Chapter 13 — STATISTICS

Theory

Mean Deviation about Mean is a measure of dispersion that indicates the average of absolute deviations of observations from their arithmetic mean.

Formula:

\[ \text{Mean Deviation about Mean} = \dfrac{1}{N}\sum |x_i - \overline{x}| \]

Key understanding:

• Mean acts as a central reference point
• Deviation measures distance from mean
• Absolute value removes negative signs
• Average of deviations gives dispersion

Solution Roadmap

• Find total number of observations \(N\)
• Compute arithmetic mean \(\overline{x}\)
• Calculate each deviation \(x_i - \overline{x}\)
• Convert into absolute deviations
• Find their sum
• Divide by \(N\)

Visual Intuition

Solution

Step 1: Write observations

\[ 4,\; 7,\; 8,\; 9,\; 10,\; 12,\; 13,\; 17 \]

Step 2: Number of observations

\[ N = 8 \]

Step 3: Compute sum of observations

\[ \sum x_i = 4 + 7 + 8 + 9 + 10 + 12 + 13 + 17 = 80 \]

Step 4: Calculate mean

\[ \overline{x} = \frac{80}{8} = 10 \]

Step 5: Compute deviations and absolute deviations

\(x_i\)	\(x_i - \overline{x}\)	\(|x_i - \overline{x}|\)
4	4 − 10 = −6	6
7	7 − 10 = −3	3
8	8 − 10 = −2	2
9	9 − 10 = −1	1
10	10 − 10 = 0	0
12	12 − 10 = 2	2
13	13 − 10 = 3	3
17	17 − 10 = 7	7
\(\sum x_i = 80\)	—	\(\sum |x_i - \overline{x}| = 24\)

Step 6: Apply formula

\[ \text{Mean Deviation} = \frac{1}{N} \sum |x_i - \overline{x}| \] \[ = \frac{24}{8} \] \[ = 3 \]

Final Answer:

\[ \boxed{3} \]

Concept Significance

• Important for CBSE board exams (3–5 mark questions)
• Strengthens base for variance and standard deviation
• Used in data interpretation for competitive exams like JEE Main
• Helps analyse consistency of data distribution

Theory

Mean Deviation about Mean measures the average distance of each observation from the arithmetic mean, ignoring the sign of deviations.

Formula:

\[ \text{Mean Deviation about Mean} = \frac{1}{N}\sum |x_i - \overline{x}| \]

Important understanding:

• Mean represents the central value of data
• Deviation shows how far each value lies from mean
• Absolute value ensures all deviations are positive
• Average of these deviations gives dispersion

Solution Roadmap

• Count number of observations \(N\)
• Compute sum of observations
• Find mean \(\overline{x}\)
• Calculate deviations \(x_i - \overline{x}\)
• Take absolute values
• Find total deviation
• Divide by \(N\)

Visual Intuition

Solution

Step 1: Write observations

\[ 38,\; 70,\; 48,\; 40,\; 42,\; 55,\; 63,\; 46,\; 54,\; 44 \]

Step 2: Number of observations

\[ N = 10 \]

Step 3: Compute sum

\[ \sum x_i = 38+70+48+40+42+55+63+46+54+44 = 500 \]

Step 4: Calculate mean

\[ \overline{x} = \frac{500}{10} = 50 \]

Step 5: Compute deviations and absolute deviations

\(x_i\)	\(x_i - \overline{x}\)	\(|x_i - \overline{x}|\)
38	38 − 50 = −12	12
70	70 − 50 = 20	20
48	48 − 50 = −2	2
40	40 − 50 = −10	10
42	42 − 50 = −8	8
55	55 − 50 = 5	5
63	63 − 50 = 13	13
46	46 − 50 = −4	4
54	54 − 50 = 4	4
44	44 − 50 = −6	6
\(\sum x_i = 500\)	—	\(\sum |x_i - \overline{x}| = 84\)

Step 6: Apply formula

\[ \text{Mean Deviation} = \frac{1}{N} \sum |x_i - \overline{x}| \] \[ = \frac{84}{10} \] \[ = 8.4 \]

Final Answer:

\[ \boxed{8.4} \]

Concept Significance

• Frequently asked in CBSE board exams with full-step marking
• Common in MCQs for JEE Main and CUET
• Strengthens foundation for standard deviation
• Helps in analysing consistency of real-life data

Theory

Mean Deviation about Median measures the average of absolute deviations of observations from the median.

