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Class 11 NCERT Solutions- Chapter 15 Statistics – Exercise 15.3

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  • Last Updated : 21 Feb, 2021
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Question 1. From the data given below state which group is more variable, A or 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

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

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

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.

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

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:

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

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                      
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.

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                      586 648
Mean of monthly wages Rs 5253 Rs 5253
Variance of the distribution of wages            100 121

(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             0 1 2 3 4
No. of matches 1 9 7 5 3

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 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.

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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