下列各式的值
(1)9! (2)
>> factorial(9)
ans =
362880
>>nchoosek(10,3)*factorial(3)
ans =
720
(3) (4)
>> nchoosek(10,3)
ans =
120
>> nchoosek(15,12)
ans =
455
2.碰运气能否通过英语四级考试
大学英语四级考试是全面检验大学生英语水平的一种综合考试,具有一定难度。这种考试包括听力、语法结构、阅读理解、写作等。除写作占15分外,其余85道为单项选择题,每道题附有A、B、C、D四个选项。这种考试方法使个别学生产生碰运气和侥幸心理,那么,靠运气能通过英语四级考试吗?
n=85;
for i=1:n
p=nchoosek(n,i)*(1/4)^i*(3/4)^(n-i);
fprintf('p(% d)=% d.\n',i,p);
end
运行结果:
p( 1)= 6.799940e-010.
p( 2)= 9.519916e-009.
p( 3)= 8.779478e-008.
p( 4)= 5.999310e-007.
p( 5)= 3.239627e-006.
p( 6)= 1.439834e-005.
.
.
.
.
p( 79)= 2.130399e-040.
p( 80)= 5.325997e-042.
p( 81)= 1.095884e-043.
p( 82)= 1.781926e-045.
p( 83)= 2.146898e-047.
p( 84)= 1.703888e-049.
p( 85)= 6.681912e-052.
第2章 随机变量及其分布
1. 随机变量X服从参数为试验次数20,概率为0.25的二项分布。
(1)生成X的概率分布;
(2)产生18个随机数(3行6列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> binocdf(0:20,20,0.25)
ans =
Columns 1 through 8
0.0032 0.0243 0.0913 0.2252 0.4148 0.6172 0.7858 0.8982
Columns 9 through 16
0.9591 0.9861 0.9961 0.9991 0.9998 1.0000 1.0000 1.0000
Columns 17 through 21
1.0000 1.0000 1.0000 1.0000 1.0000
(2)>> binornd(20,0.25,3,6)
ans =
9 8 3 4 6 6
6 3 4 5 6 2
5 6 6 4 7 4
(3)>> binoinv(0.45,20,0.25)
ans =
5
(4)>> x=0:20;y=binopdf(x,20,0.25);plot(x,y,'.')
>> x=0:0.01:20;y=binocdf(x,20,0.25);plot(x,y)
2、随机变量X服从参数为3的泊松分布。
(1)生成X的概率分布;
(2)产生21个随机数(3行7列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> poisspdf(0:20,3)
ans =
Columns 1 through 8
0.0498 0.1494 0.2240 0.2240 0.1680 0.1008 0.0504 0.0216
Columns 9 through 16
0.0081 0.0027 0.0008 0.0002 0.0001 0.0000 0.0000 0.0000
Columns 17 through 21
0.0000 0.0000 0.0000 0.0000 0.0000
(2)>> poissrnd(3,3,7)
ans =
0 3 0 4 6 3 4
3 7 0 2 3 2 2
2 7 1 3 3 8 1
(3)>> poissinv(0.45,3)
ans =
3
(4)>> x=0:10;y=poisspdf(x,3);plot(x,y,'.')
>> x=0:0.01:10;y=poisscdf(x,3);plot(x,y)
3、随机变量X服从参数为4的指数分布。
(1)求分布函数在-2,-1,0,1,2的函数值;
(2)产生16个随机数(4行4列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> expcdf(-2:2,4)
ans =
0 0 0 0.2212 0.3935
(2)>> exprnd(4,4,4)
ans =
0.4868 7.4617 4.5061 1.9785
3.5886 7.7543 3.6473 7.1617
13.2547 2.0038 3.6019 6.6833
1.1713 5.4741 3.8057 9.4312
(3)>> expinv(0.45,4)
ans =
2.3913
(4)>> x=0:10;y=exppdf(x,3);plot(x,y)
>> x=0:0.01:10;y=expcdf(x,3);plot(x,y)
4.随机变量X服从标准正态分布。
(1)求分布函数在-2,-1,0,1,2,3,4,5的函数值;
(2)产生18个随机数(3行6列);
(3)又已知分布函数F(x)=0.45,求x;
(4)在同一个坐标系画出X的概率密度和分布函数图形。
(1)>> normpdf(-2:5,0,1)
ans =
0.0540 0.2420 0.3989 0.2420 0.0540 0.0044 0.0001 0.0000
(2)>> normrnd(0,1,3,6)
ans =
0.7425 0.1798 0.3257 0.6532 -2.0516 0.9298
1.1436 -0.9833 1.2963 -0.5051 -0.4483 0.9019
-0.9147 0.3848 1.0992 -0.4760 -1.5512 0.1383
(3)>> norminv(0.45,0,1)
ans =
-0.1257
(4)>> x=-10:0.01:10;y=normpdf(x,0,1);plot(x,y)
>> x=-10:0.01:10;y=normcdf(x,0,1);plot(x,y)
5.公共汽车车门的高度是按成年男子与车门碰头的机会在0.01以下的标准来设计的。根据统计资料,成年男子的身高X服从均值为168厘米,方差为7厘米的正态分布,那么车门的高度应该至少设计为多少厘米?
