python分类频数图 python 频数分布

《概率论与数理统计》作业一，python画频率分布表

• 5.1

• 2：
• 5：
• 6：

• 5.2

• 2：
• 3:

• 5:

• 5.3

• 3:
• 4:
• 5：
• 8:
• 10:
• 13:
• 24:
• 28:

• (1)
• (2)
• (3)

• 频率分布表画图函数(按照分割区间大小/按照分组

• (1)按照分组数
• （2）按照分割区间大小

5.1

2：

总体：全体成年男子的抽烟情况
样本：50个同学调查到的全部5000名男子
总体分布：Bernoulli分布

5：

总体：某场生产的所有电容器
样本：抽出的n件产品
样本分布：
假设每个样本的分布iid,且都服从指数分布
$P(X_1=x_1,X_2=x_2,...,X_n=x_n)=\Pi_{i=1}^{n} \lambda e^{-\lambda x_i }$

6：

我认为这个结论是不合理的，因为总体是所有毕业生，但是样本是返校毕业生，工资低混的不好的毕业生不太愿意返校，抽样不随机。毕业生平均工资低于5万美金。

平均工资，平均年龄等样本数据一般有偏，样本均值不适合代表平均水平。

5.2

2：

3+4+8+3+2=20

分布函数要求右连续
$F_{20}(x)=\left\{ \begin{aligned} &0 \qquad & x< 38 \\ &\frac{3}{20} & 38\leq x< 48 \\ &\frac{7}{20} & 48\leq x< 58\\ &\frac{3}{4} &58\leq x<68\\ &\frac{9}{10}&68\leq x< 78\\ &1& x\geq 78 \end{aligned} \right.$

3:

#顺序排列
import numpy as np
import pandas as pd
t2=[909,1086,1120,999,1320,1091,1071,1081,
    1130,1336,967,1572,825,914,992,1232,950,
    775,1203,1025,1096,808,1224,1044,871,1164,971,950,866,738]
t2=np.sort(t2)#排序
print(t2.shape,t2,(np.max(t2)-np.min(t2))/6)

(30,) [ 738  775  808  825  866  871  909  914  950  950  967  971  992  999
 1025 1044 1071 1081 1086 1091 1096 1120 1130 1164 1203 1224 1232 1320
 1336 1572] 139.0

#频率分布表
#取间隔为140
t22=pd.cut(t2,6, labels=[u"(737,877]",u"(877,1017]",u"(1017,1157]",u"(1157,1297]",u"(1297,1437]",u"(1437,1577]"])
t22=t22.value_counts()
t22=pd.DataFrame(t22)
t22['分组区间'] = t22.index
t22.columns = ['频数','分组区间']
t22.reset_index(drop=True, inplace=True)  
t22['组中值'] =[807,947,1087,1227,1367,1507]
t22['频率']=t22['频数']/30
##计算累计频率
ljpl=[0]
for i in t22['频率']:
    ljpl.append(i+ljpl[-1])
t22['累计频率']=ljpl[1:]
t22=t22[['分组区间','组中值','频数','频率','累计频率']]
t22

	分组区间	组中值	频数	频率	累计频率
0	(737,877]	807	6	0.200000	0.200000
1	(877,1017]	947	8	0.266667	0.466667
2	(1017,1157]	1087	9	0.300000	0.766667
3	(1157,1297]	1227	4	0.133333	0.900000
4	(1297,1437]	1367	2	0.066667	0.966667
5	(1437,1577]	1507	1	0.033333	1.000000

#画直方图
import matplotlib.pyplot as plt  
plt.rcParams['font.family'] = 'sans-serif'
plt.rcParams['font.sans-serif'] = 'SimHei'
plt.rcParams['axes.unicode_minus'] = False

plt.hist(t2, bins=6)
plt.title('第三题直方图')

Text(0.5, 1.0, '第三题直方图')

