第2章数理统计基础习题解答

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概率论与数理统计2.第二章练习题(答案)

概率论与数理统计2.第二章练习题(答案)

第二章练习题(答案)一、单项选择题1.已知连续型随机变量X 的分布函数为⎪⎩⎪⎨⎧≥<≤+<=ππx x b kx x x F ,10,0,0)( 则常数k 和b 分别为 ( A )(A )0,1==b k π (B )π1,0b k = (C )0,21==b k π (D )π21,0==b k . 2.下列函数哪个是某随机变量的分布函数 ( A )A. f (x )={xa e −x 22a,x ≥01, x <0(a >0); B. f (x )={12cosx, 0< x <π0, 其他C. f (x )={cosx, −π2< x <π20, 其他D. f (x )={sinx, −π2< x <π20, 其他3.若函数()f x 是某随机变量X 的概率密度函数,则一定成立的是 ( C ) A. ()f x 的定义域是[0,1] B. ()f x 的值域为[0,1] C. ()f x 非负 D. ()f x 在(,)-∞+∞内连续4. 设)1,1(~N X ,密度函数为)(x f ,则有( C ) A.{}{}00>=≤X P X P B. )()(x f x f -= C. {}{}11>=≤X P X P D. )(1)(x F x F --=5. 设随机变量()16,~μN X ,()25,~μN Y ,记()41-<=μX P p ,()52+>=μY P p ,则正确的是 ( A ).(A )对任意μ,均有21p p = (B )对任意μ,均有21p p < (C )对任意μ,均有21p p > (D )只对μ的个别值有21p p = 6. 设随机变量2~(10,)X N ,则随着的增加{10}P X ( C )A.递增B.递减C.不变D.不能确定7.设F 1(x )与F 2(x )分别为随机变量X 1、X 2的分布函数,为使F (x )=aF 1(x )-bF 2(x )是某一随机变量的分布函数,在下列给定的多组数值中应取 ( A )A . a =53, b =52-; B . a =32, b =32;C . 21-=a , 23=b ; D . 21=a , 23-=b .8.设X 1与X 2是任意两个相互独立的连续型随机变量,它们的概率密度函数分别为f 1(x )和f 2(x ),分布函数分别为F 1(x )和F 2(x ),则 ( D ) (A) f 1(x )+f 2(x ) 必为某个随机变量的概率密度; (B )f 1(x )•f 2(x ) 必为某个随机变量的概率密度; (C )F 1(x )+F 2(x ) 必为某个随机变量的分布函数; (D) F 1(x ) •F 2(x ) 必为某个随机变量的分布函数。

数理统计课后答案-第二章

数理统计课后答案-第二章

2.6
已知总体
ξ 服从指数分布,概率密度为
⎧λ e − λ x ϕ ( x) = ⎨ ⎩ 0 x>0 x≤0
3
其中,参数 λ > 0 , ( X 1 , X 2 , L , X n ) 是
ξ 的样本, X 是样本均值, S 2 是样本方差,
S *2 是修正样本方差,求 EX , DX , E ( S 2 ) 和 E ( S *2 ) 。
因此有
n −1 n −1 2 n −1 n −1 2 2 )= σ , E (S y σ 。 Dξ = Dη = n n n n
⎡ n ⎤ E ⎢∑ ( X i + Yi − X − Y ) 2 ⎥ ⎣ i =1 ⎦
n n ⎡n ⎤ = E ⎢∑ ( X i − X ) 2 + 2∑ ( X i − X )(Yi − Y ) + ∑ (Yi − Y ) 2 ⎥ i =1 i =1 ⎣ i =1 ⎦
解 因为总体
ξ 服从参数为 λ 的指数分布,所以 Eξ =
1
λ
, Dξ =
1
λ2

由定理 2.1 可知
EX = Eξ =
1
λ
, DX =
1 1 Dξ n −1 n −1 , E ( S *2 ) = Dξ = 2 。 = 2 ,E ( S 2 ) = Dξ = 2 n n nλ nλ λ
2.7
设 ( X 1 , X 2 , L , X n ) 是总体
∑ (X i − X )2 =
i =1
n
S x2 。 b2
2.4 设有样本 ( X 1 , X 2 , L , X n ) , X = 本方差, μ 是常数,证明
1 n 1 n 2 X 是样本均值, S = ( X i − X ) 2 是样 ∑ ∑ i n i =1 n i =1

概率论与数理统计习题解答(第2章)

概率论与数理统计习题解答(第2章)

