问题11:板球运动员在25局比赛中得分的次数如下:
26、35、94、48、82、105、53、0、39、42、71、0、64、15、34、15、34、6、71、0、64、15、34、15、34, 67、0、42、124、84、54、48、139、64、47
(i)按升序重新排列这些行程。
(ii)确定球员,是最高分。
(iii)球员没有得分多少次?
(iv)他得分了几个世纪?
(v)他得分了50次以上?
解决方案:
(i) Arranging the runs in ascending order as:
0, 0, 0, 0, 6, 15, 15, 15, 15, 26, 34, 34, 34, 34, 35, 39, 42, 42, 47, 48, 48, 53, 54, 64, 64, 64, 67, 71, 71, 82, 84, 90, 94, 124, 139.
(ii) From the above arrangement, it is observed that,
The Highest Score out of all the runs is 139.
(iii) 3 times, the player was not successful in scoring any run.
(iv) He scored 3 centuries at all.
(v) 12 times, he scored more than 50 runs.
问题12.分配的班级大小为25,第一节课的时间间隔为200-224。有七个班级间隔。
(i)写上课间隔。
(ii)写下每个间隔的等级标记。
解决方案:
Given that,
The Class size is equal to 25,
The first class interval is 200-224
There are seven class intervals in all.
(i) These seven class interval are as follows:
200-224, 225-249, 250-274, 275-299, 300-324, 325-349, 350-374.
(ii) Class marks of the following class intervals are:
- For class interval, 200-224:
- For class interval, 225-249:
- For class interval, 250-274:
- For class interval, 300-324:
- For class interval, 325-349:
- For class interval, 350-374:
问题13:在下面的每一个中写出班级人数和班级限制:
(i)104、114、124、134、144、154和164
(ii)47、52、57、62、67、72、78、82、87、92、97、102
(iii)12.5、17.5、22.5、27.5、32.5、37.5、42.5、47.5
解决方案:
(i) 104, 114, 124, 134, 144, 154 and 164
The class size for the given data is:
114 – 104 = 10
The class limits are obtained as:
Class mark |
Lower class limit |
Upper class limit |
Class limit |
104 |
104 – 10/2 = 99 |
104 + 10/2 = 109 |
99 – 109 |
114 |
114 – 10/2 = 109 |
114 + 10/2 = 119 |
109 – 119 |
124 |
119 |
129 |
119 – 129 |
134 |
129 |
139 |
129 – 139 |
114 |
139 |
149 |
139 – 149 |
154 |
149 |
159 |
149 – 159 |
164 |
159 |
169 |
159 – 169 |
(ii) 47, 52, 57, 62, 67, 72, 78, 82, 87, 92, 97, 102
The class size for the given data is:
52 – 47 = 5
The class limits are obtained as:
Class mark |
Lower class limit |
Upper class limit |
Class limit |
47 |
47 – 5/2 = 44.5 |
47 + 5/2 = 49.5 |
44.5 + 49.5 |
52 |
49.5 |
54.5 |
49.5 + 54.5 |
57 |
54.5 |
59.5 |
54.5 – 59.5 |
62 |
59.5 |
64.5 |
59.5 – 64.5 |
67 |
64.5 |
69.5 |
64.5 – 69.5 |
72 |
69.5 |
74.5 |
69.5 – 74.5 |
77 |
74.5 |
79.5 |
74.5 – 79.5 |
82 |
79.5 |
84.5 |
79.5 – 84.5 |
87 |
84.5 |
89.5 |
84.5 – 89.5 |
92 |
89.5 |
94.5 |
89.5 – 94.5 |
97 |
94.5 |
99.5 |
94.5 – 99.5 |
102 |
99.5 |
104.5 |
99.5 – 104.5 |
(iii) 12.5, 17.5, 22.5, 27.5, 32.5, 37.5, 42.5, 47.5
The class size for the given data is:
17.5 – 12.5 = 5
The class limits are obtained as:
Class mark |
Lower-class limit |
Upper-class limit |
Class limit |
12.5 |
