题目

Problem

The discrete random variable $X$ is defined by the cumulative distribution function

$x$	1	2	3	4	5
$F(x)$	$\frac{3k}{2}$	$4k$	$\frac{15k}{2}$	$12k$	$\frac{35k}{2}$

where $k$ is a constant.

(a) Find the probability distribution of $X$ .

(3)

(b) Find $P(1.5<X<3.5)$

(2)

The random variable $Y=12-7X$

(3)

(d) Calculate $P(4X<|Y|)$

(6)

解答

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因为 $F(5)=1$ ，先求 $k$ 。概率分布由相邻 cumulative probabilities 相减得到。

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Since $F(5)=1$ ,

\begin{align*} \frac{35k}{2}=1. \end{align*}

\begin{align*} k=\frac{2}{35}. \end{align*}

The probability distribution is

\begin{array}{c|ccccc} x&1&2&3&4&5\\ \hline P(X=x)&\frac{3}{35}&\frac{5}{35}&\frac{7}{35}&\frac{9}{35}&\frac{11}{35} \end{array}

Therefore

P(1.5<X<3.5)=P(X=2)+P(X=3) =\frac{5}{35}+\frac{7}{35} =\frac{12}{35}.

Now

E(X)=\frac{1(3)+2(5)+3(7)+4(9)+5(11)}{35} =\frac{25}{7}.

Also,

E(X^2)=\frac{1^2(3)+2^2(5)+3^2(7)+4^2(9)+5^2(11)}{35} =\frac{101}{7}.

Thus

\operatorname{Var}(X)=\frac{101}{7}-\left(\frac{25}{7}\right)^2 =\frac{82}{49}.

Since $Y=12-7X$ ,

\operatorname{Var}(Y)=(-7)^2\operatorname{Var}(X) =49\cdot\frac{82}{49} =82.

For $Y=12-7X$ , check each possible value:

\begin{array}{c|ccccc} X&1&2&3&4&5\\ \hline Y&5&-2&-9&-16&-23\\ 4X&4&8&12&16&20\\ |Y|&5&2&9&16&23 \end{array}

The inequality $4X<|Y|$ is true for $X=1$ and $X=5$ .

Therefore

P(4X<|Y|) =P(X=1)+P(X=5) =\frac{3}{35}+\frac{11}{35} =\frac{2}{5}.