CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01

Download Class 11 Mathematics Concept Summaries: CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01

Review targeted revision notes for Class 11 Mathematics with the CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01. Built according to official educational guidelines for the 2026-27 academic year, these downloadable summaries for Chapter 13 Statistics support daily study and last-minute exam readiness.

Access Chapter 13 Statistics Notes and Study Material

Navigate directly to the revision notes for Chapter 13 Statistics using the digital viewer below. Each summary is structured to highlight high-weightage sections, allowing students to instantly access critical definitions and focus on core exam topics.

Arithmetic Mean

Arithmetic Mean

Arithmetic mean is defined as the sum of the values of all observations divided by the number of observations.

Arithmetic Mean in an Individual Series

(i) Direct method: \( \bar{x} = \frac{\Sigma x}{N} \), where \( \Sigma x = x_1 + x_2 + \ldots + x_n \) and N = total number of items.

 

Example 1. The marks obtained by 10 students in a subject are:
StudentsABCDEFGHIJ
Marks85605075554055704565
Calculate the arithmetic mean by the direct method.

Answer:
StudentsMarks (X)
A85
B60
C50
D75
E55
F40
G55
H70
I45
J65
N = 10Σx = 600

Arithmetic mean \( \bar{x} = \frac{\Sigma x}{N} = \frac{600}{10} \) = 60 marks

(ii) Short cut method: Under this method, any figure is assumed as the mean and deviations are calculated from this assumed mean.
\( \bar{x} = A + \frac{\Sigma d}{N} \)
where \( \bar{x} \) = arithmetic mean, A = assumed mean, d = X − A = deviation of the variable from the assumed mean, Σd = Σ(X − A) = sum of the deviations from the assumed mean and N = total number of items.

 

Example 2. Calculate the arithmetic mean of the marks (of the above 10 students) by the assumed mean method.
Answer:
StudentsMarks (X)d = X − A (A = 50)
A85+35
B60+10
C50 (A)0
D75+25
E55+5
F40−10
G55+5
H70+20
I45−5
J65+15
N = 10Σd = 100

\( \bar{x} = A + \frac{\Sigma d}{N} = 50 + \frac{100}{10} \) = 60 marks

(iii) Step deviation method: In this method, the deviations from the assumed mean are divided by a common factor (C) to get step deviations. Apply the following formula:
\( \bar{x} = A + \frac{\Sigma d'}{N} \times C \)
where d = X − A is the deviation of the variable from the assumed mean, \( d' = \frac{X - A}{C} \) is the step deviation, Σd′ = sum of the step deviations and C = common factor.
Note: The step deviation method can be used only when the deviations from the assumed mean (d) are divisible by a common factor.

 

Example 3. Calculate the arithmetic mean of the marks (of the above 10 students) by the step deviation method.
Answer:
StudentsMarks (X)d = X − A, A = 50\( d' = \frac{X - A}{C} \), C = 5
A85+35+7
B60+10+2
C50 (A)00
D75+25+5
E55+5+1
F40−10−2
G55+5+1
H70+20+4
I45−5−1
J65+15+3
N = 10Σd′ = 20

\( \bar{x} = A + \frac{\Sigma d'}{N} \times C = 50 + \frac{20}{10} \times 5 \) = 60 marks

Arithmetic Mean in a Discrete Series

In case of a discrete series (ungrouped frequency distribution), the values of the variable show repetitions, i.e., frequencies are given corresponding to different values of the variable. N = sum total of frequencies = Σf.
(i) Direct method: \( \bar{x} = \frac{\Sigma fx}{\Sigma f} \), where Σfx = sum of the products of the variable with the respective frequencies and Σf = total number of items.

 

Example 4. From the following data of the marks obtained by 60 students of a class, calculate the average marks by the direct method.
Marks203040506070
No. of students812201064

Answer:
Marks (X)No. of students (f)fx
208160
3012360
4020800
5010500
606360
704280
Σf = 60Σfx = 2460

\( \bar{x} = \frac{\Sigma fx}{\Sigma f} = \frac{2460}{60} \) = 41 marks

(ii) Short cut method: \( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} \), where A = assumed mean, d = X − A is the deviation of the variable from the assumed mean, Σfd = sum of the products of the deviations (d) with the respective frequencies (f) and Σf = total number of items.

