期末习题精选含中英文对照答案
期末习题精选含中英文对照答案 前半部分 一.填空
Part 1. Blank filling 1 . 已知P(A)=
3
1, P(B)=5
1,则
(1) 当A, B 互不相容时, P(A B)= ; (2) 当A, B 相互独立时, P(A B)= .
1 . Suppose : P(A)=
3
1, P(B)=
51, then
(1) When events A & B are mutually exclusive, P(A B)= P(A)+ P(B) =
3
1+
5
1=
15
8;
(2) When events A & B are independent, P(A B)= P(A)+ P(B)- P(A)× P(B) =
3
1+
5
1-
3
1×
5
1=
15
8-
15
1=
15
7
1’ If and only if A ∩B=Φ , events A & B are mutually exclusive;
If and only if P(AB)= P(A)× P(B) or P(A │B)=P(A) , events A & B are independent.
2. 若随机变量X 的分布律见右表
则(1) 常数c= ; (2) 数学期望EX= ;
(3)方差V ar(X)= .
注 (1) 0.2+0.5+c=1,解出c=0.3 (2)
)
=(-1)×0.2+0×0.5+2×0.3=0.4
(3) 先计算E(X 2)( 为上表第一、三行对应元素乘积之和) =1×0.2+0×0.5+4×
0.3=1.4 V ar(X)= E(X 2
)-( EX)2
=1.4-0.42
=1.4-0.16=1.24
2. The probability distribution of the random variable is shown as follows:
Then (1) the constant value c= 0.3 ; (2) expected value EX= 0.4 ;
(3)variance of X V ar(X)= 1.24 .
Part 3. If random variable X~B(4,0.3)
Question : a. P(X=2)=
b. P(X>0)=
c. EX=
d. V ar(X)=
二. 射击游戏中,连续射击三次,射中一次记10分. 某射手命中率为0.7, 以X表示该射手在游戏中射中次数, 以Y 中得分则
(1) X服从什么分布?射手期望射中次数为多少?
(2)求Y的分布律和数学期望
Part 2. In shooting game, one continue to shoot 3 times . One gets 10 points for each success. A certain shooter
hits the target 7 out of 10. Use X to stand for the times of quiver and Y for his score, then
(1)What kind of distribution does X obey? How many times does the shooter expect to hit the target?
X~ B(n,p)=B(3,0.7);EX=np=3×0.7=2.1
EY=0×0.027+10×0.189+20×0.441+30×0.343=21
二’. 射击游戏中射中可以连射,但最多射击三次,射中一次记10分. 某射手命中率为0.7, 以X表示该射手在游戏中示该射手在游戏中得分则
(1) X服从什么分布?射手期望射中次数为多少?
Part 2. In shooting game, one can continue to shoot if he hit the target and he can shoot 3 times at most. One gets 10 points for each success. A certain shooter hits the target 7 out of 10. Use X to stand for the times of quiver and Y for his score, then
EX=0×0.3+1×0.21+2×0.147+3×0.343=1.533
EY=0×0.3+10×0.21+20×0.147+30×0.343=15.33
三. 随机变量X~N(1,32
) 求(1)P{X<4.9}
(2) P{-2.9≤X ≤4.9} (3) P{X>a}=0.1,求a
注:查表6.1知标准正态变量Z 满足等式P{0≤Z<1.3}=0.4
Part 3. (15 basis) the random variable X~N(1,32) Question : (1)P{X<4.9} （2） P{-2.9≤X ≤4.9}
（3） P{X>a}=0.1, a=?
