第11类RD Sharma解决方案–第32章统计–练习32.7

问题1.工厂的两个工厂A和B显示有关工人人数和付给他们的工资的以下结果

	Plant A	Plant B
No. of workers	5000	6000
Average monthly wages	₹2500	₹2500
The variance of distribution of wages	81	100

在哪个工厂A或B中，个人工资的差异更大?

解决方案：

Variation of the distribution of wages in plant A (σ² =18)

So, Standard deviation of the distribution A (σ – 9)

Similarly, the Variation of the distribution of wages in plant B (σ² =100)

So, Standard deviation of the distribution B (σ – 10)

And, Average monthly wages in both the plants is 2500,

Since, the plant with a greater value of SD will have more variability in salary.

∴ Plant B has more variability in individual wages than plant A

问题2：班上50名学生的身高和体重的均值和标准差如下：

	Weights	Heights
Mean	63.2 Kg	63.2 inch
Standard deviation	5.6 Kg	11.5 inch

哪个显示出更多的可变性，高度或重量?

解决方案：

We observe that the average weights and height for the 50 students is same i.e. 63.2.

Therefore, the parameter with greater variance will have more variability.

Thus, height has greater variability

问题3.两种分布的变异系数分别为60％和70％，其标准偏差分别为21和16。他们的算术手段是什么?

解决方案：

Coefficient of variation = $\frac{\sigma}{x}\times100$

So, we have:

$60\% =\frac{21}{\overline{x}}\times100\Rightarrow\overline{x}=\frac{21}{.60}\times100=35\\ 70\% =\frac{16}{\overline{x}}\times100\Rightarrow\overline{x}=\frac{16}{.70}\times100=22.85\\$

∴ Means are 35 and 22.85

问题4.根据以下数据计算变异系数：

Income(in ₹):	1000 – 1700	1700 – 2400	2400 – 3100	3100 – 3800	3800 – 4500	4500 – 5200
No. of families:	12	18	20	25	35	10

解决方案：

Now,

N = 120, $\sum u_i^2f_i=321$

Mean, $\overline{X}=A+h\left(\frac{\sum u_if_i}{N}\right)\\ \overline{X}=2750+700\left(\frac{83}{120}\right)\\ =3234.17\\ Var(X)=h^2\left[\frac{1}{N}\displaystyle\sum_{i=1}^nf_iu_i^2-\left(\frac{1}{N}\sum_{i=1}^nu_if_i\right)^2\right]\\ Var(X)=490000\left[\left(\frac{321}{120}\right)-\left(\frac{83}{120}\right)^2\right]$

Variance = 1076332.64

Standard Deviation, $\sigma=\sqrt{1076332.64}\\ =1037.47$

Coefficient of variation = $\frac{1037.46}{3234.17}\times100$

= 32.08

∴ The coefficient variation is 32.08

问题5.对属于同一行业的两家公司A和B中付给工人的每周工资的分析得出以下结果：

Class	Fi	xi	$U_i=\frac{x_i-mean}{700}$	fiui	fiui2
1000 – 1700	12	1350	-2	-24	48
1700 – 2400	18	2050	-1	-18	18
2400 – 3100	20	2750	0	0	0
3100 – 3800	25	3450	1	25	25
3800 – 4500	35	4150	2	70	140
4500 – 5200	10	4850	3	30	90
	$\sum f_i=120$			$\sum u_if_i=83$	$\sum u_i^2f_i=321$

(i)哪家公司A或B支付了较大的每周工资?

(ii)哪家公司A或B的个人工资差异较大?

解决方案：

(i) Average weekly wages = $\frac{Total\ weekly\ wages}{No.\ of\ workers}$

Total weekly wages = (Average weekly wages) × (No. of workers)

Total weekly wages of Firm A = 52.5 × 586 = Rs 30765

Total weekly wages of Firm B = 47.5 × 648 = Rs 30780

Firm B pays a larger amount as Firm A

(ii) Here, 

S.D (Firm A) = 10 and S.D (Firm B) = 11

Coefficient variance (Firm A) = $\frac{10}{52.5}\times100$

= 19.04

Coefficient variance (Firm B) = $\frac{11}{47.5}\times100$

= 23.15

∴ Coefficient variance of Firm B is greater than that of Firm A, Firm B has greater variability in individual wages.

