第10类RD Sharma解决方案–第7章统计–练习7.3 |套装2

问题14：找到以下频率分布的均值：

Class interval:	25 – 29	30 – 34	35 – 39	40 – 44	45 – 49	50 – 54	55 – 59
Frequency:	14	22	16	6	5	3	4

解决方案：

Let’s consider the assumed mean (A) = 42

From the table it’s seen that,

A = 42 and h = 5

Mean = A + h x (Σf_i u_i/N)

= 42 + 5 x (-79/70)

= 42 – 79/14

= 42 – 5.643

= 36.357

问题15。对于以下分布，请使用所有合适的方法计算均值：

Class interval	Mid – value xi	di = xi – 42	ui = (xi – 42)/5	fi	fiui
25 – 29	27	-15	-3	14	-42
30 – 34	32	-10	-2	22	-44
35 – 39	37	-5	-1	16	-16
40 – 44	42	0	0	6	0
45 – 49	47	5	1	5	5
50 – 54	52	10	2	3	6
55 – 59	57	15	3	4	12
				N = 70	Σ fiui = -79

解决方案：

By direct method

Mean = (sum/N) + A

= 848/64

= 13.25

By assuming mean method

Let the assumed mean (A) = 65

Mean = A + sum/N

= 6.5 + 6.75

= 13.25

问题16.以下是2004-2005年某城市每周生活成本指数的观察结果。如果计算生活指数，则计算每周费用。

Size of item:	1 – 4	4 – 9	9 – 16	16 – 20
Frequency:	6	12	26	20

解决方案：

Let the assumed mean (A) = 1650

We have

A = 16, h = 100

Mean = A + h (sum/N)

= 1650 + (175/13)

= 21625/13

= 1663.46

问题17.下表显示了140名学生在某一篇论文的考试中获得的成绩：

Class interval	Mid value xi	Frequency fi	fixi
1 – 4	2.5	6	15
4 – 9	6.5	12	18
9 – 16	12.5	26	325
16 – 27	21.5	20	430
		N = 64	Sum = 848

使用所有三种方法计算平均分数：直接方法，假定均值偏差和捷径方法。

解决方案：

(i) Direct method:

Mean = sum/N

= 3620/140

= 25.857

(ii) Assumed mean method:

Let the assumed mean = 25

Mean = A + (sum/N)

Mean = A + (sum/N)

= 25 + (120/140)

= 25 + 0.857

= 25.857

(iii) Step deviation method:

Let the assumed mean (A) = 25

Mean = A + h(sum/N)

= 25 + 10(12/140)

= 25 + 0.857

= 25.857

问题18.以下频率分布的平均值为62.8，所有频率的总和为50。计算未命中频率f ₁和f ₂ 。

Class interval	Mid value xi	ui = (xi – A) = xi – 65	Frequency fi	fiui
1 – 4	2.5	-4	6	-25
4 – 9	6.5	0	12	0
9 – 16	12.5	6	26	196
16 – 27	21.5	15	20	300
			N = 64	Sum = 432

解决方案：

Given,

Sum of frequency = 50

5 + f₁ + 10 + f₂ + 7 + 8 = 50

f₁ + f₂ = 20

3f₁ + 3f₂ = 60        —(1) [Multiply both side by 3]

and mean = 62.8

Sum/N = 62.8

(30f₁ + 70f₂ + 2060)/50 = 62.8

30f₁ + 70f₂ = 3140 – 2060

30f₁ + 70f₂ = 1080

3f₁ + 7f₂ = 108      —(2) [divide it by 10]

Subtract equation (1) from equation (2)

3f₁ + 7f₂ – 3f₁ – 3f₂ = 108 – 60

4f₂ = 48

f₂ = 12

Put value of f₂ in equation (1)

3f₁ + 3(12) = 60

f₁ = 24/3 = 8

f₁ = 8, f₂ = 12

问题19.以下分配显示了给多层建筑物的孩子的每日零用钱。平均零用钱为18.00卢比。找出丢失的频率。

Cost of living index	Number of students	Cost of living index	Number of students
1400 – 1500	5	1700 – 1800	9
1500 – 1600	10	1800 – 1900	6
1600 – 1700	20	1900 – 2000	2

解决方案：

Given mean = 18,

Let the missing frequency be v

Mean = sum/N

18 = 752 + 20×44 + x $\frac{752+20x}{44+x}$

792 + 18x = 752 + 20x

2x = 40

x = 20

问题20.如果以下分布的均值为27。找到p的值。

Class interval	Mid value xi	di = xi – A = xi – 1650	$<strong>u_i=(\frac{x_i-1650}{100})</strong>$	Frequency fi	fiui
1400 – 1500	1450	-200	-2	5	-10
1500  – 1600	1550	-100	-1	10	-10
1600 – 1700	1650	0	0	20	0
1700 – 1800	1750	100	1	9	9
1800 – 1900	1850	200	2	6	12
1900 – 2000	1950	300	3	2	6
				N = 52	Sum = 7

