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title: 機率HW 8 110590064 資工二 劉韶軒

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# 機率HW 8 110590064 資工二 劉韶軒
## Problem 1
### 原文題目
> Suppose that 15 points are selected at random and independently from the interval (0, 1). In average, how many of them will be greater than 3/4?
### 中文題目
> 假設有15個點是獨立地從區間（0, 1）中隨機選擇的。平均而言，有多少個點會大於3/4？
### 解決方式
期望值計算公式:$E=P*Val$
$15 × \frac{1}{4} = \frac{15}{4} = 3.75$
## Problem 2
### 原文題目
> A point is selected at random on a line segment of length k. What is the probability that none of the two segments is smaller than k/3?
### 中文題目
> 在長度為k的線段上隨機選擇一個點。兩個線段都不小於k/3的機率是多少？
### 解決方式
左邊線段必須讓切斷點x>$\frac{k}{3}$
右邊線斷臂旭讓切斷點x<$\frac{2k}{3}$
所以切斷點範圍$\frac{k}{3}$<x<$\frac{2k}{3}$
所以我們可以得出x/k = $\frac{k}{3}$
## Problem 3
### 原文題目
> Let X be a random number from (0, 1). Find the density functions of Y = -ln(1 - X ) .
### 中文題目
> 設X是從（0, 1）中的隨機數。找出Y = -ln(1 - X)的概率密度函數
### 解決方式
$CDF(Y) = P(Y ≤ x) = P(-ln(1 - X) ≤ x)$
$1 - X ≥ e^{-y}$
$- X ≥ e^{-y} - 1$
$X ≥ 1 - e^{-y}$
$F_X(x) = P(X ≤ x) = x$
$F_Y(y) = P(Y ≤ y) = P(-ln(1 - X) ≤ y) = P(X ≥ 1 - e^(-y)) = 1 - (1 - e^(-y)) = e^(-y)$
$f_Y(y) = \frac{d F_Y(y)}{dy} = -e^{-y}$
## Problem 4
### 原文題目
Let Z be a standard normal random variable and α be a given constant. Find the real number x that maximizes P(x < Z < x + α).
### 中文題目
### 解決方式
$φ(z) = (1 / √(2π)) * e^{\frac{-z^2}{2}}$
$P(x < Z < x + α) = ∫_{x}^{x+α} φ(z) dz$
$d/dx P(x < Z < x + α) = d/dx ∫_{x}^{x+α} φ(z) dz= φ(x+α) * (d/dx (x+α)) - φ(x) * (d/dx)$
$φ(x+α) - φ(x) = 0$
對稱分布$φ(-z) = φ(z)$
$x + α = -x,x = -α/2$
## Problem 5
### 原文題目
> The grades for a certain exam are normally distributed with mean 67 and variance 64. What percent of students get A(≥ 90), B(80 - 90), C(70 - 80), D(60 - 70), and F(< 60)?
### 解決方式
$Z= \frac{X−μ}{σ}$
$σ = \sqrt{64}=8$
$Z_{90}=\frac{90-67}{8}=2.875$
$Z_{80}=\frac{90-67}{8}=1.625$
$Z_{70}=\frac{90-67}{8}=0.375$
$Z_{60}=\frac{90-67}{8}=-0.875$
$A(≥ 90)=1-P(Z<2.875)=0.003$
$A(80-90)=P(1.625<Z<2.875)$
$A(70-80)=P(0.375<Z<1.625)$
$A(60-70)=P(-0.875<Z<0.375)$
$A(<60)=P(Z<-0.875)$
## Problem 6
### 原文題目
Let ~ (N,$σ^2$). Prove that P(| X - μ |> kσ ) does not depend on μ or σ .
### 解決方式
Z = $\frac{X - μ}{σ}$为X的标准化值。Z均值为0
$P(|X - μ| > kμ) = P(|\frac{X - μ}{σ}| > k)$
Z = $\frac{X - μ}{σ}$
这意味着它对正态分布的均值和标准差的变化是不变的。
## Problem 7
### 原文題目
Let X ~ N(0,1) . Calculate the probability density function of Y = $sqrt{|X|}$.
### 解決方式
$P(Y≤y)=P(|X|≤\sqrt{y})=P(|X|≤y2)=P(−y^2≤X≤y^2)=P(X≤y^2)−P(X<−y^2)=Φ(y^2)−Φ(−y^2)$
$g(y)=\frac{dP(Y≤y)}{dy}=2yϕ(y2)+2yϕ(−y2)=4yϕ(y2)$
symmetry of ϕ(x) around x=0
## Problem 8
### 原文題目
> $LetX~be~an~exponential~random~variable~with~parameter~a~,~mean~E(X)~and~standard~deviation~\\σ_X .~Find~P(|X- E(X) | 2σ_X )$
### 解決方式
$E(X)=\frac{1}{a},σ_X = \frac{1}{a}$
$P(|X - E(X)| ≤ 2σ_X)。$
$f_X(x) = \frac{1}{a} * e^{\frac{-x}{a}}$
$|X - E(X)| = |X - 1/a|$
$when~X >\frac{1}{a} ,X - \frac{1}{a} > 0，|X - \frac{1}{a}| = X - \frac{1}{a}。$
$when~X ≤ \frac{1}{a} ,X - \frac{1}{a} ≤ 0，|X - \frac{1}{a}| = -(X - \frac{1}{a}) = \frac{1}{a} - X。$
$P(X - \frac{1}{a} ≤ 2σ_X) = P(X - \frac{1}{a} ≤ \frac{2}{a}) = P(X ≤ \frac{3}{a})$
$P(X ≤ 3/a) = 1 - e^{-3/a}$
$P(|X - E(X)| ≤ 2σ_X) = P(X -\frac{1}{a} ≤ 2σ_X) + P(\frac{1}{a} - X ≤ 2σ_X) = (1 - e^{\frac{-3}{a}}) + 1$
$P(|X - E(X)| ≤ 2σ_X) = 2 - e^\frac{-3}{a}$
