> Homework 1 Yerin Lee
> #1.9
> # 1. What is the smallest amount? The largest?
> Cooper<-c(15.9, 21.4, 19.9, 21.9, 20.0, 16.5, 17.9, 17.5)
> min(Cooper)
[1] 15.9
> max(Cooper)
[1] 21.9
> # 2. Find the average amount.
> mean(Cooper)
[1] 18.875
> # 3. Find the differences of the largest and smallest amounts from the mean.
> diff(Cooper)
[1] 5.5 -1.5 2.0 -1.9 -3.5 1.4 -0.4
> min(diff(Cooper))
[1] -3.5
> max(diff(Cooper))
[1] 5.5
First, I put the data for the cooper. Next, I got the minimum(the smallest amount) of the
cooper which is 15.9, maximum(the largest amount) of the cooper which is 21.9, and the
mean(average amount) of the cooper which is 18.875. To find the differences of the largest
and smallest amounts from the mean, I have to find out the differences first. The differences
of total data of Cooper is 5.5, -1.5, 2.0, -1.9, -3.5, 1.4, -0.4. Therefore, the minimum(the
smallest amount) of the cooper is -3.5, the maximum(the largest amount) of the cooper is
5.5.
> # 1.15
> # According to The Digital Bits (http://www.digitalbits.com/),
> # monthly sales (in 10,000s) of DVD players in 2003 were
> JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
79 74 161 127 133 210 99 143 249 249 368 302
> # Enter the data into a data vector dvd.
> # By slicing, form two data vectors: one containing the months with 31 days, the other the
remaining months.
> # Compare the means of these two data vectors.
> dvd<-c(79,74,161,127,133,210,9,143,249,249,368,302)
> dvd31<-c(1,3,5,7,8,10,12)
> dvd30<-c(2,4,6,9,11)
> mean(dvd31)
[1] 6.571429
> mean(dvd30)
[1] 6.4
> mean(dvd31)-mean(dvd30)
[1] 0.1714286