12.5 – 2.5 = 10 |
12.5 + 2.5 = 15 |
10 – 15 |
17.5 |
17.5 – 2.5 = 15 |
17.5 + 2.5 = 20 |
15 – 20 |
22.5 |
22.5 – 2.5 = 20 |
22.5 + 2.5 = 25 |
20 – 25 |
27.5 |
27.5 – 2.5 = 25 |
27.5 + 2.5 = 30 |
25 – 30 |
32.5 |
32.5 – 2.5 = 30 |
32.5 + 2.5 = 35 |
30 – 35 |
37.5 |
37.5 – 2.5 = 35 |
37.5 + 2.5 = 40 |
35 – 40 |
42.5 |
42.5 – 2.5 = 40 |
42.5 + 2.5 = 45 |
40 – 45 |
47.5 |
47.5 – 2.5 = 45 |
47.5 + 2.5 = 50 |
45 – 50 |
问题14.以下数据提供了40个家庭中的儿童人数:
1,2,6,5,1,5,1,1,2,2,6,2,3,4,2,0,0,4,4,4,3,2,2,0,0,1,2, 2、4、3、2、1、0、5、1、2、4、3、4、1、6、2、2
用频率分布的形式表示它。
解决方案:
The frequency distribution of the following sequence of data :
Number of children |
Tally marks |
Number of families |
0 |
5 |
|
1 |
7 |
|
2 |
12 |
|
3 |
5 |
|
4 |
6 |
|
5 |
||| |
3 |
6 |
||| |
3 |
问题15:下面是40名IX级学生在数学上的得分:
81、55、68、79、85、43、29、68、54、73、47、35、72、64、95、44、50、77、64、35、79、52、45、54、70, 83、62、64、72、92、84、76、63、43、54、38、73、68、52、54。
解决方案:
Obtaining the frequency distribution for the marks scored by 40 students of class IX are as follows:
Marks |
Tally marks |
Frequency |
20 – 30 |
| |
1 |
30 – 40 |
||| |
3 |
40 – 50 |
5 |
|
50 – 60 |
8 |
|
60 – 70 |
8 |
|
70 – 80 |
9 |
|
80 – 90 |
|||| |
4 |
90 – 100 |
|| |
2 |
Total = 40 |
问题16:IX班的30名学生的身高(厘米)如下:
155、158、154、158、160、148、149、150、153、159、161、148、157、153、157、153、157、162、159、151、154、156、152、156、160, 152、147、155、163、155、157、153。
准备一个以160 – 164作为班级间隔之一的频率分配表。
解决方案:
The frequency distribution table with 160 – 164 as one of the class intervals is as follows :
Height (in cm) |
Tally marks |
Frequency |
145 – 149 |
|||| |
4 |
150 – 154 |
9 |
|
155 – 159 |
12 |
|
160 – 164 |
6 |
|
Total = 30 |
问题17:工厂中30名工人的月工资如下:
830,835,890,810,835,836,869,845,898,890,820,860,832,833,855,845,804,808,812,840,885,835,836,878,840, 868、890、806、840、890。
解决方案:
Obtaining the frequency distribution for the monthly wages of 30 workers in a factory are as follows:
Height (in cm) |
Tally marks |
Frequency |
800 – 810 |
||| |
3 |
810 – 820 |
|| |
2 |
820 – 830 |
| |
1 |
830 – 840 |
8 |
|
840 – 850 |
5 |
|
850 – 860 |
| |
1 |
860 – 870 |
||| |
3 |
870 – 880 |
| |
1 |
880 – 890 |
| |
1 |
890 – 900 |
5 |
|
Total = 30 |
问题18. 11月份某个城市的每日最高气温(摄氏度)如下:
25.8,24.5,25.6,20.7,21.8,20.5,20.6,20.9,22.3,22.7,23.1,22.8,22.9,21.7,21.3,20.5,20.9,23.1,22.4,21.5,22.7,22.8,22.0,23.9,24.7, 22.8、23.8、24.6、23.9、21.1。
解决方案:
Obtaining the frequency distribution for the daily maximum temperatures (in degree Celsius) recorded in a certain city during the month of November are as follows:
Maximum temperature (in degree Celsius) |
Tally marks |
Frequency |
20.0 – 21.0 |
6 |
|
21.0 – 22.0 |
5 |
|
22.0 – 23.0 |
9 |
|
23.0 – 24.0 |
5 |
|
24.0 – 25.0 |
