Ch 13  ·  Q–
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Class 10 Mathematics Exercise 13.2 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.

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Q1
NUMERIC3 marks

The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Concept Used

  • Mean (Assumed Mean Method): Used for grouped data to simplify calculations.
  • \[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \]
  • Mode (Grouped Data): Most frequent class (modal class).
  • \[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]

Solution Roadmap

  1. Find class marks \(x_i\)
  2. Choose assumed mean \(a\)
  3. Compute \(d_i = x_i - a\)
  4. Find \(f_i d_i\)
  5. Apply mean formula
  6. Identify modal class
  7. Apply mode formula
  8. Compare results

Solution

Step 1: Class marks \(x_i\)

\[ x_i = \frac{\text{Upper limit + Lower limit}}{2} \]

Class \(f_i\) \(x_i\) \(d_i = x_i - 30\) \(f_i d_i\)
5-15610-20-120
15-251120-10-110
25-35213000
35-45234010230
45-55145020280
55-6556030150
Total \[ \sum f_i = 80 \] \[ \sum f_i d_i = 430 \]

Step 2: Mean Calculation

\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \] \[ = 30 + \frac{430}{80} \] \[ = 30 + 5.375 \] \[ = 35.375 \approx 35.38 \]

Mean = 35.38 years

Step 3: Mode Calculation

Modal class = 35–45 (highest frequency = 23)

\[ l = 35,\quad h = 10,\quad f_1 = 23,\quad f_0 = 21,\quad f_2 = 14 \] \[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \] \[ = 35 + \left( \frac{23 - 21}{2(23) - 21 - 14} \right) \times 10 \] \[ = 35 + \left( \frac{2}{46 - 35} \right) \times 10 \] \[ = 35 + \frac{2}{11} \times 10 \] \[ = 35 + 1.82 \] \[ = 36.82 \]

Mode = 36.82 years

Graphical Understanding (Modal Class Peak)
Age Groups Frequency
Interpretation
  • Mean age ≈ 35.38 years (average patient age)
  • Mode ≈ 36.82 years (most common age group concentration)
  • Mode > Mean ⇒ slight right skew (more patients in higher age group)
Exam Significance
  • Direct CBSE board question type (3–4 marks)
  • Tests concept clarity of grouped data
  • Frequently used in CUET, SSC, Railway exams
  • Understanding comparison (Mean vs Mode) is key for case-study questions
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1 / 6  ·  17%
Q2 →
Q2
NUMERIC3 marks

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Lifetimes (in hours) 0-20 20-40 40-60 60-80
\(\tiny\text{(Modal Class)}\)		 80-100 100-120
Frequency 10 35 52 \((f_0)\) 61 \((f_1)\) 38 \((f_2)\) 29

Concept Used

  • Mode (Grouped Data): It represents the most frequent observation.
  • For grouped data, mode is estimated using:
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]
  • \(l\): Lower limit of modal class
  • \(h\): Class width
  • \(f_1\): Frequency of modal class
  • \(f_0\): Frequency of preceding class
  • \(f_2\): Frequency of succeeding class

Solution Roadmap

  1. Identify modal class (highest frequency)
  2. Extract \(l, h, f_0, f_1, f_2\)
  3. Substitute into formula
  4. Simplify step-by-step carefully

Solution

Step 1: Identify Modal Class

Highest frequency = 61 ⇒ Modal Class = 60–80

Step 2: Determine Parameters

\[ l = 60,\quad h = 20 \] \[ \begin{aligned} f_1 &= 61,\\ f_0 &= 52,\\ f_2 &= 38 \end{aligned} \]

Step 3: Apply Mode Formula

\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]

Step 4: Substitution

\[ = 60 + \left( \frac{61 - 52}{2(61) - 52 - 38} \right) \times 20 \]

Step 5: Simplify Numerator

\[ 61 - 52 = 9 \]

Step 6: Simplify Denominator

\[ 2(61) = 122 \] \[ \begin{aligned} 122 - 52 - 38 &= 122 - 90 \\&= 32 \end{aligned} \]

