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Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.3
  • Last Updated : 21 Feb, 2021

Question 1. From the data given below state which group is more variable, A or B?

Marks10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Group A917323340109
Group B1020302543157

Solution:

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



MarksGroup A (fi)Mid-point (Xi)Yi = (xi – A)/h(Yi)2fiyifi(yi)2
10 – 20915((15 – 45)/10) = -3(-3)2 = 9-2781
20 – 301725((25 – 45)/10) = -2(-2)2 = 4-3468
30 – 403235((35 – 45)/10) = -1(-1)2 = 1-3232
40 – 503345((45 – 45)/10) = 00200
50 – 604055((55 – 45)/10) = 1 12 = 14040
60 – 701065((65 – 45)/10) = 222 = 42040
70 – 80975((75 – 45)/10) = 332 = 92781
Total150   -6342

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

Where A = 45,

and yi = (xi – 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.

MarksGroup B (fi)Mid-point XiYi = (xi – A)/h(Yi)2fiyifi(yi)2
10 – 201015((15 – 45)/10) = -3(-3)2 = 9-3090
20 – 302025((25 – 45)/10) = -2(-2)2 = 4-4080
30 – 403035((35 – 45)/10) = -1(-1)2 = 1-3030
40 – 502545((45 – 45)/10) = 00200
50 – 604355((55 – 45)/10) = 112 = 14343
60 – 70  1565((65 – 45)/10) = 222 = 43060
70 – 80775((75 – 45)/10) = 332 = 92163
Total150   -6366

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.

Question 2. From the prices of shares X and Y below, find out which is more stable in value:

X35545253565852505149
Y108107105105106107104103104101

Solution:

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                      
35108122511664
54107291611449
52105270411025
53105280911025
56106813611236
58107336411449
52104270410816
50103250010609
51104260110816
4910126360110290
Total = 510105026360110290

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.

Question 3. An analysis of monthly 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 wages earners                     586648
Mean of monthly wagesRs 5253Rs 5253
Variance of the distribution of wages           100121

(i) Which firm A or B pays larger amount as monthly wages?

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

Solution:

(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.

Question 4. The following is the record of goals scored by team A in a football session:

No. of goals scored            01234
No. of matches19753

For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

Solution:

From the given data,

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

Number of goals xiNumber of matches fifixiXi2fixi2
01000
19919
2714428
3515945
Total2550 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.

Question 5. The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

\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

Which is more varying, the length or weight?

Solution:

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.

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