概率论与数理统计习题解答(第2章)习 题 二(A )三、解答题1.一颗骰子抛两次,以X 表示两次中所得的最小点数(1) 试求X 的分布律; (2) 写出X 的分布函数.解: (1)分析:这里的概率均为古典概型下的概率,所有可能性结果共36种,如果X=1,则表明两次中至少有一点数为1,其余一个1至6点均可,共有1-612⨯C(这里12C 指任选某次点数为1,6为另一次有6种结果均可取,减1即减去两次均为1的情形,因为612⨯C 多算了一次)或1512+⨯C 种,故2{}36113615361-611212=+⨯=⨯==C C X P ,其他结果类似可得.(2)⎪⎪⎪⎪⎩⎪⎪⎪⎪⎨⎧≥<≤=+=+=+=+=<≤=+=+=+=<≤=+=+=<≤=+=<≤=<=6165}5{}4{}3{}2{}1{54 }4{}3{}2{}1{43 }3{}2{}1{32}2{}1{21}1{1 0 )(x x X P X P X P X P X P x X P X P X P X P x X P X P X P x X P X P x X P x x F ,,,,,,,⎪⎪⎪⎪⎪⎪⎪⎩⎪⎪⎪⎪⎪⎪⎪⎨⎧≥<≤<≤<≤<≤<≤<=6 165363554 363243 36273236202136111 0 x x x x x x x ,,,,,,, 2.某种抽奖活动规则是这样的:袋中放红色球及白色球各5只,抽奖者交纳一元钱后得到一次抽奖的机会,然后从袋中一次取出5只球,若5只球同色,则获奖100元,否则无奖,以X 表示某抽奖者在一次抽取中净赢钱数,求X 的分布律.3解:注意,这里X 指的是赢钱数,X 取0-1或100-1,显然{}1261299510===C X P .3.设随机变量X 的分布律为;,2,1,0,!}{>===λλ k k ak X P k为常数,试求常数a .解:因为1!0==-∞=∑λλaek ak k,所以λ-=e a .4.设随机变量X 的分布律为(1) 求X 的分布函数;(2) 求}21{≤X P ,}2523{≤<X P ,}32{≤≤x P . 解: (1)⎪⎪⎪⎩⎪⎪⎪⎨⎧≥<≤<≤-<=⎪⎪⎩⎪⎪⎨⎧≥<≤=+-=<≤--=<=3x 132432141-1x 03x 132}2{}1{21}1{-1x 0)(,,,,,,,,x x x X P X P x X P x f ,4(2) {}41121=-==⎭⎬⎫⎩⎨⎧≤X p X P 、 {}2122523===⎭⎬⎫⎩⎨⎧≤<X P X P , {}{}{}{}{}{}43323232==+=====≤≤X P X P X X P X P .5.设随机变量X 的分布律为 ,2,1,21}{===k k X P k求: (1) P {X = 偶数} (2) P {X ≥ 5} (3) P {X = 3的倍数} 解:(1) {}3121121121lim 212121222242=⎪⎪⎪⎪⎭⎫⎝⎛-⎪⎭⎫ ⎝⎛-=++++==∞→i i iX P 偶数,(2) {}{}16116151415=-=≤-=≥X P X P , (3) {}7121121121lim 21333313=-⎥⎥⎦⎤⎢⎢⎣⎡⎪⎭⎫ ⎝⎛-===∞→∞=∑i i i i X P 的倍数.6. 某公安局在长度为t 的时间间隔内收到的紧急呼救的次数X 服从参数为0.5t 的泊松分布,而与时间间隔的起点无关(时间以小时计) (1) 求某一天中午12时至下午3时没有收到紧急呼救的概率.(2) 求某一天中午12时至下午5时至少收到5一次紧急呼救的概率. 解:(1) ()()5.15.0~P t P X = {}5.10-==e X P .(2) 5.25.0=t {}{}5.21011--==-=≥e x P x P .7. 某人进行射击,每次射击的命中率为0.02,独立射击400次,试求至少击中2次的概率.解:设射击的次数为X ,由题意知().20400~,B X , {}{},98.002.0111240010400k k k kC X P X P -=∑-=≤-=≥9972.028.01!81810=-=-≈-=∑e k k K ,其中8=400×0.02.8. 设事件A 在每一次试验中发生的概率为0.3,当A 发生不少于3次时,指示灯发出信号.现进行5次独立试验,试求指示灯发出信号的概率. 解:设X 为事件A 在5次独立重复实验中出现的次数,().305~,B X 则指示灯发出信号的概率{}{})7.03.07.03.07.03.0(131********55005C C C X P X P p ++-=<-=≥=1631.08369.01=-=.9. 设顾客在某银行窗口等待服务的时间X (以分钟计)服从参数为5指数分布.某顾客在6窗口等待服务,若超过10分钟,他就离开.他一个月要到银行5次,以Y 表示他未等到服务而离开窗口的次数.写出Y 的分布律,并求P {Y ≥ 1}. 解:因为X 服从参数为5的指数分布,则51)(xex F --=,{}2)10(110-=-=>e F X P ,()25~-e B Y ,,则50,1,k ,)1()(}{5225=-==---k k k e e Ck Y P .0.516711}0{-1}1{52=--===≥-)(e Y P Y P10.设随机变量X的概率密度为⎪⎩⎪⎨⎧>≤=2||,02||,cos )(ππx x x a x f ,试求:(1) 系数a ;(2) X 落在区间)4,0(π内的概率. 解:(1) 由归一性知:⎰⎰-∞+∞-===222cos )(1ππaxdx a dx x f ,所以21=a . (2) .42|sin 21cos 21}40{404===<<⎰πππx xdx X P . 11.设连续随机变量X的分布函数为7⎪⎩⎪⎨⎧≥<≤<=1,110,0,0)(2x x Ax x x F试求:(1) 系数A ;(2) X 落在区间(0.3,0.7)内的概率;(3) X 的概率密度.解 (1)由F (x )在x =1的连续性可得)1()(lim )(lim 11F x F x F x x ==-→+→,即A=1.(2){}=<<7.03.0X P 4.0)3.0()7.0(=-F F .(3)X 的概率密度⎩⎨⎧<<='=,010,2)()(x x x F x f .12.设随机变量X 服从(0,5)上的均匀分布,求x 的方程02442=+++X Xx x有实根的概率.解:因为X 服从(0,5)上的均匀分布,所以⎪⎩⎪⎨⎧<<=其他05051)(x x f若方程24422=+++X Xx x 有实根,则3216)4(2≥--=∆X X ,即12-≤≥X X ,所以有实根的概率为{}{}53510511252152==+=-≤+≥=⎰⎰-∞-x dx dx X P X P p813.设X ~N (3,4)(1) 求};3{},2{},104{},52{>>≤<-≤<X P X P X P X P (2) 确定c 使得};{}{c X P c X P ≤=>(3) 设d 满足9.0}{≥>d X P ,问d 至多为多少?解: (1) 因为4)(3~,N X 所以 )2()5(}52{F F X P -=≤<5328.016915.08413.01)5.0()1(=-+=-Φ-Φ={})4()10(104--=≤<-F F X P9996.019998.021)5.3(21)5.3()5.3(=-⨯=-Φ=--Φ-Φ={}{}212≤-=>X P X P {}221≤≤--=X P[])2()2(1---=F F [])5.2()5.0(1-Φ--Φ-=[])5.0()5.2(1Φ-Φ-=3023.01-=6977.0={}{}313≤-=>X P X P )3(1F -=)0(1Φ-=5.01-=5.0=. (2){}{}c X P c X P ≤-=>1,则{}21=≤c X P 21)23()(=-Φ==c c F ,经查表得21)0(=Φ,即023=-c ,得3=c ;由概率密度关于9x=3对称也容易看出。

数理统计第二章课后习题参考答案

数理统计第二章课后习题参考答案

第二章 参数估计2.4 设子样1.3,0.6,1.7,2.2,0.3,1.1是来自具有密度函数()1f x ββ=;,0x β<<的总体,试用矩法估计总体均值、总体方差及参数β. 解: 1.30.6 1.7 2.20.3 1.1 1.26X μ+++++===.()()()()()()()22222222111 1.3 1.20.6 1.2 1.7 1.2 2.2 1.20.3 1.2 1.1 1.26ni i X X n σ=⎡⎤=-=-+-+-+-+-+-⎣⎦∑ ()222222210.10.60.510.90.10.4076σ=+++++==. ()()0112E X x f x dx xdx ββββ+∞-∞===⎰⎰;.令()E X X =,则12X β=,即2X β=.参数β的矩估计量为ˆ22 1.2 2.4X β==⨯=.2.6 设总体X 的密度函数为()f x θ;,1X ,2X ,…,n X 为其样本,求下列情况下θ的MLE.(iii )()()100x x e x f x ααθθαα--⎧>⎪=⎨⎪⎩,;,其它α已知解:当0i X >()12i n = ,,,时,似然函数为: ()()()()111111ni i i n n n x n x i i i i i i L f x x e x eαααθθαθθθαθα=----===∑⎛⎫=== ⎪⎝⎭∏∏∏;.()()11ln ln ln 1ln n ni i i i L n n x x αθθααθ===++--∑∑.由()1ln 0ni i L nx αθθθ=∂=-=∂∑,得θ的MLEˆθ,即1ˆnii nxαθ==∑.2.7 设总体X 的密度函数为()()1f x x ββ=+,01x <<,1X ,2X ,…,n X 为其子样,求参数β的MLE 及矩法估计。

