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NCERT Solutions for Class 11 Maths Chapter 15 : Statistics

Q: Q1. Explain the difference between Variance and Standard Deviation.

Answer: The standard deviation of a set of numbers is the distance between them and the mean. The variance is a measure of how much each point deviates from the mean on average. Variance is the average of all data points inside a group, whereas standard deviation is the square root of the variance.

Q: Q2. How many exercises are there in Statistics Class 11 NCERT Solutions?

Answer: The NCERT Solutions Class 11 Maths Chapter 15 Statistics has 34 sums separated between three exercises and one miscellaneous activity. They range from simple formula-based problems to difficult word problems. The solutions to the exercise-specific questions are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.

Q: Q3. What are the key subtopics in Statistics Class 11 Maths Chapter 15 that could be tested?

Answer: All the topics covered in NCERT Solutions Class 11 Maths Chapter 15 are equally important. Statistics is a topic that is commonly used in the financial and economic industries for planning and forecasting. They aid in the development of mathematical models and the analysis of time series data. It is not only an important lesson for exams but it is also required of students in the 12th grade. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q: Q4. Is it necessary to practice all of the questions in Statistics Class 11 NCERT Solutions?

Answer: Statistics Class 11 NCERT Solutions are well-designed to help students study basic statistical concepts easily. It thoroughly covers all essential principles, allowing students to grasp the concepts. Students must memorize multiple formulas and do complex calculations. All of the exercises, examples, and practice problems have been meticulously designed to help students gain a rapid and effective comprehension of the various topics.

Statistics Class 11 Maths Chapter 15 deals with the study of data collection for specific purposes. After that, we can make decisions based on the data by analyzing and interpreting it. Class 11 Maths Chapter 15 explains the key concepts of statistics. Class 11 Maths Chapter 15 Statistics will help students to gain a thorough understanding of this concept and its practical applications. The expert faculty of Toppr produced these solutions to assist students with their first-term exam preparations.

Statistics Class 11 Chapter 15 also covers topics such as Measures of dispersions, range, mean deviation, variance of grouped and ungrouped data, variance and standard deviation of discrete and continuous frequency distributions, and coefficient of variation. Students will quickly learn about these concepts if they practice Chapter 15 Maths Class 11 NCERT Solutions of Statistics. All of these solutions have been created with the new CBSE pattern in mind, ensuring that students have a thorough understanding of their exams. It goes over all major concepts in depth, allowing students to better understand the ideas.

Chapter 15 Maths Class 11 Questions and Answers can help you get good grades in tests and properly prepare for all of the important concepts. Several examples provided in the solutions will assist students with a better understanding of Statistics. Statistics Class 11 NCERT Solutions Chapter 15 is available below in pdf format, and a few solutions are also included in the exercises. These solutions explain the topics covered with examples so that students can easily relate to the notion being discussed.

Table of Content

Introduction

Access NCERT Solutions for Class 11 Maths Chapter 15 : Statistics

Exercise 15.1

Question 1

Find the mean deviation about the mean for the data:

4, 7, 8, 9, 10, 12, 13, 17

Medium

Solution

Verified by Toppr

M e a n (\overline{x}) = \frac{4 + 7 + 8 + 9 + 1 0 + 1 2 + 1 3 + 1 7}{8} = \frac{8 0}{8} = 10

$x_{i}$	$∣ x_{i} - \overline{x} ∣$
$4$	$6$
$7$	$3$
$8$	$2$
$9$	$1$
$10$	$0$
$12$	$2$
$13$	$3$
$17$	$7$
	$\sum ∣ x_{i} - \overline{x} ∣ = 24$

\Rightarrow

Mean deviation about mean

= \frac{2 4}{8} = 3

Question 2

Find the mean deviation about the mean for the data

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Medium

Solution

Verified by Toppr

The given data is

38, 70, 48, 40, 42, 55, 63, 46, 54, 44

Mean

\overset{x}{ˉ} = \frac{3 8 + 7 0 + 4 8 + 4 0 + 4 2 + 5 5 + 6 3 + 4 6 + 5 4 + 4 4}{1 0}

= \frac{5 0 0}{1 0} = 50

The deviations of the respective observation from the mean

\overset{x}{ˉ},

i.e.

x_{i} = \overset{x}{ˉ}

are

- 12, 20, - 2, - 10, - 8, 5, 13, - 4, 4, - 6

.

