Focus Keyword: NCERT Solutions Class 10 Math Chapter 13 Statistics
Secondary Keywords & LSI: CBSE Class 10 Maths Statistics solutions, Class 10 Maths Chapter 13 Exercise 13.1 solutions, Class 10 Maths Exercise 13.2, Class 10 Maths Exercise 13.3, mean median mode grouped data Class 10, CBSE Class 10 Maths board exam preparation
SEO Meta Description: Complete NCERT Solutions for Class 10 Math Chapter 13 Statistics. Step-by-step Mean, Median, Mode formulas, calculation tables, missing frequencies, and exam tips.
H1 Title: NCERT Solutions for Class 10 Math Chapter 13: Statistics (Complete Step-by-Step Guide)
Navigating through the CBSE Class 10 Mathematics curriculum requires an in-depth understanding of statistical frequency distributions, central tendencies, grouped data intervals, and missing frequency derivations. Chapter 13 of Class 10 Mathematics, “Statistics”, forms the foundation of demographic analysis, economic forecasting, quality control engineering, and scientific research methodology. It investigates the three primary measures of central tendency—Mean, Mode, and Median; explores the direct, assumed mean, and step-deviation methods for computing arithmetic averages; and details the quantitative determination of modal intervals and cumulative frequency medians from continuous distributions. To help students master every aspect of this high-weightage chapter, this comprehensive guide offers textbook-accurate, highly structured, and pedagogically sound responses strictly aligned with the latest CBSE evaluation standards.
Every question presented in the official NCERT textbook—ranging from foundational textbook examples to the complete question sets of Exercise 13.1, Exercise 13.2, and Exercise 13.3, along with an expanded set of 15 board-level FAQs—has been solved with exhaustive detail. Key scoring terms, systematic tabular workflows, and algebraic substitutions have been highlighted to ensure students secure maximum marks in their CBSE Board Examinations.
Chapter 13: Statistics
Master Chapter Summary & Formula Blueprint
In the rationalised NCERT Class 10 curriculum, Chapter 13 focuses entirely on grouped numerical data. Outdated topics such as cumulative frequency curves (ogives) have been removed, making computational accuracy for Mean, Mode, and Median the primary focus of board examinations.
| Measure of Central Tendency | Method / Condition | Standard Governing Formula | Core Parameters & Definitions | CBSE Marks Weightage |
|---|---|---|---|---|
| Mean ($\bar{x}$) | Direct Method | $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ | $x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}$, $f_i = \text{frequency}$ | 2 to 3 Marks |
| Mean ($\bar{x}$) | Assumed Mean Method | $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$ | $a = \text{assumed mean}$, $d_i = x_i – a$ | 3 to 4 Marks |
| Mean ($\bar{x}$) | Step-Deviation Method | $\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$ | $u_i = \frac{x_i – a}{h}$, $h = \text{class size}$ | 3 to 4 Marks |
| Mode | Grouped Data | $\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$ | $l = \text{lower limit of modal class}$, $f_1 = \text{modal freq}$, $f_0 = \text{preceding freq}$, $f_2 = \text{succeeding freq}$ | 3 to 4 Marks |
| Median | Grouped Data | $\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h$ | $l = \text{lower limit of median class}$, $cf = \text{preceding cumulative freq}$, $f = \text{median class freq}$ | 3 to 5 Marks |
| Empirical Relationship | Moderately Asymmetric Data | $3\text{ Median} = \text{Mode} + 2\text{ Mean}$ | Interlinks all 3 measures of central tendency | 1 Mark (MCQ / Fill-in) |
🧠 Examiner’s Secret: When continuous class intervals are discontinuous (e.g., $118 – 126, 127 – 135$), students must convert them into continuous boundaries before identifying the modal class or median class. Subtract $0.5$ from the lower limit and add $0.5$ to the upper limit (e.g., $117.5 – 126.5, 126.5 – 135.5$). Failing to convert boundaries results in an incorrect lower limit $l$ and a deduction of marks.
Foundational Statistical Concepts and Central Tendencies
Mean of Grouped Data
The arithmetic mean of grouped data is the weighted average value of the numerical distribution, calculated by dividing the sum of the products of class midpoints and their corresponding frequencies by the total frequency. Three Methods to Compute Mean Direct Σ(fi·xi) / Σfi Assumed Mean a + Σ(fi·di)/Σfi Step-Deviation a + h·[Σ(fi·ui)/Σfi]
When values of class mark $x_i$ and frequency $f_i$ are small, the Direct Method is practical. If numerical values are large, the Assumed Mean Method or Step-Deviation Method reduces large calculations to small deviations, minimizing calculation errors.
Mode of Grouped Data
The mode of grouped data is the value inside the modal class that occurs with the highest frequency across the entire distribution.
In grouped frequency distributions, the modal class is identified by locating the class interval with the maximum absolute frequency $f_1$. The mode is then calculated using the formula:
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
where:
- $l$ = lower limit of the modal class
- $h$ = size of the class interval (assuming equal class sizes)
- $f_1$ = 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
Median of Grouped Data
The median of grouped data is the measure of central tendency that identifies the exact middle observation, dividing the total frequency distribution into two equal halves.
To find the median:
- Construct a cumulative frequency ($cf$) column.
- Calculate $\frac{n}{2}$, where $n = \sum f_i$.
- Identify the median class, which is the class interval whose cumulative frequency is greater than and closest to $\frac{n}{2}$.
- Apply the formula:
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h$$
where:
- $l$ = lower limit of the median class
- $n$ = total number of observations ($\sum f_i$)
- $cf$ = cumulative frequency of the class preceding the median class
- $f$ = frequency of the median class
- $h$ = class size of the median class
💡 Did You Know?: The empirical formula interlinking the three measures of central tendency, $3\text{ Median} = \text{Mode} + 2\text{ Mean}$, was formulated by Karl Pearson. It applies reliably to unimodal, moderately skewed (moderately asymmetric) frequency distributions.
[👉 Also Read: Class 10 Math Chapter 12 Surface Areas and Volumes NCERT Solutions]
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 13.1
Question 1 (Page 185) [CBSE 2014, 2018, 2021]
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 $20\text{ houses}$ 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 |
Which method did you use for finding the mean, and why?