今得子样观察值为0.3,0.8,0.27,0.35,0.62及0.55,求参数β的估计值。

概率论与数理统计(经管类)第二章课后习题答案

概率论与数理统计(经管类)第二章课后习题答案

习题2.11.设随机变量X 的分布律为P{X=k}=,k=1, 2,N,求常数a.aN 解:由分布律的性质=1得∑∞k =1p kP(X=1) + P(X=2) +…..+ P(X=N) =1N*=1,即a=1aN 2.设随机变量X 只能取-1,0,1,2这4个值,且取这4个值相应的概率依次为,,求常数c.12c 34c ,58c ,716c 解:12c +34c +58c +716c =1C=37163.将一枚骰子连掷两次,以X 表示两次所得的点数之和,以Y 表示两次出现的最小点数,分别求X,Y 的分布律.注: 可知X 为从2到12的所有整数值.可以知道每次投完都会出现一种组合情况,其概率皆为(1/6)*(1/6)=1/36,故P(X=2)=(1/6)*(1/6)=1/36(第一次和第二次都是1)P(X=3)=2*(1/36)=1/18(两种组合(1,2)(2,1))P(X=4)=3*(1/36)=1/12(三种组合(1,3)(3,1)(2,2))P(X=5)=4*(1/36)=1/9(四种组合(1,4)(4,1)(2,3)(3,2))P(X=6)=5*(1/36=5/36(五种组合(1,5)(5,1)(2,4)(4,2)(3,3))P(X=7)=6*(1/36)=1/6(这里就不写了,应该明白吧)P(X=8)=5*(1/36)=5/36P(X=9)=4*(1/36)=1/9P(X=10)=3*(1/36)=1/12P(X=11)=2*(1/36)=1/18P(X=12)=1*(1/36)=1/36以上是X 的分布律投两次最小的点数可以是1到6里任意一个整数,即Y 的取值了.P(Y=1)=(1/6)*1=1/6 一个要是1,另一个可以是任何值P(Y=2)=(1/6)*(5/6)=5/36 一个是2,另一个是大于等于2的5个值P(Y=3)=(1/6)*(4/6)=1/9 一个是3,另一个是大于等于3的4个值P(Y=4)=(1/6)*(3/6)=1/12一个是4,另一个是大于等于4的3个值P(Y=5)=(1/6)*(2/6)=1/18一个是5,另一个是大于等于5的2个值P(Y=6)=(1/6)*(1/6)=1/36一个是6,另一个只能是6以上是Y 的分布律了.4.设在15个同类型的零件中有2个是次品,从中任取3次,每次取一个,取后不放回.以X 表示取出的次品的个数,求X 的分布律.解:X=0,1,2X=0时,P=C 313C 315=2235X=1时,P=C 213∗C 12C 315=1235X=2时,P=C 013∗C 22C 315=1355.抛掷一枚质地不均匀的硬币,每次出现正面的概率为,连续抛掷8次,以X 表示出现正面的次数,求23X 的分布律.解:P{X=k}=, k=1, 2, 3, 8C k 8(23)k (13)8‒k 6.设离散型随机变量X 的分布律为X -123P141214解:求P {X ≤12}, P {23<X ≤52}, P {2≤X ≤3}, P {2≤X <3}P {X ≤12}=14P {23<X ≤52}=12P {2≤X ≤3}=12+14=34P {2≤X <3}=127.设事件A 在每一次试验中发生的概率分别为0.3.当A 发生不少于3次时,指示灯发出信号,求:(1)进行5次独立试验,求指示灯发出信号的概率;(2)进行7次独立试验,求指示灯发出信号的概率.解:设X 为事件A 发生的次数,(1)P {X ≥3}=P {X =3}+P {X =4}+P {X =5}=C 35(0.3)3(0.7)2+C 45(0.3)4(0.7)1+C 55(0.3)5(0.7)0=0.1323+0.02835+0.00243=0.163(2) P{X≥3}=1‒P{X=0}‒P{X=1}‒P{X=2}=1‒C07(0.3)0(0.7)7‒C17(0.3)1(0.7)6‒C27(0.3)2(0.7)5=1‒0.0824‒0.2471‒0.3177=0.3538.甲乙两人投篮,投中的概率分别为0.6,0.7.现各投3次,求两人投中次数相等的概率.解:设X表示各自投中的次数P{X=0}=C03(0.6)0(0.4)3∗C03(0.7)0(0.3)3=0.064∗0.027=0.002P{X=1}=C13(0.6)1(0.4)2∗C13(0.7)1(0.3)2=0.288∗0.189=0.054P{X=2}=C23(0.6)2(0.4)1∗C23(0.7)2(0.3)1=0.432∗0.441=0.191P{X=3}=C33(0.6)3(0.4)0∗C33(0.7)3(0.3)0=0.216∗0.343=0.074投中次数相等的概率= P{X=0}+P{X=1}+P{X=2}+P{X=3}=0.3219.有一繁忙的汽车站,每天有大量的汽车经过,设每辆汽车在一天的某段时间内出事故的概率为0.0001.在某天的该段时间内有1000辆汽车经过,问出事故的次数不小于2的概率是多少?(利用泊松分布定理计算)解:设X表示该段时间出事故的次数,则X~B(1000,0.0001),用泊松定理近似计算=1000*0.0001=0.1λP{X≥2}=1‒P{X=0}‒P{X=1}=1‒C01000(0.0001)0(0.9999)1000‒C11000(0.0001)1(0.9999)999=1‒e‒0.1‒0.1e‒0.1=1‒0.9048‒0.0905=0.004710.一电话交换台每分钟收到的呼唤次数服从参数为4的泊松分别,求:(1)每分钟恰有8次呼唤的概率;(2)每分钟的呼唤次数大于10的概率.解: (1) P{X=8}=P{X≥8}‒P{X≥9}=0.051134‒0.021363=0.029771(2) P{X>10}=P{X≥11}=0.002840习题2.21.求0-1分布的分布函数.解:F(x)={0, x<0q, 0≤x<11,x≥12.设离散型随机变量X的分布律为:3 OF 18X -123P0.250.50.25求X 的分布函数,以及概率,.P {1.5<X ≤2.5} P {X ≥0.5}解:當x <‒1時,F (x )=P {X ≤x }=0;當‒1≤x <2時,F (x )=P {X ≤x }=P {X =‒1}=0.25;當2≤x <3時,F (x )=P {X ≤x }=P {X =‒1}+P {X =2}=0.25+0.5=0.75;當x ≥3時,F (x )=P {X ≤x }=P {X =‒1}+P {X =2}+P {X =3}=0.25+0.5+0.25=1;则X 的分布函数F(x)为:F (x )={0, x <‒10.25, ‒1≤x <20.75, 2≤x <31, x ≥3P {1.5<X ≤2.5}=F (2.5)‒F (1.5)=0.75‒0.25=0.5 P {X ≥0.5}=1‒F (0.5)=1‒0.25=0.753.设F 1(x),F 2(x)分别为随机变量X 1和X 2的分布函数,且F(x)=a F 1(x)-bF 2(x)也是某一随机变量的分布函数,证明a-b=1.证: F (+∞)=aF (+∞)‒bF (+∞)=1,即a ‒b =14.如下4个函数,哪个是随机变量的分布函数:(1)F 1(x )={0, x <‒212, ‒2≤x <02, x ≥0(2)F 2(x )={0, x <0sinx, 0≤x <π1, x ≥π(3)F 3(x )={0, x <0sinx, 0≤x <π21, x ≥π2(4)F 4(x )={0, x <0x +13, 0<x <121, x ≥125.设随机变量X 的分布函数为F(x) =a+b arctanx ,‒∞<x <+∞,求(1)常数a,b;(2) P {‒1<X ≤1}解: (1)由分布函数的基本性质 得:F (‒∞)=0,F (+∞)=1{a +b ∗(‒π2)=0a +b ∗(π2)=1of backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, fullof humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy5 OF 18解之a=, b=121π(2)P {‒1<X ≤1}=F (1)‒F (‒1)=a +b ∗π4‒(a +b ∗‒π4)=b ∗π2=12(将x=1带入F(x) =a+b arctanx )注: arctan 为反正切函数,值域(), arctan1=‒π2,π2 π46.设随机变量X 的分布函数为F (x )={0, x <1lnx, 1≤x <e1, x ≥e求P {X ≤2},P {0<X ≤3},P {2<X ≤2.5}解: 注: P {X ≤2}=F(2)=ln2 F(x)=P {X ≤x }P {0<X ≤3}=F (3)‒F (0)=1‒0=1;P {2<X ≤2.5}=F (2.5)‒F (2)=ln2.5‒ln2=ln2.52=ln1.25习题2.31.设随机变量X 的概率密度为:f (x )={acosx, |x |≤π20, 其他.求: (1)常数a; (2);(3)X 的分布函数F(x).P {0<X <π4}解:(1)由概率密度的性质∫+∞‒∞f (x )dx =1,∫π2‒π2acosxdx =a sinx |π2‒π2=asin π2‒asin (‒π2)=asin π2+asin π2=a +a =1A =12(2)P {0<X <π4}=(12)sin(π4)‒(12)sin (0)=12∗22+12∗0=24一些常用特殊角的三角函数值正弦余弦正切余切0010不存在π/61/2√3/2√3/3√3π/4√2/2√2/211of backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, full of humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy(3)X 的概率分布为:F (x )={0, x <‒π212(1+sinx ), ‒π2≤x <π21, x ≥π2 2.