Step 7: Final Calculation

\[ \text{Mode} = 60 + \left( \frac{9}{32} \right) \times 20 \] \[ = 60 + \frac{180}{32} \] \[ = 60 + 5.625 \] \[ = 65.625 \]

Modal Lifetime = 65.625 hours

Graphical Insight (Modal Class Peak)
Lifetimes (hours) Frequency
Interpretation
  • Most components last around 65.6 hours
  • Peak concentration lies in 60–80 hours interval
  • Useful in predicting typical product performance
Exam Significance
  • Very common CBSE board question (Mode of grouped data)
  • Direct formula-based 3-mark question
  • Tests accuracy in substitution and simplification
  • Important for CUET, SSC, Railway exams
← Q1
2 / 6  ·  33%
Q3 →
Q3
NUMERIC3 marks

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (₹) 1000-1500 1500-2000 2000-2500 2500-3000 3000-3500 3500-4000 4000-4500 4500-5000
Families 24 40 33 28 30 22 16 7

Concept Used

  • Mode: Most frequent expenditure group.
  • Mean (Step Deviation Method): Efficient for large values.
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \] \[ \overline{x} = a + \left( \frac{\sum f_i u_i}{\sum f_i} \right) \times h \]

Solution Roadmap

  1. Find class marks \(x_i\)
  2. Choose assumed mean \(a\)
  3. Compute \(u_i = \frac{x_i - a}{h}\)
  4. Find \(f_i u_i\)
  5. Calculate Mean
  6. Identify modal class
  7. Apply mode formula

Solution

Step 1: Class Marks

\[ x_i = \frac{\text{Upper + Lower}}{2} \]
Class \(f_i\) \(x_i\) \(u_i=\frac{x_i-2750}{500}\) \(f_i u_i\)
1000-1500241250-3-72
1500-2000401750-2-80
2000-2500332250-1-33
2500-300028275000
3000-3500303250130
3500-4000223750244
4000-4500164250348
4500-500074750428
Total \[ \sum f_i = 200 \] \[ \sum f_i u_i = -35 \]

Step 2: Mean Calculation

\[ \overline{x} = 2750 + \left( \frac{-35}{200} \right) \times 500 \] \[ = 2750 - \frac{35}{200} \times 500 \] \[ = 2750 - 87.5 \] \[ = 2662.5 \]

Mean Monthly Expenditure = ₹2662.50

Step 3: Mode Calculation

Modal class = 1500–2000 (highest frequency = 40)

\[ l=1500,\quad h=500,\quad f_1=40,\quad f_0=24,\quad f_2=33 \] \[ \text{Mode} = 1500 + \left( \frac{40-24}{2(40)-24-33} \right) \times 500 \] \[ = 1500 + \left( \frac{16}{80-57} \right) \times 500 \] \[ = 1500 + \frac{16}{23} \times 500 \] \[ = 1500 + 347.83 \] \[ = 1847.83 \]

Modal Expenditure = ₹1847.83

Graphical Insight (Distribution Shape)
Expenditure (₹) Families
Interpretation
  • Most families spend around ₹1848 (modal value)
  • Average spending is higher (₹2662.5)
  • Mean > Mode ⇒ right-skewed distribution (few high spenders pulling average up)
Exam Significance
  • Highly important CBSE board question (combined Mean + Mode)
  • Tests step deviation efficiency
  • Common in case-study questions
  • Important for CUET, SSC, Banking exams
← Q2
3 / 6  ·  50%
Q4 →
Q4
NUMERIC3 marks

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Students per Teacher States / U.T.
15-203
20-258
25-309
30-3510
35-403
40-450
45-500
50-552

Concept Used

  • Mean (Assumed Mean Method)
  • Mode (Grouped Data)
\[ \overline{x} = a + \frac{\sum f_i d_i}{\sum f_i} \] \[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]

Solution Roadmap

  1. Find class marks \(x_i\)
  2. Choose assumed mean \(a\)
  3. Compute \(d_i = x_i - a\)
  4. Find \(f_i d_i\)
  5. Compute mean
  6. Identify modal class
  7. Apply mode formula

