NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

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The task of creating textbooks for all classes and subjects of the CBSE has been delegated to the National Council of Educational Research and Training, or NCERT. NCERT textbooks are accepted as the required reading for all exams by CBSE and other State Boards and institutions. The NCERT textbooks are designed to give learners the all-encompassing knowledge they need for their complete mental growth, helping them score great results.

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NCERT Solutions for Class 11 are accessible in the Extramarks for all topics. It is the one place where students can find every NCERT solution they need. Along with the other exercises in the chapter, the NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2 are provided.Examples and visuals are provided when needed to help explain the solutions more clearly. Thanks to the subject-matter experts at Extramarks, the NCERT Solutions for Class 11 Maths, Chapter 15, Exercise 15.2 are prepared in accordance with CBSE regulations.Students can familiarise themselves with the Extramarks pattern used in all of their NCERT Solutions by using NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2.

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NCERT Solutions Class 11 by Extramarks

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H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

One can access the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 from the Extramarks official website or mobile application. It is available every time one downloads it, as per their convenience.H3 – NCERT Solutions for Class 11 Maths Chapters

Extramarks provides NCERT Solutions for all chapters of Class 11 Mathematics in both Hindi and English. The NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2 in English are available on this page.Below are also the links to the different chapters of Class 1 Mathematics.

NCERT Solutions for Class 11 Mathematics Chapters

Chapter 1 – Sets
Chapter 2 – Relations and Functions
Chapter 3 – Trigonometric Functions
Chapter 4 – Principle of Mathematical Induction
Chapter 5 – Complex Numbers and Quadratic Equations
Chapter 6 – Linear Inequalities
Chapter 7 – Permutations and Combinations
Chapter 8 – Binomial Theorem
Chapter 9 – Sequences and Series
Chapter 10 – Straight Lines
Chapter 11 – Conic Sections
Chapter 12 – Introduction to 3D Geometry
Chapter 13 – Limits and Derivatives
Chapter 14 – Mathematical Reasoning
Chapter 15 – Statistics
Chapter 16 – Probability

The different topics covered under the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 are given below:

Limitations of mean deviation
Variance and standard deviation
Standard deviation
Standard deviation of a discrete frequency distribution
Standard deviation of a continuous frequency distribution
Evaluating variance and standard deviation using the shortcut method

H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

The NCERT Class 11 Chapter 15 has other exercises as well. Below is the breakdown of the different exercises and the number of questions in each one of them.

Exercise 15.1 – 12 questions and their solutions
Exercise 15.2 -10 questions and their solutions
Exercise 15.3 -5 questions and their solutions
Miscellaneous exercise – 7 questions and their solutions

H3 – All Topics Under NCERT Class 11 Maths for Exercise 15.2

Class 11 Mathematics NCERT Solution Chapter 15 – Statistics (Term I)

Statistics is an important chapter in advanced Mathematics. Students will discover some dispersion metrics and computation techniques for both grouped and ungrouped data in this chapter. Measures of dispersion, mean deviation, range, variance, and standard deviation of grouped and ungrouped data, as well as the study of frequency distributions, are only a few of the various ideas covered in this chapter. Students can easily get a better score after they are familiar with the themes and strategies for answering the questions.

The NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 deals with all these topics in a comprehensive yet easy-to-understand format.

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

These answers come with thorough explanations and examples drawn from actual scenarios to make them easier to recall. Students can look up NCERT Solutions for Class 11 Maths Chapter 15 Exercise 15.2 and other solutions for reference.

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The solutions are presented in clear, easy-to-understand language. As a result, students experience less stress as they prepare for upcoming exams.For ease of use by students, all the questions are answered in the NCERT Solutions For Class 11 Maths Chapter 15 Exercise 15.2 in a simple step-by-step manner.

Q.1 Find the mean and variance for each of the data in Exercies 1 to 5.

