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Statistics Class 11th Notes - Free NCERT Class 11 Maths Chapter 15 notes - Download PDF

Statistics Class 11th Notes - Free NCERT Class 11 Maths Chapter 15 notes - Download PDF

Edited By Komal Miglani | Updated on Jul 25, 2025 09:45 AM IST

Statistics isn’t tough when you have the right notes. We are surrounded by data, from the scores of a Football match to the marks obtained by a student in an examination; data is everywhere. Ever thought about how we can knit all these data together so that they all make perfect sense? Well, this is where statistics come into the picture. From NCERT Class 11 Maths, the chapter Statistics contains the concepts of Measures of Dispersion, Grouped and Ungrouped Data, Variance and Standard Deviation, etc. These concepts will help the students grasp more advanced Statistics concepts easily and enhance their problem-solving ability in real-world applications. These NCERT notes for class 11 Maths help to make learning uncomplicated, simple, and easy in a stress-free environment.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Suggested: JEE Main: high scoring chaptersPast 10 year's papers

This Story also Contains
  1. Statistics Class 11 Notes PDF download: Free PDF Download
  2. Statistics Class 11 Notes
  3. Statistics: Previous Year Question and Answer
  4. Importance of NCERT Class 11 Maths Chapter 13 Notes:
  5. NCERT Class 11 Notes Chapter Wise

This article on NCERT notes for Class 11 Maths Chapter 13 Statistics provides well-structured NCERT notes to help students easily grasp the concepts of Statistics. Students who want to revise the key topics of Statistics quickly will find this article very useful. It will also significantly boost the exam preparation of students. These NCERT Class 11 Maths Chapter 13 notes on Statistics are made by the Subject Matter Experts of Careers360 following the latest CBSE syllabus, ensuring that students can grasp the basic concepts effectively. For syllabus, notes, and PDF, refer to this link: NCERT.

Statistics Class 11 Notes PDF download: Free PDF Download

Students who wish to access the Statistics Class 11 Maths notes can click on the link below to download the entire notes in PDF.

Download PDF

Statistics Class 11 Notes

Careers360 experts have curated these Statistics Class 11 Notes to help students revise quickly and confidently.

Statistics

Collecting and analysing data in different forms is called statistics.

Measure of dispersion

Measuring the difference or the variations in the values is called dispersion measurement.

Range

The difference between the highest and the lowest values gives us the range.

Mean

Mean is generally the ratio between the sum of observations to the number of observations.

Denoted by x.
Mathematical representation: 1ni=1nxi
n denotes the number of observations.
xi denotes ith observations.

Median

The middlemost term in all the observations, after arranging them in either ascending or descending order, denotes the Median.

For the odd number of observations :

Median =(n+12)th observation
For an even number of observations:
It is the mean of (n2) th and (n+12)th observation.

Mode

The observation that repeats most frequently is the mode of the data.

Mean Deviation for ungrouped data

Let n be the observation: x1,x2,x3,,xn

Mean Deviation about Mean

MD(x¯)=|xix|n

Here, x is the mean.
Steps to calculate the mean deviation about a mean:
Step 1: Calculate the mean value x¯ using the formula.

1ni=1nxi

Step 2: Find the value of deviation from xi to x.

Step 3: Find the value of the mean of absolute value from the formula specified above under the definition.

Mean deviation about Median

MD(M)=|xiM|n
Here, M is the median.

Mean Deviation for grouped data

It can be of two types:

  1. Discrete distribution

  2. Continuous distribution

Mean Deviation about the mean for a discrete distribution

Let n be the observation: x1,x2,x3,,xn, and the frequencies are f1,f2,f3,.......,fn respectively.

MD(x¯)=fi|xix¯|fi=fi|xix¯|N
Here, xi and fj are the respective observations and frequencies.
N is the sum of frequencies.

Mean Deviation about median for a discrete distribution

MD(M)=fi|xiM|fi=MD(x¯)=fi|xiM|N
Here, M denotes the median.

Mean Deviation about the mean for a continuous distribution

Let n be the observation: x1,x2,x3,.xn, and the frequencies are f1,f2,f3,..fn respectively.

MD(x¯)=fi|xix¯|fi=MD(x¯)=fi|xix¯|N
 Step deviation:x=a+i=1nfidiNh
Where we have di=(xiA)h
Here, a is the assumed mean, h is the interval between the upper and lower class frequencies

Mean Deviation about median for continuous distribution

MD(M)=fi|xiM|fi=MD(x¯)=fi|xiM|N
Here, M denotes median.

