MATH 2411 Homework 2

Q1

a

$$\begin{array}{rl}{F}_Y(x)& =P(Y\le x)\\ & =P(\max (X_1,X_2,\dots ,X_n)\le x)\\ & =P(X_1\le x,X_2\le x,\dots X_n\le x)\\ & =P(X_1\le x)P(X_2\le x)\dots P(X_n\le x)\\ & =(F(x){)}^n\end{array}$$ $$\begin{array}{rl}{f}_Y(x)& =\frac{d}{dx}(F(x{)}^n)\\ & =n(F(x){)}^{n-1}f(x)\end{array}$$

b

$$\begin{array}{rl}E(Y)& ={\int }_0^{T}y{f}_Y(x)dy\\ & ={\int }_0^{T}yn(F(x){)}^{n-1}f(x)dy\\ & =n(F(x){)}^{n-1}f(x){\left[\frac{y^2}{2}\right]}_0^T\\ & =\frac{n{T}^2}{2}f(x)(F(x){)}^{n-1}\end{array}$$

Q2

$$\begin{array}{rl}& P(\text{Error})=\frac{1}{10^5}\\ & P(\text{Success})=1-\frac{1}{10^5}\end{array}$$

Assumes that

• The outcome of each transmission is either error or success
• The transmissions are identical
• The transmissions are independent

Denote the number of errors in these $n$ transmissions by the r.v. $X\sim B(10^5,\frac{1}{10^5})$

$$\begin{array}{rl}P(X>3)& =1-P(X=0)-P(X=1)-P(X=2)-P(X=3)\\ & =1-\left(\genfrac{}{}{0ex}{}{10^5}{0}\right){\left(1-\frac{1}{10^5}\right)}^{10^5}\\ & -\left(\genfrac{}{}{0ex}{}{10^5}{1}\right)\left(\frac{1}{10^5}\right){\left(1-\frac{1}{10^5}\right)}^{10^5-1}\\ & -\left(\genfrac{}{}{0ex}{}{10^5}{2}\right){\left(\frac{1}{10^5}\right)}^2{\left(1-\frac{1}{10^5}\right)}^{10^5-2}\\ & -\left(\genfrac{}{}{0ex}{}{10^5}{3}\right){\left(\frac{1}{10^5}\right)}^3{\left(1-\frac{1}{10^5}\right)}^{10^5-3}\\ & =0.01898754\end{array}$$

1 - pbinom(3, 10**5, 1/10**5)

Q3

qnorm(0.99, mean=175, sd=8)

From R we get the height of a person at the 99th percentile is 193.6108 cm

The minimum door height would be 193.6108 + 5 = 198.6108 cm

Q4

$$Z=\frac{3.8\times 10^6-4.4\times 10^6}{3\times 10^5}=-2$$

Let $X$ denotes the lifetimes of chips, $X\sim N(4.4\times 10^6,(3\times 10^5{)}^2)$

$$p(X>3.8\times 10^6)=P(Z>-2)=0.9772499$$

1 - pnorm(-2)

Let $Y$ denotes the number of chips whose lifetimes are more than $3.8\times 10^6$ hours, $Y\sim B(100,0.9772499)$

$$P(Y\ge 96)=1-P(Y\le 95)=0.9214286$$

1 - pbinom(95, 100, 1 - pnorm(-2))

Q5

a

Let $X$ denotes the annual rainfall, $X\sim N(40,8.7^2)$

$$P(X>42)=0.4090905$$

1 - pnorm(42, mean=40, sd=8.7)

Let $Y$ denotes that the rainfall is over 42 inches in each year, we can see that the outcomes is either true or false, and we assumes that the annual rainfalls are independent and identical between years, $Y\sim B(3,0.4090905)$

$$P(Y=3)=0.4090905^3=0.06846338$$

b

Assumes the annual rainfalls are independent between years

$$E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)$$ $$\mathrm{Var}(X_1+X_2+X_3)=\mathrm{Var}(X_1)+\mathrm{Var}(X_2)+\mathrm{Var}(X_3)$$ $$\begin{array}{rl}P(X_1+X_2+X_3>126)& =P\left(Z>\frac{126-(3\times 40)}{\sqrt{3\times 8.7^2}}\right)\\ & =0.3452515\end{array}$$

m <- 3 * 40
v <- 3 * 8.7**2
s <- sqrt(v)
 
1 - pnorm(126, mean=m, sd=s)
# 1 - pnorm((126 - m) / s) # Alternative