NCERT Solutions for Exercise 15.3 Class 11 Maths Chapter 15 Statistics

NCERT solutions for Class 11 Maths chapter 15 exercise 15.3 has a total of 5 questions with some questions having subpart also. Questions from exercise 15.3 mainly deal with the comparison between two sets of data in which we have to compare the variation and consistency among the given sets. Exercise 15.3 Class 11 Maths uses the same concept, hence if the initial few questions are done properly, the whole exercise can be tackled easily.

Solving NCERT Solutions for Class 11 Maths chapter 15 exercise 15.3 is highly recommended as it is easy as well as scoring from the exam point of view. Sometimes, direct questions are asked from this exercise as well in Boards. Also, students can refer to the following exercise of NCERT for further information about upcoming exercises.

• Statistics Exercise 15.1

• Statistics Exercise 15.2

• Statistics Miscellaneous Exercise

  Statistics Class 11 Chapter 15 -Exercise: 15.3

  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

  
  

  Answer:

  The group having a higher coefficient of variation will be more variable.

  Let the assumed mean, A = 45 and h = 10

  For Group A

  Marks	Group A $f_i$	Midpoint $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$ $= \frac{x_i-45}{10}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
  10-20	9	15	-3	9	-27	81
  20-30	17	25	-2	4	-34	68
  30-40	32	35	-1	1	-32	32
  40-50	33	45	0	0	0	0
  50-60	40	55	1	1	40	40
  60-70	10	65	2	4	20	40
  70-80	9	75	3	9	27	81
  	$\sum{f_i}$ =N = 150				$\sum f_iy_i$ = -6	$\sum f_iy_i ^2$ =342

  Mean,

  $\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6$

  We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

  $\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}$

  $\therefore \sigma = \sqrt{227.84} = 15.09$

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

  C.V.(A) = $\frac{15.09}{44.6}\times100 = 33.83$

  Similarly,

  For Group B

  Marks	Group A $f_i$	Midpoint $x_i$	$\dpi{100} y_i = \frac{x_i-A}{h}$ $= \frac{x_i-45}{10}$	$y_i^2$	$f_iy_i$	$f_iy_i^2$
  10-20	10	15	-3	9	-30	90
  20-30	20	25	-2	4	-40	80
  30-40	30	35	-1	1	-30	30
  40-50	25	45	0	0	0	0
  50-60	43	55	1	1	43	43
  60-70	15	65	2	4	30	60
  70-80	7	75	3	9	21	72
  	$\sum{f_i}$ =N = 150				$\sum f_iy_i$ = -6	$\sum f_iy_i ^2$ =375

  Mean,

  $\overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6$

  We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2$

  $\\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}$

  $\therefore \sigma = \sqrt{249.84} = 15.80$

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

  C.V.(B) = $\frac{15.80}{44.6}\times100 = 35.42$

  Since C.V.(B) > C.V.(A)

  Therefore, 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

  
  

  Answer:

  X( $x_i$ )	Y( $y_i$ )	$x_i^2$	$y_i^2$
  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	2401	10201
  =510	= 1050	=26360	=110290

  For X,

  Mean , $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51$

  Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]$

  $\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{35} = 5.91$

  C.V.(X) = $\frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58$

  Similarly, For Y,

  Mean , $\overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105$

  Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]$

  $\\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{4} = 2$

  C.V.(Y) = $\frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904$

  Since C.V.(Y) < C.V.(X)

  Hence Y is more stable.

  Question:3(i) 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 wage earners	$586$	$648$
  Mean of monthly wages	$Rs\hspace {1mm} 5253$	$Rs\hspace {1mm} 5253$
  Variance of the distribution of wages	$100$	$121$

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

  Answer:

  Given, Mean of monthly wages of firm A = 5253

  Number of wage earners = 586

  Total amount paid = 586 x 5253 = 30,78,258

  Again, Mean of monthly wages of firm B = 5253

  Number of wage earners = 648

  Total amount paid = 648 x 5253 = 34,03,944

  Hence firm B pays larger amount as monthly wages.

  Question:3(ii) 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 wage earners	$586$	$648$
  Mean of monthly wages	$Rs\hspace {1mm} 5253$	$Rs\hspace {1mm} 5253$
  Variance of the distribution of wages	$100$	$121$

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

  Answer:

  Given, Variance of firm A = 100

  Standard Deviation = $\sigma_A = \sqrt{Variance}= \sqrt{100} = 10$

  Again, Variance of firm B = 121

  Standard Deviation = $\sigma_B = \sqrt{Variance}= \sqrt{121} = 11$

  Since $\sigma_B>\sigma_A$ , firm B has 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?

  Answer:

  No. of goals scored $x_i$	Frequency $f_i$	$x_i^2$	$f_ix_i$	$f_ix_i^2$
  0	1	0	0	0
  1	9	1	9	9
  2	7	4	14	28
  3	5	9	15	45
  4	3	16	12	48
  	$\sum{f_i}$ =N = 25		$\sum f_ix_i$ = 50	$\sum f_ix_i ^2$ =130

  For Team A,

  Mean,

  $\overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i =\frac{50}{25}= 2$

  We know, Variance, $\sigma^2 = \frac{1}{N^2}\left [N\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]$

  $\\ \implies \sigma^2 = \frac{1}{(25)^2}\left [25(130) - (50)^2 \right ] \\ \\ = \frac{750}{625} =1.2$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}$

  $\therefore \sigma = \sqrt{1.2} = 1.09$

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

  For Team B,

  Mean = 2

  Standard deviation, $\sigma$ = 1.25

  C.V.(B) = $\frac{\sigma}{\overline x}\times100 = \frac{1.25}{2}\times100 = 62.5$

  Since C.V. of firm B is more than C.V. of A.

  Therefore, 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:

  $\sum_{i=1}^{50}x_i=212, \sum _{i=1}^{50}x_i^2=902.8,\sum _{i=1}^{50}y_i=261,\sum _{i=1}^{50}y_i^2=1457.6$
  Which is more varying, the length or weight?

  Answer:

  For lenght x,

  Mean, $\overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24$

  We know, Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]$

  $\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28$

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

  For weight y,

  Mean,

  Mean, $\overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22$

  We know, Variance, $\sigma^2 = \frac{1}{n^2}\left [n\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]$

  $\\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036$

  We know, Standard Deviation = $\sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37$

  C.V.(y) = $\frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24$

  Since C.V.(y) > C.V.(x)

  Therefore, weight is more varying.