5:

t5=[5954,5022,14667,6582,6870,1840,2662,4508,
   1208,3852,618,3008,1268,1978,7963,2048,
   3077,993,353,14263,1714,11127,6926,2047,
   714,5923,6006,14267,1697,13867,4001,2280,
   1223,12579,13588,7315,4538,13304,1615,8612]
t5=np.sort(t5)
print(t5.shape,t5)

(40,) [  353   618   714   993  1208  1223  1268  1615  1697  1714  1840  1978
  2047  2048  2280  2662  3008  3077  3852  4001  4508  4538  5022  5923
  5954  6006  6582  6870  6926  7315  7963  8612 11127 12579 13304 13588
 13867 14263 14267 14667]

(14667-353)/1700

8.42

ran=[]
for i in range(10):ran.append(352+i*1700)

lable=[]
for i in range(9):
    lable.append('('+str(ran[i])+','+str(ran[i+1])+']')
lable

['(352,2052]',
 '(2052,3752]',
 '(3752,5452]',
 '(5452,7152]',
 '(7152,8852]',
 '(8852,10552]',
 '(10552,12252]',
 '(12252,13952]',
 '(13952,15652]']

t55=pd.cut(t5,ran, labels=lable)
t55=t55.value_counts()
t55=pd.DataFrame(t55)
t55['分组区间'] = t55.index
t55.columns = ['频数','分组区间']
t55.reset_index(drop=True, inplace=True)  
#组中值
zzz=[]
for i in range(9):
    zzz.append(ran[i]+1700/2)
t55['组中值'] =zzz
t55['频率']=t55['频数']/40
##计算累计频率
ljpl=[0]
for i in t55['频率']:
    ljpl.append(i+ljpl[-1])
t55['累计频率']=ljpl[1:]
t55=t55[['分组区间','组中值','频数','频率','累计频率']]
t55

	分组区间	组中值	频数	频率	累计频率
0	(352,2052]	1202.0	14	0.350	0.350
1	(2052,3752]	2902.0	4	0.100	0.450
2	(3752,5452]	4602.0	5	0.125	0.575
3	(5452,7152]	6302.0	6	0.150	0.725
4	(7152,8852]	8002.0	3	0.075	0.800
5	(8852,10552]	9702.0	0	0.000	0.800
6	(10552,12252]	11402.0	1	0.025	0.825
7	(12252,13952]	13102.0	4	0.100	0.925
8	(13952,15652]	14802.0	3	0.075	1.000

plt.hist(t5, bins=ran)
plt.title('第五题直方图')

Text(0.5, 1.0, '第五题直方图')

5.3

3:

$\bar{y}=3\bar{x}-4$

$s_y^2=\frac{1}{n-1}\sum_{i}(y_i-\bar{y})^2=\frac{1}{n-1}\sum_{i}(3x_i-4-(3\bar{x}-4))^2=\frac{1}{n-1}\sum_{i}9(x_i-\bar{x})^2=9s_x^2$

4:

pf:
$(n+1)\bar{x_{n+1}}-(n+1)\bar{x_n}=x_{n+1}-\bar{x_n}$
左右同时除以n+1即得所证

pf:
$ns_{n+1}^2-(n-1)s_{n}2=\sum_{i=1}^{{n+1}(x_i-\bar{x}_{n+1})}2-\sum_{i=1}^{{n}(x_i-\bar{x}_n)}2
=x_{n+1}^{2-2(\sum_{i=1}}{n+1}x_i \bar{x}{n+1}-\sum{i=1}^{n}x_i \bar{x}{n})+((n+1)\bar{x}{n+1}^2-n\bar{x}_n2)=x_{n+1}^{2-2[x_{n+1}\bar{x}_{n+1}-\sum_{i=1}}{n}x_i(\bar{x}{n+1}-\bar{x}{n})]+((n+1)\bar{x}{n+1}^2-n\bar{x}_n2)=x{n+1}^{2-2[x_{n+1}\bar{x}_{n+1}-\frac{n}{n+1}(x_{n+1}-\bar{x}_n)\bar{x}_n]+((n+1)\bar{x}_{n+1}}2-n\bar{x}_n^2)
$
把$\bar{x}_{n+1}$带入上一条证明中的$\bar{x}_n+\frac{1}{n+1}(x_{n+1}-\bar{x}_n)$
可得$n s_{n+1}^2-(n-1)s_{n}^2=\frac{n}{n+1}(x_{n+1}-\bar{x}_{n})^2$
两边同时除以n即为所求

