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# NCERT Solutions for Class 11 Maths Chapter 15: Statistics - Exercise 15.3

Last updated date: 10th Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 15 Statistics

Free PDF download of NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.3 (Ex 15.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 11 Maths Chapter 15 Statistics Exercise 15.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

 Class: NCERT Solutions for Class 11 Subject: Class 11 Maths Chapter Name: Chapter 15 - Statistics Exercise: Exercise - 15.3 Content-Type: Text, Videos, Images and PDF Format Academic Year: 2024-25 Medium: English and Hindi Available Materials: Chapter WiseExercise Wise Other Materials Important QuestionsRevision Notes
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## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

1. From the data given below, 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

Ans: First we have to find the standard deviation of group A.

Standard deviation of A is calculate as follows,

 Marks Group A$\left( {{f_i}} \right)$ Mid-point$\left( {{x_i}} \right)$ ${y_i} = \dfrac{{{x_i} - 45}}{{10}}$ ${y_i}^2$ ${f_i}{y_i}$ ${f_i}{y_i}^2$ 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 81 150 $- 6$ 342

Here,

$h = 10$

$N = 150$

$A = 45$

${\text{Mean = }}A + \dfrac{{\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} }}{N} \times h$

${\text{Mean = }}45 + \dfrac{{\left( { - 6} \right)}}{{150}} \times 10$

${\text{Mean = }}45 - 0.4$

${\text{Mean = }}44.6$

Calculating standard deviation,

${\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} - {{\left( {\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} } \right)}^2}} \right]$

${\sigma ^2} = \dfrac{{100}}{{22500}}\left[ {150 \times 342 - {{\left( { - 6} \right)}^2}} \right]$

${\sigma ^2} = \dfrac{1}{{225}}\left( {51264} \right)$

$\sigma = \sqrt {227.84}$

$\sigma = 15.09$

$\sigma = \sqrt {227.84}$

$\sigma = 15.09$

Now we have to find the standard deviation of group B.

Standard deviation of B is calculate as follows,

 Marks Group A$\left( {{f_i}} \right)$ Mid-point$\left( {{x_i}} \right)$ ${y_i} = \dfrac{{{x_1} - 45}}{{10}}$ ${y_i}^2$ ${f_i}{y_i}$ ${f_i}{y_i}^2$ 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 63 150 $- 6$ 366

${\text{Mean = }}A + \dfrac{{\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} }}{N} \times h$

${\text{Mean = }}45 + \dfrac{{\left( { - 6} \right)}}{{150}} \times 10$

${\text{Mean = }}45 - 0.4$

${\text{Mean = }}44.6$

Calculating standard deviation,

${\sigma ^2} = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} - {{\left( {\sum\limits_{i = 1}^{i = 7} {{x_i}{f_i}} } \right)}^2}} \right]$

${\sigma ^2} = \dfrac{{100}}{{22500}}\left[ {150 \times 366 - {{\left( { - 6} \right)}^2}} \right]$

${\sigma ^2} = \dfrac{1}{{225}}\left( {54846} \right)$

${\sigma ^2} = 243.84$

$\sigma = \sqrt {243.84}$

$\sigma = 15.61$

As the mean of both the groups is same, therefore, the group with greater standard deviation is more variable.

Therefore, group B is more variable.

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

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

Ans: To compare the stability of both shares we have to calculate the standard deviation of both shares.

Coefficient of variance i.e. C.V. of share X can be calculated as,

The prices of shares X are,

$35,54,52,53,56,58,52,50,51,49$

Total number of observation is 10

Mean,

$\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{x_i}}$

$\overline x = \dfrac{1}{{10}} \times 510$

$\overline x = 51$

 ${x_i}$ $\left( {{x_i} - \overline x } \right)$ ${\left( {{x_i} - \overline x } \right)^2}$ 35 $- 16$ 256 54 3 9 52 1 1 53 2 4 56 5 25 58 7 49 52 1 1 50 $- 1$ 1 51 0 0 49 $- 2$ 4 350

Variance of shares of X can be calculated as,

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \overline x } \right)}^2}}$

${\sigma ^2} = \dfrac{1}{{10}} \times 350$

${\sigma ^2} = 35$

$\sigma = \sqrt {35}$

$\sigma = 5.91$

$C.V.{\text{ of shares X }} = \dfrac{\sigma }{{\overline x }} \times 100$

$C.V.{\text{ of shares X}} = \dfrac{{5.91}}{{51}} \times 100$

$C.V.{\text{ of shares X}} = 11.58$

Coefficient of variance i.e. C.V. of share Y can be calculated as,

The prices of shares Y are,

$108,107,105,105,106,107,104,103,104,101$

Total number of observation is 10

Mean,

$\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{x_i}}$

$\overline x = \dfrac{1}{{10}} \times 1050$

$\overline x = 105$

 ${y_i}$ $\left( {{y_i} - \overline y } \right)$ ${\left( {{y_i} - \overline y } \right)^2}$ 108 3 9 107 2 4 105 0 0 105 0 0 106 1 1 107 2 4 104 $- 1$ 1 103 $- 2$ 4 104 $- 1$ 1 101 $- 4$ 16 40

