第10类RD Sharma解决方案–第7章统计–练习7.3 |套装1

问题1.下表列出了城市中体力劳动者的家庭总支出(以卢比计)的分布。

Expenditure (in rupees) (x)	Frequency (fi)	Expenditure (in rupees) (xi)	Frequency (fi)
100 – 150	24	300 – 350	30
150 – 200	40	350 – 400	22
200 – 250	33	400 – 450	16
250 – 300	28	450 – 500	7

找出每个家庭的平均支出(以卢比为单位)。

解决方案：

Let the assumed mean (A) = 275

It’s seen that A = 275 and h = 50

So,

Mean = A + h x (Σf_i u_i/N)

= 275 + 50 (-35/200)

= 275 – 8.75

= 266.25

问题2。作为他们的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本地200所房屋中植物数量的数据。找出每所房子的平均植物数量。

Class interval	Mid value (xi)	di = xi – 275	ui = (xi – 275)/50	Frequency fi	fiui
100 – 150	125	-150	-3	24	-72
150 – 200	175	-100	-2	40	-80
200 – 250	225	-50	-1	33	-33
250 – 300	275	0	0	28	0
300 – 350	325	50	1	30	30
350 – 400	375	100	2	22	44
400 – 450	425	150	3	16	48
450 – 500	475	200	4	7	28
				N = 200	Σ fiui = -35

您使用哪种方法求平均值，为什么?

解决方案：

From the given data,

To find the class interval we know that,

Class marks (x_i) = (upper class limit + lower class limit)/2

Now, let’s compute x_i and f_ix_i by the following

Here,

Mean = Σ f_iu_i/N

= 162/ 20

= 8.1

Thus, the mean number of plants in a house is 8.1

We have used the direct method as the values of class mark x_i and f_i is very small.

问题3.考虑以下工厂工人的日工资分配

Number of plants:	0 – 2	2 – 4	4 – 6	6 – 8	8 – 10	10 – 12	12 – 14
Number of house:	1	2	1	5	6	2	3

用适当的方法求出工厂工人的平均日工资。

解决方案：

Let us assume mean (A) = 150

It’s seen that,

A = 150 and h = 20

So,

Mean = A + h x (Σf_i u_i/N)

= 150 + 20 x (-12/50)

= 150 – 24/5

= 150 = 4.8

= 145.20

问题4。30名妇女在医院接受了医生检查，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数。

Number of plants	Number of house (fi)	xi	fixi
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
Total	N = 20		Σ fiui = 162

解决方案：

Using the relation (x_i) = (upper class limit + lower class limit)/ 2

And, class size of this data = 3

Let the assumed mean (A) = 75.5

So, let’s calculate d_i, u_i, f_iu_i as following:

From table, it’s seen that

N = 30 and h = 3

So, the mean = A + h x (Σf_i u_i/N)

= 75.5 + 3 x (4/30

= 75.5 + 2/5

= 75.9

Therefore, the mean heart beats per minute for those women are 75.9 beats per minute.

问题5.找到以下每个频率分布的均值：

Daily wages (in ₹)	100 – 120	120 – 140	140 – 160	160 – 180	180 – 200
Number of workers:	12	14	8	6	10

解决方案：

Let’s consider the assumed mean (A) = 15

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σf_i u_i/N)

= 15 + 6 x (3/40)

= 15 + 0.45

= 15.45

问题6.找到以下频率分布的均值：

Class interval	Mid value xi	di = xi – 150	ui = (xi – 150)/20	Frequency fi	fiui
100 – 120	110	-40	-2	12	-24
120 – 140	130	-20	-1	14	-14
140 – 160	150	0	0	8	0
160 – 180	170	20	1	6	6
180 – 200	190	40	2	10	20
				N= 50	Σ fiui = -12

解决方案：

Let’s consider the assumed mean (A) = 100

From the table it’s seen that,

A = 100 and h = 20

Mean = A + h x (Σf_i u_i/N)

= 100 + 20 x (61/100)

= 100 + 12.2

= 112.2

问题7：找到以下频率分布的均值：

Number of heart beats 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：

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf_i u_i/N)

= 20 + 8 x (7/40)

= 20 + 1.4

= 21.4

问题8：找到以下频率分布的均值：

Number of heart beats per minute	Number of women (fi)	xi	di = xi – 75.5	ui = (xi – 755)/h	fiui
65 – 68	2	66.5	-9	-3	-6
68 – 71	4	69.5	-6	-2	-8
71 – 74	3	72.5	-3	-1	-3
74 – 77	8	75.5	0	0	0
77 – 80	7	78.5	3	1	7
80 – 83	4	81.5	6	2	8
83 – 86	2	84.5	9	3	6
	N = 30				Σ fiui = 4

