Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).
Comprehensive Concepts & Principles
Percentiles and Quartiles forms an essential building block in CBSE Class 11 Applied Mathematics (Measures of Dispersion and Percentiles). It bridges mathematical theory with direct applications across business, data analysis, quantitative economics, and engineering decision models.
Core Focus Areas:
Calculation of Quartiles (Q1, Q2, Q3), Deciles, and Percentiles
Percentile Rank formula and interpretation in comparative scoring and standardized testing
Mathematical Framework & Key Formulas
Master the governing relationships and definitions established in this exercise:
$$Q_k = Value of (k(N+1)4)th term$$
$$P_k = Value of (k(N+1)100)th term$$
$$Percentile Rank = B + 0.5EN × 100$$
Worked Benchmark Examples & Classroom Lesson Plan
Example 1 (Quartiles, Interquartile Range & Coefficient of QD): The monthly sales incentives (in thousands ₹) earned by $15$ marketing executives in a firm are recorded as:
$$12, 15, 18, 20, 22, 25, 28, 30, 32, 35, 38, 40, 42, 45, 50$$
Calculate: (i) First Quartile ($Q_1$), (ii) Median ($Q_2$), (iii) Third Quartile ($Q_3$), (iv) Interquartile Range ($IQR$), and (v) Coefficient of Quartile Deviation.
Show Step-by-Step Solution
Example 2 (Classroom Case: Percentile Rank from Ungrouped Test Marks): A class of $25$ students took a standardized business statistics assessment. Their scores arranged in ascending order are:
$$42, 45, 48, 52, 55, 56, 58, 60, 62, 65, 68, 70, 72, 72, 74, 76, 78, 80, 82, 85, 88, 90, 92, 95, 98$$
Calculate the Percentile Rank ($PR$) of a student who scored $72$. Interpret what this percentile rank means.
Show Step-by-Step Solution
CBSE Standard Lesson Formula for Percentile Rank:
$$PR = \frac{B + 0.5E}{N} \times 100$$
where:
$B$ = Number of scores strictly below the given score ($X = 72$)
$E$ = Number of scores exactly equal to the given score ($X = 72$)
$N$ = Total number of observations in the dataset ($25$)
Step 1: Count $B$ and $E$
• Scores strictly below $72$: $\{42, 45, 48, 52, 55, 56, 58, 60, 62, 65, 68, 70\} \implies B = 12$
• Scores equal to $72$: $\{72, 72\} \implies E = 2$
• Total scores $N = 25$
Step 2: Substitute into formula
$$PR = \frac{12 + (0.5 \times 2)}{25} \times 100$$
$$PR = \frac{12 + 1}{25} \times 100 = \frac{13}{25} \times 100 = 52\%$$ Educational Interpretation:
The student's Percentile Rank is $52^{\text{nd}}$ percentile ($PR = 52$). This indicates that the student scored as high as or higher than $52\%$ of all test-takers in the class.
Example 3 (Decile and Percentile Interpolation): Using the $15$-executive sales incentive dataset from Example 1:
$$12, 15, 18, 20, 22, 25, 28, 30, 32, 35, 38, 40, 42, 45, 50$$
Calculate: (i) The $6^{\text{th}}$ Decile ($D_6$) and (ii) The $65^{\text{th}}$ Percentile ($P_{65}$).
Show Step-by-Step Solution
(ii) Calculating $65^{\text{th}}$ Percentile ($P_{65}$):
$$\text{Position of } P_{65} = \frac{65(N + 1)}{100} = \frac{65 \times 16}{100} = \frac{1040}{100} = 10.4^{\text{th}} \text{ term}$$
Interpolate between $10^{\text{th}}$ term ($35$) and $11^{\text{th}}$ term ($38$):
$$P_{65} = \text{Term}_{10} + 0.4 \times (\text{Term}_{11} - \text{Term}_{10})$$
$$P_{65} = 35 + 0.4 \times (38 - 35) = 35 + 0.4 \times 3 = 35 + 1.2 = 36.2$$
So, $P_{65} = ₹36,200$.
CHECK YOUR PROGRESS 10.4
Graded textbook exercises based directly on the CBSE Applied Mathematics support material for Percentiles and Quartiles. Solve each question independently before toggling the step-by-step solutions.
