stats_calvin's_midterm

1.) MTBE

$$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{98}{201}=\sim 0.48756$$

2.)

$$P(A)=\frac{\text{Sum of A}}{\text{Sum of all}}=\frac{120}{223}=\sim .5381$$

3.) Given:

$P(\text{positive}|\text{users})=.95$ $P(\text{negative}|\text{non-users})=.89$

---

8.) $P(Y\le 1.1)$

we know that $f(y)=\{\begin{array}{ll}\frac{2}{3}{y}^2& [0,2]\\ 0& \text{elsewhere}\end{array}$

[Graph of f(y) from 0 to 2]

$F(1.1)=\frac{2}{9}(1.1{)}^3$

9.) $F(y)=\{\begin{array}{ll}\frac{2}{9}{y}^3& [0,2]\end{array}$ $\int \frac{2}{3}{y}^{2}dy\to d=1$ $\frac{2}{9}{y}^3{|}_0^2$ $F(y)=\frac{2}{9}{y}^3[0,2]$

$F(3)-F(1)=0.875$

10.) $P(8\le Y\le 13)=P(13)-P(8)$

$=\text{dnorm}(13,12,4)-\text{dnorm}(8,12,4)$ $=0.0562$

11.) Guessed.

12.) Birthday Problem

$P(A)=1-P(\bar{A})$ $P(\bar{A})=\frac{365(365-1)\dots (365-k+1)}{365^k}$

$=1-\frac{\exp (\text{lchoose}(365,k)+1\text{factorial}(k))}{k\log (365)}$

Thus birthday(20) = .4114

---

4.)

$\text{epagas}\leftarrow \text{read.csv}(\text{"EFMEHS.csv"})$ \text{z\_epagas} \leftarrow (\text{epagas} \ \text{MPG} - \text{mean}(\text{epagas} $ \text{MPG}))/ \text{sd}(\text{epagas} $ \text{MPG})$

$\text{outlier\_epagas}\leftarrow \text{subset}(\text{epagas, abs(z\_epagas)}\ge 2$ $\&\text{ abs(z\_epagas)}<=3)$

$\min (\text{outlier\_epagas})$ $=300$

5.) Given Standard Deviation is 3.

[Drawing of a normal distribution bell curve] $\sim 99.9\%$ $99\%$

6.)

$\text{mean}-\text{sd}\cdot 2\text{, }\text{mean}+\text{sd}\cdot 2\text{ : }6$ [Drawing of L and U on a normal distribution] To find U: U = \text{mean}(\text{epagas} \ \text{MPG})+ \text{sd}(\text{epagas} $ \text{MPG})= 39.4119$

7.)

$$f(y)=\{\begin{array}{ll}c{y}^2& [0,2]\\ 0& \text{elsewhere}\end{array}$$ $$\begin{array}{c}f(y)=\frac{3}{8}{y}^2\\ =c\frac{y^3}{3}{|}_0^2\\ {\int }_0^{2}c{y}^{2}dy=1\\ \therefore c=\frac{3}{8}\\ =c\frac{(2{)}^3}{3}-0=\\ \frac{8}{3}c=1\end{array}$$