M=l+N2Cfh

Here, l is the lowest boundary value
N is the sum of frequencies
C is the value of the cumulative frequency of the selected class
f is the selected frequency
h is the interval between the upper and lower class frequencies.
To overcome a few limitations of the mean deviation, we use the concepts called variance and standard deviation.

Variance

It is the square of the Standard deviation.

σ2=1n|xix¯|2σ2= variance x¯= mean n= sum of frequencies 

Standard Deviation

It is the square root of variance.

σ=1n|xix¯|2

Standard deviation of discrete frequencies

σ=1Nfi|xix¯|2

Standard deviation of continuous frequencies

σ=1NNfixi2(fixi)2

Another formula:

σ=hNNfidi2(fidi)2
Here, h is the width of the class

di=(xiA)h, where A is assumed mean 

Coefficient of variance

CV= Standard deviation  Mean ×100CV=σx¯×100

Note: When the CV is the coefficient of variance is greater than we have greater variance than that of a smaller variance.

With this topic, we conclude the NCERT Class 11 Statistics notes.

Statistics: Previous Year Question and Answer

Question 1:

If xi’s are the midpoints of the class intervals of grouped data, fi’s are the corresponding frequencies and x¯ is the mean, then Σ(fixix¯) is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Solution:
Mean: It is the average of the given numbers/observations. It is easy to calculate the mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean (x¯)=fixin
By cross multiplication, we get,
fixi=x¯n(1)
(fixix¯)=fixix¯
=nx¯x¯ (from equation (1))
=nx¯nx¯
= 0

Hence, the correct answer is 0.

Question 2:

The abscissa of the point of intersection of the less-than-type and the more-than-type cumulative frequency curves of a grouped data gives its
(A) mean
(B) median
(C) mode
(D) all three above

Solution:
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarise categorical variables.
If we make a graph of less than type and of more than type grouped data and find the intersection point, then the value at the abscissa is the median of the grouped data.
Hence, option (B) is correct.

Question 3:

For the following distribution:

Class

0-5

5-10

10-15

15-20

20-25

Frequency

10

15

12

20

9

The sum of the lower limits of the median class and the modal class is
(A) 15
(B) 25
(C) 30
(D) 35

Solution:
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However, mostly we use frequency distribution to summarise categorical variables.

Class

Frequency

Cumulative Frequency (C.F)

0-5

10

10

5-10

15

(10+15=25)

10-15

12

(25+12=37)

15-20

20

(37+20=57)

20-25

9

(57+9=66)

N=66


N2=662=33 which ies in the class 10-15.
Hence, the median class is 10 – 15.
The class with the maximum frequency is the modal class, which is 15 – 20
The lower limit of the median class = 10
The lower limit of the modal class = 15
Sum = 10 + 15 = 25

Hence, the correct answer is 25.

Importance of NCERT Class 11 Maths Chapter 13 Notes:

NCERT Class 11 Maths Chapter 13 Notes play a vital role in helping students grasp the core concepts of the chapter easily and effectively, so that they can remember these concepts for a long time. Some important points of these notes are:

  • Effective Revision: These notes provide a detailed overview of all the important theorems and formulas, so that students can revise the chapter quickly and effectively.
  • Clear Concepts: With these well-prepared notes, students can understand the basic concepts effectively. Also, these notes will help the students remember the key concepts by breaking down complex topics into simpler and easier-to-understand points.
  • Time Saving: Students can look to save time by going through these notes instead of reading the whole lengthy chapter.
  • Exam Ready Preparation: These notes also highlight the relevant contents for various exams, so that students can get the last-minute minute very useful for exams.

NCERT Class 11 Notes Chapter Wise

For students' preparation, Careers360 has gathered all Class 11 Maths NCERT Notes here for quick and convenient access.

NCERT Notes Subject-Wise

Given below are some subject-wise links for the NCERT Notes for class 11.

Subject-Wise NCERT Exemplar Solutions

After finishing the textbook exercises, students can use the following links to check the NCERT exemplar solutions for a better understanding of the concepts.

Subject-Wise NCERT Solutions

Students can also check these well-structured, subject-wise solutions.

NCERT Books and Syllabus

Students should always analyse the latest CBSE syllabus before making a study routine. The following links will help them check the syllabus. Also, here is access to more reference books.

Articles

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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