问题6.以下是班上男孩和女孩的权重分布的一些细节：

	Firm A	Firm B
No. of wage earners	586	648
Average weekly wages	₹52.5	₹47.5
The variance of the distribution of wages	100	121

哪个分布更易变?

解决方案：

Given:

S.D (Boys) is 3 and S.D (Girls) is 2

Coefficient variance (Boys) = $\frac{3}{60} \times 100$

= 5

Coefficient variance (Girls) = $\frac{2}{45} \times 100$

= 4.4

∴ Coefficient variance of Boys is greater than Coefficient variance of girls, and then the distribution of weights of boys is more variable than that of girls.

问题7.下面是数学，物理和化学三个学科的50个班级的学生获得的分数的平均值和标准偏差：

	Boys	Girls
Number	100	50
Mean weight	60 Kg	45 Kg
Variance	9	4

解决方案：

In order to compare the variability of marks in Math, Physics and Chemistry.

We have to calculate their coefficient of variation.

Let σ₁, σ₂ and σ₃ denote the standard deviation of marks in Math, Physics and Chemistry respectively. Further, Let $\overline{X_1},\ \overline{X_2}\ and\ \overline{X_3}$ be the mean scores in Math, Physics and Chemistry respectively.

We have 

$\overline{X_1}=42\ \ \ \ \ \overline{X_2}=32\ \ \ \ \ \overline{X_3}=40.9$

⇒ σ₁  = 12       σ₂  = 15         σ₃  = 20

Now,

Coefficient of variation in Maths = $\frac{\sigma_1}{X_1}\times100=\frac{12}{42}\times100=28.57$

Coefficient of variation in Physics = $\frac{\sigma_2}{X_2}\times100=\frac{15}{32}\times100=46.88$

Coefficient of variation in Chemistry = $\frac{\sigma_3}{X_3}\times100=\frac{20}{40.9}\times100=48.90$

Clearly, coefficient of variation in marks is greatest in Chemistry and lowest in Math.

So, marks in chemistry show highest variability and marks in maths show lowest variability.

问题8.从下面给出的数据可以看出，哪个组的变量更大，G ₁或G ₂ ?

Subject            Mean	Mathematics          42	Physics             32	Chemistry              40.9
Standard deviation	12	15	20

解决方案：

Let’s first find the coefficient of variable for group G₁

Here, N = 150, A = 45, $\sum f_iu_i=-6,\ \sum f_iu_i^2=342$  and h = 10

∴ Mean = $\overline{x}=A+h\left(\frac{1}{N}\sum f_iu_i\right)\\ \overline{x}=45+10\left(\frac{-6}{150}\right)=44.6\\ Var(x)=h^2\left[\frac{1}{N}\sum f_iu_i^2-\left(\frac{1}{N}\sum f_iu_i\right)^2\right]=100\left[\frac{342}{150}-\left(\frac{-6}{150}\right)^2\right]=227.84\\ S.D.=\sqrt{Var(x)}=\sqrt{227.84}=15.09$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times100=\frac{15.09}{44.6}\times100=33.83$

Now, lets find the coefficient of variable for group G₂

CI                   f               10 – 20              10      20 – 30              20	x                 u=(x – A)/h              15                        -3       25                        -2	fu              u2            -30               9      -40               4	fu2             90        80
30 – 40              30      40 – 50              25	35                        -1       45                         0	-30               1         0                0	30         0
50 – 60              43	55                         1	43               1	43
60 – 70              15      70 – 80               7	65                         2       75                         3	30               4        21               9	60        63
150		-6	366