解决方案：

Let’s consider the assumed mean (A) = 15

Class interval	Mid – value xi	di = xi – 15	ui = (xi – 15)/6	fi	fiui
0 – 6	3	-12	-2	7	-14
6 – 12	9	-6	-1	5	-5
12 – 18	15	0	0	10	0
18 – 24	21	6	1	12	12
24 – 30	27	12	2	6	12
				N = 40	Σ fiui = 5

From the table it’s seen that,

A = 15 and h = 6

Mean = A + h x (Σf_i u_i/N)

= 15 + 6 x (5/40)

= 15 + 0.75

= 15.75

问题9.找到以下频率分布的均值：

Class interval:	0 – 6	6 – 12	12 – 18	18 – 24	24 – 30
Frequency:	6	8	10	9	7

解决方案：

Let’s consider the assumed mean (A) = 25

Class interval	Mid – value xi	di = xi – 25	ui = (xi – 25)/10	fi	fiui
0 – 10	5	-20	-2	9	-18
10 – 20	15	-10	-1	12	-12
20 – 30	25	0	0	15	0
30 – 40	35	10	1	10	10
40 – 50	45	20	2	14	28
				N = 60	Σ fiui = 8

From the table it’s seen that,

A = 25 and h = 10

Mean = A + h x (Σf_i u_i/N)

= 25 + 10 x (8/60)

= 25 + 4/3

= 79/3 = 26.333

问题10：找到以下频率分布的均值：

Class interval	Mid – value xi	di = xi – 15	ui = (xi – 15)/6	fi	fiui
0 – 6	3	-12	-2	6	-12
6 – 12	9	-6	-1	8	-8
12 – 18	15	0	0	10	0
18 – 24	21	6	1	9	9
24 – 30	27	12	2	7	14
				N = 40	Σ fiui = 3

解决方案：

Let’s consider the assumed mean (A) = 20

Class interval	Mid – value xi	di = xi – 20	ui = (xi – 20)/8	fi	fiui
0 – 8	4	-16	-2	5	-10
8 – 16	12	-4	-1	9	-9
16 – 24	20	0	0	10	0
24 – 32	28	4	1	8	8
32 – 40	36	16	2	8	16
				N = 40	Σ fiui = 5

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf_i u_i/N)

= 20 + 8 x (5/40)

= 20 + 1

= 21

问题11：找到以下频率分布的均值：

Class interval:	50 – 70	70 – 90	90 – 110	110 – 130	130 – 150	150 – 170
Frequency:	18	12	13	27	8	22

解决方案：

Let’s consider the assumed mean (A) = 20

Class interval	Mid – value xi	di = xi – 20	ui = (xi – 20)/8	fi	fiui
0 – 8	4	-16	-2	5	-12
8 – 16	12	-8	-1	6	-8
16 – 24	20	0	0	4	0
24 – 32	28	8	1	3	9
32 – 40	36	16	2	2	14
				N = 20	Σ fiui = -9

From the table it’s seen that,

A = 20 and h = 8

Mean = A + h x (Σf_i u_i/N)

= 20 + 6 x (-9/20)

= 20 – 72/20

= 20 – 3.6

= 16.4

问题12：找到以下频率分布的均值：

Class interval	Mid – value xi	di = xi – 100	ui = (xi – 100)/20	fi	fiui
50 – 70	60	-40	-2	18	-36
70 – 90	80	-20	-1	12	-12
90 – 110	100	0	0	13	0
110 – 130	120	20	1	27	27
130 – 150	140	40	2	8	16
150 – 170	160	60	3	22	66
				N= 100	Σ fiui = 61

解决方案：

Let’s consider the assumed mean (A) = 60

Class interval	Mid – value xi	di = xi –60	ui = (xi – 60)/20	fi	fiui
10 – 30	20	-40	-2	5	-10
30 – 50	40	-20	-1	8	-8
50 – 70	60	0	0	12	0
70 – 90	80	20	1	20	20
90 – 110	100	40	2	3	6
110 – 130	120	60	3	2	6
				N = 50	Σ fiui = 14

From the table it’s seen that,

A = 60 and h = 20

Mean = A + h x (Σf_i u_i/N)

= 60 + 20 x (14/50)

= 60 + 28/5

= 60 + 5.6

= 65.6

问题13：找到以下频率分布的均值：

Class interval:	0 – 8	8 – 16	16 – 24	24 – 32	32 – 40
Frequency:	6	7	10	8	9

解决方案：

Let’s consider the assumed mean (A) = 50

Class interval	Mid – value xi	di = xi – 50	ui = (xi – 50)/10	fi	fiui
25 – 35	30	-20	-2	6	-12
35 – 45	40	-10	-1	10	-10
45 – 55	50	0	0	8	0
55 – 65	60	10	1	12	12
65 – 75	70	20	2	4	8
				N = 40	Σ fiui = -2

From the table it’s seen that,

A = 50 and h = 10

Mean = A + h x (Σf_i u_i/N)

= 50 + 10 x (-2/40)

= 50 – 0.5

= 49.5