设随机变量X 的概率密度为f (x )=ae ‒|x |, ‒∞<x <+∞,求: (1)常数a; (2); (3)X 的分布函数. P {0≤X ≤1}解:(1),即a=∫+∞‒∞f(x)dx =∫0‒∞ae x dx +∫+∞ae ‒x dx =a +a =112(2)P {0≤X ≤1}=F (1)‒F (0)=12(1‒e ‒1)(3)X 的分布函数F (x )={12e x, x ≤01‒12e ‒x, x >03.求下列分布函数所对应的概率密度:(1)F 1(x )=12+1πarctanx , ‒∞<x <+∞;解:(柯西分布)f 1(x )=1π(1+x 2)(2)F 2(x )={1‒e ‒x 22, x >00, x ≤0π/3√3/21/2√3√3/3π/210不存在0π-1不存在7 OF 18解:(指数分布) f 2(x )={x e ‒x 22, x >00, x ≤0(3)F 3(x )={0, x <0sinx , 0≤ x ≤π21, x >π2解: (均匀分布)f 3(x )={cosx , 0≤ x ≤π20, 其他4.设随机变量X 的概率密度为f (x )={x, 0≤x <12‒x, 1≤ x <20, 其他.求: (1); (2)P {X ≥12} P {12<X <32}.解:(1)P {X ≥12}=1‒F (12)=1‒1222=1‒18=78(2)(2)P {12<X <32}=F(32)‒F(12)=(2∗32‒1‒3222)‒(3222)=345.设K 在(0,5)上服从均匀分布,求方程(利用二次式的判别式)4x 2+4Kx +K +2=0有实根的概率.解: K~U(0,5)f (K )={15 , 0≤x ≤50, 其他方程式有实数根,则Δ≥0,即(4K)2‒4∗4∗(K +2)=16K 2‒16(K +2)≥02≤K ≤‒1故方程有实根的概率为:P {K ≤‒1}+P {K ≥2}=∫5215dx =0.66.设X ~ U(2,5),现在对X 进行3次独立观测,求至少有两次观测值大于3的概率.解:P {K >3}=1‒F (3)=1‒3‒25‒2=23至少有两次观测值大于3的概率为:C 23(23)2(13)1+C 33(23)3(13)0=20277.设修理某机器所用的时间X 服从参数为λ=0.5(小时)指数分布,求在机器出现故障时,在一小时内可以修好的概率.解: P {X ≤1}=F (1)=1‒e‒0.58.设顾客在某银行的窗口等待服务的时间X(以分计)服从参数为λ=的指数分布,某顾客在窗口等待159 OF 18服务,若超过10分钟,他就离开.他一个月要到银行5次,以Y 表示他未等到服务而离开窗口的次数.写出Y 的分布律,并求P {Y ≥1}.解:“未等到服务而离开的概率”为P {X ≥10}=1‒F (10)=1‒(1‒e‒15∗10)=e ‒2P {Y =k }=C k 5(e ‒2)k(1‒e ‒2)5‒k , (k =0,1,2,3,4,5)Y 的分布律:Y 012345P0.4840.3780.1180.0180.0010.00004P {Y ≥1}=1‒P {Y =0}=1‒0.484=0.5169.设X ~ N(3,),求:22(1);P {2<X ≤5}, P {‒4<X ≤10}, P {|X |>2}, P {X >3}(2).常数c,使P {X >c }=P {X ≤c }解: (1)P {2<X ≤5}=Φ(5‒32)‒Φ(2‒32)=Φ(1)‒[1‒Φ(12)]=0.8413‒(1‒0.6915)=0.5328P {‒4<X ≤10}=Φ(10‒32)‒Φ(‒4‒32)=Φ(3.5)‒[1‒Φ(3.5)]=0.9998‒0.0002=0.9996 P {|X |>2}= 1‒P {‒2≤X ≤2}=1‒[Φ(2‒32)‒Φ(‒2‒32)]=1‒(0.3085‒0.0062)=0.6977P {X >3}= P {X ≥3}=1‒Φ(3‒32)=1‒Φ(0)=1‒0.5=0.5(2)P {X >c }=P {X ≤c }P {X >c }=1‒P {X ≥c }P {X >c }+P {X ≥c }=1Φ(c ‒32)+Φ(c ‒32)=1Φ(c ‒32)=0.5经查表,即C=3c ‒32=010.设X ~ N(0,1),设x 满足P {|X |>x }<0.1.求x 的取值范围.解:P {|X |>x }<0.12[1‒Φ(x )]<0.1‒Φ(x )<‒1920Φ(x )≥1920Φ(x )≥0.95经查表当 1.65时x ≥Φ(x )≥0.95即 1.65时x ≥P {|X |>x }<0.111.X ~ N(10,),求:22(1)P {7<X ≤15};(2)常数d,使P {|X ‒10|<d }<0.9.解: (1)P {7<X ≤15}=Φ(15‒102)‒Φ(7‒102)=Φ(2.5)‒[1‒Φ(1.5)]=0.9938‒0.0668=0.927(2)P {|X ‒10|<d }=P {10‒d <X <10+d }<0.9=Φ(10+d ‒102)‒Φ(10‒d ‒102)<0.9=Φ(d2)<0.95经查表,即d=3.3d2=1.6512.某机器生产的螺栓长度X(单位:cm)服从正态分布N(10.05,),规定长度在范围10.050.12内 0.062±为合格,求一螺栓不合格的概率.解:螺栓合格的概率为:P {10.05‒0.12<X <10.05+0.12}=P {9.93<X <10.17}=Φ(10.17‒10.050.06)‒Φ(9.93‒10.050.06)=Φ(2)‒[1‒Φ(2)]=0.9772∗2‒1=0.9544螺栓不合格的概率为1-0.9544=0.045613.测量距离时产生的随机误差X(单位:m)服从正态分布N(20,).进行3次独立测量.求:402(1)至少有一次误差绝对值不超过30m 的概率;(2)只有一次误差绝对值不超过30m的概率.解:(1)绝对值不超过30m的概率为:P{‒30<X<30}=Φ(30‒2040)‒Φ(‒30‒2040)=Φ(0.25)‒[1‒Φ(1.25)]=0.4931至少有一次误差绝对值不超过30m的概率为:1−C 03(0.4931)0(1‒0.4931)3=1‒0.1302=0.8698(2)只有一次误差绝对值不超过30m的概率为:C13(0.4931)1(1‒0.4931)2=0.3801习题2.41.设X的分布律为X-2023P0.20.20.30.3求(1)的分布律.Y1=‒2X+1的分布律; (2)Y2=|X|解: (1)的可能取值为5,1,-3,-5.Y1由于P{Y1=5}=P{‒2X+1=5}=P{X=‒2}=0.2P{Y1=1}=P{‒2X+1=1}=P{X=‒2}=0.2P{Y1=‒3}=P{‒2X+1=‒3}=P{X=2}=0.3P{Y1=‒5}=P{‒2X+1=‒5}=P{X=3}=0.3从而的分布律为:Y1X-5-315Y10.30.30.20.2(2)的可能取值为0,2,3.Y2由于P{Y2=0}=P{|X|=0}=P{X=0}=0.2P{Y2=2}=P{|X|=0}=P{X=‒2}+P{X=2}=0.2+0.3=0.5P{Y2=3}=P{|X|=3}=P{X=3}=0.3从而的分布律为:Y2X023Y20.20.50.32.设X的分布律为X-1012P0.20.30.10.411 OF 18求Y=(X‒1)2的分布律.解:Y的可能取值为0,1,4.由于P{Y=0}=P{(X‒1)2=0}=P{X=1}=0.1P{Y=1}=P{(X‒1)2=1}=P{X=0}+P{X=2}=0.7P{Y=4}=P{(X‒1)2=4}=P{X=‒1}=0.2从而的分布律为:YX014Y0.10.70.23.X~U(0,1),求以下Y的概率密度:(1)Y=‒2lnX; (2)Y=3X+1; (3)Y=e x.解: (1) Y=g(x)=‒2lnX, 值域為(0,+∞),X=ℎ(y)=e‒Y2, ℎ'(y)=12e‒Y2 f Y(y)=f x(ℎ(y))| ℎ'(y)|=1∗12e‒Y2=12e‒Y2.即f Y(y)={12e‒Y2, y>0,0, y≤0(2) Y=g(x)=3X+1,值域為(‒∞,+∞), X=ℎ(y)=Y‒13, ℎ'(y)=13f Y(y)=f x(ℎ(y))| ℎ'(y)|=1∗13=13即f Y(y)={13, 1< y<4,0, 其他注: 由X~U(0,1),,当X=0时,Y=3*0+1=1; ,当X=1时,Y=3*1+1=4 Y=3X+1(3) Y=g(x)=e x, X=ℎ(y)=lny, ℎ'(y)=1yf Y(y)=f x(ℎ(y))| ℎ'(y)|=1∗1y=1y即f Y(y)={1y, 0< y<e,0, 其他注: ,当X=0时,; ,当X=1时,Y=e0=0 Y=e1=e4.设随机变量X的概率密度为f X(x)={32x2, ‒1<x<00, 其他.of backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, fullof humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy13 OF 18求以下Y 的概率密度:(1)Y=3X; (2) Y=3-X; (3)Y =X 2.解: (1) Y=g(x)=3X,X =ℎ(y )=Y 3, ℎ'(y)=13f Y (y )=f x (ℎ(y ))| ℎ'(y)|=Y 26∗13=Y218即f Y (y )={Y 218, ‒3< y <0,0, 其他(2)Y=g(x) =3-X, X=h(y) =3-Y,-1ℎ'(y)=f Y (y )=f x (ℎ(y ))| ℎ'(y)|=32∗(3‒Y)2+1=3(3‒Y)22即f Y (y )={3(3‒Y)22, 3< y <4,0, 其他(3), X=h(y)=,Y =g(x)=X 2Y ℎ'(y)=12Y,即f Y (y )=f x (ℎ(y ))| ℎ'(y)|=3Y 22∗1 2Y=3Y4f Y (y )={3Y4, 0< y <1,0, 其他5.设X 服从参数为λ=1的指数分布,求以下Y 的概率密度:(1)Y=2X+1; (2)(3) Y =e x; Y =X 2.