remark:这道题说明随着抽样样本的增加可逐次计算样本 均值与方差

5：

pf:
$\bar{x}=\frac{1}{m+n}\sum_{i}^{m+n}x_{i}=\frac{\sum_{j=1}^{m}x_{j}^{2}+\sum_{i=1}^{n}x_{i}^{1}}{m+n}=\frac{n\bar{x}_1+m\bar{x}_2}{m+n}$

其中
$x_{j}^1$表示容量为n的样本中的样本的取值
$x_{i}^2$表示容量为m的样本中的样本的取值

pf:

$s^2=\frac{\sum_{i=1}^{n}(x_{i}^1-\bar{x} )^2+\sum_{i=1}^{m}(x_i^2-\bar{x})^2}{m+n-1}$

$=\frac{\sum_{i=1}^{n}(x_{i}^1-\frac{n\bar{x}_1+m\bar{x}_2}{m+n} )^2+\sum_{i=1}^{m}(x_i^2-\frac{n\bar{x}_1+m\bar{x}_2}{m+n}}{m+n-1}$

$=\frac{\sum_{i=1}^n(x_i^1-\bar{x}_1+\frac{m(\bar{x}_1-\bar{x}_2)^2}{m+n})^2}{m+n-1}+\frac{\sum_{i=1}^m(x_i^2-\bar{x}_2+\frac{n(\bar{x}_1-\bar{x}_2)^2}{m+n})^2}{m+n-1}$

$=\frac{(n-1)s_1^2+(m-1)s_2^2+\frac{mn(\bar{x}_1-\bar{x}_2)^2}{m+n}}{m+n-1}$

由上式记得所求。

8:

$E(\bar{x})=E(\frac{\sum_{i=1}^n x_n}{n})=0$

$Var(\bar{x})=\frac{1}{n^2}\sum_{i=1}^{n} Var(x_i)=\frac{1}{n}Var(x_i)$

$Var(x_i)=E(x_i^2)=\frac{1}{2}\int_{-1}{1}x^2 dx=\frac{1}{3}$

$Var(\bar{x})=\frac{1}{3n}$

10:

$\sum_{i<j}(x_i-x_j)^2=\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}((x_i-\bar{x})+(\bar{x}-x_j))^2=\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}(x_i-\bar{x})^2+(x_j-\bar{x})^2-2(x_ix_j+\bar{x}^2)=\frac{1}{2}\sum_{i=1}^{n}\sum_{j=1}^{n}[(x_i-\bar{x})^2+(x_j-\bar{x})^2]=n(n-1)s^2$

13:

由正态分布的再生性
$\bar{x}_1\sim N(\mu,\frac{\sigma^2}{n}),\bar{x}_2\sim N(\mu,\frac{\sigma^2}{n})$
$\bar{\mu}=\bar{x}_1-\bar{x}_2,\quad \bar{\mu}\sim N(0,\frac{2\sigma^2}{n})$
记$\phi$为标准正态分布的分布函数
解$P(|\bar{\mu}>\sigma|)\leq 0.01\rightarrow 2\phi(\frac{\sigma}{\sigma \sqrt{\frac{2}{n}}})-1$得$n\geq 14$

24:

