Ch 13  ·  Q–
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Class 10 Mathematics Exercise 13.1 NCERT Solutions Olympiad Board Exam

Chapter 13 — Statistics

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

📋9 questions
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Q1
NUMERIC3 marks

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of houses 1 2 1 5 6 2 3

Concept & Theory

  • In grouped data, each class interval is represented by its class mark (midpoint).
  • \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Mean (Direct Method) is calculated as: \[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]
  • This method is suitable when class marks are small and calculations are manageable.

Solution Roadmap

  1. Find class marks \(x_i\) for each class interval.
  2. Multiply frequency \(f_i\) with class mark \(x_i\).
  3. Compute \(\sum f_i\) and \(\sum f_i x_i\).
  4. Apply mean formula.

Visualization of Class Marks

1 3 5 7 9 11 13

Solution

Step 1: Calculate class marks \(x_i\)

Class Interval \(f_i\) \(x_i\) \(f_i x_i\)
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339
Total \(\sum f_i = 20\) \(\sum f_i x_i = 162\)

Step 2: Apply mean formula

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \]

Step 3: Substitute values

\[ \overline{x} = \frac{162}{20} \]

Step 4: Simplify

\[ \overline{x} = 8.1 \]
Mean number of plants per house = 8.1

Method Justification

Direct Method is used because class marks are small and calculations are simple, making it efficient and less error-prone.

Exam Significance

  • Very frequently asked in CBSE Board Exams (2–4 marks).
  • Tests understanding of grouped data and mean calculation.
  • Important base for Assumed Mean and Step Deviation methods.
  • Useful in aptitude sections of competitive exams (SSC, Banking).
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1 / 9  ·  11%
Q2 →
Q2
NUMERIC3 marks

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in ₹) 500-520 520-540 540-560 560-580 580-600
Number of workers 12 14 8 6 10

Concept & Theory

  • When class intervals are large, Assumed Mean Method simplifies calculations.
  • Class mark is calculated as: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation is taken from assumed mean \(a\): \[ d_i = x_i - a \]
  • Mean formula becomes: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Find class marks \(x_i\).
  2. Select a convenient assumed mean \(a\).
  3. Compute deviations \(d_i = x_i - a\).
  4. Calculate \(f_i d_i\).
  5. Find \(\sum f_i\) and \(\sum f_i d_i\).
  6. Substitute into mean formula.

Visualization of Assumed Mean Concept

a = 550 510 530 570 590

Solution

Step 1: Calculate class marks \(x_i\)

\[ x_i = 510,\; 530,\; 550,\; 570,\; 590 \]

Step 2: Choose assumed mean

\[ a = 550 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i = x_i - 550\) \(f_i d_i\)
500-52012510-40-480
520-54014530-20-280
540-560855000
560-580657020120
580-6001059040400
Total \(\sum f_i = 50\) \(\sum f_i d_i = -240\)

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute values

\[ \overline{x} = 550 + \frac{-240}{50} \]

Step 6: Simplify

\[ \overline{x} = 550 - 4.8 \] \[ \overline{x} = 545.2 \]
Mean daily wages = ₹ 545.2

Method Justification

Assumed Mean Method is used because class intervals are large, making direct multiplication tedious. This method reduces computational effort and minimizes arithmetic errors.

Exam Significance

  • Highly important for CBSE Board Exams (3–4 marks question).
  • Frequently tested with method selection (Direct vs Assumed Mean).
  • Builds foundation for Step Deviation Method.
  • Useful in Data Interpretation sections of SSC, Banking, and Railways exams.
← Q1
2 / 9  ·  22%
Q3 →
Q3
NUMERIC3 marks

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency \(f\).

Daily pocket allowance (in ₹) 11-13 13-15 15-17 17-19 19-21 21-23 23-25
Number of children 7 6 9 13 f 5 4

Concept & Theory

  • When mean is given and frequency is missing, we form an equation using mean formula.
  • Using Assumed Mean Method: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]
  • If assumed mean \(a = \overline{x}\), then: \[ \sum f_i d_i = 0 \]
  • This simplifies calculation significantly.

Solution Roadmap

  1. Find class marks \(x_i\).
  2. Choose assumed mean \(a = 18\).
  3. Compute deviations \(d_i = x_i - a\).
  4. Compute \(f_i d_i\).
  5. Use condition \(\sum f_i d_i = 0\).
  6. Solve for \(f\).

