第10类NCERT解决方案–第14章统计–练习14.1

问题1.作为环境意识计划的一部分，一群学生进行了一项调查，他们收集了以下有关当地20所房屋中植物数量的数据。找出每所房子的平均植物数量。

Number of Plants	0-2	2-4	4-6	6-8	8-10	10-12	12-14
Number of houses	1	2	1	5	6	2	3

您使用哪种方法求平均值，为什么?

解决方案：

Step 1: Let us find out the Class Mark (x_i) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

 $\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

No.of Plants (Class Interval)	No. of Houses (Frequency) (fi)	Class Mark (xi)	fi * xi
0-2	1	1	1
2-4	2	3	6
4-6	1	5	5
6-8	5	7	35
8-10	6	9	54
10-12	2	11	22
12-14	3	13	39
	Sum: ∑ fi = 20		Sum: ∑ fixi = 162

Now, after creating this table we will be able to find the mean very easily – 

$\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

= 16

= 8.1

Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.

问题2。请考虑以下工厂的50名工人的日工资分配。

Daily Wages (in ₹)	500-520	520-540	540-560	560-580	580-600
Number of Workers	12	14	8	6	10

用适当的方法求出工厂工人的平均日工资。

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.

u_i = (x_i – A)/h 

=> u_i = (xi – 150)/20

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

$\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Daily wages (Class interval)	Number of workers frequency (fi)	Mid-point (xi)	ui = (xi – 150)/20	fiui
100-120	12	110	-2	-24
120-140	14	130	-1	-14
140-160	8	150	0	0
160-180	6	170	1	6
180-200	10	190	2	20
Total	Sum ∑fi = 50			Sum ∑fiui = -12

So, the formula to find out the mean is:

Mean = $\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

           = 150 + (20 × -12/50) 

           = 150 – 4.8

           = 145.20

Thus, mean daily wage of the workers = Rs. 145.20.

问题3。以下分布显示了本地儿童的每日零用钱。平均零用钱为18卢比。找到缺失的频率f。

Daily pocket allowance (in ₹)	11-13	13-15	15-17	17-19	19-21	21-23	23-25
Number of children	7	6	9	13	f	5	4

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middle-Class Mark as our Assumed Mean(A).

Step 3: Now we will apply the general formula to calculate the mean

$\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class interval	Number of children (fi)	Mid-point (xi)	fixi
11-13	7	12	84
13-15	6	14	84
15-17	9	16	144
17-19	13	18 = A	234
19-21	f	20	20f
21-23	5	22	110
23-25	4	24	96
Total	∑ fi = 44 + f		Sum ∑fixi = 752 + 20f

The mean formula is

Mean = $\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

          = (752 + 20f)/(44 + f)

Now substitute the values and equate to find the missing frequency (f)

⇒ 18 = (752 + 20f)/(44 + f)

⇒ 18(44 + f) = (752 + 20f)

⇒ 792 + 18f = 752 + 20f

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f

⇒ f = 20

So, the missing frequency, f = 20.

问题4。30名妇女在医院接受医生检查，每分钟的心跳数记录并总结如下。选择合适的方法，找出这些女性每分钟的平均心跳。

Number of heartbeats per minute	65-68	68-71	71-74	74-77	77-80	80-83	83-86
Number of Women	2	4	3	8	7	4	2

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.

d_i = (x_i – A) 

=> d_i = (x_i – 75.5)

Step 3: Now we will apply the Assumed Mean Formula to calculate the mean

$\large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class Interval	Number of women (fi)	Mid-point (xi)	di = (xi – 75.5)	fidi
65-68	2	66.5	-9	-18
68-71	4	69.5	-6	-24
71-74	3	72.5	-3	-9
74-77	8	75.5 = A	0	0
77-80	7	78.5	3	21
80-83	4	81.5	6	24
83-86	2	84.5	9	18
	Sum ∑fi = 30			Sum ∑fiui = 12

Mean = $\large \overline{x}=A+\frac{\sum f_id_i}{\sum f_i}$

= 75.5 + (12/30)

= 75.5 + 2/5

= 75.5 + 0.4

= 75.9

Therefore, the mean heartbeats per minute for these women is 75.9

问题5.在零售市场上，水果摊贩正在出售装在包装盒中的芒果。这些盒子里装着不同数量的芒果。以下是根据箱子数量分配的芒果。

Number of Mangoes	50-52	53-55	56-58	59-61	62-64
Number of Boxes	15	110	135	115	25

找到装在包装箱中的芒果的平均数量。您选择了哪种求平均值的方法?

解决方案：

Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add  0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula 

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.