解: (1) Y=g(x)=2X+1,X =ℎ(y )=Y ‒12, ℎ'(y )=12X 的概率密度为:f X (x )={λe ‒λx, x >0,0, x ≤0f Y (y )=f x (ℎ(y ))| ℎ'(y)|=λe ‒λ∗Y ‒12∗12=12e ‒Y ‒12即f Y (y )={12e ‒Y ‒12, y >00, 其他(2)Y =g (x )=e x , X =ℎ(y )=lnY,ℎ'(y )= 1Y注意是绝对值 ℎ'(y)of backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, full of humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happyf Y (y )=f x (ℎ(y ))| ℎ'(y)|=e‒lnY∗1Y =1e lnY ∗1Y =1Y ∗1Y =1Y 2即f Y (y )={1Y2, y >10, 其他(3)Y =g (x )=X 2,X =ℎ(y )=Y , ℎ'(y )=12Y,,f Y (y )=f x (ℎ(y ))| ℎ'(y)|=e ‒Y∗12Y=12Ye ‒Y即f Y (y )={12Ye ‒Y, y >00, 其他6.X~N(0,1),求以下Y 的概率密度:(1) Y =|X |; (2)Y =2X 2+1解: (1) Y =g (x )=|X |, X =ℎ(y )=±Y, ℎ'(y )=1f X (x )=12πσe‒(x ‒μ)22σ2‒∞<x <+∞当X=+Y 时:f Y (y )=f x (ℎ(y ))| ℎ'(y)|=12πe‒y 22当X=-Y 时: f Y (y )=f x (ℎ(y ))| ℎ'(y)|=12πe ‒y 22故f Y (y )=12πe ‒y 22+12πe‒y 22=22πe ‒y 22=42πe‒y 22=2πe ‒y 22f Y (y )={2πe ‒y 22, y >00, y ≤0(2)Y =g (x )=2X 2+1, X =ℎ(y )=Y ‒12,ℎ'(y )=12Y ‒12永远大于0.e x 当x>0是,>1e xof backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, fullof humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy15 OF 18f Y (y )=f x (ℎ(y ))| ℎ'(y)|=12πe‒(Y ‒12)22∗12Y ‒12=12π(y ‒1)e‒y ‒14即f Y (y )={12π(y ‒1)e ‒y ‒14, y >10, y ≤1自测题一,选择题1,设一批产品共有1000件,其中有50件次品,从中随机地,有放回地抽取500件产品,X 表示抽到次品的件数,则P{X=3}= C .A. B.C. D.C 350C 497950C 5001000A 350A 497950A 5001000C 3500(0.05)3(0.95)497 35002.设随机变量X~B(4,0.2),则P{X>3}= A .A. 0.0016B. 0.0272C. 0.4096D. 0.8192解:P{X>3}= P{X=4}= (二项分布)C 44(0.2)4(1‒0.2)03.设随机变量X 的分布函数为F(x),下列结论中不一定成立的是D .A. B. C. D. F(x) 为连续函数F (+∞)=1 F (‒∞)=00≤F (x )≤14.下列各函数中是随机变量分布函数的为 B .A. B.F 1(x )=11+x 2, ‒∞<x <+∞F 2(x )={0, x ≤0x 1+x , x >0C.D.F 3(x )=e ‒x, ‒∞<x <+∞F 4(x )=34+12πarctanx, ‒∞<x <+∞5.设随机变量X 的概率密度为 则常数a= A .f (x )={a x 2, x >100, x ≤10A. -10B.C.D. 10解: F(x) =‒15001500∫+∞‒∞a x2dx =‒ax =16.如果函数是某连续型随机变量X 的概率密度,则区间[a,b]可以是 C f (x )={x, a<x <b0, 其他A. [0, 1]B. [0, 2]C. D. [1, 2][0,2]不晓得为何课后答案为Dof backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, fullof humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy7.设随机变量X 的取值范围是[-1,1],以下函数可以作为X 的概率密度的是 A A. B. {12, ‒1< x <10, 其他{2, ‒1< x <10, 其他C.D. {x, ‒1< x <10, 其他{x 2, ‒1< x <10, 其他8.设连续型随机变量X 的概率密度为 则= B .f (x )={x2, 0< x <20, 其他P{‒1≤ X ≤1}A. 0 B. 0.25 C. 0.5 D. 1解:P {‒1≤ X ≤1}=∫1‒1x2dx =x 24|1‒1=149.设随机变量X~U(2,4),则= A . (需在区间2,4内)P{3< x <4}A. B. P{2.25< x <3.25}P{1.5< x <2.5}C. D. P{3.5< x <4.5}P{4.5< x <5.5}10. 设随机变量X 的概率密度为 则X~ A .f (x )=122πe ‒(x ‒1)28A. N (-1, 2)B. N (-1, 4)C. N (-1, 8)D. N (-1, 16)11.已知随机变量X 的概率密度为fx(x),令Y=-2X,则Y 的概率密度fy(y)为 D .A.B.C.D. 2f X (‒2y)f X (‒y2)12f X(‒y2)12f X (y 2)二,填空题1.已知随机变量X 的分布律为X 12345P2a0.10.3a0.3则常数a= 0.1 .解:2a+0.1+0.3+a+0.3=12.设随机变量X 的分布律为X 123P162636记X 的分布函数为F(x)则F(2)=.解: 1216+263.抛硬币5次,记其中正面向上的次数为X,则=.P{ X ≤4}3132解:P { X ≤4}=1‒P { X =5}=1‒C 55(12)5(12)自己算的结果是12f X(‒y2)17 OF 184.设X 服从参数为λ(λ>0)的泊松分布,且,则λ= 2 .P { X =0}=12P { X =2}解:分别将.P { X =0},P { X =2}帶入P k =P { X =k }=λk k!e ‒λ5.设随机变量X 的分布函数为F (x )={0, x <a0.4, a ≤x <b1, x ≥b其中0<a<b,则= 0.4.P {a2<X <a +b 2}解:P { a 2<X <a +b 2}=F (a +b 2)‒F (a 2)=0.4‒0=0.46.设X 为连续型随机变量,c 是一个常数,则= 0.P { X =c }7. 设连续型随机变量X 的分布函数为F (x )={13e x, x <013(x +1), 0≤x <21, x ≥2则X 的概率密度为f(x),则当x<0是f(x)=.13e x 8. 设连续型随机变量X 的分布函数为其中概率密度为f(x),F (x )={1‒e ‒2x , x >00, x ≤0则f(1)= .2e ‒29. 设连续型随机变量X 的概率密度为其中a>0.要使,则常数a=f (x )={12a, ‒a < x <a 0, 其他P { X >1}=13 3 .解:P { X >1}=1‒P { X ≤1}=13,P { X ≤1}=23=12a10.设随机变量X~N(0,1),为其分布函数,则= 1 .Φ(x)Φ(x )+Φ(‒x)11.设X~N ,其分布函数为为标准正态分布函数,则F(x)与之间的关系是(μ,σ2)F (x ),Φ(x)Φ(x)=.F (x )Φ(x ‒μσ)12.设X~N(2,4),则= 0.5 .P { X ≤2}13.设X~N(5,9),已知标准正态分布函数值,为使,则Φ(0.5)=0.6915P { X <a }<0.6915常数a< 6.5. 解:, F (a )=Φ(a ‒μσ)=a ‒53a ‒53<0.514. 设X~N(0,1),则Y=2X+1的概率密度= .f Y (y )122πe‒(Y ‒1)28解:Y =g (x )=2X +1, X =ℎ(y )=Y ‒12,ℎ'(y )=12f Y (y )=f x (ℎ(y ))| ℎ'(y)|=12πe‒(Y ‒12)22∗12=122πe‒(Y ‒1)28三.袋中有2个白球3个红球,现从袋中随机地抽取2个球,以X 表示取到红球的数,求X 的分布律.解: X=0,1,2当X=0时,P { X =0}=C 03∗C 22C 25=110当X=1时,P { X =1}=C 13∗C 12C 25=610当X=2时,P { X =2}=C 23∗C 02C 25=310X 的分布律为:X 012P110610310四.设X 的概率密度为求: (1)X 的分布函数F(x);(2).f (x )={|x|, ‒1≤ x ≤10, 其他 P { X <0.5},P { X >‒0.5}解: (1)当x <-1时. F(x)=0;;当‒1≤x <0时,F(x)=∫x‒1‒x dx =‒x 22|x ‒1=12‒x 22当0≤x <1时,F (x )=1‒ 1∫xx dx =1‒x 22|1x =12+x 22当x ≥1时. F(x)=1F (X )={0, X <‒112‒x22, ‒1≤X <012+x22, 0≤X <11, X ≥1(2)P { X <0.5}=F (0.5)=12+0.522=58;P { X >‒0.5}=1‒F (‒0.5)=1‒(12‒0.522)=58五.