>> norminv(1-0.01,168,7)
ans =
184.2844
第3章 随机变量的数字特征
1、若 ,求 。
>> [M,V]=binostat(10,0.5)
M =
5
V =
2.5000
2、若 ,求 。
>> [M,V]=poisstat(4)
M =
4
V =
4
3、若随机变量X服从期望为1,标准差为5的正态分布,求 。
>> [M,V]=normstat(1,5)
M =
1
V =
25
4.设随机变量 的概率密度为:
,求 。
>> syms x;
>> f1=2*x+1;
>> f2=4-x;
>> Ex=int(x*f1,0,2)+int(x*f1,2,4);
>> Ex2=int(x^2*f1,0,2)+int(x^2*f2,2,4);
>> Dx=Ex2-Ex^2
Ex =
152/3
Dx =
-22876/9
5.设有标着1,2,…,9号码的9只球放在一个盒子中,从其中有放回地取出4只球,重复取100次,求所得号码之和X的数学期望及其方差。
>>x=1:9;
>>Ex=100*4*sum(x)/9
>>Dx=100*4*(sum(x.^2)/9-(sum(x)/9)^2)
Ex =
20##
Dx =
2.6667e+003
6.假定国际市场上每年对我国某种出口商品需求量 是随机变量(单位:吨),它服从[2000, 4000]上的均匀分布。如果售出一吨,可获利3万元,而积压一吨,需支付保管费及其它各种损失费用1万元,问应怎样决策才能使收益最大?
syms x y;
ita1=3*y;
ita2=3*x-(-1)*(y-x);
phix=1/2000;
Eita=simplify(int((ita2)*(phix),x,2000,y)+int(ita1*phix,x,y,4000))
dif=diff(Eita,y)
y=solve(dif)
E=eval(Eita)
Eita =
- y^2/2000 + 5*y - 20##
dif =
5 - y/1000
y =
5000
E =
10500
7.某厂生产的某种型号的细轴中任取20个,测得其直径数据如下(单位:mm):
13.26,13.63,13.13,13.47,13.40,13.56,13.35,13.56,13.38,13.20,
13.48,13.58,13.57,13.37,13.48,13.46,13.51,13.29,13.42,13.69
求以上数据的样本均值与样本方差。
>> A=[13.26 13.63 13.13 13.47 13.40 13.56 13.35 13.56 13.38 13.20 13.48 13.58 13.57 13.37 13.48 13.46 13.51 13.29 13.42 13.69];
>> Ex=mean(A)
>> Dx=var(A)
Ex =
13.4395
Dx =
0.0211
8.将一枚硬币重复掷n次,并以X,Y分别表示出现正面和反面的次数.求X和Y的相关系数。
>> syms n;
>> px=[n/2 n/2];
>> x=[0 1];
>> py=[n/2 n/2];
>> y=[0 1];
>> corrcoef(x,y)
ans =
1.0000 1.0000
1.0000 1.0000
9.设某小型水电站一天的供电量X(kWh)在[100,200]上均匀分布,而当地人们的需求量Y在[100,250]上均匀分布。设水电站每供电1kWH有利润0.2元;若需求量超过供电量,则水电站可以从电网上取得附加电量来补充,每供电1kWH有利润0.1元。求该水电站在一天内利润的数学期望。
syms x y;
ita1=0.2*y;
ita2=0.2*x-0.1*(y-x);
phix=1/100;
phiy=1/150;
Eita=simplify(int((ita2)*(phiy),x,2000,y)+int(ita1*phix,x,y,4000));
dif=diff(Eita,y);
y=solve(dif);
E=eval(Eita)
E =
27200/3
第4章 大数定理和中心极限定理
1. 在次品率为 的大批产品中,任意抽取300件产品。利用中心极限定理计算抽取的产品中次品件数在(40,60)的概率。
>> R=binornd(300,1/6,1,1000);
>> pro=sum(R>40&R<60)/1000
pro =
0.8570
2. 在天平上重复独立地称一重为a(单位:g)的物品,各次称得的结果 都服从正态分布 。若以 表示 次称得结果的算术平均值,为使
是少要称多少次?分别用切比雪夫不等式和独立同分布的中心极限定理求解.