Answer:
Step 1: Construct Calculation Table (Direct Method)
We use the Direct Method because the numerical values of both the class marks $x_i$ and frequencies $f_i$ are small.
| Class Interval (Plants) | Frequency ($f_i$) | Class Mark ($x_i = \frac{\text{Upper} + \text{Lower}}{2}$) | $f_i x_i$ |
|---|---|---|---|
| 0 – 2 | 1 | 1 | 1 |
| 2 – 4 | 2 | 3 | 6 |
| 4 – 6 | 1 | 5 | 5 |
| 6 – 8 | 5 | 7 | 35 |
| 8 – 10 | 6 | 9 | 54 |
| 10 – 12 | 2 | 11 | 22 |
| 12 – 14 | 3 | 13 | 39 |
| Total | $\sum f_i = 20$ | $\sum f_i x_i = 162$ |
Step 2: Formula and Step-by-Step Substitution
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$$
$$\bar{x} = \frac{162}{20} = 8.1\text{ plants}$$
Final Answer:
The mean number of plants per house is $8.1\text{ plants}$. We chose the Direct Method because the numerical values of $x_i$ and $f_i$ are small, making direct multiplication straightforward.
Question 2 (Page 185) [CBSE 2015, 2019, 2023]
Consider the following distribution of daily wages of $50\text{ workers}$ of a factory.
| Daily wages (in ₹) | 500 – 520 | 520 – 540 | 540 – 560 | 560 – 580 | 580 – 600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Step 1: Construct Calculation Table (Assumed Mean Method)
Class size $h = 20$. Let assumed mean $a = 550$ (central class mark).
Deviation $d_i = x_i – a = x_i – 550$.
$u_i = \frac{d_i}{h} = \frac{x_i – 550}{20}$.
| Class Interval (₹) | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i – 550$ | $u_i = \frac{d_i}{20}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 500 – 520 | 12 | 510 | -40 | -2 | -24 |
| 520 – 540 | 14 | 530 | -20 | -1 | -14 |
| 540 – 560 | 8 | 550 ($a$) | 0 | 0 | 0 |
| 560 – 580 | 6 | 570 | 20 | 1 | 6 |
| 580 – 600 | 10 | 590 | 40 | 2 | 20 |
| Total | $\sum f_i = 50$ | $\sum f_i u_i = -12$ |
Step 2: Formula and Step-by-Step Substitution (Step-Deviation Method)
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 550 + \left(\frac{-12}{50}\right) \times 20$$
$$\bar{x} = 550 – \frac{240}{50} = 550 – 4.8 = 545.2$$
Final Answer:
The mean daily wage of the workers is ₹$545.20$.
Question 3 (Page 186) [CBSE 2013, 2017, 2020 Standard, 2024]
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹$18$. 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 |
Answer:
Step 1: Construct Calculation Table
Given mean $\bar{x} = 18$. Choose assumed mean $a = 18$.
Deviation $d_i = x_i – 18$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i – 18$ | $f_i d_i$ |
|---|---|---|---|---|
| 11 – 13 | 7 | 12 | -6 | -42 |
| 13 – 15 | 6 | 14 | -4 | -24 |
| 15 – 17 | 9 | 16 | -2 | -18 |
| 17 – 19 | 13 | 18 ($a$) | 0 | 0 |
| 19 – 21 | $f$ | 20 | 2 | $2f$ |
| 21 – 23 | 5 | 22 | 4 | 20 |
| 23 – 25 | 4 | 24 | 6 | 24 |
| Total | $\sum f_i = 44 + f$ | $\sum f_i d_i = 2f – 40$ |
Sum of negative deviations: $-42 – 24 – 18 = -84$.
Sum of positive deviations: $20 + 24 = 44$.
Net sum: $44 – 84 + 2f = 2f – 40$.
Step 2: Apply Assumed Mean Formula
$$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$
$$18 = 18 + \frac{2f – 40}{44 + f}$$
Subtract $18$ from both sides:
$$0 = \frac{2f – 40}{44 + f}$$
$$2f – 40 = 0 \implies 2f = 40 \implies f = 20$$
Final Answer:
The missing frequency $f$ is $20$.
Question 4 (Page 186) [CBSE 2012, 2016, 2022 Term-2]
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
| Number of heart beats 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 |
Answer:
Step 1: Construct Calculation Table (Step-Deviation Method)
Class size $h = 3$. Let assumed mean $a = 75.5$.
$u_i = \frac{x_i – 75.5}{3}$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i – 75.5}{3}$ | $f_i u_i$ |
|---|---|---|---|---|
| 65 – 68 | 2 | 66.5 | -3 | -6 |
| 68 – 71 | 4 | 69.5 | -2 | -8 |
| 71 – 74 | 3 | 72.5 | -1 | -3 |
| 74 – 77 | 8 | 75.5 ($a$) | 0 | 0 |
| 77 – 80 | 7 | 78.5 | 1 | 7 |
| 80 – 83 | 4 | 81.5 | 2 | 8 |
| 83 – 86 | 2 | 84.5 | 3 | 6 |
| Total | $\sum f_i = 30$ | $\sum f_i u_i = 4$ |
$$\sum f_i u_i = (-6 – 8 – 3) + (7 + 8 + 6) = -17 + 21 = 4$$
Step 2: Formula and Substitution
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 75.5 + \left(\frac{4}{30}\right) \times 3 = 75.5 + \frac{12}{30} = 75.5 + 0.4 = 75.9$$
Final Answer:
The mean heart beats per minute for the women is $75.9$.
Question 5 (Page 186) [CBSE 2015, 2018, 2020 Standard]
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers 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?
Answer:
Step 1: Convert Discontinuous to Continuous Class Intervals
The given class intervals are inclusive (discontinuous): $50-52, 53-55$.
The gap between the upper limit of one class and the lower limit of the next is $53 – 52 = 1$.
Adjustment factor $= \frac{1}{2} = 0.5$.
Continuous intervals: $49.5 – 52.5, 52.5 – 55.5$, etc. Class size $h = 3$.
