设随机变量X的概率密度为，求X的数学期望E(X)与方差D(X)．
【答案】：EX=∫(0,1)x*3x^2dx=3/4EX^2=∫(0,1)x^2*3x^2dx=3/5
所以DX=EX^2-(EX)^2=3/5-(3/4)^2=3/80
绛旓細瑙ｏ細濡傚浘1锛孎(x)锛漃(X鈮)锛屽湪0鈮锛1鏃讹紝姹俋鐨勫垎甯冨嚱鏁帮紝姹傜殑鏄疿鈮鐨姒傜巼,鑰屼笉鏄眰X鈮1鐨勬鐜囷紝鍥犳绉垎鑼冨洿涓0鍒皒銆
绛旓細銆愮瓟妗堛戯細EX=鈭(0,1)x*3x^2dx=3/4 EX^2=鈭(0,1)x^2*3x^2dx=3/5 鎵浠X=EX^2-(EX)^2=3/5-(3/4)^2=3/80
绛旓細f(x)=3x³/胃³=3x³/6=x³/2銆2銆佷簬鏄垎甯冨嚱鏁颁负F(x)=鈭(-鈭,x) f(x)dx=鈭(-鈭,x) x³/2dx= x^4/8銆備护x^4/8=1锛屾垜浠彲浠ユ眰寰x鐨鍙栧艰寖鍥存槸锛-鈭,4鈭8) (4娆℃牴鍙8锛夈
绛旓細A=2銆璁鹃殢鏈哄彉閲廥鍏锋湁姒傜巼瀵嗗害fX(x)锛-鈭<x<鈭烇紝鐢辫鍑芥暟g(x)澶勫鍙涓旀亽鏈塯'(x)>0锛堟垨鎭掓湁g'(x)<0锛夛紝鍒橸=g(X)鏄繛缁瀷闅忔満鍙橀噺銆傜Н鍑烘潵鐨勫嚱鏁版槸ax^3/4,绉垎鍊兼槸a*1^3/4-a*0^3/4=a/4銆傚f(x)=Ax鍦0鍒1涓婄Н鍒嗭紝 寰楀埌0.5A=1锛岃В寰桝=2銆
绛旓細鍏堢敱姒傜巼姹鍑哄弬鏁颁负2锛屽啀鐢卞叕寮忔眰鍑烘湡鏈涗笌鏂瑰樊銆侲X=鈭紙0锛1锛x*3x^2dx=3/4 EX^2=鈭紙0锛1锛墄^2*3x^2dx=3/5 鎵浠X=EX^2-锛圗X锛塣2=3/5-锛3/4锛塣2=3/80 鍗曠函鐨勮姒傜巼瀵嗗害 娌℃湁瀹為檯鐨勬剰涔夛紝瀹冨繀椤绘湁纭畾鐨勬湁鐣屽尯闂翠负鍓嶆彁銆傚彲浠ユ妸姒傜巼瀵嗗害鐪嬫垚鏄旱鍧愭爣锛屽尯闂寸湅鎴愭槸妯潗鏍囷紝...
绛旓細璁捐繛缁瀷闅忔満鍙橀噺X鐨勬鐜囧瘑搴︿负 姹:(1)甯告暟a;(2)X鐨勫垎甯冨嚱鏁? 璁捐繛缁瀷闅忔満鍙橀噺X鐨勬鐜囧瘑搴︿负姹:(1)甯告暟a;(2)X鐨勫垎甯冨嚱鏁般... 璁捐繛缁瀷闅忔満鍙橀噺X鐨勬鐜囧瘑搴︿负姹:(1)甯告暟a;(2)X鐨勫垎甯冨嚱鏁般 灞曞紑 鎴戞潵绛 鍒嗕韩 鏂版氮寰崥 QQ绌洪棿 2涓洖绛 #鐑# 鎴愬勾浜虹殑鎶戦儊鏄惁澶у鍥犱负娌￠挶? 42娓╂煍...
绛旓細銆愮瓟妗堛戯細瑙ｏ細锛1锛塮(x)=ae^(x) ,x鈮0 ;f(x)=ae^(-x),x锛0 鐢姒傜巼瀵嗗害鍑芥暟鐨勬ц川寰 鈭玜e^xdx(绉垎鍖洪棿涓鸿礋鏃犵┓鍒0)=1/2 寰梐=1/2 锛2锛塅(x)=(1/2) (e^x),x鈮0 F(x)=1-(1/2)e^(-x),x锛0 浠ｅ叆P{0鈮鈮1}=F(1)-F(0)=(1/2)(1-1/e)鎴栬匬{0鈮鈮...
绛旓細P{|x|<1} 涓1/2銆傚垎甯冨嚱鏁癋(x)涓1/2+arctanx锛寈灞炰簬锛-蟺/2锛屜/2锛夈傝f(x)鍦▁杩炵画锛屾鐜囧瘑搴鍑芥暟灏辨槸姒傜巼鍒嗗竷鍑芥暟鐨勫鏁帮紝浠庡鏁扮殑瑙掑害鐞嗚В姒傜巼瀵嗗害鍑芥暟姣旇緝瀹规槗鐞嗚В锛屾鐜囧瘑搴﹀嚱鏁拌〃鐜扮殑鏄鐜囧垎甯冨嚱鏁板湪鏌愪竴鐐圭殑鍙樺寲鐜囷紝鍒嗗竷鍑芥暟鏄竴涓疮绉嚱鏁帮紝鏄鍑芥暟锛屾墍浠ヨ繖涓彉鍖栧彧鏄鍔犵殑蹇參鐨...
绛旓細绠鍗曡绠椾竴涓嬪嵆鍙紝绛旀濡傚浘鎵绀
绛旓細鏍规嵁姒傜巼瀵嗗害鍑芥暟鐨勫畾涔夛紝绉垎銆0,2銆 锛x/2+c锛塪x=1.鎵浠ワ紙x^2/4+cx锛夈0,2銆=1銆傛墍浠1+2c=1.鎵浠=0.___绗簩闂瓼锛坸锛=0(x<=0鏃)F锛坸锛=绉垎銆0,x銆 锛坸/2锛=x^2/4.(0<x<=2)F(x)=1.(x>2)鈥斺斺旂涓夐棶P锛坸<o.5锛=F(0.5)=0.5^2/4=0.0625.涓嶆噦鍐...
