问题1.您对“统计”一词的理解是什么：
(i)单数形式(ii)复数形式?
解决方案：
The word statistics can be used in multiple ways, both singularly and plurally.
(i) In singular form, statistics can be used collectively as the science of collection, presentation, analysis, and interpretation of numerical data.
(ii) In plural form, statistics, however, implies numerical facts or observations that are collected together to serve a definite purpose.
问题2：描述统计的一些基本特征。
解决方案：
Some fundamental characteristics of statistics are:
(1) Statistics are always collected with a definite purpose.
(2) Statistics are only expressed in quantitative and not qualitative terms.
(3) Statistics can be classified into various segments and often are comparable in an experiment.
问题3.什么是(i)主要数据(ii)次要数据?主数据或辅助数据中的哪一个更可靠，为什么?
解决方案：
(i) Primary data: The raw data collected without using any of the existing resources of the data by an investigator with a definite plan is called the primary data.
(ii)Secondary data: The data collected after using information from existing resources such as other published or unpublished sources is called secondary data.
Primary data is more reliable and relevant because it is raw in nature, collected without any prior information on the existing topic. It is original in nature, gathered by some individuals and/or the research bodies, whereas the secondary data are gathered by an individual or an institution for some purpose by taking help from existing resources and are used by someone else in another context.
问题4.为什么要将数据分组?
解决方案：
The data obtained in original form without any prior resources on the same, is called raw data. It is essentially difficult to capture information from the raw data, which is difficult to understand. Grouping of data increases the ease of understanding and interpretation. Also, it helps to simulate calculations on commodities which helps in describing and analyzing the data.
问题5.解释以下术语的含义：
(i)变化
(ii)班级间隔
(iii)班级规模
(iv)商标
(v)频率
(vi)班级限制
(vii)正确的班级限制
解决方案：
(i) Variate: Any character that can vary from one individual to another is called variate.
(ii) Class-interval: In the data of each group into which raw data is considered.
(iii) Class-size: The difference between the upper limit and lower limit of the specified class.
(iv) Class-mark: The middle value of the selected class is marked as the class-mark.
The following formula is used in its calculation :
Class-mark = (Upper limit + Lower limit) /2
(v) Frequency: The number of observations corresponding to a given class is known as frequency.
(vi) Class limits: Each class is bounded by two figures, called the class limits. The figure on the right side is called the upper limit while figure on the left side of the class is called the lower limit.
(vii)True class limits:
If classes are inclusive. For example 5-9, 10-14, 15-19, and so on
Then, the true lower limit of class = Lower limit of class – 0.5
Also, the true upper limit of class = Upper limit of class + 0.5
In this case, the True limits of the class are 5-9 are 4.5 and 9.5
But, if classes are exclusive like 20-30, 30-40, 40-50, etc., then class limits and true class limits are equivalent.
问题6.一组十个学生的年龄如下。年龄记录以年和月为单位：
8 – 6、9 – 0、8 – 4、9 – 3、7 – 8、8 – 11、8 – 7、9 – 2、7 – 10、8 – 8
(i)最低年龄是多少?
(ii)最高年龄是多少?
(iii)确定范围?
解决方案：
The given ages of ten students of a group are as follows:
8 – 6, 9 – 0, 8 – 4, 9 – 3, 7 – 8, 8 – 11, 8 – 7, 9 – 2, 7 – 10, 8 – 8.
(i) The Lowest age in the given group = 7 years 8 months
(ii) The Highest age in the given group = 9 years 3 months
(iii) Range = The Highest age – The Lowest age
Substituting the values,
Range = (9 years 3 months) – (7 years 8 months)
= 1 year 7 months
问题7.六个朋友的每月零用钱如下：
45卢比，30卢比，40卢比，50卢比，25卢比，45卢比。
(i)最高零用钱是多少?
(ii)最低零用钱是多少?
(iii)范围是多少?
(iv)零用钱的金额按升序排列。
解决方案：
The monthly pocket money of six friends are given below in the form of a sequence:
Rs 45, Rs 30, Rs 40, Rs 50, Rs 25, Rs 45
(i) The Highest pocket money given = Rs 50
(ii) The Lowest pocket money given = Rs 25
(iii) Range = The Highest pocket money – The Lowest pocket money
Substituting the values, we get,
= Rs 50 – Rs 25
= Rs 25
(iv) Arranging the amounts of pocket money in ascending order is: Rs 25, Rs 30, Rs 40, Rs 45, Rs 45,Rs 50.