Step 2: Construct Calculation Table (Step-Deviation Method)
Let assumed mean $a = 57$. $u_i = \frac{x_i – 57}{3}$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i – 57}{3}$ | $f_i u_i$ |
|---|---|---|---|---|
| 49.5 – 52.5 | 15 | 51 | -2 | -30 |
| 52.5 – 55.5 | 110 | 54 | -1 | -110 |
| 55.5 – 58.5 | 135 | 57 ($a$) | 0 | 0 |
| 58.5 – 61.5 | 115 | 60 | 1 | 115 |
| 61.5 – 64.5 | 25 | 63 | 2 | 50 |
| Total | $\sum f_i = 400$ | $\sum f_i u_i = 25$ |
$$\sum f_i u_i = (-30 – 110) + (115 + 50) = -140 + 165 = 25$$
Step 3: Formula and Step-by-Step Substitution
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 57 + \left(\frac{25}{400}\right) \times 3 = 57 + \frac{75}{400} = 57 + \frac{3}{16} = 57 + 0.1875 \approx 57.19$$
Final Answer:
The mean number of mangoes kept in a packing box is $57.19$. We used the Step-Deviation Method because the frequencies were large and this method simplified calculations.
Question 6 (Page 186) [CBSE 2014, 2019 Set-2]
The table below shows the daily expenditure on food of $25\text{ households}$ 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.
Answer:
Step 1: Construct Calculation Table (Step-Deviation Method)
Class size $h = 50$. Let assumed mean $a = 225$.
$u_i = \frac{x_i – 225}{50}$.
| Class Interval (₹) | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i – 225}{50}$ | $f_i u_i$ |
|---|---|---|---|---|
| 100 – 150 | 4 | 125 | -2 | -8 |
| 150 – 200 | 5 | 175 | -1 | -5 |
| 200 – 250 | 12 | 225 ($a$) | 0 | 0 |
| 250 – 300 | 2 | 275 | 1 | 2 |
| 300 – 350 | 2 | 325 | 2 | 4 |
| Total | $\sum f_i = 25$ | $\sum f_i u_i = -7$ |
$$\sum f_i u_i = (-8 – 5) + (2 + 4) = -13 + 6 = -7$$
Step 2: Formula and Step-by-Step Substitution
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 225 + \left(\frac{-7}{25}\right) \times 50 = 225 – 7 \times 2 = 225 – 14 = 211$$
Final Answer:
The mean daily expenditure on food is ₹$211$.
Question 7 (Page 187) [CBSE 2017, 2023 Set-1]
To find out the concentration of $\text{SO}_2$ in the air (in parts per million, i.e., $\text{ppm}$), the data was collected for $30\text{ localities}$ in a certain city and is presented below:
| Concentration of $\text{SO}_2$ (in ppm) | 0.00 – 0.04 | 0.04 – 0.08 | 0.08 – 0.12 | 0.12 – 0.16 | 0.16 – 0.20 | 0.20 – 0.24 |
|---|---|---|---|---|---|---|
| Frequency | 4 | 9 | 9 | 2 | 4 | 2 |
Find the mean concentration of $\text{SO}_2$ in the air.
Answer:
Step 1: Construct Calculation Table (Direct Method)
| Class Interval (ppm) | Frequency ($f_i$) | Class Mark ($x_i$) | $f_i x_i$ |
|---|---|---|---|
| 0.00 – 0.04 | 4 | 0.02 | 0.08 |
| 0.04 – 0.08 | 9 | 0.06 | 0.54 |
| 0.08 – 0.12 | 9 | 0.10 | 0.90 |
| 0.12 – 0.16 | 2 | 0.14 | 0.28 |
| 0.16 – 0.20 | 4 | 0.18 | 0.72 |
| 0.20 – 0.24 | 2 | 0.22 | 0.44 |
| Total | $\sum f_i = 30$ | $\sum f_i x_i = 2.96$ |
Step 2: Formula and Step-by-Step Substitution
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{2.96}{30} = \frac{296}{3000} \approx 0.09867\text{ ppm}$$
Rounding to three decimal places:
$$\bar{x} \approx 0.099\text{ ppm}$$
Final Answer:
The mean concentration of $\text{SO}_2$ in the air is $0.099\text{ ppm}$.
Question 8 (Page 187) [CBSE 2016, 2020 Standard]
A class teacher has the following absentee record of $40\text{ students}$ 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 | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Answer:
Notice: The class sizes ($h$) are unequal ($6, 4, 4, 6, 8, 10, 2$). Therefore, the Step-Deviation method cannot be applied directly with a single uniform class size. We use the Direct Method or Assumed Mean Method.
Step 1: Construct Calculation Table (Assumed Mean Method)
Let assumed mean $a = 17$ (from midpoint of $14-20$).
$d_i = x_i – 17$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $d_i = x_i – 17$ | $f_i d_i$ |
|---|---|---|---|---|
| 0 – 6 | 11 | 3 | -14 | -154 |
| 6 – 10 | 10 | 8 | -9 | -90 |
| 10 – 14 | 7 | 12 | -5 | -35 |
| 14 – 20 | 4 | 17 ($a$) | 0 | 0 |
| 20 – 28 | 4 | 24 | 7 | 28 |
| 28 – 38 | 3 | 33 | 16 | 48 |
| 38 – 40 | 1 | 39 | 22 | 22 |
| Total | $\sum f_i = 40$ | $\sum f_i d_i = -181$ |
Sum of negative products: $-154 – 90 – 35 = -279$.
Sum of positive products: $28 + 48 + 22 = 98$.
Net sum: $98 – 279 = -181$.
Step 2: Formula and Step-by-Step Substitution
$$\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$$
$$\bar{x} = 17 + \left(\frac{-181}{40}\right) = 17 – 4.525 = 12.475 \approx 12.48\text{ days}$$
Final Answer:
The mean number of days a student was absent is $12.48\text{ days}$ (or $12.475\text{ days}$).
Question 9 (Page 187) [CBSE 2013, 2018, 2024]
The following table gives the literacy rate (in percentage) of $35\text{ cities}$. 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 |
Answer:
Step 1: Construct Calculation Table (Step-Deviation Method)
Class size $h = 10$. Let assumed mean $a = 70$.
