题目
设随机变量X～N(60，32)，求分位数x1，x2，使X分别落在区间(-∞，x1)，(x1，x2)，(x2，+∞)内的概率之比为3：4：5。
设随机变量X～N(60，32)，求分位数x1，x2，使X分别落在区间(-∞，x1)，(x1，x2)，(x2，+∞)内的概率之比为3：4：5。
题目解答
答案
解析
步骤 1：确定概率比例
设X落在 $(-\infty ,{x}_{1})$ ,(x1,x2), $({x}_{2},+\infty )$ 内的概率分别是P1 、P2和p3,由题意 ${p}_{1}+{p}_{2}+{p}_{3}=1$ 且 .${P}_{1}:{P}_{2}:{P}_{3}=3:4:5$ 由此计算得 ${p}_{1}=\dfrac {3}{12}=\dfrac {1}{4}$, ${P}_{2}=\dfrac {4}{12}=\dfrac {1}{3}$ . ${P}_{3}=\dfrac {5}{12}$
步骤 2：计算x1
则 $P(x\lt {x}_{1})=\Phi (\dfrac {{x}_{1}-60}{3})={p}_{1}=\dfrac {1}{4}=0.25$ $\Phi (\dfrac {60-{x}_{1}}{3})=1-\Phi (\dfrac {{x}_{1}-60}{3})=1-\dfrac {1}{4}=0.75$ 即 $\dfrac {60-{x}_{1}}{3}=0.68$, 故 ${x}_{1}=60-3\times 0.68=57.96$
步骤 3：计算x2
又 P|X> x2}=1-P|X≤x2|=1-ϕ((x2-)/360 =P3=5/12, $(\dfrac {{x}_{2}-60}{3})=1-\dfrac {5}{12}=\dfrac {7}{12}=0.5833$ 则 $\dfrac {{x}_{2}-60}{3}=0.21$, 故 ${x}_{2}=60+3\times 0.21=60.63$
设X落在 $(-\infty ,{x}_{1})$ ,(x1,x2), $({x}_{2},+\infty )$ 内的概率分别是P1 、P2和p3,由题意 ${p}_{1}+{p}_{2}+{p}_{3}=1$ 且 .${P}_{1}:{P}_{2}:{P}_{3}=3:4:5$ 由此计算得 ${p}_{1}=\dfrac {3}{12}=\dfrac {1}{4}$, ${P}_{2}=\dfrac {4}{12}=\dfrac {1}{3}$ . ${P}_{3}=\dfrac {5}{12}$
步骤 2：计算x1
则 $P(x\lt {x}_{1})=\Phi (\dfrac {{x}_{1}-60}{3})={p}_{1}=\dfrac {1}{4}=0.25$ $\Phi (\dfrac {60-{x}_{1}}{3})=1-\Phi (\dfrac {{x}_{1}-60}{3})=1-\dfrac {1}{4}=0.75$ 即 $\dfrac {60-{x}_{1}}{3}=0.68$, 故 ${x}_{1}=60-3\times 0.68=57.96$
步骤 3：计算x2
又 P|X> x2}=1-P|X≤x2|=1-ϕ((x2-)/360 =P3=5/12, $(\dfrac {{x}_{2}-60}{3})=1-\dfrac {5}{12}=\dfrac {7}{12}=0.5833$ 则 $\dfrac {{x}_{2}-60}{3}=0.21$, 故 ${x}_{2}=60+3\times 0.21=60.63$