Visualization of Balance Concept

Mean = 18 -ve +ve Total negative = Total positive

Solution

Step 1: Calculate class marks \(x_i\)

\[ x_i = 12,\; 14,\; 16,\; 18,\; 20,\; 22,\; 24 \]

Step 2: Take assumed mean

\[ a = 18 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
11-13712-6-42
13-15614-4-24
15-17916-2-18
17-19131800
19-21\(f\)202\(2f\)
21-23522420
23-25424624
Total \(\sum f_i = 44 + f\) \(\sum f_i d_i = 2f - 40\)

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute given mean \( \overline{x} = 18 \)

\[ 18 = 18 + \frac{2f - 40}{44 + f} \]

Step 6: Simplify equation

\[ 18 - 18 = \frac{2f - 40}{44 + f} \] \[ 0 = \frac{2f - 40}{44 + f} \]

Step 7: Numerator must be zero

\[ 2f - 40 = 0 \] \[ 2f = 40 \] \[ f = 20 \]
Missing frequency \(f = 20\)

Method Justification

Assumed Mean Method is most efficient here because the given mean equals the assumed mean, reducing the problem to a simple linear equation.

Exam Significance

  • Very important CBSE Board question type (case-based or 3–4 marks).
  • Tests conceptual clarity of mean and deviations.
  • Common in competitive exams under Data Interpretation.
  • Strengthens equation formation skills in statistics.
← Q2
3 / 9  ·  33%
Q4 →
Q4
NUMERIC3 marks

Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 7 4 2

Concept & Theory

  • When class intervals are uniform and mid-values are manageable, Assumed Mean Method is efficient.
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation: \[ d_i = x_i - a \]
  • Mean formula: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Compute class marks \(x_i\).
  2. Select central class mark as assumed mean.
  3. Find deviations \(d_i\).
  4. Calculate \(f_i d_i\).
  5. Apply mean formula.

Visualization of Central Tendency

a = 75.5 Lower Higher Balanced deviations around mean

Solution

Step 1: Calculate class marks \(x_i\)

\[ x_i = 66.5,\; 69.5,\; 72.5,\; 75.5,\; 78.5,\; 81.5,\; 84.5 \]

Step 2: Choose assumed mean

\[ a = 75.5 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
65-68266.5-9-18
68-71469.5-6-24
71-74372.5-3-9
74-77875.500
77-80778.5321
80-83481.5624
83-86284.5918
Total \(\sum f_i = 30\) \(\sum f_i d_i = 12\)

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute values

\[ \overline{x} = 75.5 + \frac{12}{30} \]

Step 6: Simplify

\[ \overline{x} = 75.5 + 0.4 \] \[ \overline{x} = 75.9 \]
Mean heartbeats per minute = 75.9

Method Justification

Assumed Mean Method is suitable because class intervals are equal and values are moderately large, making deviation-based calculation faster and systematic.

Exam Significance

  • Frequently appears in CBSE Board Exams (3–4 marks).
  • Tests accuracy in decimal handling and deviation concept.
  • Important for medical/statistical data interpretation.
  • Relevant for competitive exams involving grouped data analysis.
← Q3
4 / 9  ·  44%
Q5 →
Q5
NUMERIC3 marks

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes 50-52 53-55 56-58 59-61 62-64
Number of boxes 15 110 135 115 25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Concept & Theory

  • The given classes are not continuous. We must convert them into continuous form.
  • Adjustment: \[ \begin{aligned} \text{New Lower Limit} &= \text{Lower Limit} - 0.5,\\ \text{New Upper Limit} &= \text{Upper Limit} + 0.5 \end{aligned} \]
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Using Assumed Mean Method: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Convert classes into continuous form.
  2. Compute class marks \(x_i\).
  3. Choose assumed mean \(a\).
  4. Find deviations \(d_i\).
  5. Calculate \(f_i d_i\).
  6. Apply mean formula.