Variance of shares of Y can be calculated as,

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{{\left( {{y_i} - \overline y } \right)}^2}}$

${\sigma ^2} = \dfrac{1}{{10}} \times 40$

${\sigma ^2} = 4$

$\sigma = \sqrt 4$

$\sigma = 2$

$C.V.{\text{ of shares X }} = \dfrac{\sigma }{{\overline x }} \times 100$

$C.V.{\text{ of shares X}} = \dfrac{4}{{105}} \times 100$

$C.V.{\text{ of shares X}} = 1.9$

On comparing the $C.V.$ of both X and Y,

$C.V.$of X is greater than $C.V.$ of Y.

Therefore, the price of Y is more stable than the price of X.

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 amounts as monthly wages ?

Ans: Monthly wages of firm A is Rs. 5253.

Total number of wage earners in firm A is 586.

Therefore, total amount paid in firm A is ${\text{Rs}}{\text{. }}5253 \times 586 = {\text{ Rs}}{\text{. }}3078258$ .

Monthly wages of firm B is Rs. 5253.

Total number of wage earners in firm B is 648.

Therefore, total amount paid in firm B is ${\text{Rs}}{\text{. }}5253 \times 648 = {\text{ Rs}}{\text{. }}3403944$ .

Therefore, firm B pays a larger amount as monthly wages.

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

Ans: Variance of distribution $\left( {{\sigma ^2}} \right)$ of wages of firm A is 100.

So, the standard deviation of the wages in firm A is

${\sigma ^2} = 100$

$\sigma = \sqrt {100}$

$\sigma = 10$

Variance of distribution $\left( {{\sigma ^2}} \right)$ of wages of firm B is 121.

So, the standard deviation of the wages in firm B is

${\sigma ^2} = 121$

$\sigma = \sqrt {121}$

$\sigma = 11$

Since, the standard deviation of firm B is greater than firm B,

Therefore, firm B has greater variability in the individual wages.

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

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

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

Ans: We have to compare team A and B, to see which team is more consistent.

For that we will calculate standard deviation of team A.

Calculation of Standard Deviation of Team A:

 No. goals scored$\left( {{x_i}} \right)$ No of matches$\left( {{f_i}} \right)$ ${f_i}{x_i}$ ${x_i}^2$ ${f_i}{x_i}^2$ 0 1 0 0 0 1 9 9 1 9 2 7 14 4 28 3 5 15 9 45 4 3 12 16 48 25 130

Mean,

$\overline x = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{f_i}{x_i}}$

$\overline x = \dfrac{1}{{25}} \times 50$

$\overline x = 2$

Standard deviation of A is,

$\sigma = \dfrac{1}{N}\sqrt {N\sum {{x_i}{f_i}} - {{\left( {\sum {{x_i}{f_i}} } \right)}^2}}$

$\sigma = \dfrac{1}{{25}}\sqrt {25 \times 130 - {{\left( {50} \right)}^2}}$

$\sigma = \dfrac{1}{{25}}\sqrt {750}$

$\sigma = 1.09$

Standard deviation of team A is $1.09$ .

Standard deviation of team B is $1.25$ .

Since the mean of both teams is the same i.e. 2.

Standard deviation of team B is greater than team A.

Therefore, team A is more consistent.

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:

$\sum\limits_{i = 1}^{50} {{x_i}} = 212,\sum\limits_{i = 1}^{50} {{x_i}^2} = 902.8,\sum\limits_{i = 1}^{50} {{y_i}} = 261,\sum\limits_{i = 1}^{50} {{y_i}^2} = 1457.6$

Which is more varying, the length or weight?

Ans: Calculating standard deviation and coefficient of length.

$\sum\limits_{i = 1}^{50} {{x_i}} = 212,\sum\limits_{i = 1}^{50} {{x_i}^2} = 902.8$

Here, we have given that,

$N = 50$

Therefore, mean will be

$\overline x = \dfrac{{\sum\limits_{i = 1}^{50} {{x_i}} }}{N}$

$\overline x = \dfrac{{212}}{{50}}$

$\overline x = 4.24$

Now, calculating variance,

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - \overline x } \right)}^2}}$

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {\left( {{x_i}^2 + {{\overline x }^2} - 2{x_i}.\overline x } \right)}$

${\sigma ^2} = \dfrac{1}{N}\left( {\sum\limits_{i = 1}^{50} {{x_i}^2} + \sum\limits_{i = 1}^{50} {{{\left( {\overline x } \right)}^2}} - \sum\limits_{i = 1}^{50} {2{x_i}.\overline x } } \right)$

${\sigma ^2} = \dfrac{1}{{50}}\left( {902.8 + 17.97 - 2 \times 4.24 \times 212} \right)$

${\sigma ^2} = \dfrac{1}{{50}}\left( {3.54} \right)$

${\sigma ^2} = 0.07$

$\sigma = \sqrt {0.07}$

$\sigma = 0.26$

So, the standard deviation of length is $0.26$ .

