1) 0
2) 1+4 or 2+3 = 5
2*(P(1 and 4) or P(2 and 3)) = 2*(P(1)*P(4) + P(2)*P(3)) = 2*(1/6*1/6 + 1/6*1/6) = 2*2/36 = 1/9 = 0.11
3) possible combination that adds up to 12 is 6+6
P(6 and 6) = 1/6*1/6 = 1/36 = 0.028
No, there people who live below the poverty line and speak a foreign language at home so the two are not disjoint.
14.6 - 4.2 = 10.4% Americans live below poverty line and speak only English at home
14.6 + 20.7 = 35.3% Americans are below poverty line or speak a foreign language at home.
20.7 - 4.2 = 16.5% speak a foreign language at home and above poverty line. 100 - 16.5 = 84.5% are above poverty line and only speak English
P(below poverty line) * P(foreign language) = (0.146 x 0.207) = 0.030
P(below poverty and foreign language) = 0.042
The two are not independent
library(VennDiagram)
## Loading required package: grid
## Loading required package: futile.logger
#venndiagram
venn.plot <- draw.pairwise.venn(14.6, 20.7, 4.2, c("Below poverty line", "Foreign language"))
grid.draw(venn.plot)
144 / 204 = 0.71
78 / 114 = 0.68
19 / 54 = 0.35
11/36 = 0.305
P(H and P) = 0.295 x 0.71 = 0.209
0.298 x 0.76 = 0.227
0.76 x 0.295 = 0.224
Revenue with 0 bags = 0 Revenue from ppl with 1 bag = 34 * 25 = $850 Revenue from ppl with 2 bags = 12 * 50 = $600
total revenue = $1450 Revenue per person (the mean) = $14.50 (xi - )^2*P(xi) x1 = 0, P(x1) = 0.54; (-14.50)^2*0.54 = 113.54 x2 = $25, P(x2) = 0.34; (25 - 14.50)^2*0.34 = 37.485 x3 = $50, P(x3) = 0.12; (50 - 14.50)^2*0.12 = 151.23
SD = (113.54 + 37.54 + 151.23) = (302.31) = 17.386
sqrt(302.31)
## [1] 17.38706
income <- c("$1 - $9,999 or loss",
"$10,000 to $14,999",
"$15,000 to $24,999",
"$25,000 to $34,999",
"$35,000 to $49,999",
"$50,000 to $64,000",
"$65,000 to $74,999",
"$75,000 to $99,999",
"$100,000 or more")
fr_total <- c( 2.2, 4.7, 15.8,18.3, 21.2, 13.9, 5.8, 8.4, 9.7)
income_df <- data.frame(income, fr_total)
income_df
## income fr_total
## 1 $1 - $9,999 or loss 2.2
## 2 $10,000 to $14,999 4.7
## 3 $15,000 to $24,999 15.8
## 4 $25,000 to $34,999 18.3
## 5 $35,000 to $49,999 21.2
## 6 $50,000 to $64,000 13.9
## 7 $65,000 to $74,999 5.8
## 8 $75,000 to $99,999 8.4
## 9 $100,000 or more 9.7
barplot(income_df$fr_total, xlab = income_df$income)
fr_less_than_fifty <- ( 2.2 + 4.7 + 15.8 + 18.3 + 21.2)
fr_less_than_fifty
## [1] 62.2
men <- (96420486 * 59 )/ 100
women <- (96420486 * 41) / 100
men
## [1] 56888087
women
## [1] 39532399
total_lt_fifty <- (96420486 * fr_less_than_fifty) / 100
total_lt_fifty
## [1] 59973542
f_lt_fifty <- (total_lt_fifty * 41)/100
f_lt_fifty
## [1] 24589152
(62.2 * 41)/100
## [1] 25.502
f2_lt_fifty <- (women * 71.8) / 100
The distribution here is right skewed
fr_less_than_fifty = 2.2 + 4.7 + 15.8 + 18.3 + 21.2) = 62.2 %
Assuming that the ratio of males to females is similar across all categories,
Probability of women making less than 50K = 62.2 * 41 = 25.5%
71.8% of women making less than 50K is 28,384,262, however based on the assumption made in c) total women in this category would be 24,589,152 which is less than 71.8%, hence the assumption is incorrect.