$u_i = \frac{x_i – 70}{10}$.
| Class Interval (%) | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i – 70}{10}$ | $f_i u_i$ |
|---|---|---|---|---|
| 45 – 55 | 3 | 50 | -2 | -6 |
| 55 – 65 | 10 | 60 | -1 | -10 |
| 65 – 75 | 11 | 70 ($a$) | 0 | 0 |
| 75 – 85 | 8 | 80 | 1 | 8 |
| 85 – 95 | 3 | 90 | 2 | 6 |
| Total | $\sum f_i = 35$ | $\sum f_i u_i = -2$ |
$$\sum f_i u_i = (-6 – 10) + (8 + 6) = -16 + 14 = -2$$
Step 2: Formula and Step-by-Step Substitution
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$$
$$\bar{x} = 70 + \left(\frac{-2}{35}\right) \times 10 = 70 – \frac{20}{35} = 70 – \frac{4}{7} = 70 – 0.57 = 69.43%$$
Final Answer:
The mean literacy rate is $69.43%$.
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 13.2
Question 1 (Page 191) [CBSE 2012, 2017, 2020 Standard]
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 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
Part I: Finding the Mode
- Identify the maximum frequency:
The maximum frequency is $23$, corresponding to the class interval $35 – 45$.
Therefore, the Modal Class is $35 – 45$. - Identify modal parameters:
- Lower limit of modal class, $l = 35$
- Class size, $h = 10$
- Frequency of modal class, $f_1 = 23$
- Frequency of preceding class, $f_0 = 21$
- Frequency of succeeding class, $f_2 = 14$
- Apply Mode Formula:
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
$$\text{Mode} = 35 + \left(\frac{23 – 21}{2(23) – 21 – 14}\right) \times 10$$
$$\text{Mode} = 35 + \left(\frac{2}{46 – 35}\right) \times 10 = 35 + \frac{20}{11} = 35 + 1.82 = 36.82\text{ years}$$
Part II: Finding the Mean (Step-Deviation Method)
Let assumed mean $a = 30$, $h = 10$.
| Class Interval | Frequency ($f_i$) | Class Mark ($x_i$) | $u_i = \frac{x_i – 30}{10}$ | $f_i u_i$ |
|---|---|---|---|---|
| 5 – 15 | 6 | 10 | -2 | -12 |
| 15 – 25 | 11 | 20 | -1 | -11 |
| 25 – 35 | 21 | 30 ($a$) | 0 | 0 |
| 35 – 45 | 23 | 40 | 1 | 23 |
| 45 – 55 | 14 | 50 | 2 | 28 |
| 55 – 65 | 5 | 60 | 3 | 15 |
| Total | $\sum f_i = 80$ | $\sum f_i u_i = 43$ |
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 30 + \left(\frac{43}{80}\right) \times 10 = 30 + \frac{43}{8} = 30 + 5.375 = 35.38\text{ years}$$
Interpretation:
- Mode = $36.82\text{ years}$: The maximum number of patients admitted were of age approximately $36.82\text{ years}$.
- Mean = $35.38\text{ years}$: On average, the age of a patient admitted to the hospital was $35.38\text{ years}$.
Question 2 (Page 191) [CBSE 2015, 2019 Set-1]
The following data gives the information on the observed lifetimes (in hours) of $225\text{ electrical components}$:
| Lifetimes (in hours) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
|---|---|---|---|---|---|---|
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Answer:
Step 1: Identify Modal Class and Parameters
- The maximum class frequency is $61$, occurring in the interval $60 – 80$.
- Modal Class = $60 – 80$
- Lower limit of modal class, $l = 60$
- Class size, $h = 20$
- $f_1 = 61$ (modal frequency)
- $f_0 = 52$ (preceding frequency)
- $f_2 = 38$ (succeeding frequency)
Step 2: Formula and Step-by-Step Substitution
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
$$\text{Mode} = 60 + \left(\frac{61 – 52}{2(61) – 52 – 38}\right) \times 20$$
$$\text{Mode} = 60 + \left(\frac{9}{122 – 90}\right) \times 20 = 60 + \left(\frac{9}{32}\right) \times 20$$
Simplify $\frac{9 \times 20}{32} = \frac{9 \times 5}{8} = \frac{45}{8} = 5.625$:
$$\text{Mode} = 60 + 5.625 = 65.625\text{ hours}$$
Final Answer:
The modal lifetime of the components is $65.625\text{ hours}$.
Question 3 (Page 191) [CBSE 2014, 2018, 2023 Set-2]
The following data gives the distribution of total monthly household expenditure of $200\text{ families}$ of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:
| Expenditure (in ₹) | Number of families |
|---|---|
| 1000 – 1500 | 24 |
| 1500 – 2000 | 40 |
| 2000 – 2500 | 33 |
| 2500 – 3000 | 28 |
| 3000 – 3500 | 30 |
| 3500 – 4000 | 22 |
| 4000 – 4500 | 16 |
| 4500 – 5000 | 7 |
Answer:
Part I: Finding the Mode
- Maximum frequency = $40$, belonging to class $1500 – 2000$.
- Modal Class = $1500 – 2000$
- $l = 1500$, $h = 500$, $f_1 = 40$, $f_0 = 24$, $f_2 = 33$
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
$$\text{Mode} = 1500 + \left(\frac{40 – 24}{2(40) – 24 – 33}\right) \times 500$$
$$\text{Mode} = 1500 + \left(\frac{16}{80 – 57}\right) \times 500 = 1500 + \frac{8000}{23} = 1500 + 347.83 = 1847.83$$
Part II: Finding the Mean (Step-Deviation Method)
Let assumed mean $a = 2750$, $h = 500$.
| Class Interval | $f_i$ | $x_i$ | $u_i = \frac{x_i – 2750}{500}$ | $f_i u_i$ |
|---|---|---|---|---|
| 1000 – 1500 | 24 | 1250 | -3 | -72 |
| 1500 – 2000 | 40 | 1750 | -2 | -80 |
| 2000 – 2500 | 33 | 2250 | -1 | -33 |
| 2500 – 3000 | 28 | 2750 ($a$) | 0 | 0 |
| 3000 – 3500 | 30 | 3250 | 1 | 30 |
| 3500 – 4000 | 22 | 3750 | 2 | 44 |
| 4000 – 4500 | 16 | 4250 | 3 | 48 |
| 4500 – 5000 | 7 | 4750 | 4 | 28 |
| Total | $\sum f_i = 200$ | $\sum f_i u_i = -35$ |
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h = 2750 + \left(\frac{-35}{200}\right) \times 500 = 2750 – \frac{175}{2} = 2750 – 87.5 = 2662.5$$
Final Answer:
The modal monthly expenditure is ₹$1847.83$, and the mean monthly expenditure is ₹$2662.50$.