PS: From form 6.1, we know that standard normal distribution Z satisfy P{0≤Z<1.3}=0.4 Answer: (1)P{X<4.9}=P{319.431-<-X } =P{Z<1.3}=0.5+0.4=0.9
(2)P{-2.9≤X ≤4.9}=P{3
19.43
13
1
9.2-≤-≤--X } =P{-1.3≤Z ≤1.3}==2×0.4=0.8
(3) P{X>a}=1.0}3
13
1{
=->-a X P ,
查表P{标准正态>1.3}=0.5-P{0≤标准正态≤1.3}=0.5-0.4=0.1,9.4,3.13
1==-∴
a a
三’总体X,E(X)=2,V ar(X)=4,抽样X 1,X 2,…X 36.求 (1)P{X <2.43} (2)P{│X -2│<0.433}
注:查表6.1知标准正态变量Z 满足等式P{0≤Z<1.3}=0.4
Part 3’ A simple random sample of X 1,X 2,…X 36 is selected from a population X with E(X)=2,V ar(X)=4. Question : (1)P{X <2.43} (2)P{│X -2│<0.433} Answer: )36
4,
2(),(~2
N n
N X =σμ
(1)P{X <2.43}=P{
3.1622
43.26
22=-<
-X }= P{Z<1.3}=0.5+0.4=0.9
(2)P{│X -2│≤0.433}=P{6
2433.06
226
2433.0≤
-≤
-X } =P{-1.3≤Z ≤1.3}==2×0.4=0.8
.三” 样本容量为100的简单随机抽样,总体的比例p=0.4, 求(1)样本比例p 的近似分布 (2)P{p <46.37%}
Part3” A simple random sample of size 100 is selected from a population with p=0.4. Question
(1) Show approximate sampling distribution of p (easy to compute) (2) What is the probability P{p <46.37%} Answer (1) p ~N(p,
)1(1p p n
-)=N(0.4,
6.04.0100
1??)=N(0.4,0.0492
)
(2) P{p <46.37%}=}049
.04
.04637.0049
.04.0{
-<
-p P =P{z<1.3}=0.5+0.4=0.9
PS: From form 6.1, we know that standard normal distribution Z satisfy P{0≤Z<1.3}=0.4
四’考察样本数据值27 25 20 34 20 30 25 35 20 15 .
1) 计算样本均值、最小值、1/4分位数、中位数、3/4分位数、最大值、众数 2) 计算极差、样本方差、样本标准差、变异系数
3) 计算35百分位数
解:排序①15 ②20 ③20 ④20 ⑤25 ⑥25 ⑦27 ⑧30 ⑨34 ⑩35 2) 3) 1) 样本均值∑==
n
i i
x
n
x 1
1=25.1、 最小值15、
1/4分位数20(i=(25/100)×10=2.5取整进一得i=3,故25百分位数为3号数据20)、
中位数25(样本容量偶数取(⑤25+ ⑥25)/2=25)、
3/4分位数30(i=(75/100)×10=7.5取整进一得i=8,故25百分位数为8号数据30)、最大值35、众数20 2) 极差x
x
i
n
i i
n
i Range ≤≤≤≤-=11min
max
=20、
样本方差()∑-=-=
n
i x
x s i n 1
2
2
1
1=42.767、
样本标准差s
s 2
=
=6.5396、变异系数=?=
100x
s 26.054
3) i=(35/100)×10=3.5取整进一得i=4,所以35百分位数为4号数据20
Part 4’ Consider the sample with data values of 27 25 20 34 20 30 25 35 20 and 15. Compute the statistics
1) Sample mean ，min, first quartile(Q1)，median, third quartile(Q3),max, mode 2) Range,sample variance,standard deviation,coefficient of variation
3) 35th percentile
Solution: Data in order: ①15 ②20 ③20 ④20 ⑤25 ⑥25 ⑦27 ⑧30 ⑨34 ⑩3
1) Sample mean=25.1，min=15, first quartile(Q1)=20，median=25, third quartile(Q3)=30,max=35, mode=22) Range=20,sample variance=42.767,standard deviation=6.5396,coefficient of variation=26.054 3) i=(35/100)×10=3.5,i=[3.5]+1=4, 35th percentile=④20
后半部分
五.
设置信水平α=0.05,总体标准差估计为9.65,问用样本均值
估计未知总体均值μ要求误差小于2,样本容量应为多少? 解: 代入公式
96.12/05.0=z 即可
取9043
.892/2
65
.996
.1)(2
2
2
2
2
2
==?=
?=n n E z σα
Part 5.
