11类NCERT解决方案-第15章统计-练习15.3

问题1.从下面给出的数据中可以看出，哪个组的变量更大，A还是B?

Marks	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70	70 – 80
Group A	9	17	32	33	40	10	9
Group B	10	20	30	25	43	15	7

解决方案：

For comparing the variability or dispersion of two series, we calculate the coefficient of variance for each series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

Co-efficient of variation (C.V.) = (σ/ x̅) × 100

Where, σ = standard deviation, x̅ = mean

For Group A

$Mean,\ \overline{x}=A+\frac{\displaystyle\sum_{i=1}^af_iy_i}{N}\times h$

Where A = 45,

and y_i = (x_i – A)/h

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

Then, variance σ² = $\frac{h^2}{N^2}[N\sum f_iy_i^2-(\sum f_iy_i)^2]$

σ² = (10²/150²) [150(342) – (-6)²]

= (100/22500) [51,300 – 36]

= (100/22500) × 51264

= 227.84

Hence, standard deviation = σ = √227.84

= 15.09

∴ C.V for group A = (σ/ x̅) × 100

= (15.09/44.6) × 100

= 33.83

Now, for group B.

Mean, $\overline{x}=A+\frac{\displaystyle\sum_{i=1}^af_iy_i}{N}\times h$

Where A = 45,

h = 10

So, x̅ = 45 + ((-6/150) × 10)

= 45 – 0.4

= 44.6

Then, Variance σ² = $\frac{h^2}{N^2}[N\sum f_iy_i^2-(\sum f_iy_i)^2]$

σ² = (10²/150²) [150(366) – (-6)²]

= (100/22500) [54,900 – 36]

= (100/22500) × 54,864

= 243.84

Hence, standard deviation = σ = $\sqrt{243.84}$

= 15.61

∴ C.V for group B = (σ/ x̅) × 100

= (15.61/44.6) × 100

= 35

By comparing C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.

问题2.从下面的X和Y股的价格中，找出哪个更稳定：

Marks	Group A (fi)	Mid-point (Xi)	Yi = (xi – A)/h	(Yi)2	fiyi	fi(yi)2
10 – 20	9	15	((15 – 45)/10) = -3	(-3)2 = 9	-27	81
20 – 30	17	25	((25 – 45)/10) = -2	(-2)2 = 4	-34	68
30 – 40	32	35	((35 – 45)/10) = -1	(-1)2 = 1	-32	32
40 – 50	33	45	((45 – 45)/10) = 0	02	0	0
50 – 60	40	55	((55 – 45)/10) = 1	12 = 1	40	40
60 – 70	10	65	((65 – 45)/10) = 2	22 = 4	20	40
70 – 80	9	75	((75 – 45)/10) = 3	32 = 9	27	81
Total	150				-6	342

解决方案：

From the given data,

Let us make the table of the given data and append other columns after calculations.

X (xi)	Y (yi)	Xi2	Yi2
35	108	1225	11664
54	107	2916	11449
52	105	2704	11025
53	105	2809	11025
56	106	8136	11236
58	107	3364	11449
52	104	2704	10816
50	103	2500	10609
51	104	2601	10816
49	101	26360	110290
Total = 510	1050	26360	110290

We have to calculate Mean for x,

Mean x̅ = $\sum \frac{x_i}{n}$

Where, n = number of terms

= 510/10

= 51

Then, Variance for x = $\frac{1}{n^2}[N\sum x_i^2-(\sum x_i)^2]$

= (1/10²)[(10 × 26360) – 510²]

= (1/100) (263600 – 260100)

= 3500/100

= 35

WKT Standard deviation = $\sqrt{Variance}$

= √35

= 5.91

So, co-efficient of variation = (σ/ x̅) × 100

= (5.91/51) × 100

= 11.58

Now, we have to calculate Mean for y,

Mean $\overline{y}=\sum \frac{y_i}{n}$

Where, n = number of terms

= 1050/10

= 105

Then, Variance for y = $\frac{1}{n^2}[N\sum y_i^2-(\sum y_i)^2]$

= (1/10²)[(10 × 110290) – 1050²]

= (1/100) (1102900 – 1102500)

= 400/100

= 4

WKT Standard deviation = $\sqrt{Variance}$

= √4

= 2

So, co-efficient of variation = (σ/ x̅) × 100

= (2/105) × 100

= 1.904

By comparing C.V. of X and Y.

C.V of X > C.V. of Y

So, Y is more stable than X.

