设随机变量X服从区间[0,5]上的均匀分布,计算概率P{2X+4≤10}= 设随机变量X的概率密度为f（x）=2x，0＜x＜10，现对X...
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F_Y(y)=P{Y<=y}
y<=-4, F_Y(y)=P{Y<=y}=0; \u56e0\u4e3a\u6b64\u65f6{Y<=y}\u4e0d\u53ef\u80fd;
y>=5, F_Y(y)=P{Y<=y}=1; \u56e0\u4e3a\u6b64\u65f6{Y<=y}\u5fc5\u7136;
-4<y<=-3, F_Y(y)=P{Y<=y}=P{X^2-2X-3<=y}=P{(X-1)^2-4<=y}=P{1-\u6839\u53f7\u4e0b4+y<=X<=1+\u6839\u53f7\u4e0b4+y}
=F_X(1+\u6839\u53f7\u4e0b4+y)-F_X(1-\u6839\u53f7\u4e0b4+y)=(\u6839\u53f7\u4e0b4+y)/2;
-3<y<5, F_Y(y)=P{Y<=y}=P{X^2-2X-3<=y}=P{(X-1)^2-4<=y}=P{0<X<=1+\u6839\u53f7\u4e0b4+y}
=F_X(1+\u6839\u53f7\u4e0b4+y)-F_X(0)=F_X(1+\u6839\u53f7\u4e0b4+y)=(1+\u6839\u53f7\u4e0b4+y)/4.
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随机变量X服从区间[0,5]上的均匀分布 ---> f(x)=1/5, 0<x<5.2X+4≤10 --> X≤3
所以, P{2X+4≤10} = P(X≤3) = (3)(1/5) = 3/5
P{2X+4≤10}=P{X≤3}=F(3)=(3-0)/(5-0)=3/5
绛旓細P{2X+4鈮10}=P{X鈮3}=F(3)=(3-0)/(5-0)=3/5
绛旓細16-4X鈮0 X鈮4 鎵浠ユ鐜囦负4/10=2/5
绛旓細P{X鈮3}=F(3)=(3-0)/(5-0)=3/5
绛旓細瑙ｏ細P(X<=1/2)=1/4 Y~(3,1/4)P(Y=1)=C(3,1)1/4*(3/4)^2=27/64 濡傛湁鎰忚锛屾杩庤璁猴紝鍏卞悓瀛︿範锛涘鏈夊府鍔╋紝璇烽変负婊℃剰鍥炵瓟锛
绛旓細闅忔満鍙橀噺X鏈嶄粠鍖洪棿銆1,5銆涓婄殑鍧囧寑鍒嗗竷 浠庤屾柟绋嬫湁瀹炴牴鐨勬鐜嘝=锛5-2锛/锛5-1锛=0.75 渚嬪锛氭柟绋媥2+2Yx+1=0鏈夊疄鏍 鍒欌柍=锛2Y锛塪u²-4=4Y²-4鈮0 瑙ｅ緱Y鈮1鎴朰鈮-1 鍙圷鈭堬紙0,5锛夋墍浠鈮1鐨勬鐜囦负锛5-1锛/5=4/5 鏁呮柟绋媥2+2Yx+1=0鏈夊疄鏍圭殑姒傜巼涓4/5 ...
绛旓細璁＄畻杩囩▼濡備笅锛氭牴鎹鎰忓彲鐭 鏂圭▼4x^2+4Xx+X+2=0鐨勪袱涓牴瀹炴牴閮芥槸瀹炴牴 鎵浠モ柍=4X*4X-4*4锛圶+2锛>=0 X>=2鎴朮<=-1 鍥犱负X鍦紙0锛5锛変笂鏈嶄粠鍧囧寑鍒嗗竷 鎵浠灞炰簬[2锛5)鎵浠ヤ袱涓牴瀹炴牴閮芥槸瀹炴牴鐨勬鐜嘝=锛5-2锛/锛5-0锛=3/5 ...
绛旓細搴旇娌￠敊 瑕佹眰鏂圭▼鏈夊疄鏍癸紝鍒欐弧瓒砨^2-4ac鈮0 鍗筹紙4X锛塣2-4脳4(X+2)鈮0 瑙ｆ柟绋嬬粍寰梄鈮-1鎴朮鈮2 鍙堚埖X鈭堬紙0锛5锛夆埓浣挎柟绋嬫湁瀹炴牴鐨刋鍙栧艰寖鍥翠负[2锛5锛夆埓P=锛5-2锛/5=0.6
绛旓細x)=1/5锛寈鈭(0,5)銆乫(x)=0锛寈涓哄叾瀹冿紱Y鐨勫瘑搴﹀嚱鏁癴(y)=5e^(-5y)锛寉鈭(0,鈭)銆乫(y)=0锛寉涓哄叾瀹冿紝鑰岋紝X銆乊鐩镐簰鐙珛锛屸埓(X,Y)鐨勫瘑搴﹀嚱鏁癴(x,y)=f(x)*f(y)=e^(-5y)锛寈鈭(0,5)銆亂鈭(0,鈭)銆乫(x,y)=0锛寈∉(0,5)銆亂∉(0,鈭)銆備緵鍙傝冦
绛旓細0.52x+锛118-x锛*0.33=53
绛旓細21銆璁鹃殢鏈哄彉閲廥鏈嶄粠鍖洪棿[0,5]涓婄殑鍧囧寑鍒嗗竷,鍒橮{X鈮3}= 0.6 銆22銆佽(X,Y)鐨勫垎甯冨緥涓:鍒 YX -1 1 20 a 1 a= 7/30 銆23銆佽X~N(-1,4),Y~N(1,9)涓擷涓嶻鐩镐簰鐙珛,鍒橷+Y~ N(0, 13) 銆24銆佽浜岀淮闅忔満鍙橀噺(X,Y)姒傜巼瀵嗗害涓篺(x,y)= 鍒檉x(x)= 銆25銆佽闅忔満鍙橀噺X鍏锋湁鍒嗗竷 = ...