>> syms a;
>> h=norminv(0.95,a,0.2)
h =
a + 2963104872567065/9007199254740992
>> times = 1000;
>> R = normrnd(a,0.2,times,1);
>> pro = sum (R<h)/times
3. 设个零件的重量都是随机变量,他们相互独立且服从相同的分布,其数学期望为0.5kg,均方差为0.1kg,问5000只零件的总重量超过2510kg的概率是多少?
>> R = normrnd(0.5,0.1,1,5000);
>> pro = sum (R*5000>2510)/5000
pro =
0.4972
4. 学校图书馆阅览室共有880个座位,学校共有12000名学生。已知每天晚上每个学生到阅览室去自习的概率为8%。
(1)求阅览室晚上座位不够用的概率;
(2)若要以80%的概率保证晚上去阅览室自习的学生都有座位,阅览室还需要增添多少个座位?
(1)>> R=binornd(12000,0.08,1,10000);
>> pro = sum (R>880)/10000
pro =
0.9965
(2) >> n=binoinv(0.8,12000,0.08)-880
n =
105
5. 有一批钢材,其中80%的长度不小于3m,现从钢材中随机抽出100根,试用中心极限定理求小于3m的钢材不超过30根的概率。
>> R=binornd(3,0.8,1,1000);
>> pro = sum (R<3)/1000
pro =
0.5040
6. 一生产线生产的产品成箱包装,每箱的重量是随机的。假设每箱平均重50kg,标准差为5kg,若用最大载重量为5t的汽车承运,试利用中心极限定理说明每辆车最多可以装多少箱,才能保证不超载的概率大于0.977。
>> n=norminv(0.977,50*100,5*100)
n =
5.9977e+003
7.对同一目标进行300次独立射击,设每次射击时的命中率均为0.44,试求300次射击最可能命中几次?其相应的概率是多少?试用matlab进行模拟,观察试验与理论结果的差异。
>> n=300*0.44
n =
132
>> R=binornd(300,0.44,1,1000);
>> p = sum (R<n)/1000
p =
0.4750
第二篇:成都理工大学数学实验概率论与数理统计分册习题参考答案
数学实验 题与参考答案
概率论与数理统计分册习
第1章 古典概型
1.求下列各式的值
(1)9! (2)P10 >> factorial(9) ans =
362880
3
3
>>nchoosek(10,3)*factorial(3) ans = 720
12
(3)C10 (4)C15 >> nchoosek(10,3) ans = 120
2.碰运气能否通过英语四级考试
大学英语四级考试是全面检验大学生英语水平的一种综合考试,具有一定难度。这种考试包括听力、语法结构、阅读理解、写作等。除写作占15分外,其余85道为单项选择题,每道题附有A、B、C、D四个选项。这种考试方法使个别学生产生碰运气和侥幸心理,那么,靠运气能通过英语四级考试吗? n=85;
for i=1:n
p=nchoosek(n,i)*(1/4)^i*(3/4)^(n-i); fprintf('p(% d)=% d.\n',i,p); end
>> nchoosek(15,12) ans = 455
运行结果:
p( 1)= 6.799940e-010. p( 2)= 9.519916e-009.