 

Example 5. Calculate the arithmetic mean of the marks (of the above 60 students) by the short cut method.
Answer:
MarksNo. of students (f)d = X − A (A = 40)fd
208−20−160
3012−10−120
40 (A)2000
5010+10+100
606+20+120
704+30+120
Σf = 60Σfd = +60

\( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} = 40 + \frac{60}{60} \) = 41 marks

(iii) Step deviation method: In this method, the values of the deviations (d) are divided by a common factor (C) to ease the calculation process.
\( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C \)
where d′ = step deviations (deviations from the assumed mean divided by the common factor) and Σfd′ = sum of the products of the step deviations (d′) with the respective frequencies (f).

 

Example 6. Calculate the arithmetic mean of the marks (of the above 60 students) by the step deviation method.
Answer:
Marks XNo. of students fd = X − A (A = 40)\( d' = \frac{X - A}{C} \) (C = 10)fd′
208−20−2−16
3012−10−1−12
40 (A)20000
5010+10+1+10
606+20+2+12
704+30+3+12
Σf = 60Σfd′ = +6

\( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C = 40 + \frac{6}{60} \times 10 \) = 41 marks

Arithmetic Mean in a Continuous Series

In the case of a continuous series (grouped frequency distribution), the values of the variable are grouped in various class intervals along with their respective frequencies. In a continuous series, the mid-points of the various class intervals are used to replace the class intervals. Once it is done, there is no difference between a continuous series and a discrete series.
In a continuous series also, the following three methods are used to calculate the arithmetic mean: (i) direct method, (ii) short cut method, (iii) step deviation method.

 

Example 7. The following table gives the marks in English secured by 30 students of a class in their weekly test. Calculate the average marks by the direct method.
Marks0 – 55 – 1010 – 1515 – 2020 – 25
No. of students286104

Answer:
Marks XNo. of students fMid-value (m)fm
0 – 522.55
5 – 1087.560
10 – 15612.575
15 – 201017.5175
20 – 25422.590
Σf = 30Σfm = 405

\( \bar{x} = \frac{\Sigma fm}{\Sigma f} = \frac{405}{30} \) = 13.50 marks

 

Example 8. Calculate the average marks (of the above 30 students) by the short cut method, \( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} \).
Answer:
Marks XNo. of students fMid-value md = m − A (A = 12.5)fd
0 – 522.5−10−20
5 – 1087.5−5−40
10 – 15612.5 (A)00
15 – 201017.5+5+50
20 – 25422.5+10+40
Σf = 30Σfd = +30

\( \bar{x} = A + \frac{\Sigma fd}{\Sigma f} = 12.5 + \frac{30}{30} \) = 13.50 marks

Step deviation method: When the class intervals for all the classes in a continuous series are of the same magnitude (width), the short cut method can be further simplified by the step deviation method: \( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C \)

 

Example 9. Calculate the average marks (of the above 30 students) by the step deviation method.
Answer:
Marks XNo. of students fMid-value md = m − A (A = 12.5)\( d' = \frac{m - A}{C} \) (C = 5)fd′
0 – 522.5−10−2−4
5 – 1087.5−5−1−8
10 – 15612.5 (A)000
15 – 201017.5+5+1+10
20 – 25422.5+10+2+8
Σf = 30Σfd′ = +6

Average marks \( \bar{x} = A + \frac{\Sigma fd'}{\Sigma f} \times C = 12.5 + \frac{6}{30} \times 5 \) = 13.50 marks

Median

Median

Median may be defined as the middle value in the data set when its elements are arranged in a sequential order, that is, in either ascending or descending order of magnitude.

Median in an Individual Series

Median (Me) = size of \( \left(\frac{N + 1}{2}\right) \)th item
(i) In case of an odd number of items: Median = middle item of the distribution.
(ii) In case of an even number of items: Median = average of the two middle items.

 

Example 10. Marks of five students are 42, 40, 60, 55 and 50. Calculate the median.
Answer:
Step 1: Marks arranged in ascending order: 40, 42, 50, 55, 60.
Step 2: Me = size of \( \left(\frac{N + 1}{2}\right) \)th item = size of \( \left(\frac{5 + 1}{2}\right) \)th item = size of 3rd item
Median = 50

Median in a Discrete Series

In a discrete series, the values of the variable are given along with their frequencies. Cumulative frequencies are found and Me = size of \( \left(\frac{N + 1}{2}\right) \)th item.

