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HW5(习题二、三)

约 687 个字 预计阅读时间 3 分钟

B23

Question

设某群体的BMI(体重指数)值(单位：kg/m²) \(X\sim N(22.5,2.5^2)\) .医学研究发现身体肥胖者患高血压的可能性增大：当 \(X\le 25\) 时，患高血压的概率为10%;当 \(25<X\le 27.5\) 时，患高血压的概率为15%;当 \(X>27.5\) 时，患高血压的概率为30%.

(1)从该群体中随机选出1人，求他患高血压的概率；

(2)若他患有高血压，求他的BMI值超过25的概率；

(3)随机独立地选出3人，求至少有1人患高血压的概率.

Answer
\[P\{|x-\mu| < \sigma\} = 0.6826,P\{|x-\mu| < 2\sigma\} = 0.9544,P\{|x-\mu| < 3\sigma\} = 0.9974\]
\[P(X \le 25) = \frac{0.6826}{2} + 0.5 = 0.8413\]
\[P(25 < X \le 27.5) = \frac{0.9544 - 0.6826}{2} = 0.1359\]
\[P(X > 27.5) = 1 - \frac{0.9974}{2} = 0.0228\]
\[\therefore P(\text{患高血压}) = 0.8413 \times 0.1 + 0.1359 \times 0.15 + 0.0228 \times 0.3 = 0.111355\]

  1. \(P_2 = \frac{P(25 < X \le 27.5)\times 0.15 + P(X > 27.5)\times 0.3}{P(\text{患高血压})} = \frac{5445}{22271} \approx 0.2445\)

  2. \(P_3 = 1 - (1 - P(\text{患高血压}))^3 = 1 - (1 - 0.111355)^3 \approx 0.2982\)

B33

Question

已知随机变量 \(X\) 的密度函数为

\[f(x) = \begin{cases}c(4-x^2), & -1<x<2 \\ 0, &\text{其他}\end{cases}\]

(1)求常数 \(c\) 的值；

(2)设 \(Y = 3X\) ，求 \(Y\) 的密度函数；

(3)设 \(Z = |X|\) ，求 \(Z\) 的分布函数及密度函数.

Answer

1.\(\int_{-1}^{2}c(4-x^2)dx = 1 \Rightarrow c = \frac{1}{9}\)

  1. \(F_Y(y) = P(Y \le y) = P(3X \le y) = P(X \le \frac{y}{3}) = F_X(\frac{y}{3})\)

两边都对 \(y\) 求导，得 \(f_Y(y) = \frac{1}{3}f_X(\frac{y}{3}) = \begin{cases}\frac{1}{27}[4 - (\frac{y}{3})^2], & -3 < y < 6 \\ 0, &其他\end{cases}\)

  1. \(F_Z(z) = P(Z \le z) = P(|X| \le z) = P(-z \le X \le z) = F_X(z) - F_X(-z)\)

两边都对 \(z\) 求导，得 \(f_Z(z) = f_X(z) + f_X(-z) = \begin{cases}\frac{2}{9}[4 - z^2], & 0 \le z < 1 \\ \frac{1}{9}[4 - z^2], & 1 \le z < 2 \\ 0, &\text{其他}\end{cases}\)

B39

Question

设随机变量 \(X \sim N(0,1)\) ，记 \(Y = e^X,\ Z = ln|X|.\)

(1)求 \(Y\) 的密度函数；

(2)求 \(Z\) 的密度函数；

Answer
  1. \(F_Y(y) = P(Y \le y) = P(e^X \le y) = P(X \le ln y) = F_X(ln y)\)

两边都对 \(y\) 求导，得 \(f_Y(y) = \begin{cases}\frac{1}{y}f_X(ln y) = \frac{1}{y}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(ln y)^2},&y>0 \\ 0, &\text{其他} \end{cases}\)

  1. \(F_Z(z) = P(Z \le z) = P(ln|X| \le z) = P(-e^z \le X \le e^z) = F_X(e^z) - F_X(-e^z)\)

两边都对 \(z\) 求导，得 \(f_Z(z) = e^z \cdot f_X(e^z) + e^z \cdot f_X(-e^z) = \frac{1}{2\sqrt{\pi}}e^{z-\frac{1}{2}e^{2z}}, -\infty < z < +\infty\)

A5

Question

设二维离散型随机变量 \((X,Y)\) 的联合分布律为

X\Y 0 1
0 0.3 a
1 b 0.2

且已知事件 \(\{ X = 0\}\) 与事件 \(\{X + Y = 1\}\) 相互独立， 求常数 \(a,b\) 的值.

Answer
\[\begin{cases} a+b = 0.5 \\ (0.3+a)(a+b) = a \end{cases} \rightarrow \begin{cases} a = 0.3 \\ b = 0.2 \end{cases}\]

A8

Question

随机变量 \(X,Y\) 的概率分布律分别为

X 0 1
P 0.4 0.6
Y 0 1 2
P 0.2 0.5 0.3

且已知 \(P\{X=0,Y=0\} = P\{X=1,Y=2\} = 0.2\)

(1)写出 \((X,Y)\) 的联合分布律；

(2)写出给定 \(\{X=0\}\) 条件下 \(Y\) 的条件分布律；

Answer
X\Y 0 1 2
0 0.2 0.1 0.1
1 0 0.4 0.2
\[P\{Y=0|X=0\} = 0.5\]
\[P\{Y=1|X=0\} = 0.25\]
\[P\{Y=2|X=0\} = 0.25\]

A9

Question

将一枚均匀的骰子抛2次，记 \(X\) 为第一次出现的点数，\(Y\) 为两次点数的最大值.

(1)求 \((X,Y)\) 的联合分布律及边际分布律；

(2)写出给定 \(\{Y=6\}\) 的条件下 \(X\) 的条件分布律；

Answer

1.

X\Y 1 2 3 4 5 6 \(\{X = i\}\)
1 \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{6}\)
2 0 \(\frac{1}{18}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{6}\)
3 0 0 \(\frac{1}{12}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{6}\)
4 0 0 0 \(\frac{1}{9}\) \(\frac{1}{36}\) \(\frac{1}{36}\) \(\frac{1}{6}\)
5 0 0 0 0 \(\frac{5}{36}\) \(\frac{1}{36}\) \(\frac{1}{6}\)
6 0 0 0 0 0 \(\frac{1}{6}\) \(\frac{1}{6}\)
\(\{Y = i\}\) \(\frac{1}{36}\) \(\frac{1}{12}\) \(\frac{5}{36}\) \(\frac{7}{36}\) \(\frac{1}{4}\) \(\frac{11}{36}\)
X\Y 1 2 3 4 5 6
\(P\{X=i\|Y=6\}\) \(\frac{1}{11}\) \(\frac{1}{11}\) \(\frac{1}{11}\) \(\frac{1}{11}\) \(\frac{1}{11}\) \(\frac{1}{11}\)

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