Skip to content
Related Articles

Related Articles

Improve Article

Class 11 RD Sharma Solutions – Chapter 32 Statistics – Exercise 32.7

  • Last Updated : 03 Mar, 2021

Question 1. Two plants A and B of a factory show the following results about the number of workers and the wages paid to them

 

       Plant A      

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

     Plant B    



No. of workers                         

5000

6000

Average monthly wages

₹2500

₹2500

The variance of distribution of wages       

81

100

In which plant A or B is there greater variability in individual wages?

Solution:



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

Question 2. The means and standard deviations of heights and weights of 50 students in a class are as follows:

 

    Weights   

   Heights   



Mean

63.2 Kg

63.2 inch

Standard deviation              

5.6 Kg

11.5 inch

Which shows more variability, heights or weights?

Solution:

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

Question 3. The coefficient of variation of two distribution are 60% and 70%, and their standard deviations are 21 and 16 respectively. What is their arithmetic means?

Solution:

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

Question 4. Calculate coefficient of variation from the following data:

Income(in ₹):     

1000 – 1700

1700 – 2400

2400 – 3100

3100 – 3800

3800 – 4500



4500 – 5200

No. of families:    

12

18

20

25

35

10

Solution:

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

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

Question 5. An analysis of the weekly wages paid to workers in two firms A and B, belonging to the same industry gives the following results:

 

       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

(i) Which firm A or B pays out the larger amount as weekly wages?

(ii) Which firm A or B has greater variability in individual wages?

Solution:

(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.

Question 6. The following are some particulars of the distribution of weights of boys and girls in a class:

 Boys         Girls       
Number 10050
Mean weight               60 Kg45 Kg
Variance94

Which of the distributions is more variable?

Solution:

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.

Question 7. The mean and standard deviation of marks obtained by 50 students of a class in three subjects, mathematics, physics and chemistry are given below:

Subject           

Mean

        Mathematics         



42

           Physics            

32

          Chemistry             

40.9

Standard deviation     

12

15

20

Solution:

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.

Question 8. From the data given below state which group is more variable, G1 or G2?

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 G21020        302543        157

Solution:

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

       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

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

Question 9. Find the coefficient of variation for the following data:

Size (in cms): 10 – 1515 – 2020 – 2525 – 3030 – 3535 – 40
No. of items: 2820352015

Solution:



        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

Question 10. From the prices of shares X and Y given below: find out which is more stable in value:

X:35545253565852505149
Y:108107105105106107104103104101

Solution:

         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.




My Personal Notes arrow_drop_up
Recommended Articles
Page :