 

Example 11. Calculate the median from the following data:
Size (X)Frequency (f)
32
41
53
67
74

Answer:
Size (X)Frequency (f)Cumulative frequency
322
413
536
6713
7417

N = Σf = 17
Me = size of \( \left(\frac{N + 1}{2}\right) \)th item = size of \( \left(\frac{17 + 1}{2}\right) \)th item = size of 9th item
The 9th item falls in the cumulative frequency 13, which corresponds to size 6.
∴ Me = 6

Median in a Continuous Series

Steps to calculate the median:
Step 1: Arrange the data in ascending or descending order.
Step 2: Calculate the cumulative frequencies (c.f.).
Step 3: Find the median item as: Me = size of \( \left(\frac{N}{2}\right) \)th item.
Step 4: Find the c.f. which is either equal to or just greater than this.
Step 5: Find the class corresponding to the c.f. equal to \( \frac{N}{2} \) or just greater than this. This class is called the median class.
Step 6: \( Me = l_1 + \frac{\frac{N}{2} - c.f.}{f} \times i \)
where \( l_1 \) = lower limit of the median class, c.f. = cumulative frequency of the class preceding the median class, f = frequency of the median class and i = width of the median class.

 

Example 12. Find the median of the following data:
X0 – 1010 – 2020 – 3030 – 4040 – 50
f34279

Answer:
Xfc.f.
0 – 1033
10 – 2047
20 – 3029
30 – 40716
40 – 50925

N = Σf = 25
Median item = \( \frac{N}{2} = \frac{25}{2} \) = 12.5th item. The 12.5th item lies in the group 30 – 40.
\( l_1 \) = 30, c.f. = 9, f = 7, i = 10
By applying the formula, \( Me = l_1 + \frac{\frac{N}{2} - c.f.}{f} \times i = 30 + \frac{12.5 - 9}{7} \times 10 = 30 + 5 = 35 \)
Median = 35

Mode

Mode

Mode is the value occurring most frequently in a set of observations and around which the other items of the set cluster most densely.

Mode in an Individual Series

(A) Mode by way of observation:
Step 1: Arrange the data in ascending or descending order.
Step 2: The item which occurs most often in the series is the ‘mode’.

 

Example 13. From the heights of 15 students, calculate the value of mode.
Height (in inches): 52, 50, 66, 70, 66, 72, 71, 66, 60, 67, 69, 67, 48, 60, 65.

Answer:
By arranging the series in ascending order we get:
48, 50, 52, 60, 60, 65, 66, 66, 66, 67, 67, 69, 70, 71, 72.
By observation, the height 66 inches occurs most often, therefore the mode (Z) is 66 inches.

(B) Mode by converting an individual series into a discrete series: The data are written as a frequency table and the value with the maximum frequency is the mode.

 

Example 14. Calculate the value of mode from the data of heights of the above 15 students by converting the data into a discrete series.
Answer:
Height (in inches)Frequency
481
501
521
602
651
663
672
691
701
711
721

The height of 66 inches has the maximum frequency (3), therefore the modal height (Z) is 66.
Mode = 66 inches

Mode in a Discrete Series

There are two methods to determine the mode in a discrete series:
(i) Mode by observation
(ii) Mode by grouping method

 

Example 15. Find the mode of the following series by observation:
Daily wages (in Rs.)100110120130140150
No. of persons2481054

Answer:
By inspection, we can see that 130 occurs most frequently (10 times) in the series.
Hence the mode is Rs. 130.

Steps of the grouping method:
Step 1: Prepare a grouping table with six columns. Column I contains the given frequencies.
Step 2: In Column II, add the frequencies in two’s starting from the first; in Column III, add them in two’s leaving the first frequency.
Step 3: In Column IV, add the frequencies in three’s starting from the first; in Column V, in three’s leaving the first; and in Column VI, in three’s leaving the first two.
Step 4: Prepare an analysis table: for each column, mark (✓) the values of the variable that make up the maximum total of that column. The value with the largest number of marks is the mode.

 

Example 16. Calculate the value of mode from the above data of daily wages by the grouping method.
Answer:
Grouping table
Wages (in Rs.)FrequencyIn two’sIn three’s
Column IColumn IIColumn IIIColumn IVColumn VColumn VI
10022+4 = 62+4+8 = 14
11044+8 = 124+8+10 = 22
12088+10 = 188+10+5 = 23
1301010+5 = 1510+5+4 = 19
14055+4 = 9
1504

In the analysis table, we mark (✓) the values having the maximum frequencies in each column of the grouping table.
Analysis table
Column No.100110120130140150
I✓
II✓✓
III✓✓
IV✓✓✓
V✓✓✓
VI✓✓✓
Total13631

Since the value 130 has occurred the maximum number of times, i.e. 6, the modal wage is Rs. 130.
Mode = Rs. 130

Mode in a Continuous Series

In a continuous series, the mode lies in a particular class or group, which is called the modal class. The following two methods are used in determining the mode:
(i) Observation method:
Step 1: Determine the modal class, i.e., the class with the highest frequency.
Step 2: Determine the exact value of the mode by the following formula:
\( Mo = l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \)
where Mo = mode, \( l_1 \) = lower limit of the modal class, \( f_1 \) = frequency of the modal class, \( f_0 \) = frequency of the class preceding the modal class, \( f_2 \) = frequency of the class succeeding the modal class and i = class interval (width) of the modal class.

