设随机变量X~B(100,0.2),应用中心极限定理可得P{X≥30}≈多少（Φ（2.5）=0.9938） 大学理工科专业都要学高等数学吗？有哪些专业不学？

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1.
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2.
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3.
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4.
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解 E(X)=100*0.2=20,D(X)=100*0.2*0.8=16

于是由中心极限定理得

璁鹃殢鏈哄彉閲弜~b(10,0.2),y~e(0.5),涓攛涓巠浜掔浉鐙珛,鍒檈(4x-3y)=?
绛旓細浣犲ソ锛闅忔満鍙橀噺X~B(10,0.2),Y~e(0.5),鍒欐湁EX=10*0.2=2锛孍Y=1/0.5=2锛屾墍浠(4X-3Y)=4EX-3EY=2銆傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒

璁鹃殢鏈哄彉閲廥~B(1000,0.1),璇曟眰姒傜巼p(110,<X<=120)
绛旓細浣跨敤涓績鏋侀檺瀹氱悊锛寈鐢1000涓簩椤瑰垎甯冨彲浠ヨ繎浼艰涓烘槸姝ｆ佸垎甯冧簡锛屼綘绠楀嚭浠栫殑鏈熸湜鏂瑰樊 鐒跺悗鎶妜鍖栦负鏍囧噯姝ｅお鍒嗗竷灏卞ソ楗夸簡

1.璁鹃殢鏈哄彉閲廥~B(10,0.2),鍒橢(X)= ? 2.璁鹃殢鏈哄彉閲廥~U(1.7),鍒橠
绛旓細1銆丒(X)=np=10路0.2=2 2銆丏(X)=(b-a)^2/12 =36/12 =3

璁鹃殢鏈哄彉閲弜~b(10,0.1),鍒檝ar(x+1)=?
绛旓細1.9銆x鏈嶄粠浜岄」鍒嗗竷B(10锛0.1)锛屾牴鎹叕寮廍X=nP=10*0.1=1DX=nP(1-P)=10*0.1*0.9=0.9=E(X-EX)^2=E(X^2-2x+1)=0.9E(X^2)=0.9+2EX-1=0.9+2-1=1.9E(5X^2+3)=5E(X^2)+3=5*1.9+3=12.5銆侲(X^2-2x+1)=E(X^2)-2EX+1=0.9 绉诲悜鏁寸悊锛孍(X^2)...

璁鹃殢鏈哄彉閲廥~B(10,0.4),鍒橷
绛旓細E(X) =np = 10(0.4)=4 D(X)=np(1-p) = 10(0.4)(0.6) = 2.4 D(X) =E(X^2) -[E(X)]^2 2.4 =E(X^2) -16 E(X^2)=18.4

姒傜巼璁洪鐩(濉┖棰,涓嶉渶瑕佽繃绋,鏈夌瓟妗堝氨濂)鏁戞晳瀛╁瓙!
绛旓細(1)璁句簨浠禔銆丅鐩镐簰鐙珛锛孭锛圓UB锛=0.6锛孭锛圓锛=0.4锛屽垯 P(AUB)=P(A)+P(B)-P(A)P(B)0.6=0.4+P(B)-0.4P(B)0.6P(B) = 0.2 P(B) = 1/3 (2)璁鹃殢鏈哄彉閲廥~B锛400锛0.1锛夛紝D(X)=np(1-p) = 400(0.1)(1-0.1) = 36 (3)X~U锛1锛4锛塜鐨勫瘑搴﹀嚱鏁 f(...

璁鹃殢鏈哄彉閲廥涓嶻鐩镐簰鐙珛,X~B(1,0.3),Y~U(-1,1),璁癦=X+Y銆傝瘯姹俍鐨勬鐜...
绛旓細Z=X+Y鐨勬鐜囧瘑搴﹀嚱鏁颁负锛歡(y)=鈭玆 p(x)f(y-x)dx銆=0 y鈮0銆俫(y)=鈭玆 p(x)f(y-x)dx=0 y鈮0銆傗埆[0,y]e^(x-y)dx=1-e^(-y) 0<y鈮1銆俍鐨勬鐜囧瘑搴︼細鈭玔0,1]e^(x-y)dx=e^(1-y)-e^(-y) y>1銆

璁鹃殢鏈哄彉閲弜~b(3,0.4)鍒檖(x鈮1)
绛旓細浣犲ソ锛乆鏄簩椤瑰垎甯冿紝鎵浠(X鈮1)=1-P(X<1)=1-P(X=0)=1-0.6^3=0.784銆傜粡娴庢暟瀛﹀洟闃熷府浣犺В绛旓紝璇峰強鏃堕噰绾炽傝阿璋紒

璁鹃殢鏈哄彉閲弜~b(3,0.2),涓旈殢鏈哄彉閲廦=X(3-X)/2,鍒橮(Y=0)=
绛旓細P(Y=0)=P(X=3)+P(X=0)=0.2^3+0.8^3=0.52婊℃剰璇峰ソ璇勫摝锛屾垜姒傜巼璁鸿冨畬浜嗭紝鍢诲樆锛岀浣犺冭瘯椤哄埄鍟

璁鹃殢鏈哄彉閲廥~B(3, 0.4),, 鍒橮(Y=1)鐨勫间负
绛旓細X~B(3,0.4)锛屽垯P(X=1)=C(3,1)(0.4)(0.6^2)=0.432 P(X鈮1)=1-P(X=0)=1-C(2 0)*(p^0)(1-p)^2=1-(1-p)^2=3/4 p=1/2 p(y鈮1)=1-P(Y=0)=1-C(3 0)*(p^0)(1-p)^3=7/8

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