Problem 1Discrete Frequency Distribution • Quartiles & QD
The daily wages earned by $60$ casual workers in an industrial park are distributed as follows:
Daily Wages ($X$ in ₹)
$200$
$250$
$300$
$350$
$400$
$450$
$500$
No. of Workers ($f$)
$3$
$8$
$12$
$15$
$11$
$7$
$4$
Calculate:
(i) First Quartile ($Q_1$)
(ii) Median ($Q_2$)
(iii) Third Quartile ($Q_3$)
(iv) Quartile Deviation ($QD$) and its relative coefficient.
Show Step-by-Step Solution
Step 1: Construct the Cumulative Frequency Table ($N = 60$):
• $X = 200, f = 3 \implies cf = 3$
• $X = 250, f = 8 \implies cf = 11$
• $X = 300, f = 12 \implies cf = 23$
• $X = 350, f = 15 \implies cf = 38$
• $X = 400, f = 11 \implies cf = 49$
• $X = 450, f = 7 \implies cf = 56$
• $X = 500, f = 4 \implies cf = 60$
(i) First Quartile ($Q_1$):
$$\text{Position of } Q_1 = \frac{N + 1}{4} = \frac{61}{4} = 15.25^{\text{th}} \text{ term}$$
The first cumulative frequency greater than or equal to $15.25$ is $23$, which corresponds to $X = ₹300$.
Thus, $Q_1 = ₹300$.
(ii) Median ($Q_2$):
$$\text{Position of } Q_2 = \frac{2(N + 1)}{4} = \frac{122}{4} = 30.5^{\text{th}} \text{ term}$$
Cumulative frequency $\ge 30.5$ is $38$, corresponding to $X = ₹350$.
Thus, $Q_2 = ₹350$.
(iii) Third Quartile ($Q_3$):
$$\text{Position of } Q_3 = \frac{3(N + 1)}{4} = \frac{183}{4} = 45.75^{\text{th}} \text{ term}$$
Cumulative frequency $\ge 45.75$ is $49$, corresponding to $X = ₹400$.
Thus, $Q_3 = ₹400$.
(i) Lower Quartile ($Q_1$):
$$\frac{N}{4} = \frac{50}{4} = 12.5$$
The $cf$ just exceeding $12.5$ is $20$, so the $Q_1$ class is $20 - 30$.
Here $L = 20, \; cf = 6, \; f = 14, \; h = 10$.
$$Q_1 = L + \left(\frac{\frac{N}{4} - cf}{f}\right) \times h = 20 + \left(\frac{12.5 - 6}{14}\right) \times 10$$
$$Q_1 = 20 + \frac{6.5 \times 10}{14} = 20 + \frac{65}{14} \approx 20 + 4.64 = ₹24,640$$ (ii) Upper Quartile ($Q_3$):
$$\frac{3N}{4} = \frac{150}{4} = 37.5$$
The $cf$ just exceeding $37.5$ is $45$, so the $Q_3$ class is $40 - 50$.
Here $L = 40, \; cf = 36, \; f = 9, \; h = 10$.
$$Q_3 = L + \left(\frac{\frac{3N}{4} - cf}{f}\right) \times h = 40 + \left(\frac{37.5 - 36}{9}\right) \times 10$$
$$Q_3 = 40 + \frac{1.5 \times 10}{9} = 40 + \frac{15}{9} \approx 40 + 1.67 = ₹41,670$$ (iii) Interquartile Range ($IQR$):
$$IQR = Q_3 - Q_1 = 41.67 - 24.64 = 17.03 \text{ (in thousands ₹) } = ₹17,030$$
Problem 3Standardized Assessment • Percentile Rank
In an all-India competitive management aptitude test, $80$ candidates appeared at an examination center. Scores were compiled in ascending order.
A student named Aryan scored $68$ marks. In the tabulated score sheet, exactly $54$ candidates scored strictly less than $68$ marks, while $4$ candidates scored exactly $68$ marks.
(i) Calculate Aryan's Percentile Rank ($PR$) using the standardized testing formula.
(ii) Interpret what this percentile rank signifies.
(iii) If candidates in the $85^{\text{th}}$ percentile and above qualify for the interview round, did Aryan qualify?
Part (ii) Educational Interpretation:
A percentile rank of $70$ indicates that Aryan performed equal to or better than $70\%$ of all examinees in the assessment group.
Part (iii) Qualification Status:
Since the cutoff requires $PR \ge 85$ and Aryan's $PR = 70$, Aryan did not qualify for the interview round.
