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2888.9 2916.2 2939.5 2968.3 2986.7 3008.4 3035.3 3059 3078.4 3096.8 3125.2 3157.6 3186 3215.2 3257.8 3296 3330.6 3366.2 3402.9 3426.1 3461.3 3488 3509.5 3536 3561 3593.2 3620.4 3630.4 3643.1 3672.5 3692.2 3719.4 3744.1 3768.3 3803.5 3838.9 3860.4 3879.8 3905.5 3932.4 3959.7 3980.7 4012.8 4037.7 4065 4086.4 4106.9 4137.5 4166.3 4177.8 4176.4 4189.8 4218 4235.9 4237.9 4264.6 4295.5 4327.8 4340.1 4340.2 4375.3 4405.2 4433.3 4472 4507.4 4525.5 4562.5 4581.6 4591 4614 4643.3 4674.6 4687.4 4703.2 4728.3 4757.1 4765.2 4785.4 4810.1 4830.2 4843.3 4861.1 4875.6 4897.3 4901.5 4900.4 4914.6 4930.2 4917 4936.1 4942.3 4951.1 4975.6 4973.5 4963.4 4974.8 5001.8 5013.4 5007.9 4985.6 4967.1 4988.9 4999.8 4988.3 4975.5 4981.1 4993.4 4992.9 4994.1 5014.4 5028.6 5025.4 5021.7 5026.9 5026.6
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R Code
library(MASS) library(car) par1 <- as.numeric(par1) if (par2 == '0') par2 = 'Sturges' else par2 <- as.numeric(par2) x <- as.ts(x) #otherwise the fitdistr function does not work properly r <- fitdistr(x,'normal') print(r) bitmap(file='test1.png') myhist<-hist(x,col=par1,breaks=par2,main=main,ylab=ylab,xlab=xlab,freq=F) curve(1/(r$estimate[2]*sqrt(2*pi))*exp(-1/2*((x-r$estimate[1])/r$estimate[2])^2),min(x),max(x),add=T) dev.off() bitmap(file='test3.png') qqPlot(x,dist='norm',main='QQ plot (Normal) with confidence intervals') grid() dev.off() load(file='createtable') a<-table.start() a<-table.row.start(a) a<-table.element(a,'Parameter',1,TRUE) a<-table.element(a,'Estimated Value',1,TRUE) a<-table.element(a,'Standard Deviation',1,TRUE) a<-table.row.end(a) a<-table.row.start(a) a<-table.element(a,'mean',header=TRUE) a<-table.element(a,r$estimate[1]) a<-table.element(a,r$sd[1]) a<-table.row.end(a) a<-table.row.start(a) a<-table.element(a,'standard deviation',header=TRUE) a<-table.element(a,r$estimate[2]) a<-table.element(a,r$sd[2]) a<-table.row.end(a) a<-table.end(a) table.save(a,file='mytable.tab')
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