date()
## [1] "Thu Sep 12 01:58:13 2013"
Due Date: September 12, 2013 Total Points: 30
1 The following values are the annual number hurricanes that have hit the United States since 1990. Follow instructions and answer questions by typing the appropriate R commands.
a. Enter the data into R. (2)
Hurricanes = c(0, 1, 1, 1, 0, 2, 2, 1, 3, 3, 0, 0, 1, 2, 6, 6, 0, 1, 3, 0, 1)
b. How many years are there? (2)
length(Hurricanes)
## [1] 21
c. What is the total number of hurricanes over all years? (2)
sum(Hurricanes)
## [1] 34
2 Answer the following questions by typing the appropriate R commands.
a. Create a vector of numbers starting with 0 and ending with 25. (2)
VD1 = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25)
b. What is the length of this vector? (2)
length(VD1)
## [1] 25
c. Create a new vector from the original vector by subtracting the mean value over all numbers in the vector. (2)
VD2 = VD1 - mean(VD1)
3 Suppose you keep track of your mileage each time you fill up. At your last 8 fill-ups the mileage was
65311 65624 65908 66219 66499 66821 67145 67447
a. Enter these numbers into a vector called miles. (2)
miles = c(65311, 65624, 65908, 66219, 66499, 66821, 67145, 67447)
b. Use the function diff() to determine the number of miles between fill-ups. (2)
diff(miles)
## [1] 313 284 311 280 322 324 302
c. What is the maximum, minimum, and mean number of miles between fill-ups? (3)
max(diff(miles))
## [1] 324
min(diff(miles))
## [1] 280
mean(diff(miles))
## [1] 305.1
4 Create the following sequences using the seq() and rep() functions as appropriate.
a. “a”, “a”, “a”, “a” (2)
rep("a", 5)
## [1] "a" "a" "a" "a" "a"
b. The odd numbers in the interval from 1 to 100 (2)
seq(1, 100, by = 2)
## [1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45
## [24] 47 49 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91
## [47] 93 95 97 99
c. 1, 1, 1, 2, 2, 2, 3, 3, 3 (2)
rep(c(1, 2, 3), c(3, 3, 3))
## [1] 1 1 1 2 2 2 3 3 3
d. 1, 1, 1, 2, 2, 3 (2)
rep(c(1, 2, 3), c(3, 2, 1))
## [1] 1 1 1 2 2 3
e. 1, 2, 3, 4, 5, 4, 3, 2, 1 (3) Hint: Use the c() function.
c(seq(1, 5), seq(4, 1))
## [1] 1 2 3 4 5 4 3 2 1