||| |
3 |
25.0 – 26.0 |
|| |
2 |
Total = 30 |
问题19.根据以下关于在工厂工作的28名工人的月工资(卢比)数据,使用相同的班级间隔来构建频率表,其中一个班级间隔为210 – 230(不包括230)
220、268、258、242、210、268、272、242、311、290、300、320、319、304、302、218、306、292、254、278、210、240、280、316、306, 215、256、236。
解决方案:
The following table depicts the frequency distribution on the monthly wages (in rupees) of 28 laborers working in a factory :
Monthly wages (in rupees) |
Tally marks |
Frequency |
210 – 230 |
|||| |
4 |
230 – 250 |
|||| |
4 |
250 – 270 |
5 |
|
270 – 290 |
||| |
3 |
290 – 310 |
7 |
|
310 – 330 |
5 |
|
Total = 28 |
问题20.在某些北极地区记录的每日最低摄氏温度如下:
-12.5,-10.8,-18.6,-8.4,-10.8,-4.2,-4.8,-6.7,-13.2,-11.8,-2.3,-1.2,-2.6、0、2.4、0、3.2、2.7、3.4 ,0,-2.4、0、3.2、2.7、3.4、0,-2.4,-5.8,-8.9,-14.6,-12.3,-11.5,-7.8,-2.9。
将它们表示为频率分布表,以-19.9至-15作为第一类间隔。
解决方案:
Frequency distribution with lower limit included and upper limit excluded is depicted taking the first class interval as -19.9 to -15 :
Temperature |
Tally marks |
Frequency |
-19.9 to -15 |
|| |
2 |
-15 to -10.1 |
7 |
|
-10.1 to -5.2 |
5 |
|
-5.2 to -0.3 |
|||| |
4 |
-0.3 to -4.6 |
17 |
|
Total = 35 |
问题21:八类的30名学生的血型记录如下:
A,B,O,O,AB,O,A,O,B,A,O,B,A,O,O,A,AB,O,A,A,A,O,O,AB,B,A, O,B,A,B,O
以频率分布表的形式表示此数据。在这些学生中找出哪个是最普通的,哪个是最稀有的血型。
解决方案:
Given that,
9 students out of the given blood groups have blood group A, 6 as B, 3 as AB and 12 as O
The frequency table representing the data along with the corresponding frequency is as follows:
Blood group | Number of students |
A |
9 |
B |
6 |
AB |
3 |
O |
12 |
Total |
30 |
Since, 12 students have their blood group as O, which is the maximum, therefore the most common blood group is O. Also, 3 students have their blood group as AB, which implies that the rarest blood group is AB.
问题22.将三枚硬币扔了30次。每次记下出现的磁头数量如下:
0、1、2、2、1、2、3、1、3、0
1,3,1,1,2,2,0,1,2,1
3,0,1,1,2,2,3,2,2,0
为上述数据准备一个频率分布表。
解决方案:
The frequency distribution table for the data of the three coins being tossed 30 times can be computed as follows:
Number of heads |
Frequency |
0 |
6 |
1 |
10 |
2 |
9 |
3 |
5 |
Total=30 |
问题23。30名儿童被问到他们在前一周看电视节目的小时数。结果发现如下:
1、6、2、3、5、12、5、8、4、8
10、3、4、12、2、8、15、1、17、6
3、2、8、5、9、6、8、7、14、2
(i)为该数据制作一个频率分布表,以班级宽度5和班级间隔之一为5 – 10。
(ii)每周有15个或更多小时的时间有多少儿童看电视。
解决方案:
(i) The following class intervals are 0 – 5, 5 – 10, 10 – 15 constructed.
The grouped frequency distribution table taking into account these class intervals is as follows:
Hours |
Number of children |
0 – 5 |
10 |
5 – 10 |
13 |
10 – 15 |
5 |
15 – 20 |
2 |
Total=30 |
(ii) The number of children who watched TV for at least 15 hours a week is 2, which implies that the number of children in the class interval 15 – 20 is equivalent to 2.