Question 4 (Page 192) [CBSE 2016, 2020 Standard]
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.
| Number of students per teacher | 15 – 20 | 20 – 25 | 25 – 30 | 30 – 35 | 35 – 40 | 40 – 45 | 45 – 50 | 50 – 55 |
|---|---|---|---|---|---|---|---|---|
| Number of states/U.T. | 3 | 8 | 9 | 10 | 3 | 0 | 0 | 2 |
Answer:
Part I: Finding the Mode
- Maximum frequency = $10$ in class $30 – 35$.
- Modal Class = $30 – 35$
- $l = 30$, $h = 5$, $f_1 = 10$, $f_0 = 9$, $f_2 = 3$
$$\text{Mode} = 30 + \left(\frac{10 – 9}{2(10) – 9 – 3}\right) \times 5 = 30 + \left(\frac{1}{20 – 12}\right) \times 5 = 30 + \frac{5}{8} = 30.625 \approx 30.6$$
Part II: Finding the Mean
Let assumed mean $a = 32.5$, $h = 5$.
| Class Interval | $f_i$ | $x_i$ | $u_i = \frac{x_i – 32.5}{5}$ | $f_i u_i$ |
|---|---|---|---|---|
| 15 – 20 | 3 | 17.5 | -3 | -9 |
| 20 – 25 | 8 | 22.5 | -2 | -16 |
| 25 – 30 | 9 | 27.5 | -1 | -9 |
| 30 – 35 | 10 | 32.5 ($a$) | 0 | 0 |
| 35 – 40 | 3 | 37.5 | 1 | 3 |
| 40 – 45 | 0 | 42.5 | 2 | 0 |
| 45 – 50 | 0 | 47.5 | 3 | 0 |
| 50 – 55 | 2 | 52.5 | 4 | 8 |
| Total | $\sum f_i = 35$ | $\sum f_i u_i = -23$ |
$$\bar{x} = 32.5 + \left(\frac{-23}{35}\right) \times 5 = 32.5 – \frac{23}{7} = 32.5 – 3.286 \approx 29.2$$
Interpretation:
Most states have a student-teacher ratio of $30.6$, while on average, the ratio across all states is $29.2$.
Question 5 (Page 192) [CBSE 2013, 2017]
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
| Runs scored | 3000 – 4000 | 4000 – 5000 | 5000 – 6000 | 6000 – 7000 | 7000 – 8000 | 8000 – 9000 | 9000 – 10000 | 10000 – 11000 |
|---|---|---|---|---|---|---|---|---|
| Number of batsmen | 4 | 18 | 9 | 7 | 6 | 3 | 1 | 1 |
Find the mode of the data.
Answer:
Step 1: Identify Modal Parameters
- Maximum frequency = $18$, corresponding to class $4000 – 5000$.
- Modal Class = $4000 – 5000$
- $l = 4000$, $h = 1000$, $f_1 = 18$, $f_0 = 4$, $f_2 = 9$
Step 2: Formula and Step-by-Step Substitution
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
$$\text{Mode} = 4000 + \left(\frac{18 – 4}{2(18) – 4 – 9}\right) \times 1000$$
$$\text{Mode} = 4000 + \left(\frac{14}{36 – 13}\right) \times 1000 = 4000 + \frac{14000}{23} = 4000 + 608.695 \approx 4608.7\text{ runs}$$
Final Answer:
The mode of the data is $4608.7\text{ runs}$.
Question 6 (Page 192) [CBSE 2015, 2019 Set-3, 2024]
A student noted the number of cars passing through a spot on a road for $100\text{ periods}$ each of $3\text{ minutes}$ 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 | 20 | 11 | 15 | 8 |
Answer:
Step 1: Identify Modal Parameters
- Maximum frequency = $20$, belonging to class $40 – 50$.
- Modal Class = $40 – 50$
- $l = 40$, $h = 10$, $f_1 = 20$, $f_0 = 12$, $f_2 = 11$
Step 2: Formula and Step-by-Step Substitution
$$\text{Mode} = l + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times h$$
$$\text{Mode} = 40 + \left(\frac{20 – 12}{2(20) – 12 – 11}\right) \times 10$$
$$\text{Mode} = 40 + \left(\frac{8}{40 – 23}\right) \times 10 = 40 + \frac{80}{17} = 40 + 4.706 \approx 44.7\text{ cars}$$
Final Answer:
The mode of the data is $44.7\text{ cars}$.
Step-by-Step Solutions: NCERT Class 10 Mathematics Exercise 13.3
Question 1 (Page 197) [CBSE 2012, 2016, 2020 Standard]
The following frequency distribution gives the monthly consumption of electricity of $68\text{ consumers}$ of a locality. Find the median, mean and mode of the data and compare them.
| Monthly consumption (in units) | 65 – 85 | 85 – 105 | 105 – 125 | 125 – 145 | 145 – 165 | 165 – 185 | 185 – 205 |
|---|---|---|---|---|---|---|---|
| Number of consumers | 4 | 5 | 13 | 20 | 14 | 8 | 4 |
Answer:
Step 1: Construct Calculation Table with Cumulative Frequency
| Class Interval | Frequency ($f_i$) | Cumulative Freq ($cf$) | Midpoint ($x_i$) | $u_i = \frac{x_i – 135}{20}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 65 – 85 | 4 | 4 | 75 | -3 | -12 |
| 85 – 105 | 5 | 9 | 95 | -2 | -10 |
| 105 – 125 | 13 | 22 | 115 | -1 | -13 |
| 125 – 145 | 20 | 42 | 135 ($a$) | 0 | 0 |
| 145 – 165 | 14 | 56 | 155 | 1 | 14 |
| 165 – 185 | 8 | 64 | 175 | 2 | 16 |
| 185 – 205 | 4 | 68 | 195 | 3 | 12 |
| Total | $n = 68$ | $\sum f_i u_i = 7$ |
Part I: Finding the Median
- $n = 68 \implies \frac{n}{2} = \frac{68}{2} = 34$.