Visualization of Continuous Conversion

49.5 52.5 55.5 58.5 61.5 64.5 a = 57

Solution

Step 1: Convert to continuous classes

\[\begin{aligned} 49.5-52.5,\\ 52.5-55.5,\\ 55.5-58.5,\\ 58.5-61.5,\\ 61.5-64.5 \end{aligned} \]

Step 2: Calculate class marks

\[ x_i = 51,\; 54,\; 57,\; 60,\; 63 \]

Step 3: Choose assumed mean

\[ a = 57 \]

Step 4: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
49.5-52.51551-6-90
52.5-55.511054-3-330
55.5-58.51355700
58.5-61.5115603345
61.5-64.525636150
Total \(\sum f_i = 400\) \(\sum f_i d_i = 75\)

Step 5: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 6: Substitute values

\[ \overline{x} = 57 + \frac{75}{400} \]

Step 7: Simplify

\[ \overline{x} = 57 + 0.1875 \] \[ \overline{x} = 57.1875 \approx 57.19 \]
Mean number of mangoes per box = 57.19

Method Justification

Assumed Mean Method is chosen because frequencies are large and evenly distributed, making calculations efficient and less time-consuming.

Exam Significance

  • Important CBSE Board question involving class conversion + mean.
  • Tests conceptual clarity of continuous class intervals.
  • Frequently appears in case-based questions.
  • Highly relevant for competitive exams with data grouping concepts.
← Q4
5 / 9  ·  56%
Q6 →
Q6
NUMERIC3 marks

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in ₹) 100-150 150-200 200-250 250-300 300-350
Number of households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Concept & Theory

  • When class intervals are equal and values are large, Step Deviation Method is most efficient.
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation: \[ d_i = x_i - a \]
  • Step deviation: \[ u_i = \frac{d_i}{h} \]
  • Mean formula: \[ \overline{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right)\times h \]

Solution Roadmap

  1. Find class marks \(x_i\).
  2. Select assumed mean \(a\).
  3. Compute deviations \(d_i\).
  4. Divide by class width \(h\) to get \(u_i\).
  5. Calculate \(f_i u_i\).
  6. Apply step deviation formula.

Visualization of Step Deviation

a = 225 -2h -h +h +2h Equal class width (h = 50)

Solution

Step 1: Calculate class marks

\[ x_i = 125,\; 175,\; 225,\; 275,\; 325 \]

Step 2: Choose assumed mean

\[ a = 225,\quad h = 50 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Step 4: Compute step deviations \(u_i = \frac{d_i}{h}\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(u_i\) \(f_i u_i\)
100-1504125-100-2-8
150-2005175-50-1-5
200-25012225000
250-30022755012
300-350232510024
Total \(\sum f_i = 25\) \(\sum f_i u_i = -7\)

Step 5: Apply mean formula

\[ \overline{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right)\times h \]

Step 6: Substitute values

\[ \overline{x} = 225 + \left( \frac{-7}{25} \right)\times 50 \]

Step 7: Simplify

\[ \overline{x} = 225 + (-14) \] \[ \overline{x} = 211 \]
Mean daily expenditure = ₹ 211

Method Justification

Step Deviation Method is used because class width is constant and large, allowing simplification of calculations by reducing values to smaller integers.

Exam Significance

  • Very important for CBSE Board Exams (4 marks typical).
  • Tests efficiency in choosing correct method.
  • Commonly asked in case-study based questions.
  • Highly useful in competitive exams involving grouped data analysis.
← Q5
6 / 9  ·  67%
Q7 →
Q7
NUMERIC3 marks

To find out the concentration of \(SO_2\) in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO\(_2\) Frequency
0.00-0.044
0.04-0.089
0.08-0.129
0.12-0.162
0.16-0.204
0.20-0.242

Find the mean concentration of \(SO_2\) in the air.

Concept & Theory

  • For grouped continuous data with small decimal values, Assumed Mean Method is efficient.
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation: \[ d_i = x_i - a \]
  • Mean formula: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Compute class marks \(x_i\).
  2. Choose a convenient assumed mean.
  3. Calculate deviations \(d_i\).
  4. Compute \(f_i d_i\).
  5. Apply mean formula.