Suppose there is a 95% confidence interval with std.dev of population σ=9.65,
We want to estimate population mean with sample mean and the margin of error is less then 2, please give the sample size. Answer:
五’设置信水平α=0.05,用样本比例p =35%估计未知总体比例p,要求误差小于5 %, 问样本容量n 为多少? 解:
Part 5’ Suppose the confidence level α=0.05, We want to estimate the proportion p of a population with sample p and the margin of error is less then 5%, please give the sample size.n
Answer:
96
.12
/05.0=z
90
43
.892/2
65
.996
.1)(2
2
2
2
2
2
==?=
?=
n n E
z σ
α96.12/05.0=z 350
6
.34965
.035.0)
1(2/05
.096
.1)
(2
2
2
2
==??=
-??=
n p p n E
z α96
.12
/05.0=z 3506
.34965
.035.0)
1(2/05
.096
.1)
(2
2
2
2
==??=
-??=
n p p n E
z α
假设检验四步骤
1)根据问题提出H0和H1:
问题μ=μ0? H0: μ=μ0 (H1: μ≠μ0) 问题μ>μ0? H1: μ>μ0 (H0:μ
≤μ0)
问题μ<μ0? H1: μ<μ
0 (H0:μ≥μ0)
2)根据已知条件构造合适的统计量: σ已知Z 统计量n
x z /
σμ
-
=
σ未知t 统计量n
s x t /
μ
-=
比例抽样z 统计量n
p z p
p
p /
)1(0
-
-
=
3)拒绝域: 问题μ=μ0?双侧,分界线±)(2
统计量z Z
α
或))(1(2
统计量t n t -±α
问题μ<μ0?左侧,(-∞,-Z α)(z 统计量)或(-∞,-t α(n-1))(t 统计量) 问题μ>μ0?右侧,(Z α,+∞) (z 统计量)或(t α(n-1), +∞) (t 统计量)
4) 由2)的统计量值结合3)拒绝域下结论
六.某车间用一台包装机包装糖. 包得的袋装糖重是一个随机变量(记为X)，它服从正态分布. 当机器正常时，其均值为0.5公斤， 某日开工为检验包装机是否正常，随机地抽取它所包装的糖9袋，称得净重为(公斤)： 0.49 0.50 0.47 0.51 0.49 0.46 0.47 0.51 0.49,(设袋重的标准差为0.015) (1)袋重的均值是0.5公斤吗?(α=0.05) (2)给出袋重均值的95%置信区间 PS: From form 6.1, Z 0.025=1.96,
Z 0.05=1.645,Fromform8.3,t 0.05(8)=1.86,t 0.05(9)=1.83,t 0.025(8)=2.31,t 0.025(9)=2.26
解(1)1)H0:μ=0.5, H1:μ≠0.5;
2)z=
23
/015.05.049.0/5
.0-=-=-n
x σ
3)α=0.05, 从表6.1知 Z 0.025=1.96,拒绝域(-∞,-1.96)∪(1.96,+∞) 4)-2∈拒绝域 所以拒绝H0认为μ≠0.5
(2) 袋重均值的95%置信区间 3
015.096.149.096
.1?
±=±n
x σ,or (0.4802 0.4998)
六.A workshop is doing sugar packaging with a packaging machine. Weight of every sugar
pack is a random variable, marked as X, which obeys normal distribution. When the machine works normally, the mean of all packs is 0.5kg. In order to test if the machine works well, pick up 9 packs randomly, which weighs 0.49 0.50 0.47 0.51 0.49 0.46 0.47 0.51 0.49, (suppose std.dev. of package bag weighs 0.015)
(1)Is the mean of the packs 0.5kg? (α=0.05)
(2)Work out 95% confidence interval for the mean of pack weights. 解:H0:μ=0.5, H1:μ≠0.5;
z=
23
/015.05.049.0/5
.0-=-=-n
x σ
α=0.05,From form 6.1,we know that Z 0.025=1.96, rejected area (-∞,-1.96)∪(1.96,+∞)
-2∈rejected area,we reject H0, Conclusion:μ≠0.5 95% confidence interval for the mean of pack weights
3
015.096.149.02
05.0?
