2.6 Dice Rolls

  1. 0, this is not possible
  2. Total number of combinations: 36
    Options to throw a sum of 5:
    1 and 4 = 1/36
    2 and 3 = 1/36
    4 and 1 = 1/36
    3 and 2 = 1/36
    (1/36)+(1/36)+(1/36)+(1/36) = 0.11111111 or 11.11%
  3. 6 and 6 = 1/36 = 0.02777778 or 2.78%
(1/36)+(1/36)+(1/36)+(1/36)
## [1] 0.1111111
1/36
## [1] 0.02777778

2.8 Poverty and language

  1. No, they are not disjoint. Individuals can fall into both categories.
  2. See below for venn diagram
  3. 10.4-4.2 = 6.2%
  4. 16.5 + 4.2 + 10.4 = 47%
  5. 100 - 47 = 53%
  6. No, these events are most likely dependent.
# 2.8 b
#install.packages('VennDiagram')
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
grid.newpage()
draw.pairwise.venn(area1 = 14.6, cross.area = 4.2, area2 = 20.7,
  category = c("Below the Poverty Line", "Language Other Than English"),
  lty = rep("blank", 2),
  fill = c("aquamarine", "yellow"),
  alpha = rep(0.25, 2),
  cat.pos = c(0, 0),
  cat.dist = rep(0.025, 2))

## (polygon[GRID.polygon.1], polygon[GRID.polygon.2], polygon[GRID.polygon.3], polygon[GRID.polygon.4], text[GRID.text.5], text[GRID.text.6], text[GRID.text.7], text[GRID.text.8], text[GRID.text.9])
14.6-4.2
## [1] 10.4
20.7-4.2
## [1] 16.5
10.4-4.2
## [1] 6.2

2.20 Assortative mating

  1. (78/114) * (78/108) = 0.494152
  2. 78/204 = 0.09313725
  3. 19/204 = 0.3518519
    11/204 = 0.05392157
  4. The eye colors of male respondents and their partners are dependent. There is a much higher probability that male respondents will choose individuals with the same eye color as them when compared to all of the others.
78/204
## [1] 0.3823529
(78/114) * (78/108)
## [1] 0.494152
19/54
## [1] 0.3518519
11/36
## [1] 0.3055556

2.30 Books on a bookshelf

  1. (28/95) * (59/94) = 0.1849944
  2. (72/95) * (28/94) = 0.2257559
  3. (72/95) * (28/95) = 0.2233795
  4. The answers are simple, because as you increase the sample size, effects like this seem to lessen. If the sample size were say, 5, and we compared the probabilities with and without replacement, the result would be significantly higher.
(28/95) * (59/94)
## [1] 0.1849944
(72/95) * (28/94)
## [1] 0.2257559

2.38 Baggage fees

  1. See paper for probability table and calculations. SD(X) = 16.67363188
  2. Assumptions: The passengers are independent from one another.
    The airline could expect 120 * 15.7 = 1884 in baggage fee revenue from 120 passengers, with a standard deviation of 182.65
    120*278.01 = 33361.20
    sqrt(33361.2) = 182.6504859

2.44 Income and gender

58% males, 41% females
a) This distribution is positive/right skewed. The median exists somewhere in the 35,000-49,000 range.
b) 21.2 + 18.3 + 15.8 + 4.7 + 2.2 = 62.2
c) 0.622 * 0.41 = 0.25502
We assume that income and gender are independent.
d) The statement that 71.8% of females make less than 50,000 per year invalidates the assumption we made in part c. Income and gender are dependent.