题目
计算器在进行加法时,将每个加数舍人最靠近它的整数,设所有舍入误-|||-差相互独立且在 (-0.5,0.5) 上服从均匀分布.-|||-(1)将1500个数相加,问误差总和的绝对值超过15的概率是多少?-|||-(2)最多可有几个数相加使得误差总和的绝对值小于10的概率不小于-|||-0.9.0?
题目解答
答案
解析
步骤 1：定义随机变量
设第k个加数的舍入误差为 ${X}_{k}(k=1,2,\cdots ,1500)$ ,已知Xk在 (-0.5,0.5) 上服从均匀分布,故知 $E({X}_{k})=0,D({X}_{k})=\dfrac {1}{12}$.
步骤 2：计算误差总和的分布
记 $X=\sum _{k=1}^{1500}{X}_{k}$, 由中心极限定理,当n充分大时有近似公式 $P\{ \dfrac {X-0}{\sqrt {1500}\sqrt {1/12}}\leqslant x\} \approx \Phi (x)$.
步骤 3：计算误差总和的绝对值超过15的概率
$P\{ |X|\gt 15\} =1-P\{ |X|\leqslant 15\} =1-P\{ -15\leqslant X\leqslant 15\} $ $=1-P\{ \dfrac {-15}{\sqrt {1500}\sqrt {1/12}}\leqslant \dfrac {X}{\sqrt {1500}\sqrt {1/12}}\leqslant \dfrac {15}{\sqrt {1500}\sqrt {1/12}}\} $ $=1-[ \Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})-\Phi (\dfrac {-15}{\sqrt {1500}\sqrt {1/12}})] $ $=1-[ 2\Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})-1] $ $=2[ 1-\Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})] $ $=2[ 1-\Phi (1.342)] $ $=2[ 1-0.9099] =0.1802$.
步骤 4：计算最多可有几个数相加使得误差总和的绝对值小于10的概率不小于0.90
设最多有n个数相加,使误差总和 $Y=\sum _{k=1}^{n}{X}_{k}$ 符合要求,即要确定n,使 $P\{ |Y|\lt 10\} \geqslant 0.90$. 由中心极限定理,当n充分大时有近似公式 $P\{ \dfrac {Y-0}{\sqrt {n}\sqrt {1/12}}\leqslant x\} \approx \Phi (x)$, 于是 $P\{ |Y|\lt 10\} =P\{ -10\lt Y\lt 10\} $ $=P\{ \dfrac {-10}{\sqrt {n}\sqrt {1/12}}\lt \dfrac {Y}{\sqrt {n}\sqrt {1/12}}\lt \dfrac {10}{\sqrt {n}\sqrt {1/12}}\} $ $=2\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})-1$ 因而n需满足 $2\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})-1\geqslant 0.90$ , 亦即n需满足 $\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})\geqslant 0.95=\Phi (1.645)$ , 即n应满足 $\dfrac {10}{\sqrt {n}\sqrt {1/12}}\geqslant 1.645$ , 由此得 $n\leqslant 443.4\dot {5}$ . 因n为正整数,因而所求的n为443.故最多只能有443个数加在一起,才能使得误差总和的绝对值小于10的概率不小于0.90.
设第k个加数的舍入误差为 ${X}_{k}(k=1,2,\cdots ,1500)$ ,已知Xk在 (-0.5,0.5) 上服从均匀分布,故知 $E({X}_{k})=0,D({X}_{k})=\dfrac {1}{12}$.
步骤 2：计算误差总和的分布
记 $X=\sum _{k=1}^{1500}{X}_{k}$, 由中心极限定理,当n充分大时有近似公式 $P\{ \dfrac {X-0}{\sqrt {1500}\sqrt {1/12}}\leqslant x\} \approx \Phi (x)$.
步骤 3：计算误差总和的绝对值超过15的概率
$P\{ |X|\gt 15\} =1-P\{ |X|\leqslant 15\} =1-P\{ -15\leqslant X\leqslant 15\} $ $=1-P\{ \dfrac {-15}{\sqrt {1500}\sqrt {1/12}}\leqslant \dfrac {X}{\sqrt {1500}\sqrt {1/12}}\leqslant \dfrac {15}{\sqrt {1500}\sqrt {1/12}}\} $ $=1-[ \Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})-\Phi (\dfrac {-15}{\sqrt {1500}\sqrt {1/12}})] $ $=1-[ 2\Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})-1] $ $=2[ 1-\Phi (\dfrac {15}{\sqrt {1500}\sqrt {1/12}})] $ $=2[ 1-\Phi (1.342)] $ $=2[ 1-0.9099] =0.1802$.
步骤 4：计算最多可有几个数相加使得误差总和的绝对值小于10的概率不小于0.90
设最多有n个数相加,使误差总和 $Y=\sum _{k=1}^{n}{X}_{k}$ 符合要求,即要确定n,使 $P\{ |Y|\lt 10\} \geqslant 0.90$. 由中心极限定理,当n充分大时有近似公式 $P\{ \dfrac {Y-0}{\sqrt {n}\sqrt {1/12}}\leqslant x\} \approx \Phi (x)$, 于是 $P\{ |Y|\lt 10\} =P\{ -10\lt Y\lt 10\} $ $=P\{ \dfrac {-10}{\sqrt {n}\sqrt {1/12}}\lt \dfrac {Y}{\sqrt {n}\sqrt {1/12}}\lt \dfrac {10}{\sqrt {n}\sqrt {1/12}}\} $ $=2\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})-1$ 因而n需满足 $2\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})-1\geqslant 0.90$ , 亦即n需满足 $\Phi (\dfrac {10}{\sqrt {n}\sqrt {1/12}})\geqslant 0.95=\Phi (1.645)$ , 即n应满足 $\dfrac {10}{\sqrt {n}\sqrt {1/12}}\geqslant 1.645$ , 由此得 $n\leqslant 443.4\dot {5}$ . 因n为正整数,因而所求的n为443.故最多只能有443个数加在一起,才能使得误差总和的绝对值小于10的概率不小于0.90.