题目
一家银行的服务包括人工服务和自助-|||-服务,在一天中,X表示接受人工服务所-|||-花费的时间,Y表示自助服务做花费的-|||-时间,二维随机变量X,Y联合概率密度-|||-为 f(x,y)= = (x+{y)^2),0leqslant xleqslant 1,0leqslant yleqslant 1 0, =-|||-输入答案 __ 分数
题目解答
答案
解析
步骤 1：确定积分区域
根据题目要求，我们需要计算 $P\{ 0\leqslant X\leqslant \dfrac {1}{4},0\leqslant Y\leqslant \dfrac {1}{4}\}$，即在 $0\leqslant x\leqslant \dfrac {1}{4}$ 和 $0\leqslant y\leqslant \dfrac {1}{4}$ 区域内，联合概率密度函数的积分值。
步骤 2：计算二重积分
根据联合概率密度函数 $f(x,y)=\dfrac {6}{5}(x+{y}^{2})$，我们需要计算在上述区域内的二重积分，即
$P\{ 0\leqslant X\leqslant \dfrac {1}{4},0\leqslant Y\leqslant \dfrac {1}{4}\} ={\int }_{0}^{\dfrac {1}{4}}{\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dydx$。
步骤 3：计算内层积分
首先计算内层积分 ${\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dy$，得到
${\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dy = \dfrac {6}{5}x{\int }_{0}^{\dfrac {1}{4}}dy + \dfrac {6}{5}{\int }_{0}^{\dfrac {1}{4}}{y}^{2}dy = \dfrac {6}{5}x\cdot \dfrac {1}{4} + \dfrac {6}{5}\cdot \dfrac {1}{3}\cdot {(\dfrac {1}{4})}^{3} = \dfrac {3}{10}x + \dfrac {1}{160}$。
步骤 4：计算外层积分
然后计算外层积分 ${\int }_{0}^{\dfrac {1}{4}}(\dfrac {3}{10}x + \dfrac {1}{160})dx$，得到
${\int }_{0}^{\dfrac {1}{4}}(\dfrac {3}{10}x + \dfrac {1}{160})dx = \dfrac {3}{10}{\int }_{0}^{\dfrac {1}{4}}xdx + \dfrac {1}{160}{\int }_{0}^{\dfrac {1}{4}}dx = \dfrac {3}{10}\cdot \dfrac {1}{2}\cdot {(\dfrac {1}{4})}^{2} + \dfrac {1}{160}\cdot \dfrac {1}{4} = \dfrac {3}{320} + \dfrac {1}{640} = \dfrac {1}{320}$。
根据题目要求，我们需要计算 $P\{ 0\leqslant X\leqslant \dfrac {1}{4},0\leqslant Y\leqslant \dfrac {1}{4}\}$，即在 $0\leqslant x\leqslant \dfrac {1}{4}$ 和 $0\leqslant y\leqslant \dfrac {1}{4}$ 区域内，联合概率密度函数的积分值。
步骤 2：计算二重积分
根据联合概率密度函数 $f(x,y)=\dfrac {6}{5}(x+{y}^{2})$，我们需要计算在上述区域内的二重积分，即
$P\{ 0\leqslant X\leqslant \dfrac {1}{4},0\leqslant Y\leqslant \dfrac {1}{4}\} ={\int }_{0}^{\dfrac {1}{4}}{\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dydx$。
步骤 3：计算内层积分
首先计算内层积分 ${\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dy$，得到
${\int }_{0}^{\dfrac {1}{4}}\dfrac {6}{5}(x+{y}^{2})dy = \dfrac {6}{5}x{\int }_{0}^{\dfrac {1}{4}}dy + \dfrac {6}{5}{\int }_{0}^{\dfrac {1}{4}}{y}^{2}dy = \dfrac {6}{5}x\cdot \dfrac {1}{4} + \dfrac {6}{5}\cdot \dfrac {1}{3}\cdot {(\dfrac {1}{4})}^{3} = \dfrac {3}{10}x + \dfrac {1}{160}$。
步骤 4：计算外层积分
然后计算外层积分 ${\int }_{0}^{\dfrac {1}{4}}(\dfrac {3}{10}x + \dfrac {1}{160})dx$，得到
${\int }_{0}^{\dfrac {1}{4}}(\dfrac {3}{10}x + \dfrac {1}{160})dx = \dfrac {3}{10}{\int }_{0}^{\dfrac {1}{4}}xdx + \dfrac {1}{160}{\int }_{0}^{\dfrac {1}{4}}dx = \dfrac {3}{10}\cdot \dfrac {1}{2}\cdot {(\dfrac {1}{4})}^{2} + \dfrac {1}{160}\cdot \dfrac {1}{4} = \dfrac {3}{320} + \dfrac {1}{640} = \dfrac {1}{320}$。