Q1. 6, 7, 10, 12, 13, 4, 8, 12
Q2. First n natural numbers.
Q3. First 10 multiples of 3.
Q4.

Xi	6	10	14	18	24	28	30
fi	2	4	7	12	8	4	3

Q5.

Xi	92	93	97	98	102	104	109
fi	3	2	3	2	6	3	3

Ans.

$\begin{array}{l} Mean (\bar{x}) = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} \\ = \frac{72}{8} \\ = 9 \end{array}$

$\begin{array}{l} \begin{matrix} x_{i} & (x_{i} - \bar{x}) & {(x_{i} - \bar{x})}^{2} \\ 6 & 6 - 9 = - 3 & 9 \\ 7 & 7 - 9 = - 2 & 4 \\ 10 & 10 - 9 = 1 & 1 \\ 12 & 12 - 9 = 3 & 9 \\ 13 & 13 - 9 = 4 & 16 \\ 4 & 4 - 9 = - 5 & 25 \\ 8 & 8 - 9 = - 1 & 1 \\ 12 & 12 - 9 = 3 & 9 \\ Total & 74 \end{matrix} \\ Variance (σ^{2}) = \frac{\sum {(x_{i} - \bar{x})}^{2}}{N} \\ = \frac{74}{8} \\ = 9 .25 \end{array}$

$\begin{array}{l} Mean of n natural numbers = \frac{\sum_{i = 1}^{n} x_{i}}{n} \\ = \frac{(\frac{n (n + 1)}{2})}{n} = \frac{n + 1}{2} \\ Variance (σ^{2}) = \frac{\sum_{i = 1}^{n} {x_{i}}^{2} - {(\bar{x})}^{2}}{n} \end{array}$

$\begin{array}{l} = \frac{(1^{2} + 2^{2} + 3^{2} + . .. + n^{2})}{n} - {(\frac{n + 1}{2})}^{2} \\ = \frac{1}{n} \times \frac{n (n + 1) (2 n + 1)}{6} - (\frac{n^{2} + 2 n + 1}{4}) \\ Variance (σ^{2}) = \frac{2 n^{2} + 3 n + 1}{6} - (\frac{n^{2} + 2 n + 1}{4}) \\ = \frac{4 n^{2} + 6 n + 2 - 3 n^{2} - 6 n - 3}{12} \\ = \frac{n^{2} - 1}{12} \end{array}$

$\begin{array}{l} Mean of 10 multiples of 3 \\ = \frac{(\begin{array}{l} 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 \\ + 27 + 30 \end{array})}{10} \\ = \frac{165}{10} = 16 .5 \\ Variance (σ^{2}) = \frac{\sum_{i = 1}^{n} {x_{i}}^{2} - {(\bar{x})}^{2}}{n} \\ = \frac{(\begin{array}{l} 3^{2} + 6^{2} + 9^{2} + 12^{2} + 15^{2} + 18^{2} \\ + 21^{2} + 24^{2} + 27^{2} + 30^{2} \end{array})}{10} - {(16 .5)}^{2} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{3465}{10} - 272 .25 \\ = 346 .5 - 272 .25 \\ ∴ Variance (σ^{2}) = 74 .25 \end{array}$

$\begin{array}{l} \begin{matrix} x_{i} & f_{i} & f_{i} x_{i} & (x_{i} - \bar{x}) & {(x_{i} - \bar{x})}^{2} & f_{i} {(x_{i} - \bar{x})}^{2} \\ 6 & 2 & 12 & 6 - 19 = - 13 & 169 & 338 \\ 10 & 4 & 40 & 10 - 19 = - 9 & 81 & 324 \\ 14 & 7 & 98 & 14 - 19 = - 5 & 25 & 175 \\ 18 & 12 & 216 & 18 - 19 = - 1 & 1 & 12 \\ 24 & 8 & 192 & 24 - 19 = 5 & 25 & 200 \\ 28 & 4 & 112 & 28 - 19 = 9 & 81 & 324 \\ 30 & 3 & 90 & 30 - 19 = 11 & 121 & 363 \\ Total & 40 & 760 & 1736 \end{matrix} \\ Mean (\bar{x}) = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{n} \\ = \frac{760}{40} \\ = 19 \end{array}$