Solution

Step 1: Class Marks

\[ x_i = \frac{\text{Upper + Lower}}{2} \]
Class \(f_i\) \(x_i\) \(d_i = x_i - 32.5\) \(f_i d_i\)
15-20317.5-15-45
20-25822.5-10-80
25-30927.5-5-45
30-351032.500
35-40337.5515
40-45042.5100
45-50047.5150
50-55252.52040
Total \[ \sum f_i = 35 \] \[ \sum f_i d_i = -115 \]

Step 2: Mean Calculation

\[ \overline{x} = 32.5 + \frac{-115}{35} \] \[ = 32.5 - \frac{115}{35} \] \[ = 32.5 - 3.2857 \] \[ = 29.21 \]

Mean ≈ 29.21 students per teacher

Step 3: Mode Calculation

Modal class = 30–35 (highest frequency = 10)

\[ l=30,\quad h=5,\quad f_1=10,\quad f_0=9,\quad f_2=3 \] \[ \text{Mode} = 30 + \left( \frac{10-9}{2(10)-9-3} \right) \times 5 \] \[ = 30 + \left( \frac{1}{20-12} \right) \times 5 \] \[ = 30 + \frac{5}{8} \] \[ = 30.625 \approx 30.63 \]

Mode ≈ 30.63 students per teacher

Graphical Insight
Students per Teacher States
Interpretation
  • Most common ratio ≈ 30–31 students per teacher
  • Average ratio ≈ 29 students per teacher
  • Mode > Mean ⇒ slightly positively skewed distribution
  • Indicates a few states have higher student load
Exam Significance
  • Very important CBSE question (Mean + Mode combined)
  • Interpretation part often asked in case-study questions
  • Tests conceptual clarity of skewness
  • Useful for CUET, SSC, UPSC CSAT data interpretation
← Q3
4 / 6  ·  67%
Q5 →
Q5
NUMERIC3 marks

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs Scored Number of Batsmen
3000-40004 \((f_0)\)
4000-500018 \((f_1)\)
5000-60009 \((f_2)\)
6000-70007
7000-80006
8000-90003
9000-100002
10000-110001

Concept Used

  • Mode (Grouped Data): Represents the most frequent class interval.
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]

Solution Roadmap

  1. Identify modal class
  2. Extract \(l, h, f_0, f_1, f_2\)
  3. Substitute into formula
  4. Simplify step-by-step

Solution

Step 1: Identify Modal Class

Highest frequency = 18 ⇒ Modal Class = 4000–5000

Step 2: Determine Parameters

\[ l = 4000,\quad h = 1000 \] \[ f_1 = 18,\quad f_0 = 4,\quad f_2 = 9 \]

Step 3: Apply Formula

\[ \text{Mode} = 4000 + \left( \frac{18 - 4}{2(18) - 4 - 9} \right) \times 1000 \]

Step 4: Simplify Numerator

\[ 18 - 4 = 14 \]

Step 5: Simplify Denominator

\[ 2(18) = 36 \] \[ 36 - 4 - 9 = 23 \]

Step 6: Final Calculation

\[ \text{Mode} = 4000 + \frac{14}{23} \times 1000 \] \[ = 4000 + \frac{14000}{23} \] \[ = 4000 + 608.70 \] \[ = 4608.70 \]

Mode ≈ 4608.7 runs

Graphical Insight (Modal Peak)
Runs Batsmen
Interpretation
  • Most batsmen have scored around 4600 runs
  • Peak concentration lies in 4000–5000 runs range
  • Distribution shows gradual decline after peak
Exam Significance
  • Classic CBSE question on Mode (3 marks)
  • Tests correct identification of modal class
  • Very common in sports-data case studies
  • Important for CUET, SSC exams
← Q4
5 / 6  ·  83%
Q6 →
Q6
NUMERIC3 marks

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Number of Cars 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 7 14 13 12 \((f_0)\) 20 \((f_1)\) 11 \((f_2)\) 15 8

Concept Used

  • Mode (Grouped Data): Most frequent observation interval.
\[ \text{Mode} = l + \left( \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \right) h \]

Solution Roadmap

  1. Identify modal class
  2. Determine \(l, h, f_0, f_1, f_2\)
  3. Substitute values carefully
  4. Simplify step-by-step

