R code:
x=c(10,10,16,18,18,18)
mean(x)#AM
(prod(x))^(1/length(x))#GM
median(x)# median
names(table(x))[table(x)==max(table(x))] # mode
Output:
> mean(x)#AM
[1] 15
> (prod(x))^(1/length(x))#GM
[1] 14.50963
> median(x)# median
[1] 17
> names(table(x))[table(x)==max(table(x))] # mode
[1] "18"
R code:
x=c(10,10,16,18,18,18)
var(x)#variance
sd(x)#standard deviation
(sd(x)/mean(x))*100# coefficient of variation
Output:
> var(x)#variance
[1] 15.6
> sd(x)#standard deviation
[1] 3.949684
> (sd(x)/mean(x))*100# coefficient of variation
[1] 26.33122
C.
Q1=First quartile=2nd observation+0.25*(3rd observation-2nd observation)=10+0.5*(10-10)=11.5
Q3=3rd quartile=5th observation+0.75*(6th observation-5th observation)=18+0.75*(18-18)=18
Inter quartile range=Q3-Q1=18-11.5=6.5
R code:
x=c(10,10,16,18,18,18)
quantile(x,0.25)#Q1
quantile(x,0.75)#Q3
quantile(x,0.75)-quantile(x,0.25)#Inter quartile range
Output:
> quantile(x,0.25)#Q1
25%
11.5
> quantile(x,0.75)#Q3
75%
18
> quantile(x,0.75)-quantile(x,0.25)#Inter quartile range
75%
6.5
D.
x=c(10,10,16,18,18,18)
boxplot(x,main="boxplot of hourly wages")
E.
R code:
x=c(10,10,16,18,18,18)
f=c(2,1,3)
pct=round(f/sum(f)*100)
lbls= paste(pct) # add percents to labels
lbls= paste(lbls,"%",sep="") # ad % to labels
pie(f,main="boxplot of hourly wages",labels = lbls)
F.
R code:
x=c(10,10,16,18,18,18)
hist(x,main="boxplot of hourly wages")
Descriptive Statistics: The Old-Fashioned Way (and with 2. In this problem, calculate the descriptive statistics below...
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