Exercise 10.4 Revision Worksheet
Printable classroom and homework worksheet for Percentiles and Quartiles. Covers objective drills, intermediate calculations, and high-yield applications.
Section A: Conceptual & Objective Drill3 Questions • 1 Mark Each
Q1. Deciles to Percentiles Equivalence1 Mark
State the exact corresponding Percentile ($P_k$) for: (i) Seventh Decile ($D_7$), and (ii) Median ($Q_2$).
Show Model Solution
Model Solution:
(i) $D_7 = P_{70}$ ($70^{\text{th}}$ percentile), since $D_k = P_{10k}$.
(ii) $Q_2 = D_5 = P_{50}$ ($50^{\text{th}}$ percentile), which corresponds to the Median.
Q2. Dimensionality of Relative Dispersion1 Mark
Why is the Coefficient of Quartile Deviation regarded as a relative measure of dispersion, whereas Quartile Deviation is an absolute measure?
Show Model Solution
Model Solution:
Quartile Deviation $QD = \frac{Q_3 - Q_1}{2}$ is expressed in the same physical units as the variable, making it absolute. The Coefficient of $QD = \frac{Q_3 - Q_1}{Q_3 + Q_1}$ is a ratio of differences to sums of the same unit, resulting in a pure, dimensionless number suitable for comparing datasets with different units or means.
Q3. Outlier Sensitivity Comparison1 Mark
Why is Quartile Deviation preferred over Standard Deviation when analyzing economic variables such as income distribution and real estate prices?
Show Model Solution
Model Solution:
Income and real estate distributions are heavily positively skewed with extreme right-tail outliers. Standard deviation squares deviations, amplifying extreme outlier distortions. In contrast, Quartile Deviation is based solely on the interquartile range ($Q_1$ to $Q_3$), covering the middle $50\%$ of observations and completely ignoring extreme outliers.
Section B: Short Numerical Problems2 Questions • 2 Marks Each
Q4. Ungrouped Data Partition Drill2 Marks
The weights (in kg) of $9$ athletes in a training academy are recorded as:
$$45, 52, 48, 60, 55, 62, 50, 58, 65$$
Calculate the First Quartile ($Q_1$), Third Quartile ($Q_3$), and Interquartile Range ($IQR$).
For a moderately skewed distribution, the First Quartile is $Q_1 = 35$ and the Coefficient of Quartile Deviation is $0.20$. Determine: (i) The Third Quartile ($Q_3$), and (ii) The Quartile Deviation ($QD$).
Section C: High-Yield Applied Word Problems1 Question • 4 Marks
Q6. Tech Enterprise Salary Spread Modeling4 Marks
The human resources team of a fintech enterprise analyzed monthly payroll disbursements for $100$ junior software analysts:
Monthly Salary (₹ in '000)
$30 - 40$
$40 - 50$
$50 - 60$
$60 - 70$
$70 - 80$
Number of Analysts ($f$)
$15$
$35$
$25$
$15$
$10$
Determine:
(i) First Quartile salary ($Q_1$). [1 Mark]
(ii) Median salary ($Q_2$). [1 Mark]
(iii) Third Quartile salary ($Q_3$). [1 Mark]
(iv) Quartile Deviation and economic interpretation of the central $50\%$ salary band. [1 Mark]
Show Model Solution
(i) First Quartile ($Q_1$) [1 Mark]:
$$\frac{N}{4} = \frac{100}{4} = 25 \implies Q_1 \text{ class is } 40 - 50 \; (L = 40, cf = 15, f = 35, h = 10)$$
$$Q_1 = 40 + \left(\frac{25 - 15}{35}\right) \times 10 = 40 + \frac{100}{35} \approx 40 + 2.86 = ₹42,860$$ (ii) Median ($Q_2$) [1 Mark]:
$$\frac{N}{2} = 50 \implies Q_2 \text{ class is } 40 - 50 \; (L = 40, cf = 15, f = 35, h = 10)$$
$$Q_2 = 40 + \left(\frac{50 - 15}{35}\right) \times 10 = 40 + \left(\frac{35}{35}\right) \times 10 = 40 + 10 = ₹50,000$$ (iii) Third Quartile ($Q_3$) [1 Mark]:
$$\frac{3N}{4} = 75 \implies Q_3 \text{ class is } 50 - 60 \; (L = 50, cf = 50, f = 25, h = 10)$$
$$Q_3 = 50 + \left(\frac{75 - 50}{25}\right) \times 10 = 50 + \left(\frac{25}{25}\right) \times 10 = 50 + 10 = ₹60,000$$ (iv) QD & Economic Interpretation [1 Mark]:
$$QD = \frac{Q_3 - Q_1}{2} = \frac{60 - 42.86}{2} = \frac{17.14}{2} = ₹8,570$$
Interpretation: The middle $50\%$ of junior analysts earn between $₹42,860$ and $₹60,000$, with a central salary dispersion of $₹8,570$ around the median.