Solution

Step 1: Identify Modal Class

Highest frequency = 20 ⇒ Modal class = 40–50

Step 2: Parameters

\[ l = 40,\quad h = 10 \] \[ f_1 = 20,\quad f_0 = 12,\quad f_2 = 11 \]

Step 3: Apply Formula

\[ \text{Mode} = 40 + \left( \frac{20 - 12}{2(20) - 12 - 11} \right) \times 10 \]

Step 4: Simplify Numerator

\[ 20 - 12 = 8 \]

Step 5: Simplify Denominator

\[ 2(20) = 40 \] \[ 40 - 12 - 11 = 17 \]

Step 6: Final Calculation

\[ \text{Mode} = 40 + \frac{8}{17} \times 10 \] \[ = 40 + \frac{80}{17} \] \[ = 40 + 4.7059 \] \[ = 44.7059 \approx 44.71 \]

Mode ≈ 44.71 cars

Graphical Insight (Traffic Peak)
Number of Cars Frequency
Interpretation
  • Most frequent traffic flow ≈ 45 cars per 3 minutes
  • Indicates peak traffic density around 40–50 range
  • Useful for traffic planning and congestion analysis
Exam Significance
  • Standard CBSE 3-mark question (Mode of grouped data)
  • Tests accuracy in substitution and simplification
  • Common real-life data interpretation question
  • Important for CUET, SSC exams
← Q5
6 / 6  ·  100%
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Chapter Complete!

All 6 solutions for Statistics covered.

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NCERT · Class X · Chapter 13

Statistics — Exercise 13.2

Measures of Central Tendency for Grouped Data  |  Mean · Mode · Median  |  Interactive AI Learning Engine

📐
Core Formulas — Exercise 13.2

All three measures of central tendency for grouped (continuous) frequency distributions, as studied in NCERT Class X Chapter 13.

Mean · Method 1
Direct Method
x̄ = Σ(fᵢxᵢ) / Σfᵢ
Where xᵢ = class mark (midpoint) = (lower + upper) / 2,
fᵢ = frequency of the i-th class,
Σfᵢ = n = total number of observations.

Best used when class marks and frequencies are small integers.
Mean · Method 2
Assumed Mean Method
x̄ = A + Σ(fᵢdᵢ) / Σfᵢ
Where A = assumed mean (choose a central class mark),
dᵢ = xᵢ − A = deviation of i-th class mark from A.

Reduces computation significantly when values are large.
Mean · Method 3
Step Deviation Method
x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h
Where uᵢ = (xᵢ − A) / h = step deviation,
h = class width (must be uniform),
A = assumed mean.

Most efficient for equal class intervals. uᵢ are small integers.
Mode
Grouped Data Mode
Mo = l + [(f₁−f₀) / (2f₁−f₀−f₂)] × h
Where l = lower boundary of modal class,
f₁ = frequency of modal class,
f₀ = frequency of class before modal class,
f₂ = frequency of class after modal class,
h = class width.
Median
Grouped Data Median
Md = l + [(n/2 − cf) / f] × h
Where l = lower boundary of median class,
n = Σfᵢ = total frequency,
cf = cumulative frequency before median class, f = frequency of median class,
h = class width.
Key Relationship
Empirical Formula
Mode = 3·Median − 2·Mean
This empirical (observed) relationship holds approximately for moderately skewed distributions.

Useful to: verify answers, find a missing measure when two are known, or check if a distribution is roughly symmetric.
📍
Key Concepts at a Glance
Class Mark (xᵢ)
Midpoint of a class interval.
xᵢ = (Lower limit + Upper limit) / 2
For class 10–20: xᵢ = 15
Modal Class
The class interval with the highest frequency in the distribution. Mode lies inside this class.
Median Class
Class where cumulative frequency first becomes ≥ n/2. Median lies inside this class.
Cumulative Frequency
Running total of frequencies up to (but not including) the current class. Used to locate the median class.
🧮
Interactive Step-by-Step Solver

Enter your grouped frequency distribution data below. The solver will generate a complete, exam-style step-by-step solution.