±=±
n
x Z
σ
,or (0.4802 0.4998)
PS: From form 6.1,
Z 0.025=1.96,Z 0.05=1.645,Fromform8.3,t 0.05(8)=1.86,t 0.05(9)=1.83,t 0.025(8)=2.31,t 0.025(9)=2.26
六’.某车间用一台包装机包装糖. 包得的袋装糖重是一个随机变量(记为X)，它服从正态分布. 某日开工检验包装机，随机地抽取它所包装的糖9袋，称得净重为(公斤)： 0.49
0.51 0.53 0.56 0.49 0.59 0.52 0.54 0.57
(1)袋重的均值大于0.51公斤吗?(α=0.05) (2)给出袋重均值的下方有界的95%置信区间 提示:由样本计算得样本均值x =0.53，样本标准差s=0.04
Part 6 ’ A workshop is doing sugar packaging with a packaging machine. Weight of every sugar pack is a random variable, marked as X, which obeys normal distribution. In order to
test if the machine works well, pick up 9 packs randomly, which weighs 0.49 0.51 0.53 0.56
0.49 0.59 0.52 0.54 0.57, ( sample mean x =0.53, sample std.dev. s=0.04)
(1)Is the mean of the packs >0.51kg? (α=0.05)
(2)Work out 95% confidence interval (low bound only) for the mean of pack weights. PS: From form 6.1, Z 0.025=1.96,
Z 0.05=1.645,Fromform8.3,t 0.05(8)=1.86,t 0.05(9)=1.83,t 0.025(8)=2.31,t 0.025(9)=2.26 答案(1) 1°H0:μ≤0.51 H1:μ>0.51
2°5.19
/
04.051.053.0/
=-=
-=
n
s x t μ
3°右侧拒绝域(t α(n-1), +∞)=(t 0.05(8), +∞)=(1.86, +∞) 4°1.5?(1.86, +∞),故接受H0,认为μ≤0.51 (2) 5052.03
04.086.153.0)1(=?
-=?
--n
s n x t α
下方有界95%置信区间为(0.5052, +∞)
七.填空:
表为单总体假设检验的若干种情况和相应代号
Part 7 .Fill in the blanks:
The below form shows some conditions and their no. in accordance for one sample
Take the given question and its terms in consideration, use the correct no. in above form to fill in the form below
八:以下是单因素方差分析用软件分析的输出结果
The ANOV A Procedure
Dependent V ariable: y
Source DF Sum of Squares Mean Square F V alue Pr > F Model 2 0.00105333 0.0005266732.92<.0001
Error 120.00019200 0.00001600
Corrected Total 14 0.00124533
(1)在上面划线处填空
(2)问因素对指标影响显著吗?为什么?答:高度显著;Pr<0.0001
Part 8: Below output results for one way anova is analysed by software
The ANOV A Procedure
Dependent V ariable: y
Source DF Sum of Squares Mean Square F V alue Pr > F Model 2 0.00105333 0.0005266732.92<.0001
Error 120.00019200 0.00001600
Corrected Total 14 0.00124533
(3)Give answers on the above lines
(4)Does the factor have great effect on depedent variable and why?
Answer: obviously affect;Pr<0.0001
九.给以下自变量(indepedent variable)x与因变量(depedent variable)y的数据表,求线性回归方程(linear regression equation).
b=20.6/10=2.06 a=y-b*x=2.02-2.06×1=-0.04
回归方程为y=-0.04+2.06x
Part 9. According to the below data sheet which shows indepedent variable x & depedent
b=20.6/10=2.06 a=y-b*x=2.02-2.06×1=-0.04
linear regression equation is y=-0.04+2.06x