2
p( 3)= 8.779478e-008.
p( 4)= 5.999310e-007.
p( 5)= 3.239627e-006.
p( 6)= 1.439834e-005.
p( 7)= 5.416520e-005.
p( 8)= 1.760369e-004.
p( 9)= 5.020311e-004.
p( 10)= 1.271812e-003.
p( 11)= 2.890482e-003.
p( 12)= 5.941547e-003.
p( 13)= 1.112136e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 14)= 1.906518e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 15)= 3.008062e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 16)= 4.386757e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 17)= 5.935025e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 18)= 7.473735e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 3
and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 19)= 8.784916e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 20)= 9.663408e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 21)= 9.970183e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 22)= 9.668056e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 23)= 8.827356e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 24)= 7.601334e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 25)= 6.182418e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 4
and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 26)= 4.755706e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 27)= 3.464033e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 28)= 2.391832e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and is
only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 29)= 1.567063e-002.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 30)= 9.750612e-003.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 31)= 5.766491e-003.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 32)= 3.243651e-003.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 5
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 33)= 1.736500e-003.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 34)= 8.852745e-004.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 35)= 4.299905e-004.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 36)= 1.990697e-004.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 37)= 8.787760e-005.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 38)= 3.700110e-005.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 39)= 1.486369e-005.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 6
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 40)= 5.697747e-006.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 41)= 2.084542e-006.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 42)= 7.279352e-007.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 43)= 2.426451e-007.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 44)= 7.720525e-008.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 45)= 2.344752e-008.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 46)= 6.796382e-009.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 7
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 47)= 1.879850e-009.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 48)= 4.960716e-010.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 49)= 1.248616e-010.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 50)= 2.996678e-011.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 51)= 6.855145e-012.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 52)= 1.494070e-012.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 53)= 3.100900e-013.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 8
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 54)= 6.125235e-014.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 55)= 1.150802e-014.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 56)= 2.055003e-015.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 57)= 3.485093e-016.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 58)= 5.608195e-017.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 59)= 8.554874e-018.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 60)= 1.235704e-018.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 9
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 61)= 1.688120e-019.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 62)= 2.178220e-020.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 63)= 2.650744e-021.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 64)= 3.037310e-022.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 65)= 3.270950e-023.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 66)= 3.303990e-024.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 67)= 3.123174e-025.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 10
and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 68)= 2.755742e-026.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 69)= 2.263170e-027.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 70)= 1.724320e-028.
Warning: Result may not be exact. Coefficient is greater than 1.000000e+015 and
is only accurate to 15 digits
> In nchoosek at 66
In Untitled at 3
p( 71)= 1.214310e-029.
p( 72)= 7.870526e-031.
p( 73)= 4.672002e-032.
p( 74)= 2.525406e-033.
p( 75)= 1.234643e-034.
p( 76)= 5.415101e-036.
p( 77)= 2.109780e-037.
p( 78)= 7.212922e-039.
p( 79)= 2.130399e-040.
p( 80)= 5.325997e-042.
p( 81)= 1.095884e-043.
p( 82)= 1.781926e-045.
p( 83)= 2.146898e-047.
p( 84)= 1.703888e-049.
p( 85)= 6.681912e-052.
第2章 随机变量及其分布
1. 随机变量X服从参数为试验次数20,概率为0.25的二项分布。 11
(1)生成X的概率分布;
(2)产生18个随机数(3行6列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> binocdf(0:20,20,0.25)
ans =
Columns 1 through 8
0.0032 0.0243 0.0913 0.8982
Columns 9 through 16
0.9591 0.9861 0.9961 1.0000
Columns 17 through 21
1.0000 1.0000 1.0000
(2)>> binornd(20,0.25,3,6)
ans =
9 8 3 4 6 6 3 4 5 6 5 6 6 4 7
(3)>> binoinv(0.45,20,0.25)
0.2252 0.9991 1.0000 6 2 4 12 0.4148 0.9998 0.6172 0.7858 1.0000 1.0000 1.0000
ans =
5
(4)>> x=0:20;y=binopdf(x,20,0.25);plot(x,y,'.')
>> x=0:0.01:20;y=binocdf(x,20,0.25);plot(x,y)
2、随机变量X服从参数为3的泊松分布。
(1)生成X的概率分布;
(2)产生21个随机数(3行7列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> poisspdf(0:20,3)
ans =
Columns 1 through 8
0.0498 0.1494 0.2240 0.2240 0.0216
Columns 9 through 16
0.0081 0.0027 0.0008 0.0002 0.0000
Columns 17 through 21
0.0000 0.0000 0.0000 0.0000 13 0.1680 0.0001 0.0000 0.1008 0.0504 0.0000 0.0000
(2)>> poissrnd(3,3,7)
ans =
0 3 0 4 6 3 4 3 7 0 2 3 2 2 2 7 1 3 3 8 1
(3)>> poissinv(0.45,3)
ans =
3
(4)>> x=0:10;y=poisspdf(x,3);plot(x,y,'.')