解决方案：

Given mean = 27

Mean = sum/N

1245 + 15p43 + p$\frac{1245+15p}{43+p}$ = 27

1245 + 15p = 1161 + 27p

12p = 84 

P =7

问题21.在零售市场上，水果贩子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。

Marks:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Number of students:	20	24	40	36	20

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：

Number of mangoes	Number of boxes
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x_i) may be obtained by using the relation

x_i = upperlimit + lowerclasslimit2$\frac{upperlimit+lowerclasslimit}{2}$ 

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d_i, u_i, f_iu_i as follows

Class interval	Frequency fi	Mid values xi	di = xi – A = xi – 25	$<strong>u_i=(\frac{x_i-25}{10})</strong>$	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	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	N = 400				Sum = 25

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

问题22.下表显示了一个地区25个家庭的日常食物支出

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	20	100
10 – 20	15	24	360
20 – 30	25	40	1000
30 – 40	35	36	1260
40 – 50	45	20	900
		N = 140	Sum = 3620

用适当的方法求出食物的平均每日支出。

解决方案：

We may calculate class mark (x_i) for each interval by using the relation

x_i = upperlimit + lowerclasslimit2$\frac{upperlimit+lowerclasslimit}{2}$

Class size = 50

Now, Taking 225 as assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Daily expenditure	Frequency f1	Mid value xi	di = xi – 225	$<strong>u_i=(\frac{x_i-225}{50})</strong>$	fiui
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
	N = 25				Sum = -7

Now we may observe that

N = 25

Sum = -7

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

问题23.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个地点的地点数据，如下所示：

Class interval	Mid value xi	ui = (xi – A)	Frequency fi	fiui
0 – 10	5	-20	20	-400
10 – 20	15	-10	24	-240
20 – 30	25	0	40	0
30 – 40	35	10	36	360
40 – 50	45	20	20	400
			N = 140	Sum = 120

求出空气_{中SO 2}的平均浓度

解决方案：

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Class size of this data = 0.04

Now taking 0.04 assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Concentration of SO2	Frequency f1	Class interval xi`	di = xi – 0.14	ui	fiui
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	N = 30				Sum = -31

From the table we may observe that 

N = 30

Sum = -31

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO₂ in the air is 0.099 ppm.

问题24.在整个学期中，班主任的缺勤记录如下，该班有40名班级学生。查找学生缺勤的平均天数。

Class interval	Mid value xi	di = xi – A = xi – 25	$<strong>u_i=(\frac{x_i-25}{10})</strong>$	Frequency fi	fiui
0 – 10	5	-20	-2	20	-40
10 – 20	15	-10	-1	24	-24
20 – 30	25	0	0	40	0
30 – 40	35	10	1	36	36
40 – 50	45	20	2	20	40
				N = 140	Sum = 12

解决方案：

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Now, taking 16 as assumed mean (a) we may

Calculate d_i and f_id_i as follows

Number of days	Number of students fi	Xi	d = xi + 10	fidi
0 – 6	11	3	-13	-143
6 – 10	10	8	-8	-280
10 – 14	7	12	-4	-28
14 – 20	7	16	0	0
20 – 28	8	24	8	32
28 – 36	3	33	17	51
30 – 40	1	39	23	23
Total	N = 40			Sum = -145

Now we may observe that

N = 40

Sum = -145

$Mean(\overline{x})=a+(sumN)\times \overline{x}=a+(\frac{sum}{N})$

= 16 + (-145/40)

= 16 – 3.625 

= 12.38

So mean number of days is 12.38 days, for which student was absent.

问题25.下表列出了35个城市的识字率(百分比)。找到平均识字率。

Class:	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	100 – 120
Frequency:	5	f1	10	f2	7	8

解决方案：

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d_i, u_i, f_iu_i as follows

Literacy rate (in %)	Number of cities (fi)	Mid value xi	di = xi – 70	ui = di – 50	fiui
45 – 55	3	50	-20	-20	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	N = 35				Sum = -2

Now we may observe that

N = 35

Sum = -2 

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%

问题21.在零售市场上，水果贩子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。

Class interval	Mid value xi	Frequency fi	fixi
0 – 20	10	5	50
20 – 40	30	fi	30fi
40 – 60	50	10	500
60 – 80	70	f2	70f2
80 – 100	90	7	630
100 – 120	110	8	880
		N = 50	Sum = 30f1 + 70f2 + 2060

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：

Number of mangoes	Number of boxes
50 – 52	15
53 – 55	110
56 – 58	135
59 – 61	115
62 – 64	25

We may observe that class internals are not continuous

There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (x_i) may be obtained by using the relation

x_i = upperlimit + lowerclasslimit2$\frac{upperlimit+lowerclasslimit}{2}$ 

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculated d_i, u_i, f_iu_i as follows