Exercise 10.4 Comprehensive Revision Sheet
Printable high-yield summary for last-minute revision and board examination preparation on Percentiles and Quartiles.
Concept Architecture & Key Takeaways
Partition Values: Statistical measures that divide an ordered dataset into equal proportional segments. Quartiles divide into $4$ parts, Deciles into $10$, and Percentiles into $100$.
Quartiles ($Q_1, Q_2, Q_3$): $Q_1$ cuts off the lower $25\%$, $Q_2$ represents the Median ($50\%$, dividing the dataset in half), and $Q_3$ cuts off the top $25\%$ ($75^{\text{th}}$ percentile).
Interquartile Range ($IQR$): Absolute difference between upper and lower quartiles: $IQR = Q_3 - Q_1$. Represents the dispersion of the middle $50\%$ of data.
Quartile Deviation ($QD$): Semi-Interquartile range: $QD = \frac{Q_3 - Q_1}{2}$. It is less affected by extreme values than Range or Standard Deviation.
Coefficient of Quartile Deviation: Relative dispersion formula: $\frac{Q_3 - Q_1}{Q_3 + Q_1}$. Used to compare consistency between distributions with different units.
Percentile Rank ($PR$): In standardized tests, $PR = \frac{B + 0.5E}{N} \times 100$, expressing the percentage of candidates scoring below or tied with a given raw score.
Essential Formulae & Identities
• Ungrouped Series: $Q_k = \text{Value of } \left(\frac{k(N + 1)}{4}\right)^{\text{th}} \text{ term}, \; k \in \{1, 2, 3\}$
Students often calculate the position $\frac{N+1}{4}$ and directly pick that index from raw, unsorted data. Partition values are defined strictly on ordered arrays. Always sort raw values in ascending order first! Also remember to use $cf$ of the preceding class in grouped data formulas.
Examiner Pro-Tip: Ungrouped vs Continuous Rank Rules
In discrete or individual series, use $\frac{k(N+1)}{4}$ to locate the term. In continuous grouped frequency tables, use $\frac{kN}{4}$ (without $+1$) to find the quartile class. Always verify that your interpolated quartile answer lies strictly within the identified class interval $[L, L+h]$!
Time: 45 Minutes Max Marks: 20 CBSE Board Blueprint
Which of the following identities correctly connects the partition values with the Median?
(A) $Q_2 = D_5 = P_{50} = \text{Median}$
(B) $Q_1 = D_2 = P_{25}$
(C) $Q_3 = D_7 = P_{75}$
(D) $D_4 = P_{45}$
Solution (A): Second quartile $Q_2$ ($50\%$) equals fifth decile $D_5$ ($5/10 = 50\%$) and fiftieth percentile $P_{50}$ ($50/100 = 50\%$), which is identically the Median.
Question 3MCQ • 1 Mark
In a batch of $50$ students taking an applied mathematics test, Ananya scored $82$ marks. In the batch, $38$ students scored strictly less than $82$, while $2$ students tied with $82$. What is Ananya's Percentile Rank ($PR$)?
Assertion (A): Quartile Deviation is an absolute measure of dispersion, whereas the Coefficient of Quartile Deviation is a relative measure. Reason (R): Quartile deviation is expressed in physical units of the variable, whereas the Coefficient of Quartile Deviation is a dimensionless ratio independent of scale units.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is not the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Solution (A): Both Assertion and Reason are true. $QD = (Q_3 - Q_1)/2$ carries the unit of $X$. Its coefficient $(Q_3 - Q_1)/(Q_3 + Q_1)$ cancels the units, creating a scale-free relative metric. R correctly explains A.
Section B: Short Answer Type I (Q5 – Q6)2 Questions • 2 Marks Each
Question 52 Marks
Calculate the Interquartile Range ($IQR$) and Quartile Deviation ($QD$) for the following ordered dataset:
$$10, 14, 18, 22, 26, 30, 34$$
In a frequency distribution, the First Quartile is $Q_1 = 40$ and the Coefficient of Quartile Deviation is $0.333$ (or $\frac{1}{3}$). Calculate the Third Quartile ($Q_3$).