The absolute values of the deviations i.e.

∣ x_{1} - \overset{x}{ˉ} ∣

are

12, 20, 2, 10, 8, 5, 13, 4, 4, 6

The required mean deviation about the mean is

M . D . = \frac{\sum _{i = 1}^{10} ∣ x _{i} - x ˉ ∣}{1 0}

= \frac{1 2 + 1 0 + 2 + 1 0 + 8 + 5 + 1 3 + 4 + 4 + 6}{1 0}

= \frac{8 4}{1 0}

= 8.4

Question 3

Find the mean deviation about the median for the data

13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

Medium

Solution

Verified by Toppr

Given: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 19

n = 12

median =

\frac{6 ^{t h} o b s + 7 ^{t h} o b s}{2}

= \frac{1 3 + 1 4}{2} = \frac{2 7}{2} = 13.5

mean deviation about median

\frac{( 1 1 0 - 1 3 . 5 1 + 1 1 1 . 1 3 . 5 1 + 1 1 1 - 1 3 . 5 1 + 1 1 2 - 1 3 . 5 1 + 1 1 3 - 1 3 . 5 1 + 1 1 3 - 1 3 . 5 1 + 1 1 4 - 1 3 . 5 1 + 1 1 6 - 1 3 . 5 + 1 1 6 - 1 3 . 5 1 + 1 1 7 - 1 3 . 5 1 + 1 1 7 - 1 3 . 5 1 + 1 1 8 - 1 3 . 5 1 )}{1 2}

\frac{3 . 5 + 2 . 5 + 2 . 5 + 1 . 5 + 0 . 5 + 0 . 5 + 0 . 5 + 2 . 5 + 2 . 5 + 3 . 5 + 3 . 5 + 4 . 5}{1 2}

= \frac{2 8}{1 2} = 2.23

Question 4

Find the mean deviation about the median for the data

36, 72, 46, 42, 60, 45, 53, 46, 51, 49

Medium

Solution

Verified by Toppr

Given =36,72,46,42,60,45,53,46,51,49

n = 10

median

= \frac{( \frac{n}{2} ) ^{t h} o b s + ( \frac{n}{2} + 1 ) ^{t h}}{2} o b s

\frac{5 ^{t h} o b s + 6 ^{t h} o b s}{2}

= \frac{4 6 + 4 9}{2} = \frac{9 5}{2} = 47.5

mean deviation above median =

\frac{ϵ ∣ x : - m ∣}{1 0}

\frac{( 1 3 6 - 4 7 . 5 1 + 1 4 2 - 4 7 . 5 1 + 1 4 5 - 4 7 . 5 1 + 1 4 6 - 4 7 . 5 1 + 1 4 6 - 4 7 . 5 1 + 1 4 9 - 4 7 - 5 1 + 1 5 1 - 4 7 . 5 1 + 1 5 3 - 4 7 . 5 1 + 1 6 0 - 4 7 . 5 1 + 1 7 2 - 4 7 . 5 1 )}{1 0}

\frac{1 1 . 5 + 5 . 5 + 2 . 5 + 1 . 5 + 1 . 5 + 1 . 5 + 3 . 5 + 5 . 5 + 1 2 . 5 + 2 4 . 5}{1 0}

\frac{7 0}{1 0} = 7

Question 5

Find the mean deviation about the mean for the data in Exercises 5 and 6.

x_{i}

5 10 15 20 25

f_{i}

7 4 6 3 5

Hard

Solution

Verified by Toppr

Question 6

Find the mean deviation about the mean for the data in

$x_{i}$	$10$	$30$	$50$	$70$	$90$
$f_{i}$	$4$	$24$	$28$	$16$	$8$

Medium

Solution

Verified by Toppr

x f f_{1} x_{1} - d m ∣ x - \overline{x}_{m} ∣ f (x - x_{m}) 1044024.2597 30247204.25102 502814015.75441 7016112035.75572 90 \underline{8} \underline{720} \underline{55.75} \underline{446} N = 8027401658 \overline{x}_{m} = \frac{2 7 4 0}{8 0} = 34.25 M . D = \frac{\sum f ∣ x - x _{m} ∣}{N} = \frac{1 6 5 8}{8 0} = 20.725

