---
# System prepended metadata

title: 機率HW8

---

機率HW8
===

## 1.
$15 \times \frac{1}{4}=3.75$

### A: ++3.75++

## 2.
$\frac{\frac{2k}{3} - \frac{k}{3}}{k-0}=\frac{1}{3}$

### A: ++$\frac{1}{3}$++

## 3.
Let $G$ and $g$ be the distribution and probability density functions of $Y$.
$G(t) = P(Y \le t) = P(-\ln(1-X) \le t) = P(X \le 1-e^{-t})=\frac{(1-e^{-t})-0}{1-0}=1-e^{-t}$

$$
G(t)=
\left\{
\begin{eqnarray}
  0 \qquad t \le 0\\
  1-e^{-t} \qquad 0 \le t \le 1 \\
  1 \qquad t \ge 1 \\
\end{eqnarray}
\right.
$$

$g(t)=G^{\prime}(t)=e^{-t}$
$$
g(t)=
\left\{
\begin{eqnarray}
  e^{-t} \qquad 0 \le t \le 1 \\
  0 \qquad otherwise \\
\end{eqnarray}
\right.
$$

## 4.
Because Z is a standard normal random variable,so the mean($\mu$) of Z is $0$,and the variance($\sigma^2$) of Z is $1$.

Let $f(x)=P(x < Z < x+\alpha)=\int^{x+\alpha}_{x}\frac{1}{\sqrt{2\pi}}e^{\frac{-y^{2}}{2}}dy=\frac{1}{\sqrt{2\pi}}\int^{x+\alpha}_{x}e^{\frac{-y^{2}}{2}}dy$

The  number x that maximizes $P(x < Z < x + α)$ appears when $f^{\prime}(x)=0$.
By the fundamental theorem of calculus,$f^{\prime}(x)=\frac{1}{\sqrt{2\pi}}(e^{\frac{-(x+\alpha)^2}{2}} - e^{\frac{-x^2}{2}})$
$f^{\prime}(x)=0$ when $x=-\frac{\alpha}{2}$

### A: $-\frac{\alpha}{2}$

## 5.
$\mu=67$ and $\sigma=8$.
$P(X \ge 90) = 1-P(X < 90) = 1-\phi(\frac{90-67}{8})=1-\phi(2.875)\simeq1-0.998=0.002$
$P(80 \le X < 90)=\phi(\frac{90-67}{8})-\phi(\frac{80-67}{8})=\phi(2.875)-\phi(1.625)\simeq 0.998-0.9484=0.0496$
$P(70 \le X < 80)=\phi(\frac{80-67}{8})-\phi(\frac{70-67}{8})=\phi(1.625)-\phi(0.375)\simeq 0.9484-0.6480=0.3004$
$P(60 \le X < 70)=\phi(\frac{70-67}{8})-\phi(\frac{60-67}{8})=\phi(0.375)-\phi(−0.875)\simeq 0.6480-0.1894=0.4586$
$P(X<60)=\phi(\frac{60-67}{8})=0.1894$

$A(\ge 90)=0.2$%
$B(80- 90)=4.96$%
$C(70- 80)=30.04$%
$D(60- 70)=45.86$%
$F(<60)=18.84$%

## 6.
$P(|X-\mu|>k\sigma)=P(X-\mu>k\sigma)+P(X-\mu<-k\sigma)=P(\frac{X-\mu}{\sigma}>k)+P(\frac{X-\mu}{\sigma}<-k) =(1-\phi(k))+(1-\phi(k))=2-2\phi(k)$

## 7.
Let $G$ and $g$ be the distribution and probability density functions of $Y$.
$G(t)=P(Y \le t) = P(\sqrt{|X|} \le t)=P(|X|\le t^2)=P(-t^2\le X\le t^2),t \ge 0$
$G(t)=P(-t^2\le X\le t^2)=\phi(t^2)-\phi(-t^2)=\phi(t^2)-(1- \phi(t^2))=2\phi(t^2)-1=2\int^{t^2}_{-\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}dx-1$

$g(t)=G^{\prime}(t)=\frac{2}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}|^{t^2}_{-\infty}=\frac{2}{\sqrt{2\pi}}e^{\frac{-t^4}{2}}$

## 8.
$P(|X-E(X)| \ge 2\sigma_{X})=P(|X-\frac{1}{\lambda}| \ge \frac{2}{\lambda})=P(X-\frac{1}{\lambda} \ge \frac{2}{\lambda})+P(X-\frac{1}{\lambda} \le  -\frac{2}{\lambda})=P(X \ge \frac{3}{\lambda})+P(X \le  -\frac{1}{\lambda})$
$P(X \ge \frac{3}{\lambda})+P(X \le  -\frac{1}{\lambda})=[1-P(X \le \frac{3}{\lambda})]+P(X \le  -\frac{1}{\lambda})=[1-(1-e^{-\lambda\frac{3}{\lambda}})]+(1-e^{\lambda\frac{1}{\lambda}})=e^-3\simeq0.049787$

### A:++0.049787++

## 9.
$P(\lfloor X+1 \rfloor =n)=P(n \le X+1 \le n+1) =P(n-1 \le X \le n)= \int^{n}_{n-1}\lambda e^{-\lambda x}dx=-e^{-\lambda x}|^{n}_{n-1}$
$-e^{-\lambda x}|^{n}_{n-1}=-e^{-\lambda n}+e^{-\lambda (n-1)}=e^{-\lambda n}(e^{\lambda}-1)$