Here, N = 150, A = 45, $\sum f_iu_i=-6,\ \sum f_iu_i^2=366$ and h = 10

∴ Mean = $\overline{x}=A+h\left(\frac{1}{N}\sum f_iu_i\right)\\ \overline{x}=45+10\left(\frac{-6}{150}\right)=44.6\\ Var(x)=h^2\left[\frac{1}{N}\sum f_iu_i^2-\left(\frac{1}{N}\sum f_iu_i\right)^2\right]=100\left[\frac{366}{150}-\left(\frac{-6}{150}\right)^2\right]=243.84\\ S.D.=\sqrt{Var(x)}=\sqrt{243.84}=15.62$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times100=\frac{15.62}{44.6}\times100=35.02$

Group G₂ is more variable

问题9.找到以下数据的变异系数：

Marks Group G1	10 – 20 9	20 – 30 30 – 40 17        32	40 – 50 33	50 – 60 60 – 70 40        10	70 – 80 9
Group G2	10	20        30	25	43        15	7

解决方案：

CI                 f                  x               10 – 15            2                12.5      15 – 20            8                17.5	u=(x – A)/h               fu                  u2                           -2                       -4                   4                 -1                       -8                   1	fu2              8            8
20 – 25           20               22.5      25 – 30           35               27.5	0                        0                   0                   1                       35                  1	0        35
30 – 35           20               32.5	2                       40                  4	80
35 – 40           15               37.5	3                       45                  9	135
100	108	266

Here, N = 100, A = 22.5, $\sum f_iu_i=108,\ \sum f_iu_i^2=266$ and h = 5

∴ Mean = $\overline{x}=A+h\left(\frac{1}{N}\sum f_iu_i\right)\\ \overline{x}=22.5+5\left(\frac{108}{100}\right)=27.90\\ Var(x)=h^2\left[\frac{1}{N}\sum f_iu_i^2-\left(\frac{1}{N}\sum f_iu_i\right)^2\right]=25\left[\frac{266}{100}-\left(\frac{108}{100}\right)^2\right]=37.34\\ S.D.=\sqrt{Var(x)}=\sqrt{37.34}=6.11$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times100=\frac{6.11}{27.90}\times100=21.9$

问题10.从下面给出的X和Y股的价格中：找出哪个值更稳定：

CI              f            10 – 20         9     20 – 30        17	x       u=(x – A)/h           15               -3      25               -2	fu         u2          -27         9     -34         4	fu2          81      68
30 – 40        32      40 – 50        33	35               -1      45                0	-32         1       0          0	32       0
50 – 60        40	55                1	40          1	40
60 – 70        10     70 – 80         9	65                2      75                3	20          4      27          9	40      81
150		-6	342

解决方案：

x                  d = (x – Mean)                 35                           -13         24                           -24	d2            169      576
52                              4         53                              5	16       25
56                              8	64
58                             10         52                              4	100       16
50                              2	4
51                              3	9
49                              1	1
480	980

∴ Mean = $\overline{x}=\frac{1}{n}\sum x_i=\frac{1}{10}[480]=48\\ Var(x)=\frac{1}{n}\left\{\sum (x_i-\overline{x})^2\right\}=\frac{1}{10}(980)=98\\ S.D(x)=\sqrt{Var(x)}=\sqrt{98}=9.9$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times100=\frac{9.9}{48}\times100=20.6$

x                  d = (x – Mean)                  35                           -13         24                           -24	d2                   169          576
52                            4          53                            5	16           25
56                            8	64
58                           10          52                            4	100           16
50                            2	4
51                            3	9
49                            1	1
480	980

∴ Mean = $\overline{x}=\frac{1}{n}\sum x_i=\frac{1}{10}[1050]=105\\ Var(x)=\frac{1}{n}\left\{\sum (x_i-\overline{x})^2\right\}=\frac{1}{10}(40)=4\\ S.D(x)=\sqrt{Var(x)}=\sqrt{4}=2$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times100=\frac{2}{105}\times100=1.90$

Since the coefficient of variation for share Y is smaller than the coefficient of variation for shares X, they are more stable.