- Cumulative frequency just greater than $34$ is $42$, belonging to class $125 – 145$.
- Median Class = $125 – 145$
- $l = 125$, $n/2 = 34$, $cf = 22$ (preceding cumulative frequency), $f = 20$, $h = 20$.
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h = 125 + \left(\frac{34 – 22}{20}\right) \times 20 = 125 + 12 = 137\text{ units}$$
Part II: Finding the Mean
$$\bar{x} = a + \left(\frac{\sum f_i u_i}{n}\right) \times h = 135 + \left(\frac{7}{68}\right) \times 20 = 135 + \frac{140}{68} = 135 + 2.06 = 137.06\text{ units}$$
Part III: Finding the Mode
Modal class = $125 – 145$ ($f_1 = 20$, $f_0 = 13$, $f_2 = 14$, $l = 125$, $h = 20$).
$$\text{Mode} = 125 + \left(\frac{20 – 13}{2(20) – 13 – 14}\right) \times 20 = 125 + \left(\frac{7}{40 – 27}\right) \times 20 = 125 + \frac{140}{13} = 125 + 10.77 = 135.77\text{ units}$$
Comparison:
Median = $137\text{ units}$, Mean = $137.06\text{ units}$, Mode = $135.77\text{ units}$. The three measures are approximately equal, indicating an almost symmetric data distribution.
Question 2 (Page 197) [CBSE 2013, 2017, 2020 Standard, 2023 Set-1, 2024]
If the median of the distribution given below is $28.5$, find the values of $x$ and $y$.
| Class interval | Frequency |
|---|---|
| 0 – 10 | 5 |
| 10 – 20 | $x$ |
| 20 – 30 | 20 |
| 30 – 40 | 15 |
| 40 – 50 | $y$ |
| 50 – 60 | 5 |
| Total | 60 |
Answer:
Step 1: Construct Cumulative Frequency Table
| Class Interval | Frequency ($f$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 0 – 10 | 5 | 5 |
| 10 – 20 | $x$ | $5 + x$ |
| 20 – 30 | 20 | $25 + x$ |
| 30 – 40 | 15 | $40 + x$ |
| 40 – 50 | $y$ | $40 + x + y$ |
| 50 – 60 | 5 | $45 + x + y$ |
| Total | $n = 60$ |
Step 2: Form Equation from Total Frequency
$$45 + x + y = 60$$
$$x + y = 60 – 45 \implies x + y = 15 \quad \text{— (Equation 1)}$$
Step 3: Identify Median Class
The given median is $28.5$, which lies strictly within the class interval $20 – 30$.
- Median Class = $20 – 30$
- Lower limit, $l = 20$
- Class size, $h = 10$
- Frequency of median class, $f = 20$
- Cumulative frequency of preceding class, $cf = 5 + x$
- $\frac{n}{2} = \frac{60}{2} = 30$
Step 4: Apply Median Formula to Solve for $x$
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h$$
$$28.5 = 20 + \left(\frac{30 – (5 + x)}{20}\right) \times 10$$
$$28.5 – 20 = \frac{25 – x}{2}$$
$$8.5 = \frac{25 – x}{2}$$
Multiply by $2$:
$$17 = 25 – x \implies x = 25 – 17 = 8$$
Step 5: Solve for $y$
Substitute $x = 8$ into Equation 1:
$$8 + y = 15 \implies y = 15 – 8 = 7$$
Final Answer:
The missing frequencies are $x = 8$ and $y = 7$.
Question 3 (Page 197) [CBSE 2015, 2019 Set-2]
A life insurance agent found the following data for distribution of ages of $100\text{ policy holders}$. Calculate the median age, if policies are given only to persons having age $18\text{ years}$ onwards but less than $60\text{ year}$.
| Age (in years) | Number of policy holders |
|---|---|
| Below 20 | 2 |
| Below 25 | 6 |
| Below 30 | 24 |
| Below 35 | 45 |
| Below 40 | 78 |
| Below 45 | 89 |
| Below 50 | 92 |
| Below 55 | 98 |
| Below 60 | 100 |
Answer:
Step 1: Convert Cumulative Frequencies into Class Intervals
The given table provides “less than” cumulative frequencies. Subtract consecutive values to obtain individual frequencies $f_i$:
| Class Interval (Age) | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 15 – 20 | 2 | 2 |
| 20 – 25 | $6 – 2 = 4$ | 6 |
| 25 – 30 | $24 – 6 = 18$ | 24 |
| 30 – 35 | $45 – 24 = 21$ | 45 |
| 35 – 40 | $78 – 45 = 33$ | 78 |
| 40 – 45 | $89 – 78 = 11$ | 89 |
| 45 – 50 | $92 – 89 = 3$ | 92 |
| 50 – 55 | $98 – 92 = 6$ | 98 |
| 55 – 60 | $100 – 98 = 2$ | 100 |
| Total | $n = 100$ |
Step 2: Identify Median Class
- $\frac{n}{2} = \frac{100}{2} = 50$.
- Cumulative frequency just greater than $50$ is $78$, which corresponds to the interval $35 – 40$.
- Median Class = $35 – 40$
- $l = 35$, $cf = 45$, $f = 33$, $h = 5$.
Step 3: Calculate Median
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h = 35 + \left(\frac{50 – 45}{33}\right) \times 5 = 35 + \frac{25}{33} = 35 + 0.758 \approx 35.76\text{ years}$$
Final Answer:
The median age of policy holders is $35.76\text{ years}$.