Question 7

Find the mean deviation from the mean for the following data:

$x_{i}$	$5$	$7$	$9$	$10$	$12$	$15$
$f_{i}$	$8$	$6$	$2$	$2$	$2$	$6$

Medium

Solution

Verified by Toppr

Computation of mean deviation about mean

$x_{i}$	$f_{i}$	$x_{i} f_{i}$	$∣ x_{i} - 9 ∣$	$\sum f_{i} ∣ x_{i} - 9 ∣$
$5$	$8$	$40$	$4$	$32$
$7$	$6$	$42$	$2$	$12$
$9$	$2$	$18$	$0$	$0$
$10$	$2$	$20$	$1$	$2$
$12$	$2$	$24$	$3$	$6$
$15$	$6$	$90$	$6$	$3$
	$N = \sum f_{i} = 26$	$\sum f_{i} x_{i} = 234$		$= \sum f_{i} ∣ x_{i} - 9 ∣ = 88$

We have,

N = \sum f_{i} = 26

, and

\sum f_{i} x_{i} = 234

∴

Mean

= \overset{ˉ}{X} = \frac{1}{N} \sum f_{i} x_{i} = \frac{2 3 4}{2 6} = 9

Mean deviation

= \frac{1}{N} \sum f_{i} ∣ x_{i} - 9 ∣ = \frac{2 3 4}{2 6} = 9

Mean deviation

= \frac{1}{N} = \sum f_{i} ∣ x_{i} - 9 = ∣ \frac{8 8}{2 6} = 3.38

Question 8

Find the mean deviation from the median for the following data:

$x_{i}$	$15$	$21$	$27$	$30$
$f_{i}$	$3$	$5$	$6$	$7$

Medium

Solution

Verified by Toppr

$x_{i}$	$f_{i}$	$C . F$	$X_{i} - 30$	$f_{i} ∣ X i - 30 ∣$
15	3	3	15	45
21	5	8	9	45
27	6	14	7	42
30	7	21	0	0
35	8	29	5	40
	$\sum f_{i} = 29$			$\sum f_{i} ∣ X_{i} - 30 ∣ = 172$

We have

N = 29

\Rightarrow \frac{N}{2} = 29

\Rightarrow N = 14.5

The cumulative frequency just greater than

\frac{N}{2}

i.e,

14.5

21

. The corresponding value ob variable is

30

. So median

= 30

Mean deviation

= \frac{1}{N} \sum f_{i} ∣ x_{i} - 30 ∣

= \frac{1 7 2}{2 9}

= 5.93

∴

The given answer is correct.

Question 9

Find the mean deviation about the mean for the data:

Income per day	Numbers of persons
0-100	4
100-200	8
200-300	9
300-400	10
400-500	7
500-600	5
600-700	4
700-800	3

Medium

Solution

Verified by Toppr

Income per day	Number of persons $(f_{i})$	Mid-point $X_{i}$	$(f_{i} X_{i})$	$∣ X_{i} - \overset{ˉ}{X} ∣$	$f_{i} ∣ X_{i} - \overset{ˉ}{X} ∣$
0-100	4	50	200	308	1232
100-200	8	150	1200	208	1664
200-300	9	250	2250	108	972
300-400	10	350	3500	8	80
400-500	7	450	3150	92	644
500-600	5	550	2750	192	960
600-700	4	650	2600	292	1168
	50		17900		7896

Here

N = i = 1 \sum 8 f_{i} = 50

\sum_{i = 1}^{8} f_{1} x_{i} = 17900

∴ \overset{ˉ}{X} = \frac{1}{N} i = 1 \sum 8 f_{i} x_{i} = \frac{1}{5 0} \times 17900 = 358

∴ M . D . (\overset{x}{ˉ}) = \frac{1}{N} i = 1 \sum 8 f_{i} ∣ x_{1} - \overset{x}{ˉ} ∣ = \frac{1}{5 0} \times 7896 = 157.92

Question 10

Find the mean deviation about the mean for the data

Height of cms	Number of boys
95-105	9
105-115	13
115-125	26
125-135	30
135-145	12
145-155	10