>> x=0:0.01:10;y=poisscdf(x,3);plot(x,y)
3、随机变量X服从参数为4的指数分布。
(1)求分布函数在-2,-1,0,1,2的函数值;
(2)产生16个随机数(4行4列);
(3)又已知分布函数F(x)=0.45,求x;
(4)画出X的分布律和分布函数图形。
(1)>> expcdf(-2:2,4)
ans =
0 0 0 0.2212 0.3935
(2)>> exprnd(4,4,4)
ans =
14
0.4868 7.4617 4.5061 1.9785
3.5886 7.7543 3.6473 7.1617
13.2547 2.0038 3.6019 6.6833
1.1713 5.4741 3.8057 9.4312
(3)>> expinv(0.45,4)
ans =
2.3913
(4)>> x=0:10;y=exppdf(x,3);plot(x,y)
>> x=0:0.01:10;y=expcdf(x,3);plot(x,y)
4.随机变量X服从标准正态分布。
(1)求分布函数在-2,-1,0,1,2,3,4,5的函数值;
(2)产生18个随机数(3行6列);
(3)又已知分布函数F(x)=0.45,求x;
(4)在同一个坐标系画出X的概率密度和分布函数图形。
(1)>> normpdf(-2:5,0,1)
ans =
0.0540 0.2420 0.3989 0.2420 0.0540 0.0044 0.0000
(2)>> normrnd(0,1,3,6)
ans =
0.7425 0.1798 0.3257 0.6532 -2.0516 0.9298
1.1436 -0.9833 1.2963 -0.5051 -0.4483 0.9019 15 0.0001
-0.9147 0.3848 1.0992 -0.4760 -1.5512 0.1383
(3)>> norminv(0.45,0,1)
ans =
-0.1257
(4)>> x=-10:0.01:10;y=normpdf(x,0,1);plot(x,y)
>> x=-10:0.01:10;y=normcdf(x,0,1);plot(x,y)
5.公共汽车车门的高度是按成年男子与车门碰头的机会在0.01以下的标准来设计的。根据统计资料,成年男子的身高X服从均值为168厘米,方差为7厘米的正态分布,那么车门的高度应该至少设计为多少厘米?
>> norminv(1-0.01,168,7)
ans =
184.2844
16
第3章 随机变量的数字特征
1、若X~b(10,0.5),求E(X),D(X)。
>> [M,V]=binostat(10,0.5)
M =
5
V =
2.5000
2、若X~?(4),求E(X),D(X)。
>> [M,V]=poisstat(4)
M =
4
V =
4
3、若随机变量X服从期望为1,标准差为5的正态分布,求E(X),D(X)。 >> [M,V]=normstat(1,5)
M =
17
1
V =
25
4.设随机变量X的概率密度为:
?2x?1?f(x)??4?x
?0?
>> syms x;
>> f1=2*x+1;
>> f2=4-x; 0?x?22?x?4,求E(X),D(X)。 其他
>> Ex=int(x*f1,0,2)+int(x*f1,2,4);
>> Ex2=int(x^2*f1,0,2)+int(x^2*f2,2,4);
>> Dx=Ex2-Ex^2
Ex =
152/3
Dx =
-22876/9
5.设有标着1,2,…,9号码的9只球放在一个盒子中,从其中有放回地取出4只球,重复取100次,求所得号码之和X的数学期望及其方差。 >>x=1:9;
>>Ex=100*4*sum(x)/9
>>Dx=100*4*(sum(x.^2)/9-(sum(x)/9)^2)
Ex =
18
2000
Dx =
2.6667e+003
6.假定国际市场上每年对我国某种出口商品需求量?是随机变量(单位:吨),它服从[2000, 4000]上的均匀分布。如果售出一吨,可获利3万元,而积压一吨,需支付保管费及其它各种损失费用1万元,问应怎样决策才能使收益最大? syms x y;
ita1=3*y;
ita2=3*x-(-1)*(y-x);
phix=1/2000;
Eita=simplify(int((ita2)*(phix),x,2000,y)+int(ita1*phix,x,y,4000)) dif=diff(Eita,y)
y=solve(dif)
E=eval(Eita)
Eita =
- y^2/2000 + 5*y - 2000
dif =
5 - y/1000
y =
5000
19
E =
10500
7.某厂生产的某种型号的细轴中任取20个,测得其直径数据如下(单位:mm):