$P(x_{(16)}>10)=1-P(x_{(16)}\leq 10)=1-P(x\leq 10)^16=0.937$
$P(x_{(1)>5})=[1-P(x\leq 5)]^{16}=0.331$

28:

(1)

pf:
$\eta\in [0,1]$
$P(\eta_{i}=t)=i\binom{n}{i}P(\eta=t)P(\eta<t)^{i-1}(1-P(\eta<t))^{n-i}$
由$P(\eta<t)=P(F(x)<t)=P(x<F^{-1}(t))=F\cdot F^{-1}(t)=t \overset{\text{F(x)连续，对t求导}}\rightarrow P(\eta=t)=1$
从而$P(\eta_{i}=t)=i\binom{n}{i} t^{i-1}(1-t)^{n-i}$
上述概率密度函数也是n个i.i.d.且服从$U[0,1]$的随机变量的次序统计量的概率密度函数。

(2)

$B(m,n)=\int_{0}^1 x^{m-1}(1-x)^{n-1}dx=\frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}$

$E(\eta_i)=n\binom{n-1}{i-1} \int_0^1 t^i(1-t)^{n-i}=i \frac{n!}{i!(n-i)!}\frac{(i)!(n-i)!}{(n+1)!}=\frac{i}{n+1}$
$Var(\eta_i)=i\frac{n!}{i!(n-i)!}int_{0}^1 t^{i-1}(1-t)^{n-i}(t-\frac{i}{n+1})^2 dt=i\frac{n!}{i!(n-i)!}[\frac{(i+1)!(n-i)!}{(n+2)!}-\frac{2i}{n+1}\frac{i!(n-1)!}{(n+1)!}+\frac{i^2}{(n+1)^2}\frac{(i-1)!(n-i)!}{n!}]=\frac{i(n-i+1)}{(n+1)^2(n+2)}$

(3)

协方差矩阵A，其中$A(1,1)=Var(\eta_i),A(2,2)=Var(\eta_j)$,从而只证明$A(1,2)=A(2,1)=cov(\eta_1,\eta_2)$
先求$\eta_1,\eta_2$的联合分布密度函数:
不妨设$i\leq j$,则$P(\eta_i=t_1,\eta_j=t_2)=\binom{n}{i-1,j-i-1,n-j}t_1^{i-1}(t_2-t_1)^{j-i-1}t_2^{j}$

$cov(\eta_1,\eta_2)$

$=E(\eta_i\eta_j)-E(\eta_i)E(\eta_j)$

$=E(\eta_i)-E(\eta_i(1-\eta_j))-E(\eta_i)E(\eta_j)$

$=\frac{i}{n+1}-\int_{0}^{1}\int_0^1 t_1(1-t_2) \cdot 2\binom{n}{i-1,j-i-1,n-j} t_1^{i-1}(t_2-t_1)^{j-i-1}(1-t_2)^{n-j} dt_1 dt_2-\frac{i}{n+1}\frac{j}{n+1}$

$=\frac{i(n+1-j)}{(n+2)(n+1)^2}$

$=\frac{a_1(1-a_2)}{n+2}$

对于上述积分:
$I=\int_{0}^1\int_0^1 \binom{n+2}{i,j-i-1,n-j+1}2 t_1^{i}(t_2-t_1)^{j-i-1}(1-t_{2})^{n-j+1}=1\\ E(\eta_1\eta_2)=\frac{i(n-j+1)}{(n+2)(n+1)}I$
关于$I$的积分:把积分对应到某种概率分布，利用概率密度函数的正则性计算积分。