Medium

Solution

Verified by Toppr

The following table is formed

Height of cms	Number of boys $f_{i}$	mid-point $X_{i}$	$f_{i} \times X_{i}$	$∣ x_{i} - \overset{x}{ˉ} ∣$	$f_{i} ∣ x_{i} - \overset{x}{ˉ} ∣$
95-105	9	100	900	25.3	227.7
105-115	13	110	1430	14.3	198.9
115-125	26	120	3120	5.3	137.8
125-135	30	130	3900	4.7	141
135-145	12	140	1680	14.7	176.4
145-155	10	150	1500	24.7	247

Here

N = i = 1 \sum 6 f_{i} = 100, i = 1 \sum 6 f_{1} x_{i} = 12530

∴ \overset{ˉ}{X} = \frac{1}{N} i = 1 \sum 6 f_{i} x_{i} = \frac{1}{1 0 0} \times 12530 = 125.3

∴ M . D . (\overset{x}{ˉ}) = \frac{1}{N} i = 1 \sum 6 f_{i} ∣ x_{1} - \overset{x}{ˉ} ∣ = \frac{1}{1 0 0} \times 1128.8 = 11.28

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NCERT Solutions for Class 11 Maths Chapter 15 : Statistics

NCERT Solutions for Class 11 Maths Chapter 15 Statistics – Brief Overview

Statistics

It is defined as the process of gathering and categorizing data, interpreting and presenting data, and analyzing data. Statistics is sometimes defined as extracting inferences from sample data gathered through trials.

Measures of Dispersion

The dispersion or scatter in data is measured based on the observations and the types of central tendency measures used.

Range - The range is defined as the difference between the maximum and minimum values in the given data set.

Range = Maximum Value – Minimum Value

Mean Deviation - The difference between the observed and expected values of a data point is defined as the mean deviation.

Mean Deviation = Sum of absolute values of deviations from 'a' / no. of observations

Mean Deviation for ungrouped data = ∑|x_i - x̅| / n

Mean Deviation for grouped data = ∑|x_i - x̅| / N; where N = ∑ f_i

Variance

The variance of data values is a measure of how far they deviate from the mean. It is denoted by σ²

Variance (σ²) = ∑(x_i - x̅)² / n

Standard Deviation

The standard deviation is a measure of statistical data distribution. It is written as the positive square root of the variance.

Standard deviation = σ

Coefficient of Variation

The coefficient of variation is a unit-independent measure of variability (denoted as C.V.). The coefficient of variation is defined as,

C.V. = ( σ / x̅ ) x 100

where σ and x̅ are the standard deviation and mean of the data

Related 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 Three Dimensional Geometry

Chapter 13 : Limits and Derivatives

Chapter 14 : Mathematical Reasoning

Chapter 16 : Probability

Frequently Asked Questions on NCERT Class 11 Maths Chapter 15 : Statistics

Q1. Explain the difference between Variance and Standard Deviation.

Answer: The standard deviation of a set of numbers is the distance between them and the mean. The variance is a measure of how much each point deviates from the mean on average. Variance is the average of all data points inside a group, whereas standard deviation is the square root of the variance.

Q2. How many exercises are there in Statistics Class 11 NCERT Solutions?

Answer: The NCERT Solutions Class 11 Maths Chapter 15 Statistics has 34 sums separated between three exercises and one miscellaneous activity. They range from simple formula-based problems to difficult word problems. The solutions to the exercise-specific questions are also accessible in PDF format, with the goal of assisting students in performing well in the yearly exam.

Q3. What are the key subtopics in Statistics Class 11 Maths Chapter 15 that could be tested?

Answer: All the topics covered in NCERT Solutions Class 11 Maths Chapter 15 are equally important. Statistics is a topic that is commonly used in the financial and economic industries for planning and forecasting. They aid in the development of mathematical models and the analysis of time series data. It is not only an important lesson for exams but it is also required of students in the 12th grade. Students can now study and stay up to date on the latest CBSE syllabus by using the NCERT Solutions, which are available in PDF format.

Q4. Is it necessary to practice all of the questions in Statistics Class 11 NCERT Solutions?

Answer: Statistics Class 11 NCERT Solutions are well-designed to help students study basic statistical concepts easily. It thoroughly covers all essential principles, allowing students to grasp the concepts. Students must memorize multiple formulas and do complex calculations. All of the exercises, examples, and practice problems have been meticulously designed to help students gain a rapid and effective comprehension of the various topics.

Related 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