 

Example 17. Find the mode of the following series:
Class interval0 – 55 – 1010 – 1515 – 2020 – 25
Frequency241567

Answer:
Class intervalFrequency
0 – 52
5 – 104 (\( f_0 \))
10 – 1515 (\( f_1 \)) modal class
15 – 206 (\( f_2 \))
20 – 257

The class 10 – 15 has the highest frequency, so it is the modal class.
Mode (Z) = \( l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \), where \( l_1 \) = 10, \( f_1 \) = 15, \( f_0 \) = 4, \( f_2 \) = 6, i = 5
\( Z = 10 + \frac{15 - 4}{2 \times 15 - 4 - 6} \times 5 = 10 + \frac{11}{20} \times 5 \) = 12.75
Mode = 12.75

(ii) Grouping method: When the frequencies are irregular or the maximum frequency is close to other frequencies, the modal class is found with the help of a grouping table and an analysis table, and then the formula is applied.

 

Example 18. From the following data, determine the mode.
Size10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 90
Frequency41025152322123

Answer:
Grouping table
Size XFrequencyIn two’sIn three’s
Column IColumn IIColumn IIIColumn IVColumn VColumn VI
10 – 2044+10 = 144+10+25 = 39
20 – 301010+25 = 3510+25+15 = 50
30 – 402525+15 = 4025+15+23 = 63
40 – 501515+23 = 3815+23+22 = 60
50 – 602323+22 = 4523+22+12 = 57
60 – 702222+12 = 3422+12+3 = 37
70 – 801212+3 = 15
80 – 903

Analysis table
Column No.10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 90
I✓
II✓✓
III✓✓
IV✓✓✓
V✓✓✓
VI✓✓✓
Total23531

It is clear that the modal class is 50 – 60 (5 marks) and the frequency of this class is 23.
Using the formula, Mode (Z) = \( l_1 + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times i \), where \( l_1 \) = 50, \( f_1 \) = 23, \( f_0 \) = 15, \( f_2 \) = 22, i = 10
\( Z = 50 + \frac{23 - 15}{2 \times 23 - 15 - 22} \times 10 = 50 + \frac{8}{9} \times 10 \) = 58.89
Mode = 58.89

Chapter 13 Statistics Concepts and Summary for Class 11 Mathematics

Download Chapter Summary: Class 11 Mathematics Chapter 13 Statistics

Access structured revision notes for Chapter 13 Statistics designed in alignment with the latest CBSE curriculum for Class 11 Mathematics. These summaries help clarify core themes and support effective daily study.

Concept Clarification for Chapter 13 Statistics

Cross-reference your comprehension with comprehensive NCERT solutions for Class 11 to ensure absolute clarity across all sub-topics in this chapter.

Additional Study Resources for Class 11 Mathematics

Follow up your revision by attempting the interactive online Mathematics MCQ test and practice worksheets for this chapter to verify your mastery. All platform resources are free to access.

FAQs

Where can I download the latest PDF for CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01?

You can download the teacher prepared revision notes for CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01 from StudiesToday.com. These notes are designed as per 2026-27 academic session to help Class 11 students get the best study material for Mathematics.

Are these Mathematics notes for Class 11 based on the 2026 board exam pattern?

Yes, our CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01 include 50% competency-based questions with focus on core logic, keyword definitions, and the practical application of Mathematics principles which is important for getting more marks in 2026 CBSE exams.

Do these Class 11 notes cover all topic-wise concepts for Mathematics?

Yes, our CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01 provide a detailed, topic wise breakdown of the chapter. Fundamental definitions, complex numerical formulas and all topics of CBSE syllabus in Class 11 is covered.

How can I use CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01 for quick last-minute revision?

These notes for Mathematics are organized into bullet points and easy-to-read charts. By using CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01, Class 11 students fast revise formulas, key definitions before the exams.

Is there any registration required to download Class 11 Mathematics notes?

No, all study resources on StudiesToday, including CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 01, are available for immediate free download. Class 11 Mathematics study material is available in PDF and can be downloaded on mobile.