已知某种类型电子组件的寿命X(单位:小时)服从指数分布,它的概率密度为f (x )={12000e ‒x 2000, x >00, x ≤0We will continue to improve the company's internal control system, and steady improvement in ability to manage and control, optimize business processes, to ensure smooth processes, responsibilities in place; to further strengthen internal controls, play a control post independent oversight role of evaluation complying with third-party responsibility; to actively make use of internal audit tools detect potential management, streamline, standardize related transactions, strengthening operations in accordance with law. Deepening the information management to ensure full communication "zero resistance". To constantly perfect ERP, and BFS++, and PI, and MIS, and SCM, information system based construction, full integration information system, achieved information resources shared; to expand Portal system application of breadth and depth, play information system on enterprise of Assistant role; to perfect daily run maintenance operation of records, promote problem reasons analysis and system handover; to strengthening BFS++, and ERP, and SCM, technology application of training, improve employees application information system of capacity and level. Humanistic care to ensure "zero." To strengthening Humanities care,continues to foster company wind clear, and gas are, and heart Shun of culture atmosphere; strengthening love helped trapped, care difficult employees; carried out style activities, rich employees life; strengthening health and labour protection, organization career health medical, control career against; continues to implementation psychological warning prevention system, training employees health of character, and stable of mood and enterprising of attitude, created friendly fraternity of Humanities environment. To strengthen risk management, ensure that the business of "zero risk". To strengthened business plans management, will business business plans cover to all level, ensure the business can control in control; to close concern financial, and coal electric linkage, and energy-saving scheduling, national policy trends, strengthening track, active should; to implementation State-owned assets method, further specification business financial management; to perfect risk tube control system, achieved risk recognition, and measure, and assessment, and report, and control feedback of closed ring management, improve risk prevention capacity. To further standardize trading, and strive to achieve "according to law, standardize and fair." Innovation of performance management, to ensure that potential employees "zero fly". To strengthen performance management, process control, enhance employee evaluation and levels of effective communication to improve performance management. To further quantify and refine employee standards ... Work, full play party, and branch, and members in "five type Enterprise" construction in the of core role, and fighting fortress role and pioneer model role; to continues to strengthening "four good" leadership construction, full play levels cadres in enterprise development in theof backbone backbone role; to full strengthening members youth work, full play youth employees in company development in the of force role; to improve independent Commission against corruption work level, strengthening on enterprise business key link of effectiveness monitored. , And maintain stability. To further strengthen publicity and education, improve the overall legal system. We must strengthen safety management, establish and improve the education, supervision, and evaluation as one of the traffic safety management mechanism. To conscientiously sum up the Olympic security controls, promoting integrated management to a higher level, higher standards, a higher level of development. Employees, today is lunar calendar on December 24, the ox Bell is about to ring, at this time of year, we clearly feel the pulse of the XX power generation company to flourish, to more clearly hear XX power generation companies mature and symmetry breathing. Recalling past one another across a railing, we are enthusiastic and full of confidence. Future development opportunities, we more exciting fight more spirited. Employees, let us together across 2013 full of challenges and opportunities, to create a green, low-cost operation, fullof humane care of a world-class power generation company and work hard! The occasion of the Spring Festival, my sincere wish that you and the families of the staff in the new year, good health, happy, happy19 OF 18一台仪器装有4个此种类型的电子组件,其中任意一个损坏时仪器便不能正常工作,假设4个电子组件损坏与否相互独立.试求: (1)一个此种类型电子组件能工作2000小时以上的概率;(2)一台仪器能正p 1常工作2000小时以上的概率.p 2解: (1)P 1=P {X ≥2000}=∫+∞200012000e‒x 2000dx=12000∗‒2000∗e‒x2000|+∞2000=‒e‒x 2000|+∞2000=0‒(‒e ‒1)=e ‒1(2)因4个电子组件损坏与否相互独立,故:P 2=P 14=(e ‒1)4=e ‒4当+∞带入‒x2000时变成负无穷大,e ‒∞=0。