${\text{C}}{\text{.V}}{\text{. = }}\dfrac{\sigma }{{\overline x }} \times 100$

${\text{C}}{\text{.V}}{\text{. = }}\dfrac{{0.26}}{{4.24}} \times 100$

${\text{C}}{\text{.V}}{\text{. = }}6.13$

So, the coefficient of variance of length is $6.13$ .

Calculating standard deviation and coefficient of variance of weigth.

$\sum\limits_{i = 1}^{50} {{y_i}} = 261,\sum\limits_{i = 1}^{50} {{y_i}^2} = 1457.6$

Here, we have given that,

$N = 50$

Therefore, mean will be

$\overline y = \dfrac{{\sum\limits_{i = 1}^{50} {{y_i}} }}{N}$

$\overline y = \dfrac{{261}}{{50}}$

$\overline y = 5.22$

Now, calculating variance,

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - \overline y } \right)}^2}}$

${\sigma ^2} = \dfrac{1}{N}\sum\limits_{i = 1}^{50} {\left( {{y_i}^2 + {{\overline y }^2} - 2{y_i}.\overline y } \right)}$

${\sigma ^2} = \dfrac{1}{N}\left( {\sum\limits_{i = 1}^{50} {{y_i}^2} + \sum\limits_{i = 1}^{50} {{{\left( {\overline y } \right)}^2}} - \sum\limits_{i = 1}^{50} {2{y_i}.\overline y } } \right)$

${\sigma ^2} = \dfrac{1}{{50}}\left( {1457.6 + 27.24 - 2 \times 5.22 \times 261} \right)$

${\sigma ^2} = \dfrac{1}{{50}}\left( {94.76} \right)$

${\sigma ^2} = 1.89$

$\sigma = \sqrt {1.89}$

$\sigma = 1.37$

So, the standard deviation of weight is $1.37$ .

${\text{C}}{\text{.V}}{\text{. = }}\dfrac{\sigma }{{\overline y }} \times 100$

${\text{C}}{\text{.V}}{\text{. = }}\dfrac{{1.37}}{{5.22}} \times 100$

${\text{C}}{\text{.V}}{\text{. = 26}}{\text{.24}}$

So, the coefficient of variance of weight is ${\text{26}}{\text{.24}}$ .

Since the coefficient of variance of weight is greater than the coefficient of variance of length.

Therefore, weight varies more than length.

## NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.3

Opting for the NCERT solutions for Ex 15.3 Class 11 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 15.3 Class 11 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 11 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 11 Maths Chapter 15 Exercise 15.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 11 Maths Chapter 15 Exercise 15.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 15: Statistics - Exercise 15.3

1. Where can I obtain the most up-to-date NCERT Solutions for Chapter 15: Introduction to Statistics (Ex. 15.3), Exercise 15.3 in Class 11 Mathematics?

You can study the range in NCERT Chapter 15, which discusses statistics. For all of the Class 11 Math chapters, NCERT Solutions are available on Vedantu, India's top e-learning website. These solutions were created specifically by Vedantu subject experts in keeping with CBSE standards. These answers are clear and entirely truthful. For a free PDF copy of these study resources, go to the Vedantu website or mobile app.

2. What is a Range?

The difference between the highest and lowest values for a given data collection is the range in statistics. For example, if the given data set is 2, 5, 8, 10, 3, the range is 10 - 2 = 8. As a result, the range may alternatively be regarded as the distance between the highest and lowest observation.

3. What is a standard deviation?

The variance's positive square root is the standard deviation. One of the basic methods used in statistical analysis is the standard deviation. Commonly abbreviated as SD, the standard deviation indicates how much a number deviates from the mean value in order to provide information about it.

4. Why should one choose Vedantu's NCERT Solutions for Class 11 Math, Chapter 15: Statistics?

The NCERT solutions for Ex. 15.3 from Class 11 Maths from Vedantu are regarded as the best option for CBSE students when it comes to exam preparation. This chapter has many exercises. The solutions to Exercise 15.3 from the Class 11 Math NCERT are available on this page in PDF format. You can download and study this solution directly from the Vedantu website or app, or you can download it as necessary.

5. Do I need to solve NCERT Solutions for Class 11 Maths, Chapter 15: Statistics (Ex. 15.3) Exercise 15.3?

Studying only chapters in a subject like mathematics is not enough; you must solve the respective exercises by yourself first. The more you practice, the better you will become at this subject. When you begin to solve problems, you will be able to identify your weak points that need to be strengthened. You can refer to Vedantu for NCERT Solutions for Class 11 Maths, Chapter 15: Statistics (Ex. 15.3) Exercise 15.3 for a better understanding and to learn various approaches to solving any problem.