HW2

Problem 2.6

If you roll a pair of fair dice, what is the probability of

$6^{2} = 36$ outcomes of ‘sums’ ranging from 2 $\to$ 12. Probability of getting of sum of 0 is $\frac{0}{36}$

There are 4 ways to get a sum of 5.

$4+1$

$1+4$

$2+3$

$3+2$

The probability of getting a sum of 5 is $\frac{4}{36} = \frac{1}{9} = 0.11$

There is one way to get a sum of 12.

$6+6$

The probability of getting a sum of 12 is $\frac{1}{36} = 0.03$

Problem 2.8

The 2010 American Community Survey estimates that 14.6% of Americans live below the poverty line, 20.7% speak a language other than English (foreign language) at home, and 4.2% fall into both categories.

They are not disjointed because they both can happen at the same time. In fact, the data says that 4.2% fall into both categories.

venn.plot <- draw.pairwise.venn(
    area1 = 14.6, 
    area2 = 20.7, 
    cross.area = 4.2, 
    category = c("% below poverty", "% speak second language"),
    fill = c("blue", "red"),
    cex = 2,
    cat.cex = 1.5,
    cat.pos = c(175, 175),
    cat.dist = 0.04
)

10.4% are below the poverty line and only speak English

26.9% are below the poverty line OR speak a second language

16.5% speak a second language 10.4% are English-only speaking so we add those together. 26.4% live below the poverty line OR speak a foreign language.

Yes, just because someone lives below the poverty line doesn’t mean they have to speak a second language.

Problem 2.20

144 males or females have blue eyes. 144 out of 204 possibilities is $\frac{144}{204} \to \frac{12}{17}$ = 71%

78 total blue eyes out of 204 possibilities. So, $\frac{78}{204} \to \frac{37}{102}$ = 38%

$\frac{19}{204}$ = 9% probability that a brown-eyed man has a blue-eyed partner.

$\frac{11}{204}$ = 5% probability that a green-eyed man has a blue-eyed partner.

The eye color of mating partners is independent. Because a male has blue eyes does not mean we can accurately determine what the eye color of the female is even though the probability of blue eyes is higher.

Problem 2.30

First draw we have a total of 95 books with 28 being hardcover. So the probability of that is $\frac{28}{95}$ = 29%.

Without replacement, the second draw attempt to pick a paperback fiction is 59 out of 94 books. The probability of that is $\frac{59}{94}$ = 63%. Using the General Multiplication Rule, we get 18%.

The probability of picking a fiction book first is $\frac{72}{95}$ and a hardback book second without replacement is $\frac{12}{94}$. Multiplying these two together we get 9.7%.

The probability of picking a fiction book first is $\frac{72}{95}$ and a hardback book second without replacement is $\frac{13}{95}$. Multiplying these two together we get 10.4%.

These are very similary because the difference between replacement and without replacement is $\frac{1}{94}$.

Problem 2.38

\[ \begin{array}{ccccc} \hline i & 1 & 2 & 3 & Total \\ \hline x_{i} & \$0 & \$25 & \$70 & - \\ P(X=x_{i}) & 0.54 & 0.34 & 0.12 & 1.00 \\ x_{i} \times P(X=x_{i}) & 0 & \$8.50 & \$8.40 & \$16.90 \\ x_{i}-\mu & -16.90 & 8.10 & 53.10 & - \\ (x_{i}-\mu)^{2} & 285.61 & 65.61 & 2819.61 & - \\ (x_{i}-\mu)^{2} \times P(X=x_{i}) & 154.23 & 22.31 & 338.35 & 514.89 \\ \hline \end{array} \]

$E(X) = (0 \times .54) + (25 \times .34) + (70 \times .12) = 16.90$ per passenger

Variance is $\sigma ^{2} = 514.89$ so the $SD(X) = 22.69$

With 120 passengers the airlines can plan to make $2,028, with a deviation of around $23.

$120 \times \$16.90 = \$2,028.00$

We’re assuming that there will not be an increase in people bringing more than two bags. This seems unlikely so the average revenue is justified.

Problem 2.44

The distribution is a bell curve centered around the income between $25,000 to $50,000.

Add the percentages that fall below the income of $50,000. 2.2 + 04.7 + 15.8 + 18.3 + 21.2 = 62.2%

25.5%. Here, we’re assuming that male and females make the same amount for the same job.

My assumption is not valid. This data source indicates that the majority of the lower incomes are female.

HW2

Chad Smith

September 16, 2017