Step 3: Since the frequency values are big, hence we are using the STEP-DEVIATION METHOD.

$\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class Interval	Number of boxes (fi)	Mid-point (xi)	di = xi – A	ui=(xi – A)/h	fiui
49.5-52.5	15	51	-6	-2	-30
52.5-55.5	110	54	-3	-1	-110
55.5-58.5	135	57 =A	0	0	0
58.5-61.5	115	60	3	1	115
61.5-64.5	25	63	6	2	50
	Sum ∑fi = 400				Sum ∑fiui = 25

Mean = $\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

= 57 + 3 * (25/400)

= 57 + 0.1875

= 57.19

Therefore, the mean number of mangoes kept in a packing box is 57.19

问题6.下表显示了一个地区25户家庭的日常食物支出。用适当的方法求出食物的平均每日支出。

Daily Expenditure (in ₹)	100-150	150-200	200-250	250-300	300-350
Number of Households	4	5	12	2	2

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.

d_i = (x_i – A) 

=> d_i = (x_i – 225)

u_i = (x_i – A)/h

=> u_i = (x_i – 225)/50

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

$\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class Interval	Number of households (fi)	Mid-point (xi)	di = xi – A	ui = di/50	fiui
100-150	4	125	-100	-2	-8
150-200	5	175	-50	-1	-5
200-250	12	225 = A	0	0	0
250-300	2	275	50	1	2
300-350	2	325	100	2	4
	Sum ∑fi = 25				Sum ∑fiui = -7

Mean = $\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

= 225 + 50 (-7/25)

= 225 – 14

= 211

Therefore, the mean daily expenditure on food is ₹211

问题7.为了找出_{空气中SO 2}的浓度(百万分之一，即ppm)，收集了某城市30个地区的数据，并在下面提供：

Concentration of SO2 (in ppm)	Frequency
0.00-0.04	4
0.04-0.08	9
0.08-0.12	9
0.12-0.16	2
0.16-0.20	4
0.20-0.24	2

求出空气_{中SO 2}的平均浓度。

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

$\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Concentration of SO2 (in ppm)	Frequency (fi)	Mid-point (xi)	fixi
0.00-0.04	4	0.02	0.08
0.04-0.08	9	0.06	0.54
0.08-0.12	9	0.10	0.90
0.12-0.16	2	0.14	0.28
0.16-0.20	4	0.18	0.72
0.20-0.24	2	0.22	0.44
	Sum ∑fi = 30		Sum ∑fixi = 2.96

The formula to find out the mean is

Mean = $\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

= 2.96/30

= 0.099 ppm

Therefore, the mean concentration of SO₂ in the air is 0.099 ppm.

问题8.在整个学期中，班主任的缺勤记录如下：40名班级学生。查找学生缺勤的平均天数。

Number of Days	0-6	6-10	10-14	14-20	20-28	28-38	38-40
Number of Students	11	10	7	4	4	3	1

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.

Step 3: Now we will apply the general formula to calculate the mean

$\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class Interval	Frequency (fi)	Mid-point (xi)	fixi
0-6	11	3	33
6-10	10	8	80
10-14	7	12	84
14-20	4	17	68
20-28	4	24	96
28-38	3	33	99
38-40	1	39	39
	Sum ∑fi = 40		Sum ∑fixi = 499

The mean formula is,

Mean = $\large \overline{x}=\frac{\sum f_ix_i}{\sum f_i}$

= 499/40

= 12.48 days

Therefore, the mean number of days a student was absent = 12.48.

问题9.下表列出了35个城市的识字率(百分比)。找到平均识字率。

Literacy rate (in %)	45-55	55-65	65-75	75-85	85-95
Number of Cities	3	10	11	8	3

解决方案：

Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula

Class Mark = (Upper class Limit + Lower Class Limit)/2

Step 2: In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.

d_i = (x_i – A) 

=> d_i = (x_i – 70)

u_i = (x_i – A)/h

=> u_i = (x_i – 70)/10

Step 3: Now we will apply the Step Deviation Formula to calculate the mean

$\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

现在，让我们看一下详细的解决方案：

Class Interval	Frequency (fi)	Class Mark(xi)	di = xi – a	ui = di/h	fiui
45-55	3	50	-20	-2	-6
55-65	10	60	-10	-1	-10
65-75	11	70 = A	0	0	0
75-85	8	80	10	1	8
85-95	3	90	20	2	6
	Sum ∑fi  = 35				Sum ∑fiui = -2

So, 

Mean = $\large \overline{x}= A + h \frac{\sum f_iu_i}{\sum f_i}$

= 70 + (-2/35) × 10

= 69.42

Therefore, the mean literacy rate = 69.42%.