Visualization of Small Deviations

a = 0.10 Small -ve Small +ve Very small deviations (decimal data)

Solution

Step 1: Calculate class marks

\[ x_i = 0.02,\; 0.06,\; 0.10,\; 0.14,\; 0.18,\; 0.22 \]

Step 2: Choose assumed mean

\[ a = 0.10 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
0.00-0.0440.02-0.08-0.32
0.04-0.0890.06-0.04-0.36
0.08-0.1290.1000
0.12-0.1620.140.040.08
0.16-0.2040.180.080.32
0.20-0.2420.220.120.24
Total \(\sum f_i = 30\) \(\sum f_i d_i = -0.04\)

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute values

\[ \overline{x} = 0.10 + \frac{-0.04}{30} \]

Step 6: Simplify

\[ \overline{x} = 0.10 - 0.001333\ldots \] \[ \overline{x} \approx 0.0987 \approx 0.099 \]
Mean concentration of \(SO_2\) = 0.099 ppm

Method Justification

Assumed Mean Method is preferred because decimal values are involved, and deviations around a central value simplify calculations significantly.

Exam Significance

  • Important CBSE Board question involving decimal data handling.
  • Tests precision and rounding skills.
  • Common in environmental/statistical case studies.
  • Useful in competitive exams involving data interpretation with decimals.
← Q6
7 / 9  ·  78%
Q8 →
Q8
NUMERIC3 marks

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40
Number of students 10 11 7 4 4 3 1

Concept & Theory

  • When class intervals are unequal, Assumed Mean Method is preferred over Step Deviation.
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation: \[ d_i = x_i - a \]
  • Mean formula: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Compute class marks \(x_i\).
  2. Choose assumed mean from central class.
  3. Calculate deviations \(d_i\).
  4. Find \(f_i d_i\).
  5. Apply mean formula.

Visualization of Unequal Class Widths

a = 17 0-6 6-10 10-14 14-20 20-28 Unequal class intervals

Solution

Step 1: Calculate class marks

\[ x_i = 3,\; 8,\; 12,\; 17,\; 24,\; 33,\; 39 \]

Step 2: Choose assumed mean

\[ a = 17 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
0-6103-14-140
6-10118-9-99
10-14712-5-35
14-2041700
20-28424728
28-383331648
38-401392222
Total \(\sum f_i = 40\) \(\sum f_i d_i = -176\)

Note: Corrected multiplication errors from original table.

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute values

\[ \overline{x} = 17 + \frac{-176}{40} \]

Step 6: Simplify

\[ \overline{x} = 17 - 4.4 \] \[ \overline{x} = 12.6 \]
Mean number of absent days = 12.6

Method Justification

Assumed Mean Method is used because class intervals are unequal, making Step Deviation unsuitable. This method ensures accuracy and systematic calculation.

Exam Significance

  • Very important conceptual question in CBSE exams (error-detection + method choice).
  • Tests understanding of unequal class intervals.
  • Frequently used in case-study based questions.
  • Relevant for competitive exams requiring analytical accuracy.
← Q7
8 / 9  ·  89%
Q9 →
Q9
NUMERIC3 marks

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %) 45-55 55-65 65-75 75-85 85-95
Number of cities 3 10 11 8 3

Concept & Theory

  • For grouped data with equal class intervals, Assumed Mean Method simplifies calculations.
  • Class mark: \[ x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
  • Deviation: \[ d_i = x_i - a \]
  • Mean formula: \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Solution Roadmap

  1. Compute class marks \(x_i\).
  2. Choose central value as assumed mean.
  3. Find deviations \(d_i\).
  4. Compute \(f_i d_i\).
  5. Apply mean formula.

Visualization of Symmetry Around Mean

a = 70 -10 +10 Balanced distribution

Solution

Step 1: Calculate class marks

\[ x_i = 50,\; 60,\; 70,\; 80,\; 90 \]

Step 2: Choose assumed mean

\[ a = 70 \]

Step 3: Compute deviations \(d_i = x_i - a\)

Class Interval \(f_i\) \(x_i\) \(d_i\) \(f_i d_i\)
45-55350-20-60
55-651060-10-100
65-75117000
75-858801080
85-953902060
Total \(\sum f_i = 35\) \(\sum f_i d_i = -20\)

Step 4: Apply mean formula

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]

Step 5: Substitute values

\[ \overline{x} = 70 + \frac{-20}{35} \]

Step 6: Simplify

\[ \overline{x} = 70 - \frac{20}{35} \] \[ \overline{x} = 70 - 0.5714 \] \[ \overline{x} \approx 69.43 \]
Mean literacy rate = 69.43%

Method Justification

Assumed Mean Method is ideal because class intervals are equal and centered around a convenient midpoint, reducing calculation complexity.