Why median is used:

• Median is less affected by extreme values (outliers)
• It represents the central position of ordered data

Formula:

\[ \text{Mean Deviation about Median} = \frac{1}{N}\sum |x_i - M| \]

Solution Roadmap

• Arrange data in ascending order
• Find median \(M\)
• Compute deviations \(x_i - M\)
• Take absolute values
• Find total deviation
• Divide by \(N\)

Visual Intuition

Solution

Step 1: Arrange data in ascending order

\[ 10,\;11,\;11,\;12,\;13,\;13,\;14,\;16,\;16,\;17,\;17,\;18 \]

Step 2: Number of observations

\[ N = 12 \]

Step 3: Find median

\[ \frac{N}{2} = \frac{12}{2} = 6 \]

Median is average of 6th and 7th terms

\[ M = \frac{13 + 14}{2} = 13.5 \]

Step 4: Compute deviations and absolute deviations

\(x_i\)	\(x_i - M\)	\(|x_i - M|\)
10	10 − 13.5 = −3.5	3.5
11	11 − 13.5 = −2.5	2.5
11	11 − 13.5 = −2.5	2.5
12	12 − 13.5 = −1.5	1.5
13	13 − 13.5 = −0.5	0.5
13	13 − 13.5 = −0.5	0.5
14	14 − 13.5 = 0.5	0.5
16	16 − 13.5 = 2.5	2.5
16	16 − 13.5 = 2.5	2.5
17	17 − 13.5 = 3.5	3.5
17	17 − 13.5 = 3.5	3.5
18	18 − 13.5 = 4.5	4.5
—	—	\(\sum |x_i - M| = 28\)

Step 5: Apply formula

\[ \text{Mean Deviation} = \frac{1}{N} \sum |x_i - M| \] \[ = \frac{28}{12} \] \[ = \frac{7}{3} = 2.33 \]

Final Answer:

\[ \boxed{2.33} \]

Concept Significance

• Important for CBSE board exams (step-based questions)
• Median-based deviation is preferred in skewed data
• Useful in statistics-based MCQs in JEE, CUET
• Foundation for advanced dispersion concepts

Theory

Mean Deviation about Median is the average of absolute deviations of observations from the median. It is preferred when data contains extreme values because median is resistant to outliers.

Formula:

\[ \text{Mean Deviation about Median} = \frac{1}{N}\sum |x_i - M| \]

Solution Roadmap

• Arrange data in ascending order
• Find median \(M\)
• Compute deviations \(x_i - M\)
• Take absolute values
• Find total deviation
• Divide by \(N\)

Visual Intuition

Solution

Step 1: Arrange data in ascending order

\[ 36,\; 42,\; 45,\; 46,\; 46,\; 49,\; 51,\; 53,\; 60,\; 72 \]

Step 2: Number of observations

\[ N = 10 \]

Step 3: Find median

\[ \frac{N}{2} = \frac{10}{2} = 5 \]

Median is average of 5th and 6th terms

\[ M = \frac{46 + 49}{2} = 47.5 \]

Step 4: Compute deviations and absolute deviations

\(x_i\)	\(x_i - M\)	\(|x_i - M|\)
36	36 − 47.5 = −11.5	11.5
42	42 − 47.5 = −5.5	5.5
45	45 − 47.5 = −2.5	2.5
46	46 − 47.5 = −1.5	1.5
46	46 − 47.5 = −1.5	1.5
49	49 − 47.5 = 1.5	1.5
51	51 − 47.5 = 3.5	3.5
53	53 − 47.5 = 5.5	5.5
60	60 − 47.5 = 12.5	12.5
72	72 − 47.5 = 24.5	24.5
—	—	\(\sum |x_i - M| = 70\)

Step 5: Apply formula

\[ \text{Mean Deviation} = \frac{1}{N} \sum |x_i - M| \] \[ = \frac{70}{10} \] \[ = 7 \]

Final Answer:

\[ \boxed{7} \]

Concept Significance

• Very important for CBSE board exams with full-step marking
• Median-based deviation is highly useful in skewed distributions
• Appears in MCQs and case-study questions (CUET, JEE Main)
• Builds conceptual bridge to advanced statistical measures

Theory

For discrete frequency distribution, Mean Deviation about Mean is calculated by considering frequencies of each observation.