Enter each class interval and its frequency. Format: lower-upper (e.g. 10-20)

💡
Tips, Tricks & Shortcuts

Exam-tested strategies to solve Chapter 13 problems faster and with fewer errors.

Tip
Choose the Right Method for Mean
Use Direct Method when numbers are small. Switch to Assumed Mean when class marks exceed 50, and use Step Deviation when all class widths are equal — it gives the simplest uᵢ values (−2, −1, 0, 1, 2).
Trick
Pick the Best Assumed Mean
Always choose the class mark of the middle class as A. This ensures your dᵢ/uᵢ values are both negative and positive, so Σfᵢuᵢ is small, reducing arithmetic errors by up to 80%.
Tip
Verify with the Empirical Formula
After computing Mean, Mode and Median separately, always check: Mode ≈ 3×Median − 2×Mean. If it doesn't hold approximately, recheck your modal/median class identification.
Trick
n/2 Shortcut for Median Class
Calculate n/2 first, then scan the cumulative frequency column from top. The median class is the first row where cf ≥ n/2. Mark it visually — this avoids re-reading the table under exam pressure.
Caution
Use Class Boundaries, Not Limits
In the Mode and Median formulas, l means the lower boundary (true lower limit) of the class, not the stated lower limit. For continuous data these are equal; but for discrete data, adjust by −0.5 if needed.
Trick
Mode Formula Denominator Check
The denominator 2f₁−f₀−f₂ must always be positive if f₁ is the highest frequency. If you get a negative denominator, you've misidentified the modal class — pick the class with the largest f.
Tip
Σfᵢ Must Equal n
Always sum your frequencies first and match with n given in the problem. If they differ, you've missed a class or mis-read a frequency — catching this early saves a whole solution rewrite.
Trick
Step Deviation uᵢ = 0 is Always A
The class mark you choose as A gets uᵢ = 0. Classes above A get positive uᵢ, classes below get negative uᵢ. A quick sanity check: if your uᵢ column doesn't straddle 0, revisit your assumed mean choice.
Tip
Table Layout Saves Marks
Draw the full working table with columns: CI | xᵢ | fᵢ | dᵢ or uᵢ | fᵢuᵢ | cf even in exams. Board examiners award method marks for each column — a complete table earns full marks even if final arithmetic slips.
Trick
Median Class: Never Use cf of That Row
In the Median formula, cf is the cumulative frequency of all classes before the median class — not including it. Students who use the cf of the median class row itself get a consistently wrong answer.
⚠️
Common Mistakes & Corrections

The most frequently lost marks in board exams — understand the mistake, then the correct approach.

🔢
Mistake 1: Using class limits instead of class marks for xᵢ
❌ For class 20–30, using xᵢ = 20 or xᵢ = 30
✅ xᵢ = (20 + 30) / 2 = 25 (midpoint / class mark)
The Direct Method requires the midpoint of each class. Using limits gives a completely wrong mean. Always compute xᵢ = (L + U) / 2 before filling the fᵢxᵢ column.
📊
Mistake 2: Including cf of the median class in the Median formula
❌ cf = cumulative frequency up to AND including median class
✅ cf = cumulative frequency of all classes BEFORE the median class
If the median class is 40–50 and cf column reads 12, 20, 29 (where 29 is median class), use cf = 20, not 29. This is the single most common error in median calculations.
🎯
Mistake 3: Wrong identification of modal class
❌ Choosing modal class as the one with highest cumulative frequency
✅ Modal class = class with the highest individual (raw) frequency fᵢ
Cumulative frequency always increases — the last class always has the highest cf. The modal class is determined by looking at fᵢ (the frequency column), not the cf column.
➗
Mistake 4: Forgetting to multiply Σfᵢuᵢ/Σfᵢ by h in Step Deviation
❌ x̄ = A + Σfᵢuᵢ / Σfᵢ (missing × h)
✅ x̄ = A + (Σfᵢuᵢ / Σfᵢ) × h
The step deviation uᵢ = (xᵢ − A)/h already divided by h — so you must multiply back by h at the end to restore the original scale. Forgetting h gives a nonsensically small answer.
🔤
Mistake 5: Mixing up f₀ and f₂ in the Mode formula
❌ f₀ = frequency after modal class; f₂ = frequency before modal class
✅ f₀ = class just BEFORE modal class; f₂ = class just AFTER modal class
In Mo = l + [(f₁−f₀)/(2f₁−f₀−f₂)] × h, the order matters because the numerator only uses f₁−f₀. Swapping f₀ and f₂ changes the answer. Always write the table in ascending order of class intervals to avoid confusion.
📏
Mistake 6: Using inconsistent h in Step Deviation with unequal widths
❌ Applying step deviation method when class widths differ (e.g. 5, 10, 10, 5)
✅ Only apply step deviation when ALL class widths are equal
The step deviation formula assumes a constant h. For unequal class widths, use the Assumed Mean method (dᵢ = xᵢ − A, no division by h). Using an inconsistent h gives a weighted mean of the wrong quantities.
⚖️
Mistake 7: Confusing n/2 and (n+1)/2 for grouped data median
❌ Using (n+1)/2 as the threshold for finding the median class
✅ For grouped data, always use n/2 (not (n+1)/2)
(n+1)/2 is used for the middle term in ungrouped/raw data. For grouped frequency distributions, the median formula uses n/2. Since n is usually even in textbook problems, the distinction is critical.
📚
Concept-Building Question Bank

