date()
## [1] "Wed Sep 11 22:50:49 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.
0 1 1 1 0 2 2 1 3 3 0 0 1 2 6 6 0 1 3 0 1
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)
v1 = seq(0, 25, 1)
b. What is the length of this vector? (2)
length(v1)
## [1] 26
c. Create a new vector from the original vector by subtracting the mean value over all numbers in the vector. (2)
v2 = v1 - mean(v1)
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", 4)
## [1] "a" "a" "a" "a"
b. The odd numbers in the interval from 1 to 100 (2)
seq(1, 100, 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(1:3, c(3, 3, 3))
## [1] 1 1 1 2 2 2 3 3 3
d. 1, 1, 1, 2, 2, 3 (2)
rep(1:3, 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, 1), seq(4, 1, -1))
## [1] 1 2 3 4 5 4 3 2 1