《概率论与数理统计》第02章习题解答.docx

《概率论与数理统计》第02章习题解答.docx
=0.2(0.2 + 0.2-0.2 x 0.2) = 0.072
P{ X = 1} = P[人(瓦U瓦)U孔A ] = 0.8(0.2 + 0.2-0.04) + 0.2 x (0.8)2
= 0.416
P{X=2} =P( £%為)=(0.8)3=0.512
3、据信有20%的美国人没有任何健康保险,现任意抽查12个美国人,以X表示15人无 任何健康保险的人数(设各人是否有健康保险是相互独立的),问X服从什么分布,写出X的分布律,并求下列情况下无任何健康保险的概率
解:(1)P{X>1}=f(x)dx=j"-(4-x2)dr = (-X- — X3)
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P{y > 0.1} = 1-F(0」)=1一0.2-0.2x0」一0.6x0= 0.774

数理统计第二章参考答案

数理统计第二章参考答案

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2.10 设总体 X ~ P() , X1,…,X n 是来自总体 X 的样本,试验证:对于任何值 [0,1] , X (1 )S *2 都是 的无偏估计量。
解:
因此, X (1 )S *2 是 的无偏估计量。 2.14 设总体 X 服从泊松分布 P() , X1,…,X n 为取自总体 X 的一个样本,求参 数 2 的无偏估计量的 R-C 下界。
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2.22 已知某种元件的寿命服从指数分布 Exp() ,抽出 12 只进行寿命试验,结果 为(单位:h): 20 640 1750 50 1110 1660 640 2410 890 970 1520 750 试求 (1) 参数 和元件的平均寿命 的置信度为 90%的置信区间; (2) 元件平均寿命 的置信度为 90%的单侧置信下限及单侧置信上限; 解:
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第2章数理统计基础习题解答

第2章数理统计基础习题解答

).
1 n λ X i − 是一个统计量. ∑ n i =1 2
B.
1 n ∑ X i − E ( X ) 是一个统计量. n i =1 1 n ∑ X i − D( X ) 是一个统计量. n i =1
C. X 1 + X 2 是一个统计量.
D.
1
4. 是(
设 ( X 1 , X 2 ,L , X n ) 是来自总体 X 的样本, X 为样本平均值,则下述结论不成立的
2
8.
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2
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1 n 1 n 1 n 2 2 2 2 , , X − X S = X − X S = ( ) ( ) ∑ i ∑ i ∑ ( X i − µ )2 , 2 3 n − 1 i =1 n i =1 n − 1 i =1
2 2
28. 设 X 1 , ⋅⋅⋅, X n 为总体 X ~ B (1, p ) 的一个样本,求 E X 和 D X ,并求样本方差
( )
( )
S2 =
1 n ( X i − X ) 2 的数学期望. ∑ n − 1 i =1
解:由性质可知, E X = E ( X ) = p
( )
D ( X )= D ( X ) / n = p (1 − p ) / n
B. 每个 X i
C. ( X 1 , X 2 ,L , X n ) 是 n 维随机变量.
D. ( X 1 , X 2 ,L , X n ) 各分量相互独立且同分布.
2. 设 ( x1 , x2 , L , xn ) 是来自总体 X 的一个样本观测值,则( A
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( X 1 , X 2 , ⋅⋅⋅, X 8 )
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; 样 本 观 测 值 为 ___(3.1, 2.6, 2.8, 3.3, 2.9, 3.2, 2.4, 量
n = ____8___ ; 样 本 均 值 的 观 测 值
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1 8 1 xi = ( 3.1 + 2.6 + L + 2.5 ) = 2.8500 ∑ 8 i =1 8 1 8 2 ( xi − x ) =0.1114 ∑ 7 i =1
E( X ) =
E( X ) = λ
和方差 D ( X ) =
D( X ) / n = λ / n
.
4
19.设总体 X
N (1, 4) , ( X 1 , X 2 , X 3 ) 是来自 X 的样本,其中 S 2 为样本方差,则
2 E ( X 12 ) E ( X 2 ) E ( X 32 ) = 125
6.26
;查表
1 1 = = 0.193 F0.05 (4,5) 5.19
.
; 若 Fα (6,3) = 14.73 , 查 表 得
α=
0.025
三.应用计算题 27. 设总体 X ~ U [ a , b] ,求样本均值的期望和方差. 解:设 ( X 1 , X 2 ,L , X n ) 是来自总体 X ~ U [ a, b] 的样本,则 由 E ( X ) = ( a + b) / 2 知, E ( X ) = E ( X ) = ( a + b) / 2 由 D ( X ) = (b − a ) / 12 知, D ( X ) = D ( X ) / n = (b − a ) / 12n
2
∑( X
i =1
n
i
− 2)
2
32 服
从的分布是
χ 2 ( n)
.
23. 设样本 X 1 , X 2 , K , X n1 来自总体 X
N ( µ1 , σ 12 ) ,样本均值和样本方差分别为:
2 N (µ2 ,σ 2 ),
X=
1 n1 1 n1 −1 2 X , S = 又设样本 Y1 , Y2 , L , Yn2 来自总体 Y ( X i − X )2 , ∑ ∑ i 1 n1 i =1 n1 − 1 i =1 1 n2
2
.
17.设 ( X 1 , X 2 , L , X n ) 为总体 X ~ N ( µ , σ ) 的一个样本,则 E ( X i ) =
µ