Class interval	Frequency fi	Mid values xi	di = xi – A = xi – 25	$<strong>u_i=(\frac{x_i-25}{10})</strong>$	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	0	0	0
58.5 – 61.5	115	60	3	1	115
61.5 – 64.5	25	63	6	2	50
Total	N = 400				Sum = 25

Now we have N

Sum = 25

Mean = A +h (sum/N)

= 57 + 3 (45/400)

= 57 + 3/16

= 57 + 0.1875

= 57.19

Clearly mean number of mangoes kept in packing box is 57.19

问题22.下表显示了一个地区25个家庭的日常食物支出

Class interval:	11 – 13	13 – 15	15 – 17	17 – 19	19 – 21	21 – 23	23 – 25
Frequency:	7	6	9	13	–	5	4

用适当的方法求出食物的平均每日支出。

解决方案：

We may calculate class mark (x_i) for each interval by using the relation

x_i = upperlimit + lowerclasslimit2$\frac{upperlimit+lowerclasslimit}{2}$

Class size = 50

Now, Taking 225 as assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Daily expenditure	Frequency f1	Mid value xi	di = xi – 225	$<strong>u_i=(\frac{x_i-225}{50})</strong>$	fiui
100 – 150	4	125	-100	-2	-8
150 – 200	5	175	-50	-1	-5
200 – 250	12	225	0	0	0
250 – 300	2	275	50	1	2
300 – 350	2	325	100	2	4
	N = 25				Sum = -7

Now we may observe that

N = 25

Sum = -7

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

225 + 50 (-7/25)

225 – 14 = 211

So, mean daily expenditure on food is Rs 211

问题23.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个地点的地点数据，如下所示：

Class interval	Mid interval xi	Frequency fi	fixi
11 – 13	12	7	84
13 – 15	14	6	88
15 – 17	16	9	144
17 – 19	18	13	234
19 – 21	20	x	20x
21 – 23	22	5	110
23 – 25	14	4	56
		N = 44 + x	Sum = 752 + 20x

求出空气_{中SO 2}的平均浓度

解决方案：

We may find class marks for each interval by using the relation

x = upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Class size of this data = 0.04

Now taking 0.04 assumed mean (x_i) we may calculate d_i, u_i, f_iu_i as follows

Concentration of SO2	Frequency f1	Class interval xi`	di = xi – 0.14	ui	fiui
0.00 – 0.04	4	0.02	-0.12	-3	-12
0.04 – 0.08	9	0.06	-0.08	-2	-18
0.08 – 0.12	9	0.10	-0.04	-1	-9
0.12 – 0.16	2	0.14	0	0	0
0.16 – 0.20	4	0.18	0.04	1	4
0.20 – 0.24	2	0.22	0.08	2	4
Total	N = 30				Sum = -31

From the table we may observe that 

N = 30

Sum = -31

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

= 0.14 + (0.04)(-31/30)

= 0.099 ppm

So mean concentration of SO₂ in the air is 0.099 ppm.

问题24.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。

Class:	0 – 10	10 – 20	20 – 30	30 – 40	40 – 50
Frequency:	8	P	12	13	10

解决方案：

We may find class mark of each interval by using the relation

x =  upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Now, taking 16 as assumed mean (a) we may

Calculate d_i and f_id_i as follows

Number of days	Number of students fi	Xi	d = xi + 10	fidi
0 – 6	11	3	-13	-143
6 – 10	10	8	-8	-280
10 – 14	7	12	-4	-28
14 – 20	7	16	0	0
20 – 28	8	24	8	32
28 – 36	3	33	17	51
30 – 40	1	39	23	23
Total	N = 40			Sum = -145

Now we may observe that

N = 40

Sum = -145

$Mean(\overline{x})=a+(sumN)\times \overline{x}=a+(\frac{sum}{N})$

= 16 + (-145/40)

= 16 – 3.625 

= 12.38

So mean number of days is 12.38 days, for which student was absent.

问题25.下表列出了35个城市的识字率(百分比)。找到平均识字率。

Class interval	Mid value xi	Frequency fi	fixi
0 – 10	5	8	40
10 – 20	15	P	152
20 – 30	25	12	300
30 – 40	35	13	455
40 – 50	45	16	450
		N = 43 + P	Sum = 1245 + 15p

解决方案：

We may find class marks by using the relation

x = upperlimit + lowerclasslimit2x = $\frac{upperlimit+lowerclasslimit}{2}$

Class size (h) for this data = 10

Now taking 70 as assumed mean (a) wrong

Calculate d_i, u_i, f_iu_i as follows

Literacy rate (in %)	Number of cities (fi)	Mid value xi	di = xi – 70	ui = di – 50	fiui
45 – 55	3	50	-20	-20	-6
55 – 65	10	60	-10	-1	-10
65 – 75	11	70	0	0	0
75 – 85	8	80	10	1	8
85 – 95	3	90	20	2	6
Total	N = 35				Sum = -2

Now we may observe that

N = 35

Sum = -2 

$Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h$

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%