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Data:
5410.4 5432.2 5452.9 5477.6 5472.5 5454.9 5446 5010.6 5395.9 5360 5336.9 5333.9 5329.6 5345.7 5353.8 5377.2 5334.1 5351.1 5001 5246.4 5230 5115.8 4972.6 5077.6 5056.9 5070.7 4799.3 5076 5021.5 5026.4 4981.9 4936.6 4901.8 4853.8 4839.2 4821.3 4840.5 4847.6 4832.3 4814.7 4806.4 4803.4 4770.3 4723.4 4667.1 4636.8 4613.2 4605.3 4590.4 4595.4 4600.1 4543.3 4596.4 4575.4 4547.9 4503.7 4446.3 4401.4 4354.3 4336.3 4300.9 4304.1 4273.2 4279.9 4243.1 4199.1 4177.6 4141.7 4088.3 4021.4 3981.2 3937.2 3893.1 3864.7 3847.8 3840.8 3828.4 3798.6 3773 3737.8 3699 3674 3648.8 3645.6 3331 3674.7 3714.5 3739.7 3759.7 3708.6 3717.3 3705.3 3612.8 3665 3670.8 3687.6 3708.2 3737.2 3748.7 3785.3 3787.1 3785.8 3749.7 3716.3 3650 3096.9 3703.2 3716 3736.9 3771.9 3704 3824.2 3733.5 3827.5 3827.6 3696.5 3675.8 3757.5 3753.3 3418.7 3772.9
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R Code
par2 <- '0' par1 <- '8' 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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Big Analytics Cloud Computing Center
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