Original, non-textbook questions organised by concept. Each has a complete step-by-step solution. Click a question to reveal its solution.

Q1 · Direct The weekly pocket money (₹) of 40 students is distributed as: 20–30: 5, 30–40: 8, 40–50: 12, 50–60: 9, 60–70: 6. Find the mean pocket money using the Direct Method. ▾
✦ Solution
1
Find Class Marks (xᵢ)
20–30: x=25  |  30–40: x=35  |  40–50: x=45  |  50–60: x=55  |  60–70: x=65
2
Compute fᵢxᵢ for each class
Class xᵢ fᵢ fᵢxᵢ
20–30 25 5 125
30–40 35 8 280
40–50 45 12 540
50–60 55 9 495
60–70 65 6 390
Total 40 1830
3
Apply Formula
x̄ = Σfᵢxᵢ / Σfᵢ = 1830 / 40
∴ Mean = ₹45.75
Q2 · Assumed Heights (cm) of 50 plants in a greenhouse: 100–120: 4, 120–140: 9, 140–160: 15, 160–180: 14, 180–200: 8. Find the mean height using the Assumed Mean Method. ▾
✦ Solution
1
Choose Assumed Mean A
Class marks: 110, 130, 150, 170, 190. Middle class mark = A = 150
2
Compute dᵢ = xᵢ − A and fᵢdᵢ
Class xᵢ fᵢ dᵢ=xᵢ−150 fᵢdᵢ
100–120 110 4 −40 −160
120–140 130 9 −20 −180
140–160 150 15 0 0
160–180 170 14 +20 +280
180–200 190 8 +40 +320
Total 50 +260
3
Apply Formula
x̄ = A + Σfᵢdᵢ/Σfᵢ = 150 + (260/50) = 150 + 5.2
∴ Mean Height = 155.2 cm
Q3 · Step Dev Monthly electricity bills (₹) of 60 households: 200–400: 7, 400–600: 13, 600–800: 18, 800–1000: 14, 1000–1200: 8. Find mean using the Step Deviation Method. ▾
✦ Solution
1
Identify h and choose A
Class width h = 200. Class marks: 300, 500, 700, 900, 1100. Choose A = 700 (middle)
2
Compute uᵢ = (xᵢ−700)/200 and fᵢuᵢ
Class xᵢ fᵢ uᵢ fᵢuᵢ
200–400 300 7 −2 −14
400–600 500 13 −1 −13
600–800 700 18 0 0
800–1000 900 14 +1 +14
1000–1200 1100 8 +2 +16
Total 60 +3
3
Apply Formula
x̄ = 700 + (3/60) × 200 = 700 + 0.05 × 200 = 700 + 10
∴ Mean Bill = ₹710
Q4 · Mode Time (in minutes) taken by commuters to reach office: 0–10: 5, 10–20: 11, 20–30: 24, 30–40: 18, 40–50: 9, 50–60: 3. Find the mode of commute time. ▾
✦ Solution
1
Identify Modal Class
Frequencies: 5, 11, 24, 18, 9, 3. Highest frequency = 24 → Modal class = 20–30
2
Identify Required Values
l = 20 (lower boundary of 20–30)
f₁ = 24 (modal class frequency)
f₀ = 11 (frequency of class before: 10–20)
f₂ = 18 (frequency of class after: 30–40)
h = 10
3
Apply Mode Formula
Mo = 20 + [(24−11) / (2×24−11−18)] × 10
= 20 + [13 / (48−29)] × 10
= 20 + (13/19) × 10
= 20 + 6.84
∴ Mode = 26.84 minutes
Q5 · Mode Marks scored by 80 students in a science test: 0–10: 3, 10–20: 8, 20–30: 19, 30–40: 26, 40–50: 15, 50–60: 9. Find the modal marks. Also verify: is the distribution positively or negatively skewed? ▾
✦ Solution
1
Identify Modal Class