13.26,13.63,13.13,13.47,13.40,13.56,13.35,13.56,13.38,13.20, 13.48,13.58,13.57,13.37,13.48,13.46,13.51,13.29,13.42,13.69 求以上数据的样本均值与样本方差。
>> A=[13.26 13.63 13.13 13.47 13.40 13.56 13.35 13.56 13.38 13.20 13.48 13.58 13.57 13.37 13.48 13.46 13.51 13.29 13.42 13.69];
>> Ex=mean(A)
>> Dx=var(A)
Ex =
13.4395
Dx =
0.0211
8.将一枚硬币重复掷n次,并以X,Y分别表示出现正面和反面的次数.求X和Y的相关系数。
>> syms n;
>> px=[n/2 n/2];
>> x=[0 1];
>> py=[n/2 n/2];
>> y=[0 1];
>> corrcoef(x,y)
ans =
20
1.0000 1.0000
1.0000 1.0000
9.设某小型水电站一天的供电量X(kWh)在[100,200]上均匀分布,而当地人们的需求量Y在[100,250]上均匀分布。设水电站每供电1kWH有利润0.2元;若需求量超过供电量,则水电站可以从电网上取得附加电量来补充,每供电1kWH有利润0.1元。求该水电站在一天内利润的数学期望。
syms x y;
ita1=0.2*y;
ita2=0.2*x-0.1*(y-x);
phix=1/100;
phiy=1/150;
Eita=simplify(int((ita2)*(phiy),x,2000,y)+int(ita1*phix,x,y,4000)); dif=diff(Eita,y);
y=solve(dif);
E=eval(Eita)
E =
27200/3
21
第4章 大数定理和中心极限定理
1. 在次品率为的大批产品中,任意抽取300件产品。利用中心极限定理计算抽取的产品中次品件数在(40,60)的概率。
>> R=binornd(300,1/6,1,1000);
>> pro=sum(R>40&R<60)/1000
pro =
0.8570
2. 在天平上重复独立地称一重为a(单位:g)的物品,各次称得的结果Xi都服从正态分布Xi~N(a,0.22)。若以n表示n次称得结果的算术平均值,为使
P{n?a?0.1}?0.95
是少要称多少次?分别用切比雪夫不等式和独立同分布的中心极限定理求解. >> syms a;
>> h=norminv(0.95,a,0.2)
h =
a + 2963104872567065/9007199254740992
>> times = 1000;
>> R = normrnd(a,0.2,times,1);
>> pro = sum (R<h)/times
3. 设个零件的重量都是随机变量,他们相互独立且服从相同的分布,其数学期望为0.5kg,均方差为0.1kg,问5000只零件的总重量超过2510kg的概率是多少?
>> R = normrnd(0.5,0.1,1,5000);
22
>> pro = sum (R*5000>2510)/5000
pro =
0.4972
4. 学校图书馆阅览室共有880个座位,学校共有12000名学生。已知每天晚上每个学生到阅览室去自习的概率为8%。
(1)求阅览室晚上座位不够用的概率;
(2)若要以80%的概率保证晚上去阅览室自习的学生都有座位,阅览室还需要增添多少个座位?
(1)>> R=binornd(12000,0.08,1,10000);
>> pro = sum (R>880)/10000
pro =
0.9965
(2) >> n=binoinv(0.8,12000,0.08)-880
n =
105
5. 有一批钢材,其中80%的长度不小于3m,现从钢材中随机抽出100根,试用中心极限定理求小于3m的钢材不超过30根的概率。
>> R=binornd(3,0.8,1,1000);
>> pro = sum (R<3)/1000
pro =
0.5040
23
6. 一生产线生产的产品成箱包装,每箱的重量是随机的。假设每箱平均重50kg,标准差为5kg,若用最大载重量为5t的汽车承运,试利用中心极限定理说明每辆车最多可以装多少箱,才能保证不超载的概率大于0.977。
>> n=norminv(0.977,50*100,5*100)
n =
5.9977e+003
7.对同一目标进行300次独立射击,设每次射击时的命中率均为0.44,试求300次射击最可能命中几次?其相应的概率是多少?试用matlab进行模拟,观察试验与理论结果的差异。
>> n=300*0.44
n =
132
>> R=binornd(300,0.44,1,1000);
>> p = sum (R<n)/1000
p =
0.4750
24