Alıştırma Seti Cevap Anahtarı

30 Jan 2016

Aşağıdaki kodlar bugünkü alıştırma setinin cevap anahtarı olarak düşünülebilir.

```{r}

#store plots in pdf, will save in Documents folder if you are on windows #and do not specify a path for the file #text results will be printed on R terminal window

pdf(“plots.pdf”,width=7,height=5) ### Exercise 1 ### ### ### q1rand=rexp(1500,1/3) par(mfrow=c(1,2)) hist(q1rand,main=’Histogram for Exercise 1’) boxplot(q1rand,main=’Boxplot for Exercise 1’)

Exercise 2

### ### #(part a) nofrandom=100 q2rand=runif(nofrandom,3,10) percentage=100*sum(q2rand>5)/nofrandom sampmean=mean(q2rand) sampmedian=median(q2rand) print(c(percentage,sampmean,sampmedian))

#(part b) #population prob=1-punif(5,3,10) popmean=0.5(3+10) popmedian=popmean #symmetry print(c(100prob,popmean,popmedian))

#(part c) nofrandom=1000 q2rand=runif(nofrandom,3,10) percentage=100sum(q2rand>5)/nofrandom sampmean=mean(q2rand) sampmedian=median(q2rand) print(c(percentage,sampmean,sampmedian)) print(c(100prob,popmean,popmedian))

Exercise 3

### ### #(part a) nofrandom=2000 q4rand=rnorm(nofrandom,8,5) #(part b) nofgreater=sum(q4rand>=9) print(nofgreater) #(part c) sampmean=mean(q4rand) sampstdev=sd(q4rand) print(c(sampmean,sampstdev)) #(part d) #population percentile25=qnorm(0.25,8,5) percentile75=qnorm(0.75,8,5) print(c(percentile25,percentile75)) #(part e) orderedsample=sort(q4rand) samplepercentile25=(orderedsample[500]+orderedsample[501])/2 samplepercentile75=(orderedsample[1500]+orderedsample[1501])/2 print(c(samplepercentile25,samplepercentile75)) #(part f) print(c(pnorm(0.789),pnorm(-0.543))) #default values imply standard normal, check help!

Exercise 3

### ### #(part a) nofdim=c(2,3) nofrandom=1000 fraction=array(0,nofdim) par(mfrow=c(1,1)) for(k in nofdim){ q3rand=runif(k*nofrandom,-1,1) q3rand=matrix(q3rand,nrow=nofrandom) #matrix form distances=apply(q3rand,1,function(x) { sqrt(sum(x^2))}) #apply to each row fraction[k]=sum(distances<1)/nofrandom } plot(c(1:nofdim),fraction,type=’l’,lty=2,lwd=2,col=2,xlab=’# of dimensions’,ylab=’fraction’,main=’Exercise 3’) points(c(1:nofdim),fraction,pch=2,cex=1)

#(part b) #fraction[2] is the ratio of the areas of circle with radius of 1 and the square with side length of 2 #fraction[3] is the ratio of the volumes of circle with radius of 1 and the cube with side length of 2 approximatedpi2D=fraction[2]4/1 approximatedpi3D=fraction[3]8/(4/3) print(c(pi,approximatedpi2D,approximatedpi3D))

Exercise 5

### ### #(part a and b) nlevel=c(2,5,10,30,50,100) nofsampling=1000 par(mfrow=c(2,3)) for(i in nlevel){ q5rand=rpois(i*nofsampling,3) q5rand=matrix(q5rand,nrow=nofsampling) #matrix form means=apply(q5rand,1,mean) hist(means,main=paste(‘n=’,i)) } #as n gets larger the distribution of mean is closer to normal distribution

#(part c) par(mfrow=c(1,1)) qqnorm(means, ylab=”Standardized Scores”, xlab=”Normal Scores”, main=”Exercise 5 Part c”) qqline(means) dev.off()

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