Highest frequency = 26 → Modal class = 30–40
f₀ = 19, f₁ = 26, f₂ = 15, l = 30, h = 10
2
Apply Formula
Mo = 30 + [(26−19)/(52−19−15)] × 10 = 30 + [7/18] × 10 = 30 + 3.89
∴ Mode ≈ 33.89 marks
3
Skewness Check
Mean (by direct method) ≈ 33.5, Median ≈ 33.5, Mode ≈ 33.89
Since Mean ≈ Median ≈ Mode, the distribution is approximately symmetric (bell-shaped).
Q6 · Median Daily wages (₹) of 70 factory workers: 200–250: 8, 250–300: 15, 300–350: 22, 350–400: 17, 400–450: 8. Find the median wage. ▾
✦ Solution
1
Build Cumulative Frequency Table
Class fᵢ cf
200–250 8 8
250–300 15 23
300–350 22 45
350–400 17 62
400–450 8 70
2
Find Median Class
n = 70 → n/2 = 35
cf column: 8, 23, 45 ← first cf ≥ 35 is 45 → Median class = 300–350
l = 300, cf = 23 (before this class), f = 22, h = 50
3
Apply Formula
Md = 300 + [(35−23)/22] × 50 = 300 + (12/22) × 50 = 300 + 27.27
∴ Median Wage = ₹327.27
Q7 · Median Ages (in years) of patients in a hospital ward: 10–20: 6, 20–30: 11, 30–40: 21, 40–50: 23, 50–60: 14, 60–70: 5. Find the median age. Also find the missing frequency if one class is unknown and total patients = 85 (with one missing). ▾
✦ Solution — Part A (All frequencies given, total = 80)
1
Cumulative Frequency Table
cf: 6 | 17 | 38 | 61 | 75 | 80     n = 80, n/2 = 40
First cf ≥ 40 is 61 → Median class = 40–50
l=40, cf=38, f=23, h=10
2
Calculate Median
Md = 40 + [(40−38)/23] × 10 = 40 + (2/23)×10 = 40 + 0.87
∴ Median Age ≈ 40.87 years
✦ Part B — Finding Missing Frequency (total = 85)
3
Set Up Equation
Let missing frequency (say for 30–40) = x
6 + 11 + x + 23 + 14 + 5 = 85 → 59 + x = 85 → x = 26
Q8 · Empirical For a frequency distribution, Mean = 54.6 and Mode = 51.8. Using the empirical formula, find the Median. Then verify using the statement: "Median divides the distribution between Mean and Mode." ▾
✦ Solution
1
Empirical Formula
Mode = 3·Median − 2·Mean
51.8 = 3·Md − 2×54.6
51.8 = 3·Md − 109.2
3·Md = 161
∴ Median = 53.67
2
Verification
Mode (51.8) < Median (53.67) < Mean (54.6) ✓
This confirms negative skewness (tail on the left) — Mean > Median > Mode is reversed, so actually this is a left-skewed distribution (longer tail to the left of Mode).
Q9 · Empirical A dataset has Median = 46.2 and Mean = 45. Find the Mode using the empirical formula. If Mean is increased by 3 due to a data entry correction, what is the new Mode? (Assume Median stays the same.) ▾
✦ Solution
1
Find Mode
Mode = 3×46.2 − 2×45 = 138.6 − 90 = 48.6
Original Mode = 48.6
2
Corrected Mean = 45 + 3 = 48. New Mode?
New Mode = 3×46.2 − 2×48 = 138.6 − 96 = 42.6
Corrected Mode = 42.6

