概率论八九章习题答案(3)

其中c、??0均为未知参数,自一批这种器件中随机地抽取n件进行寿命试验.设它们的失效时间依次为x1?x2???xn,求:(1)?与c的极大似然估计量;(2)?与c的矩估计量.

解 (1)似然函数为

??[??xi?c?]/?1?ei?1L??,c????n??0 ??[??xi?c?]/??1ei?1???n??0 nn当xi?c 其他 当xn?xn?1???x1?c 其他 lnL??,c???nln???x?c? ??ii?11n?lnL??,c?n??x. ??0,故lnL??,c?关于c单调增,故c又因为ML1?c??lnL??c,?n1n再令???2??xi?c??0,得?的极大似然估计量为

????i?1??ML?x?x1. (2) EX????tc?e??t?c?/?c??dt??????e???du???c

0?????EX2????t2c?e??t?c?/?dt??2???0c????22????edu?2??2c??c

???2??c??X ??c??X ??????n令?2,即2??2?2?21n2,解得 12?????(??c)????Xi?2??2c??c?n?Xini?1?i?1???1n22Xi?X2?Sn?Sn????n i?1????c?X?Sn 例9 设总体X的概率为

????x??1e??x当x?0f?x;????

??0 当x?0?其中??0是未知参数，??0是已知参数，试根据来自总体X的简单随机样本

?X1,X2,?,Xn?，求?的极大似然估计量.

解 设?x1,x2,?,xn?是样本?X1,X2,?,Xn?的观察值,参数?的似然函数为

L?????f?xi;???????i?1nn?x?ii?1n?1??e?xi?i?1n

对数似然方程有

dlnL???nn????xi?0 d??i?1解得 ??nn

i?x?i?1d2lnL???n又因为???0,于是?的极大似然估计量为 22d?????n?x?ii?1n

例10 假设初生婴儿(男孩)的体重服从正态分布,随机抽取12名新生婴儿,

测得其体重为:3100,2520,3000,3000,3600,3160,3560,3320,2880,2600,3400,2540 (单位:g).试以95%的置信度估计新生男婴的平均体重.

解 方差?2未知时,单个正态总体的期望?的置信度为1??的置信区间为

SS??X?tn?1,X?tn?1?????/2?/2??

nn??由题意可知, ??0.05,?/2?0.025,n?12,查表得t0.025?11??2.201.于是

X?1?3100?2520???2540??3057 12112S?(Xi?3057)2?375.3 ?11i?1因此,新生男婴的平均体重的置信度为95%的置信区间为?2820,3300?.

例11 某自动包装机包装洗衣粉,其重量服从正态分布,今随机抽查12袋测

得其重量(单位:g)分别为: 1001,1004,1004,1000,997,999,1004, 1000,996,1002,998,999.试求:(1)平均袋重?的点估计值;(2)方差?2的点估计值;(3)

?的置信度为95%的置信区间;(4) ?2的置信度为95%的置信区间;(5)若已知

?2?9，求?的置信度为95%的置信区间.

1n1?解 (1) ??X??Xi??1001?1004???999??1000.33. ni?11221n2?(2) ??S?(Xi?X)2?6.932. ?n?1i?1(3) ?2未知,单个正态总体的期望?的置信度为1??的置信区间为

SS??X?tn?1,X?tn?1?????/2?/2??

nn??由题意可知, ??0.05,?/2?0.025,n?12,查表得t0.05?11??2.201,于是

St?/2?n?1??1.673 n?的置信度为95%的置信区间是?998.577,1001.923?.

(4) ?未知, ?2的置信度为1??的置信区间为

??n?1?S2n?1?S2?????2?n?1?,?2?n?1???

1??/2??/2?22222

查表得??/2?n?1???0.025?11??21.92,?1??/2?n?1???0.975?11??3.816.故?的置信

度为95%的置信区间是?3.479,19.982?.

(5)已知?2?9,?的置信度为95%的置信区间为

????X?U,X?U?/2?/2??

nn??查表得U?/2?U0.025?1.96,

?nU?/2?1.697故?的置信度为95%的置信区间是

?998.553,1001.14?.

例12 已知某种木材横纹抗压力的实验值服从正态分布,对10个试件做横纹抗压力的试验数据如下:482,493,457,471,510,446,435,418,394,496(单位:kg/cm2).试以95%的置信度估计该木材的平均横纹抗压力,并指出估计误差限.

解 样本均值为

1X??482???496??457.5

10样本方差为

1nS?(Xi?X)2?35.2 ?n?1i?1由题意可知, 1???0.95,??0.05,t?/2?9??t0.025?9??2.26. 又因为该木材的平均横纹抗压力的置信度为95%的置信区间为

SS??X?tn?1,X?tn?1?????/2?/2??

nn???即 ?457.?5?故置信区间为?431.0,484.0?.

35.22.?2610,?457.?535.?2 62.?210?五、课本习题全解

8-1 EX??1?2??0???2??1?3???2?8?

?1?EX?2?8?

1n1n 用样本M1??Xi代替总体相应的矩?1,得到2?8???Xi?X,所以

ni?1ni?1???1?2?X? 838-2 EX??x0?2?1??xdx?? ??213?1?EX??

11n1n 用样本M1??Xi代替总体相应的矩?1,得到???Xi?X,所以

3ni?1ni?1??3X ?8-3 EX????0xx?e2?x2/2?2dx??2?

?1?EX??2?

?1n1n 用样本M1??Xi代替总体相应的矩?1,得到???Xi?X,所以

ni?12ni?1???2a?2b??2EX?2a, EX?1222?X

28-4

a?b??2?4a?3a?b??2?4a, DX?32

①列出矩估计式

???1?EX??1 ?222???2?EX?DX??EX???2??1②求解关于估计量的方程组得

??1??1 ?2 ??2??2??1① 出矩估计

用样本矩M1,M2代替相应的总体矩?1,?2，故得

?1?1?2?S?2 ?X?X, ??ini?11n1?8-5 ???Xi?X??1458?1395???1496??1476.2 ni?110???2?S2n1?221458?1476.2?????1496?1476.2???6198.56 ??10?x1?x2???xn8-6 似然函数为

L?x1,x2,?,xn;p??p?q?n??x1?x2???xn??pnx?1?p?n?nx?Ln?nxn?nx?1nx?1nx?nxp?1?p??p?1?p??n?nx??p

?p?x8-7 似然函数为

L?x1,x2,?,xn;p???p?1?p?i?1nxi?1xi?n?pn?1?p?? i?1n?n?lnL?nlnp?ln?1?p???xi?n?

?i?1?

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