Question 4 (Page 198) [CBSE 2014, 2018, 2022 Term-2]
The lengths of $40\text{ leaves}$ of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:
| Length (in mm) | Number of leaves |
|---|---|
| 118 – 126 | 3 |
| 127 – 135 | 5 |
| 136 – 144 | 9 |
| 145 – 153 | 12 |
| 154 – 162 | 5 |
| 163 – 171 | 4 |
| 172 – 180 | 2 |
Find the median length of the leaves.
Answer:
Step 1: Convert Discontinuous to Continuous Class Boundaries
Difference between $127$ and $126 = 1$. Half-difference $= 0.5$.
Subtract $0.5$ from lower limits and add $0.5$ to upper limits.
| Continuous Class Interval (mm) | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 117.5 – 126.5 | 3 | 3 |
| 126.5 – 135.5 | 5 | 8 |
| 135.5 – 144.5 | 9 | 17 |
| 144.5 – 153.5 | 12 | 29 |
| 153.5 – 162.5 | 5 | 34 |
| 162.5 – 171.5 | 4 | 38 |
| 171.5 – 180.5 | 2 | 40 |
| Total | $n = 40$ |
Step 2: Identify Median Class
- $\frac{n}{2} = \frac{40}{2} = 20$.
- Cumulative frequency just greater than $20$ is $29$, corresponding to $144.5 – 153.5$.
- Median Class = $144.5 – 153.5$
- $l = 144.5$, $cf = 17$, $f = 12$, $h = 9$.
Step 3: Calculate Median
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h = 144.5 + \left(\frac{20 – 17}{12}\right) \times 9$$
$$\text{Median} = 144.5 + \left(\frac{3}{12}\right) \times 9 = 144.5 + \frac{9}{4} = 144.5 + 2.25 = 146.75\text{ mm}$$
Final Answer:
The median length of the leaves is $146.75\text{ mm}$.
Question 5 (Page 198) [CBSE 2016, 2020 Standard]
The following table gives the distribution of the life time of $400\text{ neon lamps}$:
| Life time (in hours) | Number of lamps |
|---|---|
| 1500 – 2000 | 14 |
| 2000 – 2500 | 56 |
| 2500 – 3000 | 60 |
| 3000 – 3500 | 86 |
| 3500 – 4000 | 74 |
| 4000 – 4500 | 62 |
| 4500 – 5000 | 48 |
Find the median life time of a lamp.
Answer:
Step 1: Construct Cumulative Frequency Table
| Life Time (hours) | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 1500 – 2000 | 14 | 14 |
| 2000 – 2500 | 56 | 70 |
| 2500 – 3000 | 60 | 130 |
| 3000 – 3500 | 86 | 216 |
| 3500 – 4000 | 74 | 290 |
| 4000 – 4500 | 62 | 352 |
| 4500 – 5000 | 48 | 400 |
| Total | $n = 400$ |
Step 2: Identify Median Class
- $\frac{n}{2} = \frac{400}{2} = 200$.
- Cumulative frequency just greater than $200$ is $216$, corresponding to $3000 – 3500$.
- Median Class = $3000 – 3500$
- $l = 3000$, $cf = 130$, $f = 86$, $h = 500$.
Step 3: Calculate Median
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h = 3000 + \left(\frac{200 – 130}{86}\right) \times 500$$
$$\text{Median} = 3000 + \frac{70 \times 500}{86} = 3000 + \frac{35000}{86} = 3000 + \frac{17500}{43} \approx 3000 + 406.98 = 3406.98\text{ hours}$$
Final Answer:
The median life time of a lamp is $3406.98\text{ hours}$.
Question 6 (Page 198) [CBSE 2013, 2017, 2023 Set-3]
$100\text{ surnames}$ were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
| Number of letters | 1 – 4 | 4 – 7 | 7 – 10 | 10 – 13 | 13 – 16 | 16 – 19 |
|---|---|---|---|---|---|---|
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Answer:
Step 1: Construct Complete Evaluation Table
| Class Interval | $f_i$ | $cf$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|---|
| 1 – 4 | 6 | 6 | 2.5 | 15.0 |
| 4 – 7 | 30 | 36 | 5.5 | 165.0 |
| 7 – 10 | 40 | 76 | 8.5 | 340.0 |
| 10 – 13 | 16 | 92 | 11.5 | 184.0 |
| 13 – 16 | 4 | 96 | 14.5 | 58.0 |
| 16 – 19 | 4 | 100 | 17.5 | 70.0 |
| Total | $n = 100$ | $\sum f_i x_i = 832.0$ |
Part I: Finding the Median
- $\frac{n}{2} = 50$. Cumulative frequency greater than $50$ is $76$ (Class $7 – 10$).
- $l = 7$, $cf = 36$, $f = 40$, $h = 3$.
$$\text{Median} = 7 + \left(\frac{50 – 36}{40}\right) \times 3 = 7 + \frac{14 \times 3}{40} = 7 + \frac{42}{40} = 7 + 1.05 = 8.05\text{ letters}$$
Part II: Finding the Mean
$$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{832}{100} = 8.32\text{ letters}$$
Part III: Finding the Mode
- Modal class = $7 – 10$ ($f_1 = 40$, $f_0 = 30$, $f_2 = 16$, $l = 7$, $h = 3$).
$$\text{Mode} = 7 + \left(\frac{40 – 30}{2(40) – 30 – 16}\right) \times 3 = 7 + \left(\frac{10}{80 – 46}\right) \times 3 = 7 + \frac{30}{34} = 7 + 0.88 = 7.88\text{ letters}$$
Final Answer:
- Median number of letters = $8.05$
- Mean number of letters = $8.32$
- Modal size of surnames = $7.88$
Question 7 (Page 199) [CBSE 2015, 2018, 2021]
The distribution below gives the weights of $30\text{ students}$ of a class. Find the median weight of the students.
| Weight (in kg) | 40 – 45 | 45 – 50 | 50 – 55 | 55 – 60 | 60 – 65 | 65 – 70 | 70 – 75 |
|---|---|---|---|---|---|---|---|
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
Answer:
Step 1: Construct Cumulative Frequency Table
| Weight (kg) | Frequency ($f_i$) | Cumulative Frequency ($cf$) |
|---|---|---|
| 40 – 45 | 2 | 2 |
| 45 – 50 | 3 | 5 |
| 50 – 55 | 8 | 13 |
| 55 – 60 | 6 | 19 |
| 60 – 65 | 6 | 25 |
| 65 – 70 | 3 | 28 |
| 70 – 75 | 2 | 30 |
| Total | $n = 30$ |
Step 2: Identify Median Class
- $\frac{n}{2} = \frac{30}{2} = 15$.
