第9类RD Sharma解决方案–第22章统计数据的表格表示–练习22.2

问题1.定义累积频率分布。

解决方案：

The structure showing the distribution of cumulative frequencies over various classes is known as cumulative frequency distribution.

问题2.解释频率分布和累积频率分布之间的差异。

解决方案：

The method representing raw data which is easily understandable is called frequency distribution. The structure showing the distribution of cumulative frequencies over various classes is known as cumulative frequency distribution.

问题3：55名学生在考试中获得的分数如下：

Marks	0-5	5-10	10-15	15-20	20-25	25-30	30-35
No. of students	2	6	13	17	11	4	2

准备一个累积频率表。

解决方案：

问题4.以下是一天中360名患者在医院接受治疗的年龄：

Marks	No. of students	Marks	No. of students
0-5	2	Less than 5	2
5-10	6	Less than 10	8
10-15	13	Less than 15	21
15-20	17	Less than 20	38
20-25	11	Less than 25	49
25-30	4	Less than 30	53
30-35	2	Less than 35	55
	N = 55

构造一个累积频率分布。

解决方案：

问题5.某条街道在1.1.98至31.3.98期间的32栋房屋的水费(卢比)如下：

56、43、32、38、56、24、68、85、52、47、35、58、63、74、27、84、69、35、44、75、55、30、54、65、45， 67、95、72、43、65、35、59。

将数据制成表格并使用70-79作为分类间隔之一将其显示为累积频率表。

解决方案：

The minimum and maximum bills are Rs 24 and Rs 95 respectively. 

Range of the class interval is 95-24 = 71.

Class interval given = 70-79, with the class size being 9. 

number of classes = range/class-size = 71/9 = 7.80 which is approximately 8. 

Cumulative frequency distribution is as follows : 

问题6.图书馆不同书架上的书籍数量如下：

30、32、28、24、20、25、38、37、40、45、16、20、19、24、27、30、32、34、35、42、27、28、19、34、38， 39、42、29、24、27、22、29、31、19、27、25、28、23、24、32、34、18、27、25、37、31、24、23、43、32， 28、31、24、23、26、36、32、29、28、21

使用45-49作为最后一个课时间隔，准备一个累积频率分布表。

解决方案：

The minimum and maximum book shelves are 16 and 45 respectively. 

The range of the book shelves 45-16 = 29

We have the following class interval , 45 – 49, with the class size =4.

Number of classes = Range/Class size = 29/4 = 7.25 , which approximates to 8 nearly.

No. of books	No. of shelves  (Frequency)	Cumulative Frequency
13-17	1	1
17-21	6	7
21-25	11	18
25-29	15	33
29-33	12	45
33-37	5	50
37-41	6	56
41-45	3	59
45-49	1	50

问题7.下面给出的是显示一所学校685名学生权重的累积频率。准备频率分布表。

Age (in years)	10-20	20-30	30-40	40-50	50-60	60-70
No. of Patients	90	50	60	80	50	30

解决方案：

Weight (in kg)	No. of students	Class interval	Frequency
Below 30	24	25-30	24 – 0 =24
Below 35	78	30-35	78- 24 = 54
Below 40	183	35-40	183 – 78 = 105
Below 45	294	40-45	294 – 183 = 111
Below 50	408	45-50	408 – 294 = 114
Below 55	543	50-55	543 – 408 = 135
Below 60	621	55-60	621 – 543 = 78
Below 65	674	60-65	674 – 621 = 53
Below 70	685	65-70	685 – 674 = 11

问题8.以下累积频率分布表显示了工业状态下40家工厂的每日用电量(以kW为单位)：

Age (in years)	No. of patients	Age (in years)	No. of patients
10-20	90	Less than 20	90
20-30	50	Less than 30	140
30-40	60	Less than 40	200
40-50	80	Less than 50	280
50-60	50	Less than 60	330
60-70	30	Less than 70	360
	N = 360

(i)将此表示为频率分布表。

(ii)准备一个累积频率表。

解决方案：

(i) 

Consumption (in kW)	No. of factories	Class interval	Frequency
Below 240	1	0-240	1
Below 270	4	240-270	4-1 = 3
Below 300	8	270-300	8-4 = 4
Below 330	24	300-330	24-8 = 16
Below 360	33	330-360	33 – 24 = 9
Below 390	38	360-390	38-33 = 5
Below 420	40	390-420	40-38 = 2

(ii)

Class interval	Frequency	Consumption (in kW)	No. of factories
0-240	1	More than 0	40
240-270	3	More than 270	40-1 = 39
270-300	4	More than 270	39-3 = 36
300-330	16	More than 300	36-4 = 32
330-360	9	More than 330	32-16= 16
360-390	5	More than 360	16-9 = 7
390-420	2	More than 390	7-5=2
		More than 420	2-2 = 0
	N = 40

问题9.以下是一个累积的频率分布表，显示了当地居民的年龄：

Bills (Rs.)	No. of houses(Frequency)
16-25	1
25-34	3
34-43	5
43-52	4
52-61	7
61-70	6
70-79	3
79-88	2
88-97	1

准备频率分布表。

解决方案：

Age (in years)	No. of persons	Class interval	Frequency
Above 0	1124	0-12	1124-1026 = 98
Above 12	1026	12-24	1026-809 = 217
Above 24	809	24-36	809-427 = 382
Above 36	427	36-48	427-158= 269
Above 48	158	48-60	158-20 = 138
Above 60	20	60-72	20 -5 = 15
Above 72	5	72-84	5-3 = 2
Above 84	3	84-96	3-1 = 2
Above 96	1	96-108	1-0 = 1