问题3.对属于同一行业的两家公司A和B中付给工人的月工资的分析得出以下结果：

Marks	Group B (fi)	Mid-point Xi	Yi = (xi – A)/h	(Yi)2	fiyi	fi(yi)2
10 – 20	10	15	((15 – 45)/10) = -3	(-3)2 = 9	-30	90
20 – 30	20	25	((25 – 45)/10) = -2	(-2)2 = 4	-40	80
30 – 40	30	35	((35 – 45)/10) = -1	(-1)2 = 1	-30	30
40 – 50	25	45	((45 – 45)/10) = 0	02	0	0
50 – 60	43	55	((55 – 45)/10) = 1	12 = 1	43	43
60 – 70	15	65	((65 – 45)/10) = 2	22 = 4	30	60
70 – 80	7	75	((75 – 45)/10) = 3	32 = 9	21	63
Total	150				-6	366

(i)哪家公司A或B支付的金额更大一些?

(ii)哪家公司A或B在个人工资方面显示出更大的可变性?

解决方案：

(i) From the given table,

Mean monthly wages of firm A = Rs 5253

and Number of wage earners = 586

Then,

Total amount paid = 586 × 5253

= Rs 3078258

Mean monthly wages of firm B = Rs 5253

Number of wage earners = 648

Then,

Total amount paid = 648 × 5253

= Rs 34,03,944

So, firm B pays larger amount as monthly wages.

(ii) Variance of firm A = 100

We know that, standard deviation (σ)= √100

=10

Variance of firm B = 121

Then,

Standard deviation (σ)=√(121 )

=11

Hence the standard deviation is more in case of Firm B that means in firm B there is greater variability in individual wages.

问题4：以下是A队在足球比赛中进球的记录：

X	35	54	52	53	56	58	52	50	51	49
Y	108	107	105	105	106	107	104	103	104	101

对于B队，每场比赛的平均进球数为2，标准差为1.25球。找到哪个团队可能被认为更一致?

解决方案：

From the given data,

Let us make the table of the given data and append other columns after calculations.

Number of goals xi	Number of matches fi	fixi	Xi2	fixi2
0	1	0	0	0
1	9	9	1	9
2	7	14	4	28
3	5	15	9	45
Total	25	50		130

First we have to calculate Mean for Team A,

Mean = $\frac{\sum f_ix_i}{\sum f_i}=\frac{50}{25}=2$

Then,

Variance = $\frac{1}{N^2}[N\sum f_ix_i^2-(\sum f_ix_i)^2]\\ =\frac{1}{25^2}[25\times130-2500]=\frac{750}{625}=1.2$

We know that, standard deviation σ = $\sqrt{Variance}=\sqrt{1.2}=1.09$

Hence co-efficient of variation of team A,

$C.V._A =\frac{\sigma}{\overline{x}}\times100=\frac{1.09}{2}\times100=54.5$

For Team B

Given, x̅  = 2

Standard deviation σ = 1.25

So, coefficient of variation of Team B,

$\Rightarrow C.V._B=\frac{1.25}{2}\times100=6.25$

Since C.V. of firm B is greater

∴ Team A is more consistent.

问题5.对应于50种植物产品的长度x(以cm为单位)和重量y(以gm为单位)的平方和的总和如下：

$\displaystyle \sum_{i=1}^{50} x_i=212,\ \ \sum_{i=1}^{50}x_i^2=902.8,\ \ \sum_{i=1}^{50}y_i=261,\ \ \sum_{i=1}^{50}y_i^2=1457.6$

长度或重量哪个变化更大?

解决方案：

First we have to calculate Mean for Length x,

Mean = $\overline{x}=\frac{\sum x_i}{n}=\frac{212}{50}=4.24$

Then,

Variance = $\frac{1}{N^2}[N\sum f_ix_i^2-(\sum f_ix_i)^2]\\ =\left(\frac{1}{50^2}\right)[(50\times902.8)-212^2]\\ =\left(\frac{1}{2500}\right)(45140-44944)\\ =\frac{196}{2500}\\ =0.0784$

We know that, standard deviation σ = $\sqrt{Variance}\\ =\sqrt{0.0784}\\ =0.28$

Hence co-efficient of variation of team A,

$C.V._x=\frac{\sigma}{\overline{x}}\times100=\frac{0.28}{4.24}\times100=6.603$

Now we have to calculate mean of weight y

$\overline{y}=\sum\frac{y}{n}\\ =\frac{261}{50}\\ =5.22$

Then,

Variance = $\left(\frac{1}{N^2}\right)[(N\sum f_iy_i^2)-(\sum f_iy_i)^2]\\ =\left(\frac{1}{50^2}\right)[(50\times1457.6)-261^2]\\ =\left(\frac{1}{2500}\right)(72880-68121)\\ =\frac{4759}{2500}\\ =1.9036$

When know that, standard deviation σ = $\sqrt{Variance}\\ =\sqrt{1.9036}\\ =1.37$

So, co-efficient of variance of Team B,

$C.V._y=\frac{\sigma}{\overline{x}}\times100=\frac{1.37}{5.22}\times100=26.24$

Since C.V. of firm weight y is greater

∴ Weight is more varying.