Observation: A 3-unit increase in Mean caused a 6-unit decrease in Mode — the empirical formula amplifies Mean changes in Mode by a factor of 2.
Q10 · Mixed Runs scored by a cricketer in 50 innings: 0–20: 6, 20–40: 10, 40–60: 16, 60–80: 12, 80–100: 6. Find Mean (Step Deviation), Mode, and Median. Verify with the empirical formula. ▾
✦ Full Solution (All Three Measures)
1
Mean by Step Deviation (A=50, h=20)
Class xᵢ fᵢ uᵢ=(xᵢ−50)/20 fᵢuᵢ cf
0–20 10 6 −2 −12 6
20–40 30 10 −1 −10 16
40–60 50 16 0 0 32
60–80 70 12 +1 +12 44
80–100 90 6 +2 +12 50
50 +2
x̄ = 50 + (2/50)×20 = 50 + 0.8 → Mean = 50.8
2
Mode
Modal class = 40–60 (f=16), f₀=10, f₂=12, l=40, h=20
Mo = 40 + [(16−10)/(32−10−12)]×20 = 40 + (6/10)×20 = 40 + 12 → Mode = 52
3
Median
n/2 = 25. From cf: 6, 16, 32 → median class = 40–60, cf=16, f=16, l=40
Md = 40 + [(25−16)/16]×20 = 40 + (9/16)×20 = 40 + 11.25 → Median = 51.25
4
Empirical Verification
3·Median − 2·Mean = 3×51.25 − 2×50.8 = 153.75 − 101.6 = 52.15 ≈ 52 (Mode) ✓
All three measures verified!
Q11 · Applied A survey of 100 households shows monthly savings (₹): 500–1000: 14, 1000–1500: 22, 1500–2000: 30, 2000–2500: 24, 2500–3000: 10. A financial advisor says "More than half the households save below ₹1900." Verify this claim using the Median. ▾
✦ Solution
1
Build cf Table
cf: 14 | 36 | 66 | 90 | 100. n=100, n/2=50.
First cf ≥ 50 is 66 → Median class = 1500–2000
l=1500, cf=36, f=30, h=500
2
Compute Median
Md = 1500 + [(50−36)/30]×500 = 1500 + (14/30)×500 = 1500 + 233.33
Median = ₹1733.33
3
Verify Advisor's Claim
Median = ₹1733.33 < ₹1900. This means 50% of households save below ₹1733.33 < ₹1900.
∴ The advisor's claim is TRUE — in fact, the median is even lower than ₹1900.
🎮
Interactive Learning Modules

Six hands-on activities to strengthen your understanding through exploration, not memorisation.

⚡ Quick Quiz
Test your formula recall — 5 questions, instant feedback
🃏 Formula Flashcards
Tap the card to flip between question and answer
📊 Mean vs Median vs Mode
Adjust the slider to see how skewness affects the three measures
Left skewedSymmetricRight skewed
🔨 Build & Solve
Create your own frequency table and auto-calculate all three measures
🔍 Formula Decoder
Hover over any symbol to get its plain-language meaning
l + [(n/2 − cf) / f] × h
Hover over the coloured symbols above
l + [(f₁− f₀) / (2 f₁− f₀− f₂)] × h
■ boundary/width ■ modal freq ■ before class ■ after class
🧩 Pattern Spotter Challenge
Look at the data and answer before calculating
A distribution has these frequencies (equal class widths):

3 | 7 | 18 | 25 | 20 | 11 | 4
Without calculating: which relationship will hold?
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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Exercise 13.2 Class 10 Maths Solutions – Statistics
Exercise 13.2 Class 10 Maths Solutions – Statistics — 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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