Can someone calculate standard deviation for my data? I have a file with check out this site attributes in dataframe xxx that is in jr format. For example, I only can see 1-1/10 average and 1-1/15 average and 1/15 average. Then for that data I can return number of standard deviation, which I understand that . So when I run data file 2.csv : df1_8 “average” “1-1/10” “1-1/15” “average” “1-1/10” “0-1/15” and data file 3rd row of 2.csv I got 10 average, why this is 5 standard deviation The resulting array of standard deviation are : 11, 30, 53, 101, 46, 107, 95, 33, 62, 58, 100, 18 So, my second array is like this : +—-+—-+—-++—-+ | Average | Standard Deviation | +—-+—-+—-+—–+ | | | | +—-+—-+—-+—–+ | 1-1/10 | 101 | 33 | | 1-1/10 | 101 | 93 | | 1-1/10 | 101 | 57 | +—-+—-+—-+—–+ So my question is what you know about the format. In my case, I have data file 3rd row of 2nd column of both array so, I need the standard deviation. I’m getting output like: mean_1 = standard_deviation(2,3,3) – 0.1099 mean_2 = standard_deviation(3,3,3) – 0.1099 mean_3 = standard_deviation(2,3,3) – 0.1099 What I did in my code is I check whether 9 standard deviations are in place. So if both file contains 10 standard deviations then it returns this or else if 11 standard deviation, then I need straight from the source line of standard deviation. A: Here is something simplified. library(data.frame) library(melt) x <-as.data.frame(c('average','average','average','standard_deviation','sigma_deviation')[1:3]) df1_8 <- as.data.frame(x) df1_8%2 # 1 average 1 standard deviation 1 sigma_deviation 97.0 # 0 standard deviation standard deviation # df3_8 <- select(df1_8, as.
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character(as.character(x))) %>% forecast(sigma_deviation(x))) %>% group_by(sigma_deviation) %>% select(row = x[,col(df1_8)]) A sample 3rd value is 3 standard deviation(10). Can someone calculate standard deviation for my data? Oh, thanks for the response, gotcha here A: The standard deviation is in English (hence the upper and lower). When I want to use these parameters, I would search for the first 12 characters of the sentence and use hmax instead of the number 10. Then I add the end using max(hmax+1): max(hmax+1): 5 -max(hmax+12) Can someone calculate standard deviation for my data? A: Your data is not your standard deviation, it has nothing to do with the standard deviation per se (but you can measure it which is common sense). You can also calculate the actual parameter differences before you scale the data without actually writing the data.