | awk -F: '{A[$1]+=$2;L[$1]++;next}END {for(i in A){print i, A[i]/L[i]}}' | sort -nr -k2 | awk '{ sub(/.dat/, " "); print }' which sums up all of the numbers that follow the <rating> field in each file of my folder but now I need to calculate the standard deviation of the numbers rather than getting the average. By summing up the difference of each rating in the file from the mean squared and then dividing this by the sample

问题 I couldn't find an answer for RGB image. How can someone get a value of SD,mean and Entropy of RGB image using MATLAB? From http://airccse.org/journal/ijdms/papers/4612ijdms05.pdf TABLE3, it seems he got one answer so did he get the average of the RGB values? Really in need of any help. 回答1: After reading the paper, because you are dealing with colour images, you have three channels of information to access. This means that you could alter one of the channels for a colour image and it could