- Cumulative frequency just greater than $15$ is $19$, corresponding to interval $55 – 60$.
- Median Class = $55 – 60$
- $l = 55$, $cf = 13$, $f = 6$, $h = 5$.
Step 3: Calculate Median
$$\text{Median} = l + \left(\frac{\frac{n}{2} – cf}{f}\right) \times h$$
$$\text{Median} = 55 + \left(\frac{15 – 13}{6}\right) \times 5 = 55 + \left(\frac{2}{6}\right) \times 5 = 55 + \frac{5}{3} = 55 + 1.67 = 56.67\text{ kg}$$
Final Answer:
The median weight of the students is $56.67\text{ kg}$.
[👉 Also Read: Class 10 Math Chapter 14 Probability NCERT Solutions]
Master High-Yield Board FAQs (Rank Math Schema Ready)
What is the empirical relationship between mean, median, and mode?
The empirical relationship interlinking the three central tendencies for moderately skewed distributions is $3\text{ Median} = \text{Mode} + 2\text{ Mean}$. If any two values are known, the third can be determined algebraically: $\text{Mode} = 3\text{ Median} – 2\text{ Mean}$, or $\text{Mean} = \frac{3\text{ Median} – \text{Mode}}{2}$.
When is the step-deviation method not applicable for finding the mean?
The step-deviation method cannot be applied directly when class intervals have unequal widths ($h$). If interval widths vary (e.g., $0-6, 6-10, 10-14, 14-20$), there is no single common factor $h$ to simplify the deviations, requiring students to use either the Direct Method or Assumed Mean Method instead.
What is the difference between less than and more than cumulative frequency distributions?
In a “less than” cumulative frequency distribution, frequencies are accumulated from the lowest class to the highest, with values paired with each interval’s upper class limit. In a “more than” distribution, frequencies are accumulated from the highest class down to the lowest, with values paired with each interval’s lower class limit.
How do you convert discontinuous class intervals into continuous ones?
To convert discontinuous (inclusive) intervals like $118-126, 127-135$ into continuous boundaries, determine the gap between consecutive classes ($127 – 126 = 1$). Divide this gap by 2 ($0.5$). Subtract $0.5$ from every lower limit and add $0.5$ to every upper limit, producing continuous boundaries: $117.5-126.5, 126.5-135.5$.
Can the mode of a grouped frequency distribution be calculated if the first class has the highest frequency?
Yes. If the first class has the highest frequency, it serves as the modal class. In this situation, the frequency of the preceding class $f_0$ is taken as zero ($f_0 = 0$). Similarly, if the final class has the highest frequency, the succeeding frequency $f_2$ is set to zero ($f_2 = 0$).
Which measure of central tendency is most affected by extreme values or outliers?
The Mean is the most sensitive to extreme values because its calculation incorporates the exact magnitude of every observation in the dataset ($\sum f_i x_i$). In contrast, the Median is the most robust measure against extreme outliers because it depends solely on the positional order of observations.
How is the median class identified in grouped frequency data?
To identify the median class, calculate the total frequency $n = \sum f_i$ and compute $\frac{n}{2}$. In the cumulative frequency ($cf$) column, find the first class whose cumulative frequency is greater than or equal to $\frac{n}{2}$. That corresponding class interval is the median class.
Why is cf taken from the preceding class rather than the median class itself in the median formula?
The cumulative frequency of the preceding class ($cf$) accounts for all observations lying entirely below the lower boundary $l$ of the median class. Subtracting $cf$ from $\frac{n}{2}$ determines the remaining number of observations needed within the median class to reach the exact midpoint.
What should you do if the assumed mean value is chosen incorrectly?
Choosing a different class mark as the assumed mean $a$ does not alter the final answer. The deviations $d_i = x_i – a$ adjust automatically, and the correction factor $\frac{\sum f_i d_i}{\sum f_i}$ balances the offset. Picking the central class mark is simply a convention to keep deviations small and balanced between positive and negative values.
Can mode and median lie outside the modal and median classes?
No. By definition and algebraic construction of their respective formulas, the computed value of the mode must fall within the modal class, and the computed median must fall within the median class. If a computed result lies outside the identified interval, an arithmetic error has occurred.
What units should be written with statistical measures?
The mean, median, and mode carry the same physical units as the original observations (e.g., years, ₹, hours, $\text{mm}$, $\text{kg}$, or $\text{ppm}$). Always append the appropriate unit to the final numerical result to avoid minor deductions under CBSE marking schemes.
How do you solve for two missing frequencies when the median is provided?
To find two missing frequencies $x$ and $y$: (1) use the total frequency to set up an initial linear equation, $x + y = n – \sum f_{\text{known}}$, and (2) use the given median value to identify the median class, then substitute values into the median formula to solve for $x$. Finally, substitute $x$ into the first equation to find $y$.
Are ogives and cumulative frequency graphs included in the 2026-2027 CBSE syllabus?
No. Cumulative frequency graphs (“less than” and “more than” ogives) have been removed from the rationalised NCERT textbook. Students are evaluated solely on the numerical computation and algebraic properties of grouped Mean, Mode, and Median.
Which central tendency is represented by the point of intersection of two ogives?
Historically, the abscissa ($x$-coordinate) of the intersection point of a “less than” ogive and a “more than” ogive on the same graph represents the Median of the dataset.
What common mistake do students make when applying the mode formula?
The most common mistake is confusing the order of terms in the denominator ($2f_1 – f_0 – f_2$) or subtracting incorrectly. Another frequent error is using an incorrect class width $h$ when intervals have not been converted to continuous form. Always verify that $2f_1 > f_0 + f_2$ so that the denominator remains positive.