$Variance (σ^{2}) = \frac{\sum_{i = 1}^{n} f_{i} {(x_{i} - \bar{x})}^{2}}{n}$

$\begin{array}{l} Variance (σ^{2}) = \frac{1736}{40} \\ = 43 .4 \end{array}$

$\begin{array}{l} \begin{matrix} x_{i} & f_{i} & f_{i} x_{i} & (x_{i} - \bar{x}) & {(x_{i} - \bar{x})}^{2} & f_{i} {(x_{i} - \bar{x})}^{2} \\ 92 & 3 & 276 & 92 - 100 = - 8 & 64 & 192 \\ 93 & 2 & 186 & 93 - 100 = - 7 & 49 & 98 \\ 97 & 3 & 291 & 97 - 100 = - 3 & 9 & 27 \\ 98 & 2 & 196 & 98 - 100 = - 2 & 4 & 8 \\ 102 & 6 & 612 & 102 - 100 = 2 & 4 & 24 \\ 104 & 3 & 312 & 104 - 100 = 4 & 16 & 48 \\ 109 & 3 & 327 & 109 - 100 = 9 & 81 & 243 \\ Total & 22 & 2200 & 640 \end{matrix} \\ Mean (\bar{x}) = \frac{\sum_{i = 1}^{n} f_{i} x_{i}}{n} \\ = \frac{2200}{22} \\ = 100 \\ Variance (σ^{2}) = \frac{\sum_{i = 1}^{n} f_{i} {(x_{i} - \bar{x})}^{2}}{n} \end{array}$

$\begin{array}{l} Variance (σ^{2}) = \frac{640}{22} \\ = 29 .09 \end{array}$

Q.2 Find the mean and standard deviation using short-cut method.

Xi	60	61	62	63	64	65	66	67	68
fi	2	1	12	29	25	12	10	4	5

Ans.

$\begin{matrix} x_{i} & f_{i} & d_{i} = \frac{x_{i} - A}{1} & f_{i} d_{i} & (x_{i} - \bar{x}) & {(x_{i} - \bar{x})}^{2} & f_{i} {(x_{i} - \bar{x})}^{2} \\ 60 & 2 & - 4 & - 8 & - 4 & 16 & 32 \\ 61 & 1 & - 3 & - 3 & - 3 & 9 & 9 \\ 62 & 12 & - 2 & - 24 & - 2 & 4 & 48 \\ 63 & 29 & - 1 & - 29 & - 1 & 1 & 29 \\ 64 = A & 25 & 0 & 0 & 0 & 0 & 0 \\ 65 & 12 & 1 & 12 & 1 & 1 & 12 \\ 66 & 10 & 2 & 20 & 2 & 4 & 40 \\ 67 & 4 & 3 & 12 & 3 & 9 & 36 \\ 68 & 5 & 4 & 20 & 4 & 16 & 80 \\ Total & 100 & 0 & 286 \end{matrix}$

$\begin{array}{l} Mean (\bar{x}) = A + \frac{\sum_{i = 1}^{n} f_{i} d_{i}}{n} \\ = 64 + \frac{0}{100} = 64 \\ Variance (σ^{2}) = \frac{\sum_{i = 1}^{n} f_{i} {(x_{i} - \bar{x})}^{2}}{n} \\ = \frac{286}{100} \\ = 2 .86 \\ Standard deviation \\ = \sqrt{2 .86} \\ = 1 .69 (Approx .) \end{array}$

Q.3 Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Q7.

Classes	0-30	30-60	60-90	90-120	120-150	150-180	180-210
Frequencies	2	3	5	10	3	5	2

Q8.