Exam Significance

  • Frequently asked in CBSE Board Exams (3–4 marks).
  • Tests understanding of grouped data symmetry.
  • Important for interpreting real-world statistical data.
  • Common in competitive exams involving data interpretation.
← Q8
9 / 9  ·  100%
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AI Engine • NCERT Class X • Statistics • Exercise 13.1

Statistics Engine for Mean of Grouped Data

A fully self-contained, embeddable PHP snippet for concept building, AI-style guided solving, formulas, caution points, and interactive practice based on the ideas used in NCERT Mathematics Class X Chapter 13, Exercise 13.1.

Important Formula Tips and Tricks Interactive learning blocks Step-by-step solver
Best fit: finding mean from grouped data using direct, assumed mean, and step-deviation methods; class mark, class width, deviations, and interpretation of results.

What this engine covers

This engine is aligned to the concept set of Exercise 13.1: grouped observations, class marks, frequencies, mean by direct method, mean by assumed mean method, and mean by step-deviation method. It also includes learner support features that ordinary solution pages usually miss: guided hints, error alerts, rich practice sets, flashcards, concept quizzes, and a solver that talks through the logic.

Concept 1: Reading grouped dataIdentify lower limit, upper limit, class width, frequency, and whether the intervals are equal.
Concept 2: Class markUse midpoint of each class interval as the representative value of that class.
Concept 3: Direct methodUse when numbers are manageable and multiplication is not too bulky.
Concept 4: Assumed meanUse when class marks are larger and deviations around a central value make work simpler.
Concept 5: Step-deviationUse when all deviations share a common factor, often the class width.
Concept 6: InterpretationState the final mean with correct unit such as rupees, students, plants, marks, or heartbeats.

Formula zone

Class mark
x = (lower limit + upper limit) / 2
Mean by direct method
x̄ = Σfx / Σf
Best when numbers are smallMost straightforward table
Deviation
d = x - a
Mean by assumed mean method
x̄ = a + (Σfd / Σf)
Choose a central class markReduces large multiplication load
Step deviation
u = (x - a) / h
Mean by step-deviation method
x̄ = a + h(Σfu / Σf)
Best when d has common factorFastest for equal-width classes

Ticks, tips, and common mistakes

Tick checklist

  • Write class interval, frequency, and class mark in separate columns.
  • Check the total frequency before calculating the mean.
  • Write the unit in the final answer.
  • Choose the method that makes arithmetic easiest, not hardest.

Tips for speed

  • If class widths are equal and class marks are far from zero, try assumed mean or step-deviation.
  • Take the middle class mark as assumed mean so positive and negative deviations stay small.
  • In step-deviation, use the common class width as h.
  • Estimate the answer first; the mean should lie within the general spread of class marks.

Common mistakes

  • Using class limits directly instead of the class mark.
  • Forgetting that the midpoint is the representative value of the whole class.
  • Taking a wrong class width h in step-deviation.
  • Adding frequencies incorrectly, which spoils the final answer.
  • Not converting inclusive-looking intervals into continuous form when a question needs that treatment.
  • Missing the sign of deviation, especially for classes below the assumed mean.

AI step-by-step solver

Enter class intervals and frequencies as comma-separated values. Example classes: 0-2,2-4,4-6,6-8 and frequencies: 1,2,1,5. The engine can auto-choose a suitable method or you can force a method.

Concept-building practice bank

These are fresh, non-textbook questions built around the same concepts as Exercise 13.1. They are organised by idea so a learner sees patterns clearly and does not merely memorise one table format.

Warm-upSkill: midpoint reading

Question 1. The weekly self-study time of 40 students is grouped as 2–4, 4–6, 6–8, 8–10, 10–12 hours with frequencies 5, 9, 12, 8, 6. Prepare the mean table and find the mean study time.

Stepwise solution. Class marks are 3, 5, 7, 9, and 11. Multiply by frequencies: 15, 45, 84, 72, and 66. Then Σf = 40 and Σfx = 282, so mean = 282/40 = 7.05 hours.

Why this matters. This builds the most basic pattern: read interval, find midpoint, multiply by frequency, divide total weighted sum by total frequency.

CoreSkill: method selection

Question 2. The marks of students are grouped into 40–50, 50–60, 60–70, 70–80, 80–90 with frequencies 4, 7, 15, 10, 4. Find the mean and state which method is more convenient.

