Question

R Language

library(tidyverse)

data(diamonds)

(a) How many diamonds have a `Very Good` cut or better?

              - Note that cut is an *ordered factor* so the levels are in order.

(b) Which diamond has the highest price per carat (ppc = price / carat)? What is the value?

(c) Find the 95th percentile for diamond price.

               - Try the `quantile()` function.

(d) What proportion of the diamonds with a price above the 95th percentile and have the color `D` or `J`?

(e) What proportion of diamonds with a clarity of VS2 have a Fair cut and a table below 56.1?

(f) What is the average price per carat (ppc=price / carat) for each cut?

Answer 1

a) Find the number of diamonds with cut Very Good, Premium and Ideal

length(which(diamonds$cut == "Very Good"))

length(which(diamonds$cut=='Premium'))

length(which(diamonds$cut=='Ideal'))

length(which(diamonds$cut == "Very Good"))+ length(which(diamonds$cut=='Premium'))+length(which(diamonds$cut=='Ideal'))

Sum of all: 47424

b) Make a column named ppc and the find the diamond with max ppc (price per carat)

diamonds$ppc<-diamonds$price/diamonds$carat

subset(diamonds, ppc == max(ppc))

Output:

c) FInd quantile (95%)

price = diamonds$price 

quantile(price, c(.95))
    95% 
13107.1 

d) FInd the number of diamonds with price greater than quantile, and those having color D or J

 length(which(diamonds$price>quantile(price,c(.95)) & (diamonds$color=='D' | diamonds$color=='J')))
[1] 437

Thus the proportion is

0.008101594

e) Find the number of diamonds with cut Fair , clarity VS2 and table value less than 56.1

> length(which(diamonds$cut == "Fair" & diamonds$clarity=="VS2" & diamonds$table<56.1))
[1] 85

Thus the proportion is

0.001575825

f) Find the diamond with each cut and then mean of all prices belonging to that cut

> a = diamonds[which(diamonds$cut == "Fair"),]
> b = diamonds[which(diamonds$cut == "Good"),]
> c = diamonds[which(diamonds$cut == "Very Good"),]
> d = diamonds[which(diamonds$cut == "Premium"),]
> e = diamonds[which(diamonds$cut == "Ideal"),]

[1] 4358.758
> mean(b$ppc)
[1] 3860.028
> mean(a$ppc)
[1] 3767.256
> mean(c$ppc)
[1] 4014.128
> mean(d$ppc)
[1] 4222.905
> mean(e$ppc)
[1] 3919.7