Classes	0-10	10-20	20-30	30-40	40-50
Frequencies	5	8	15	16	6

Ans.

$\begin{matrix} Classes & f_{i} & x_{i} & u_{i} = \frac{x_{i} - A}{30} & {u_{i}}^{2} & f_{i} u_{i} & f_{i} {u_{i}}^{2} \\ 0 - 30 & 2 & 15 & - 3 & 9 & - 6 & 18 \\ 30 - 60 & 3 & 45 & - 2 & 4 & - 6 & 12 \\ 60 - 90 & 5 & 75 & - 1 & 1 & - 5 & 5 \\ 90 - 120 & 10 & 105 = A & 0 & 0 & 0 & 0 \\ 120 - 150 & 3 & 135 & 1 & 1 & 3 & 3 \\ 150 - 180 & 5 & 165 & 2 & 4 & 10 & 20 \\ 180 - 210 & 2 & 195 & 3 & 9 & 6 & 18 \\ Total & 30 & 2 & 76 \end{matrix}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} M e a n = A + \frac{\sum_{i = 1}^{n} f_{i} u_{i}}{n} \times h \\ = 105 + \frac{2}{30} \times 30 \end{array}$

$\begin{array}{l} \end{array}$ $\begin{array}{l} = 107 \\ Variance (σ^{2}) = \frac{h^{2}}{n^{2}} [n \sum_{i}^{n} (f_{i} {u_{i}}^{2}) - {(\sum_{i}^{n} f_{i} u_{i})}^{2}] \\ = \frac{{(30)}^{2}}{{(30)}^{2}} [30 \times 76 - {(2)}^{2}] \\ = 2280 - 4 = 2276 \end{array}$

$\begin{array}{l} \begin{matrix} Classes & f_{i} & x_{i} & u_{i} = \frac{x_{i} - A}{10} & {u_{i}}^{2} & f_{i} u_{i} & f_{i} {u_{i}}^{2} \\ 0 - 10 & 5 & 5 & - 2 & 4 & - 10 & 20 \\ 10 - 20 & 8 & 15 & - 1 & 1 & - 8 & 8 \\ 20 - 30 & 15 & 25 = A & 0 & 0 & 0 & 0 \\ 30 - 40 & 16 & 35 & 1 & 1 & 16 & 16 \\ 40 - 50 & 6 & 45 & 2 & 4 & 12 & 24 \\ Total & 50 & 10 & 68 \end{matrix} \\ Mean = A + \frac{\sum_{i = 1}^{n} f_{i} u_{i}}{n} \times h \\ = 25 + \frac{10}{50} \times 10 \\ = 27 \\ Variance (σ^{2}) = \frac{h^{2}}{n^{2}} [n \sum_{i}^{n} (f_{i} {u_{i}}^{2}) - {(\sum_{i}^{n} f_{i} u_{i})}^{2}] \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = \frac{{(10)}^{2}}{{(50)}^{2}} [50 \times 68 - {(10)}^{2}] \\ = \frac{1}{25} [3400 - 100] \\ = 132 \end{array}$

Q.4 Find the mean, variance and standard deviation using short-cut method

Height in cms	70-75	75-80	80-85	85-90	90-95	95-100	100-105	105-110	110-115
No. of children	3	4	7	7	15	9	6	6	3

Ans.