问题8.在下面的每一个中编写类-size：
(i)0-4、5-9、10-14
(ii)10-19、20-29、30-39
(iii)100-120、120-140、160-180
(iv)0-0.25、0.25-00.50、0.50-0.75
(v)5-5.01、5.01-5.02、5.02-5.03。
解决方案：
(i) 0-4, 5-9, 10-14
Since, the given classes are inclusive, so
True lower limit of class = Lower limit of class – 0.5
And, True upper limit of class = Upper limit of class + 0.5
True class limits are 0.5-4.5, 4.5-9.5, 9.5-14.5
Therefore, class size = 14.5 – 9.5 = 5
(ii) 10-19, 20-29, 30-39
Since, classes are inclusive, so
True lower limit of class = Lower limit of class – 0.5
And, True upper limit of class = Upper limit of class + 0.5
True class limits 19.5-19.5, 19.5-29.5, 29.5-29.5
The required class size = 39.5-29.5 = 10
(iii) 100-120, 120-140, 160-180
Here the class limits and true class limits are the same
So, the required class size = 120 – 100 = 20
(iv) 0-0.25, 0.25-00.50, 0.50-0.75
Here the class limits and true class limits are the same
So, the required class size = 0.25 – 0 = 0.25
(v) 5-5.01, 5.01-5.02, 5.02-5.03
Therefore, class limits and true class limits are the same
So, class size = 5.01 – 5.0 = 0.01.
问题9. 30名学生的数学最终成绩如下：
53，61，48，60，78，68，55，100，67，90，
75、88、77、37、84、58、60、48、62、56
44、58、52、64、98、59、70、39、50、60。
(i)按升序排列这些标记，一组为30到39，一组为40到49秒，依此类推。
现在回答以下问题：
(ii)最高分数是多少?
(iii)最低分数是多少?
(iv)范围是多少?
(v)如果合格分数是40，那么多少个不及格?
(vi)有多少人得分在75分以上?
(vii)实际未出现50到60之间的哪些观察结果?
(viii)有多少得分低于50?
解决方案：
(i) Arrange marks in ascending order:
|
Class Marks |
Observations |
Frequency |
|
30-39 |
37, 39 |
2 |
|
40-49 |
44,48,48 |
3 |
|
50-59 |
50, 52, 53, 55, 56, 58, 58, 59 |
8 |
|
60-69 |
60, 60, 60, 61, 62, 64, 67, 68 |
8 |
|
70-79 |
70, 75, 77, 78 |
4 |
|
80-89 |
84,88 |
2 |
|
90-99 |
90, 98 |
2 |
|
100-109 |
100 |
1 |
(ii) Therefore, from the table above, it is observed that, the Highest score is equal to 100,
(iii) The Lowest score Is equal to 37.
(iv) Since, the range is defined as:
Range = The Highest score – The Lowest score
= 100 – 37
= 63
(v) 2
(vi) 8
(vii) 51, 54, 57
(viii) 5
问题10.在特定的一天中，医院新生婴儿的体重(以千克为单位)如下：
2.3、2.2、2.1、2.7、2.6、3.0、2.5、2.9、2.8、3.1、2.5、2.8、2.7、2.9、2.4
(i)重新布置最大重量。
(ii)确定最高重量。
(iii)确定最低重量。
(iv)确定范围。
(v)那天有几个婴儿出生?
(vi)有多少个体重在2.5公斤以下的婴儿?
(vii)多少婴儿体重超过2.8公斤?
(viii)多少婴儿重2.8公斤?
解决方案：
(i) Arrange weights in descending order:
3.1, 3.0, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.4, 2.3, 2.2, 2.1.
(ii) The Highest weight in the given sequence= 3.1 Kg
(iii) The Lowest weight in the given sequence= 2.1 Kg
(iv) Range in the given sequence = 3.1 kg – 2.1 kg = 1.0 Kg
(v) 15 babies were born on that particular day.
(vi) 4 babies out of all weigh below 2.5 Kg.
(vii) Babies with more than 2.8 Kg weight are 4 in number.