设随机变量X~N(2,4),则P(2<X<6)=? 急！！ 设随机变量x~n(2,16)求p(3<x<6)
\u8bbe\u968f\u673a\u53d8\u91cfX~N(1,4),\u5219P{| X |>2}=\u8bbe\u968f\u673a\u53d8\u91cfX~N(1,4),\u5219P{| X |>2}=\u7684\u8ba1\u7b97\u65b9\u6cd5\u5982\u4e0b\uff1a
\u968f\u673a\u4e8b\u4ef6\u6570\u91cf\u5316\u7684\u597d\u5904\u662f\u53ef\u4ee5\u7528\u6570\u5b66\u5206\u6790\u7684\u65b9\u6cd5\u6765\u7814\u7a76\u968f\u673a\u73b0\u8c61\u3002\u4f8b\u5982\u67d0\u4e00\u65f6\u95f4\u5185\u516c\u5171\u6c7d\u8f66\u7ad9\u7b49\u8f66\u4e58\u5ba2\u4eba\u6570\uff0c\u7535\u8bdd\u4ea4\u6362\u53f0\u5728\u4e00\u5b9a\u65f6\u95f4\u5185\u6536\u5230\u7684\u547c\u53eb\u6b21\u6570\uff0c\u706f\u6ce1\u7684\u5bff\u547d\u7b49\u7b49\uff0c\u90fd\u662f\u968f\u673a\u53d8\u91cf\u7684\u5b9e\u4f8b\u3002
\u6269\u5c55\u8d44\u6599
\u4e00\u4e2a\u968f\u673a\u8bd5\u9a8c\u7684\u53ef\u80fd\u7ed3\u679c\uff08\u79f0\u4e3a\u57fa\u672c\u4e8b\u4ef6\uff09\u7684\u5168\u4f53\u7ec4\u6210\u4e00\u4e2a\u57fa\u672c\u7a7a\u95f4\u03a9(\u89c1\u6982\u7387)\u3002\u968f\u673a\u53d8\u91cfx\u662f\u5b9a\u4e49\u4e8e\u03a9\u4e0a\u7684\u51fd\u6570\uff0c\u5373\u5bf9\u6bcf\u4e00\u57fa\u672c\u4e8b\u4ef6\u03c9\u2208\u03a9\uff0c\u6709\u4e00\u6570\u503cx(\u03c9)\u4e0e\u4e4b\u5bf9\u5e94\u3002
\u4ee5\u63b7\u4e00\u9897\u9ab0\u5b50\u7684\u968f\u673a\u8bd5\u9a8c\u4e3a\u4f8b\uff0c\u5b83\u7684\u6240\u6709\u53ef\u80fd\u7ed3\u679c\u89c1\uff0c\u51716\u4e2a\uff0c\u5206\u522b\u8bb0\u4f5c\u03c91,\u03c92,\u03c93,\u03c94,\u03c95,\u03c96,\u8fd9\u65f6\uff0c\u03a9={\u03c91,\u03c92,\u03c93,\u03c94,\u03c95,\u03c96},\u800c\u51fa\u73b0\u7684\u70b9\u6570\u8fd9\u4e2a\u968f\u673a\u53d8\u91cfx\uff0c\u5c31\u662f\u03a9\u4e0a\u7684\u51fd\u6570x(\u03c9k)=k\uff0ck=1,2,\u2026\uff0c6\u3002
\u4ee5Fai(x)\u8868\u793a\u6807\u51c6\u6b63\u6001\u5206\u5e03\u7684\u5206\u5e03\u51fd\u6570\u3002
P(3<x<6)=p( (3-2)/4< (x-2)/4<(6-2)/4 )=
=Fai(1)- Fai(1/4)=. \uff08\u67e5\u8868\uff09
=0.8413- 0.5987 =0.2426.
于是：2<x<6 变成：0 <t<2
P(2<x<6) = P(0<t<2) = Φ(2) - Φ(0)
= 0.9772 - 0.5 = 0.4772
Φ为标准正态分布，P=Φ((6-2)/2)-Φ(0)= Φ(2)-Φ(0) 查表
绛旓細N锛2,4锛鏄寚x鏈嶄粠渭 = 2锛屜 = 2鐨勬鎬佸垎甯冦備篃灏辨槸璇寸洿绾縳=2鏄姝ｆ佸垎甯冨浘鍍忕殑瀵圭О杞达紝鎵浠ワ紝P锝沊鈮2锝=1/6銆傝嫢闅忔満鍙橀噺X鏈嶄粠涓涓暟瀛︽湡鏈涗负渭銆佹柟宸负蟽^2鐨勬鎬佸垎甯冿紝璁颁负N(渭锛屜僞2)銆傚叾姒傜巼瀵嗗害鍑芥暟涓烘鎬佸垎甯冪殑鏈熸湜鍊嘉煎喅瀹氫簡鍏朵綅缃紝鍏舵爣鍑嗗樊蟽鍐冲畾浜嗗垎甯冪殑骞呭害銆傚綋渭 =...
绛旓細N锛2,4锛鏄寚x鏈嶄粠渭 = 2锛屜 = 2鐨勬鎬佸垎甯 涔熷氨鏄鐩寸嚎x=2鏄姝ｆ佸垎甯冨浘鍍忕殑瀵圭О杞 鎵浠ョ洿绾縳=2宸﹀彸鐨勬鐜囦釜鍗犱竴鍗婏紝鍗砅锝沊鈮2锝=1/2
绛旓細闅忔満鍙橀噺X~N锛2锛4锛锛屾墍浠锛圶)=4,D(0.5X)=0.5*0.5D(X)=0.25D(X)=1
绛旓細闅忔満鍙橀噺X~N锛2,4锛,鎵浠锛圶)=4,D(0.5X)=0.5*0.5D(X)=0.25D(X)=1
绛旓細A 鍥犱负璁鹃殢鏈哄彉閲廥 锝N锛2锛4锛夛紝鍒D锛 X锛夌殑鍊= D锛圶锛= 锛岄堿
绛旓細DY=16銆傝В绛旇繃绋嬪涓嬶細鍥犱负 X ~ N锛2锛4锛锛涙墍浠 D(X) = 4锛汥(Y)= D(3-2x)= (-2)^2*D(X)= 4*4 =16
绛旓細璁綵=(X-2)/2 鍒欙紝Z~N(0锛1)P(X鈮0)=P(Z鈮-1)=桅(-1)=1-桅(1)=1-0.8413 =0.1587
绛旓細灏嗘鎬鍙橀噺锛X~N(2,4) 鍙樻崲鎴愭爣鍑嗘鎬佸彉閲忥細t = (x-2)/2 浜庢槸锛2<x<6 鍙樻垚锛0 <t<2 P(2<x<6) = P(0<t<2) = 桅(2) - 桅(0)= 0.9772 - 0.5 = 0.4772
绛旓細Cov(X,X+1)=Cov(X,X)=4
绛旓細杩炵画鍑芥暟锛屽湪鍗曠偣鐨勬鐜囧间竴寰嬩负0.