Stepwise solution. Class marks are 45, 55, 65, 75, 85. Since these are fairly large but evenly spaced, assumed mean or step-deviation is convenient. Take a = 65 and h = 10. Then u values are −2, −1, 0, 1, 2. Σfu = −8 − 7 + 0 + 10 + 8 = 3 and Σf = 40. So mean = 65 + 10(3/40) = 65.75 marks.

Concept note. The point is not only getting the answer but learning when arithmetic can be compressed elegantly.

IntermediateSkill: reverse reasoning

Question 3. Daily water intake of children is grouped as 1–2, 2–3, 3–4, 4–5, 5–6 litres with frequencies 3, 6, f, 5, 2. If the mean intake is 3.5 litres, find f.

Stepwise solution. Class marks are 1.5, 2.5, 3.5, 4.5, 5.5. Then Σfx = 4.5 + 15 + 3.5f + 22.5 + 11 = 53 + 3.5f and Σf = 16 + f. Given mean = 3.5, so 3.5 = (53 + 3.5f)/(16 + f). Thus 56 + 3.5f = 53 + 3.5f, which is impossible; this means the given mean is inconsistent with the table. Therefore, the learner must detect bad data instead of forcing an answer.

Concept note. This is a high-value thinking question: sometimes the correct step is to spot inconsistency.

CautionSkill: reading carefully

Question 4. A data set has intervals 0–5, 5–10, 10–20, 20–30 with frequencies 6, 8, 10, 6. Find the mean.

Stepwise solution. First note that class widths are not all equal. That does not stop us from finding the mean, but it does make step-deviation less neat. Class marks are 2.5, 7.5, 15, 25. Then Σfx = 15 + 60 + 150 + 150 = 375 and Σf = 30. So mean = 375/30 = 12.5.

Concept note. Students often think equal width is compulsory for mean. It is not; it mainly affects convenience of the method.

Interactive modules

Flashcards

Card 1

Question: What is the class mark of 17–19?

Answer: (17 + 19)/2 = 18.
Card 2

Question: When is assumed mean method more useful than direct method?

Answer: When class marks and products are large, because deviations from a central assumed value reduce computation.

Quick concept quiz

Choose the best answer and get instant feedback.

Score: 0/2
1For the class interval 50–60, the class mark is
2In step-deviation method, h usually stands for

Method selector trainer

Try these quick rules mentally:

  • Small class marks and easy products → direct method.
  • Large class marks but equal spacing → assumed mean.
  • Equal class width and easy common factor in deviations → step-deviation.
  • Unequal class widths → direct method is often the safest choice.

Micro challenge

Intervals: 10–20, 20–30, 30–40 with frequencies 3, 5, 2. Can you estimate the mean before calculating?

Since the middle class has the highest frequency, the mean should be near 25, slightly below or above depending on the pull of the outer classes. Exact value: class marks 15, 25, 35; Σfx = 45 + 125 + 70 = 240; Σf = 10; mean = 24.

Teacher-ready usage notes

  • Paste this snippet directly inside an existing PHP page; it already contains scoped CSS, markup, and JavaScript.
  • No HTML, head, body, or H1 wrapper is used, so it remains embeddable.
  • No external libraries are required; everything is self-contained.
  • You can duplicate the block on different pages because the snippet auto-generates a unique scoped ID from the file itself.
  • To adapt for another chapter, change only the text content and the built-in practice bank arrays in the script.

Pedagogical scope note: Exercise 13.1 is centred on the mean of grouped data and suitable methods for calculating it, including direct, assumed mean, and step-deviation approaches.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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Class 10 Maths Chapter 13 Ex 13.1 – Complete NCERT Solutions
Class 10 Maths Chapter 13 Ex 13.1 – Complete NCERT Solutions — Complete Notes & Solutions · academia-aeternum.com
Statistics is one of the most practical and scoring chapters in the Class X Mathematics syllabus, as it bridges numerical computation with real-world interpretation of data. The textbook exercise solutions for Chapter 13, Statistics, are designed to help learners systematically master the calculation and analysis of data presented in grouped frequency distributions. These solutions focus not only on obtaining correct numerical answers but also on developing a clear understanding of statistical…
🎓 Class 10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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