频率分布表画图函数(按照分割区间大小/按照分组

(1)按照分组数

import numpy as np
import pandas as pd

def fredistable_zushu(t,n):#t是数组，n是组数
    t=np.sort(t)
    mi=np.min(t)
    ma=np.max(t)
    ran=[]
    #不需要分割区间为整数时:cut=(ma-mi)/n
    cut=int((ma-mi)/n)+1
    for i in range(n+1):
        ran.append(mi-1+i*cut)#ran.append(mi+i*cut)直接从最小值开始
    lable=[]
    for i in range(n):
        lable.append('('+str(ran[i])+','+str(ran[i+1])+']')
    t1=pd.cut(t,ran, labels=lable)
    t1=t1.value_counts()
    t1=pd.DataFrame(t1)
    t1['分组区间'] = t1.index
    t1.columns = ['频数','分组区间']
    t1.reset_index(drop=True, inplace=True)  
    #组中值
    zzz=[]
    for i in range(n):
        zzz.append(ran[i]+float(cut)/2)
    t1['组中值'] =zzz
    t1['频率']=t1['频数']/np.shape(t)[0]
    ##计算累计频率
    ljpl=[0]
    for i in t1['频率']:
        ljpl.append(i+ljpl[-1])
    t1['累计频率']=ljpl[1:]
    t1=t1[['分组区间','组中值','频数','频率','累计频率']]
    return(t1)


t5=[5954,5022,14667,6582,6870,1840,2662,4508,
   1208,3852,618,3008,1268,1978,7963,2048,
   3077,993,353,14263,1714,11127,6926,2047,
   714,5923,6006,14267,1697,13867,4001,2280,
   1223,12579,13588,7315,4538,13304,1615,8612];
fredistable_zushu(t5,9)

	分组区间	组中值	频数	频率	累计频率
0	(352,1943]	1147.5	11	0.275	0.275
1	(1943,3534]	2738.5	7	0.175	0.450
2	(3534,5125]	4329.5	5	0.125	0.575
3	(5125,6716]	5920.5	4	0.100	0.675
4	(6716,8307]	7511.5	4	0.100	0.775
5	(8307,9898]	9102.5	1	0.025	0.800
6	(9898,11489]	10693.5	1	0.025	0.825
7	(11489,13080]	12284.5	1	0.025	0.850
8	(13080,14671]	13875.5	6	0.150	1.000

（2）按照分割区间大小

def fredistable_fenge(t,cut):#t是数组，cut是分割间隔
    t=np.sort(t)
    mi=np.min(t)
    ma=np.max(t)
    ran=[]
    n=int((ma-mi)/cut)+1
    for i in range(n+1):
        ran.append(mi-1+i*cut)#ran.append(mi+i*cut)直接从最小值开始
    lable=[]
    for i in range(n):
        lable.append('('+str(ran[i])+','+str(ran[i+1])+']')
    t1=pd.cut(t,ran, labels=lable)
    t1=t1.value_counts()
    t1=pd.DataFrame(t1)
    t1['分组区间'] = t1.index
    t1.columns = ['频数','分组区间']
    t1.reset_index(drop=True, inplace=True)  
    #组中值
    zzz=[]
    for i in range(n):
        zzz.append(ran[i]+float(cut)/2)
    t1['组中值'] =zzz
    t1['频率']=t1['频数']/np.shape(t)[0]
    ##计算累计频率
    ljpl=[0]
    for i in t1['频率']:
        ljpl.append(i+ljpl[-1])
    t1['累计频率']=ljpl[1:]
    t1=t1[['分组区间','组中值','频数','频率','累计频率']]
    return(t1)

fredistable_fenge(t5,1700)

	分组区间	组中值	频数	频率	累计频率
0	(352,2052]	1202.0	14	0.350	0.350
1	(2052,3752]	2902.0	4	0.100	0.450
2	(3752,5452]	4602.0	5	0.125	0.575
3	(5452,7152]	6302.0	6	0.150	0.725
4	(7152,8852]	8002.0	3	0.075	0.800
5	(8852,10552]	9702.0	0	0.000	0.800
6	(10552,12252]	11402.0	1	0.025	0.825
7	(12252,13952]	13102.0	4	0.100	0.925
8	(13952,15652]	14802.0	3	0.075	1.000