D( X i ) =
σ2
; E ( X i2 ) =
σ 2 + µ2
.
18. 设总体服从参数为 λ 的泊松分布 X ~ P (λ ) ,求样本 ( X 1 , X 2 ,L , X n ) 均值的期望
第 2 章数理统计基础习题解答
一.选择题 1. 设 ( X 1 , X 2 ,L , X n ) 为总体 X 的样本,则不成立的是( B ).
A. 每个 X i
(i = 1,2, L , n) 与 X 有相同的分布. (i = 1,2, L , n) 是确定的数.
B. 每个 X i
C. ( X 1 , X 2 ,L , X n ) 是 n 维随机变量.
s2 =
; 样 本 二 阶 原 点 矩 的 观 测 值 为
b2 =
1 8 2 ∑ xi = 65.7600 8 = 8.2200 8 i =1
.
15. 设总体 X ~ B (1, p ) ,其中未知参数 0 < p < 1 , ( X 1 , X 2 , L , X n ) 是 X 的样本,则
P( X 1 = x1 , X 2 = x2 ,L , X n = xn ) =
2
8.
设 ( X 1 , X 2 ,L , X n ) 是来自正态总体 X ~ N ( µ , σ ) 的简单随机样本, X 为样本均
2
值,记 S12 =
1 n 1 n 1 n 2 2 2 2 , , X − X S = X − X S = ( ) ( ) ∑ i ∑ i ∑ ( X i − µ )2 , 2 3 n − 1 i =1 n i =1 n − 1 i =1
C. xi , i = 1, 2, L , n 与 X 有相同的分布.
D. xi , i = 1, 2, L , n 与 X 有相同的数学特征.
3. 已知总体 X 服从 [0, λ ] 上的均匀分布( λ 未知) X 1 , X 2 , L X n 为 X 的样本, 则( C A.
).
1 n λ X i − 是一个统计量. ∑ n i =1 2
B.
1 n ∑ X i − E ( X ) 是一个统计量. n i =1 1 n ∑ X i − D( X ) 是一个统计量. n i =1
C. X 1 + X 2 是一个统计量.
D.
1
4. 是(
设 ( X 1 , X 2 ,L , X n ) 是来自总体 X 的样本, X 为样本平均值,则下述结论不成立的
E ( S 2 ) = D( X ) = p(1 − p) .
29.在天平上重复称一重量为 a 的物品,假设各次称量结果相互独立且都服从正态分布
N (a,0.2 2 ) .若以 X n 表示 n 次称量结果的算术平均值,要使 P{| X n − a |< 0.1} ≥ 0.95 ,求
2 2
28. 设 X 1 , ⋅⋅⋅, X n 为总体 X ~ B (1, p ) 的一个样本,求 E X 和 D X ,并求样本方差(Fra bibliotek)( )
S2 =
1 n ( X i − X ) 2 的数学期望. ∑ n − 1 i =1
解:由性质可知, E X = E ( X ) = p
( )
D ( X )= D ( X ) / n = p (1 − p ) / n
样本均值和样本方差分别为: Y =
2 S12 σ 2 立,则 2 ⋅ 2 服从 S2 σ 1
∑ Yi , S22 =
i =1
n2
1 n2 ∑ (Yi − Y )2 ,且两个样本相互独 n2 − 1 i =1
F (n1 − 1, n2 − 1)
分布.
查表得 z0.0495 = 24. 设 zα 为标准正态分布的上侧分位数, 查表得 α =

⎛ σ2 ⎞ N ⎜ µ, ⎟ n ⎠ ⎝
分布.
21. 设 ( X 1 , X 2 ,L , X n ) 是来自正态总体 X 从
N ( µ , σ 2 ) 的样本,则 Z =
n(X − µ)
σ

N (0,1)
分布.
则 22. 设总体 X ~ N (2,3 ) ,X 1 , X 2 , ⋅⋅⋅, X n 为 X 的一个简单样本,
1.65
; 若 zα = 2.31 ,
0.0104
.
25. 设 tα (n) 为 t 分 布 的 上 侧 分 位 数 , 则 查 表 得 t0.025 (5) =
2.5706
;若
tα (6) = 3.7074 ,查表得 α =
0.005
.
5
26. 设 Fα ( m, n) 为 F 分布的上侧分位数,则查表得 F0.05 (5, 4) = 得 F0.95 (5, 4) =
p i=1
∑ xi
n
n − xi (1 − p ) ∑ i =1
n
.
16 . 设 ( X 1 , X 2 , L , X n ) 为 总 体 X 的 一 个 样 本 , 则 样 本 的
r
阶原点矩为
1 n r ∑ Xi n i =1
;样本的 r 阶中心矩为
1 n ( X i − X )r ∑ n i =1
A. 正态分布.
B. 自由度为 n 的 t 分布.
C.
D. F 分布.
12. 设 ( X 1 , L , X n ) 及 (Y1 , L , Ym ) 分别取自两个相互独立的正态总体 N ( µ1 , σ 2 ) 及
S12 N ( µ 2 , σ ) 的两个样本,其样本方差分别为 S 及 S ,则统计量 F = 2 服从 F 分布的自 S2
).
1 n 1 n 2 , X S = ∑ ( X i − X )2 ,则以下结论中错误的是( B ∑ i n i =1 n − 1 i =1
B. ( X − µ ) / σ ~ N (0,1) .
A. X 与 S 2 独立.
C. ( n − 1) S 2 / σ 2 ~
χ 2 (n − 1) .
D. n ( X − µ ) / S ~ t ( n − 1) .
C ).
A. X 与
∑(X
i =1
n
n
i
− X ) 2 独立.
B. 当 i ≠ j 时, X i 与 X j 独立.
C.
∑X
i =1
n
i

∑X
i =1
2 i
独立.
D. 当 i ≠ j 时, X i 与 X j 独立.
2
5. 样本 ( X 1 , X 2 ,L , X n ) 取自概率密度为 p ( x ) 的总体,则有( A
2 S4 =
1 n ( X i − µ ) 2 ,则服从自由度为 n − 1 的 t 分布的随机变量是( A ). ∑ n i =1
A.
X −µ . S1 n
B.
X −µ S2 / n

C.
X −µ . S3 n − 1
D.
X −µ . S4 n
9. 设 ( X 1 , X 2 ,L , X n ) 是来自正态总体 X
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