Step 2: First Quartile ($Q_1$) [1 Mark]:
$$\text{Position} = \frac{N + 1}{4} = \frac{41}{4} = 10.25^{\text{th}} \text{ term}$$
Cumulative frequency $\ge 10.25$ is $11$, so $Q_1 = 20$.
Step 3: Third Quartile ($Q_3$) [1 Mark]:
$$\text{Position} = \frac{3(N + 1)}{4} = \frac{123}{4} = 30.75^{\text{th}} \text{ term}$$
Cumulative frequency $\ge 30.75$ is $34$, so $Q_3 = 40$.
Question 83 Marks
State the mathematical formula to calculate the $k^{\text{th}}$ Percentile ($P_k$) for a continuous grouped frequency distribution. Explain the exact meaning of every symbol used and illustrate with $P_{80}$.
Show Marking Scheme & Solution (3 Marks)
Step 1: Governing Formula [1 Mark]:
$$P_k = L + \left(\frac{\frac{kN}{100} - cf}{f}\right) \times h$$ Step 2: Definition of Parameters [1 Mark]:
$L$ = Lower boundary limit of the percentile class interval
$N$ = Total cumulative frequency ($\sum f$)
$cf$ = Cumulative frequency of the class immediately preceding the percentile class
$f$ = Frequency of the percentile class itself
$h$ = Class width (interval magnitude)
Step 3: Case of $P_{80}$ [1 Mark]:
To evaluate $P_{80}$, compute rank index $\frac{80N}{100} = 0.8N$, locate the class containing this cumulative count, and apply:
$$P_{80} = L + \left(\frac{0.8N - cf}{f}\right) \times h$$
Section D: Long Answer Type (Q9)1 Question • 4 Marks
Question 94 Marks
The examination scores of $60$ students in a standardized quantitative assessment are summarized in the table below:
Marks Range
$0 - 10$
$10 - 20$
$20 - 30$
$30 - 40$
$40 - 50$
Students ($f$)
$5$
$12$
$20$
$15$
$8$
Calculate:
(i) First Quartile ($Q_1$). [1.5 Marks]
(ii) Third Quartile ($Q_3$). [1.5 Marks]
(iii) Coefficient of Quartile Deviation. [1 Mark]
Show Marking Scheme & Solution (4 Marks)
Cumulative Frequency ($N = 60$):
• $0 - 10 : f = 5, cf = 5$
• $10 - 20 : f = 12, cf = 17$
• $20 - 30 : f = 20, cf = 37$
• $30 - 40 : f = 15, cf = 52$
• $40 - 50 : f = 8, cf = 60$
Section E: Case-Based / Source-Based Problem (Q10)1 Case Study • 4 Marks
Question 10: Case Study4 Marks Total
Case Background: Corporate Performance Appraisal & Incentive Tiers
A multinational technology consultancy conducts annual performance assessments for its $250$ associate software engineers. Each engineer is assigned an aggregated productivity score out of $100$. The score distribution indicates:
• Lower Quartile score ($Q_1$) = $55$
• Median productivity score ($Q_2$) = $68$
• Upper Quartile score ($Q_3$) = $82$
• $90^{\text{th}}$ Percentile score ($P_{90}$) = $91$
The compensation policy dictates that engineers scoring in the top $10\%$ receive an elite performance bonus, while engineers scoring in the bottom $25\%$ are enrolled in a structured performance enhancement program.
(i) [1 Mark]: What is the score cutoff threshold for an associate engineer to qualify for the elite performance bonus?
Show Solution for (i)
[1 Mark]: The top $10\%$ cutoff corresponds to the $90^{\text{th}}$ percentile ($P_{90}$). Therefore, an engineer must score $\ge 91$ marks to receive the elite performance bonus.
(ii) [1 Mark]: How many associate engineers will be enrolled in the performance enhancement program?
Show Solution for (ii)
[1 Mark]: The bottom $25\%$ corresponds to scores strictly below $Q_1$. The number of engineers $= 25\% \times 250 = \frac{250}{4} = 62.5 \approx 62$ or $63$ engineers (exactly one-quarter of the workforce).
(iii) [2 Marks]: Calculate:
(a) The Interquartile Range ($IQR$) of productivity scores. [1 Mark]
(b) The Coefficient of Quartile Deviation. [1 Mark]