$\begin{matrix} C l a s s e s & f_{i} & x_{i} & u_{i} = \frac{x_{i} - A}{5} & u_{i}^{2} & f_{i} u_{i} & f_{i} u_{i}^{2} \\ 70 - 75 & 3 & 72.5 & - 4 & 16 & - 12 & 48 \\ 75 - 80 & 4 & 77.5 & - 3 & 9 & - 12 & 36 \\ 80 - 85 & 7 & 82.5 & - 2 & 4 & - 14 & 28 \\ 85 - 90 & 7 & 87.5 & - 1 & 1 & - 7 & 7 \\ 90 - 95 & 15 & 92.5 = A & 0 & 0 & 0 & 0 \\ 95 - 100 & 9 & 97.5 & 1 & 1 & 9 & 9 \\ 100 - 105 & 6 & 102.5 & 2 & 4 & 12 & 24 \\ 105 - 110 & 6 & 107.5 & 3 & 9 & 18 & 54 \\ 110 - 115 & 3 & 112.5 & 4 & 16 & 12 & 78 \\ T o t a l & 60 & 6 & 254 \end{matrix}$

$Mean = A + \frac{\sum_{i = 1}^{n} f_{i} u_{i}}{n} \times h$ $\begin{array}{l} \end{array}$ $\begin{array}{l} \end{array}$

$\begin{array}{l} = 92 .5 + \frac{6}{60} \times 5 \\ = 92 .5 + 0 .5 \\ = 93 \end{array}$

$\begin{array}{l} Variance (σ^{2}) = \frac{h^{2}}{n^{2}} [n \sum_{i}^{n} (f_{i} {u_{i}}^{2}) - {(\sum_{i}^{n} f_{i} u_{i})}^{2}] \\ = \frac{{(5)}^{2}}{{(60)}^{2}} [60 \times 254 - {(6)}^{2}] \\ = \frac{25}{3600} [15240 - 36] \\ = \frac{25}{3600} \times 15204 \\ = 105 .52 \\ Standard deviation \\ = \sqrt{105 .52} \\ = 10 .27 \end{array}$

Q.5 The diameters of circles (in mm) drawn in a design are given below:

Diameters	33-36	37-40	41-44	45-48	49-52
No. of circles	15	17	21	22	25

Calculate the standard deviation and mean diameter of the circles.

Ans.

$\begin{array}{l} \begin{matrix} \begin{matrix} Classe-interva \end{matrix} & \begin{matrix} f_{i} \end{matrix} & \begin{matrix} x_{i} \end{matrix} & u_{i} = \frac{x_{i} -A}{4} & {u_{i}}^{2} & f_{i} u & f_{i} {u_{i}}^{2} \\ 32.5-36.5 & 15 & 34.5 & -2 & 4 & -30 & 60 \\ 36.5-40.5 & 17 & 38.5 & -1 & 1 & -17 & -17 \\ 40.5-44.5 & 21 & 42.5=A & 0 & 0 & 0 & 0 \\ 44.5-48.5 & 22 & 46.5 & 1 & 1 & 22 & 22 \\ 48.5-52.5 & 25 & 50.5 & 2 & 4 & 50 & 100 \\ Total & 100 & 25 & 199 \end{matrix} \\ Variance (σ^{2}) = \frac{h^{2}}{n^{2}} [n \sum_{i}^{n} (f_{i} {u_{i}}^{2}) – {(\sum_{i}^{n} f_{i} u_{i})}^{2}] \\ = \frac{{(4)}^{2}}{{(100)}^{2}} [100×199- {(25)}^{2}] \\ = \frac{1}{625} [19900-625] \\ = \frac{1}{625} ×19275 \\ = 30.84 \\ Standard deviation \\ = \sqrt{30.84} \\ = 5.55 \\ Mean = A+ \frac{\sum_{i=1}^{n} f_{i} u_{i}}{n} ×h \\ = 42.5+ \frac{25}{100} ×4 \\ = 42.5+1 \\ = 43.5 \end{array}$

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

H3 – Access NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

H3 – NCERT Solution Class 11 Maths of Chapter 15 All Exercises

H3 – All Topics Under NCERT Class 11 Maths for Exercise 15.2

H2 – NCERT Solutions for Class 11 Maths Chapter 15 Statistics Exercise 15.2

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NCERT Solutions for Class 11 Maths Chapter 15 Statistics (Ex 15.2) Exercise 15.2

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