Formula:

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \] \[ \text{Mean Deviation about Mean} = \frac{1}{\sum f_i} \sum f_i |x_i - \overline{x}| \]

Key idea:

• Each observation contributes according to its frequency
• Higher frequency means greater impact on deviation

Solution Roadmap

• Compute \(f_i x_i\)
• Find total frequency \(N = \sum f_i\)
• Calculate mean \(\overline{x}\)
• Find deviations \(|x_i - \overline{x}|\)
• Multiply by frequencies \(f_i|x_i-\overline{x}|\)
• Sum and divide by \(N\)

Visual Intuition

Solution

Step 1: Compute \(f_i x_i\)

\(x_i\)	\(f_i\)	\(f_i x_i\)
5	7	35
10	4	40
15	6	90
20	3	60
25	5	125
	\(N = 25\)	\(\sum f_i x_i = 350\)

Step 2: Calculate mean

\[ \overline{x} = \frac{350}{25} = 14 \]

Step 3: Compute deviations and weighted deviations

\(x_i\)	\(f_i\)	\(|x_i - 14|\)	\(f_i|x_i - 14|\)
5	7	9	63
10	4	4	16
15	6	1	6
20	3	6	18
25	5	11	55
			\(\sum f_i|x_i - \overline{x}| = 158\)

Step 4: Apply formula

\[ \text{Mean Deviation} = \frac{1}{25} \times 158 \] \[ = 6.32 \]

Final Answer:

\[ \boxed{6.32} \]

Concept Significance

• Very important for CBSE board exams (table-based questions)
• Common in data interpretation and MCQs (JEE, CUET)
• Builds foundation for grouped data statistics
• Develops understanding of weighted averages

Theory

For discrete frequency distribution, Mean Deviation about Mean considers the weight (frequency) of each observation.

Formula:

\[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} \] \[ \text{Mean Deviation about Mean} = \frac{1}{\sum f_i}\sum f_i |x_i - \overline{x}| \]

Key insight:

• Mean is the balancing point of distribution
• Frequencies determine influence of each value
• Higher frequency → higher contribution to deviation

Solution Roadmap

• Compute \(f_i x_i\)
• Find total frequency \(N\)
• Calculate mean \(\overline{x}\)
• Find \(|x_i - \overline{x}|\)
• Multiply by frequency
• Sum and divide by \(N\)

Visual Intuition

Solution

Step 1: Compute \(f_i x_i\)

\(x_i\)	\(f_i\)	\(f_i x_i\)
10	4	40
30	24	720
50	28	1400
70	16	1120
90	8	720
	\(N = 80\)	\(\sum f_i x_i = 4000\)

Step 2: Calculate mean

\[ \overline{x} = \frac{4000}{80} = 50 \]

Step 3: Compute deviations and weighted deviations

\(x_i\)	\(f_i\)	\(|x_i - 50|\)	\(f_i|x_i - 50|\)
10	4	40	160
30	24	20	480
50	28	0	0
70	16	20	320
90	8	40	320
			\(\sum f_i|x_i - \overline{x}| = 1280\)

Step 4: Apply formula

\[ \text{Mean Deviation} = \frac{1280}{80} \] \[ = 16 \]

Final Answer:

\[ \boxed{16} \]

Concept Significance

• Important for CBSE board exams with table-based solutions
• Frequently appears in MCQs (JEE Main, CUET)
• Strengthens concept of weighted averages
• Helps in real-life data dispersion analysis

Theory

For discrete frequency distribution, Mean Deviation about Median is computed using frequencies and median as the central value.

Formula:

\[ \text{Mean Deviation about Median} = \frac{1}{\sum f_i} \sum f_i |x_i - M| \]

Why median is used:

• Median is stable and not affected by extreme values
• Useful when data is skewed

Solution Roadmap

• Compute cumulative frequency (cf)
• Find median position \(\frac{N}{2}\)
• Locate median value \(M\)
• Compute \(|x_i - M|\)
• Multiply by frequencies
• Sum and divide by \(N\)

Visual Intuition

Solution

Step 1: Compute cumulative frequencies

\(x_i\)	\(f_i\)	\(cf\)
5	8	8
7	6	14
9	2	16
10	2	18
12	2	20
15	6	26
	\(N = 26\)

Step 2: Find median position

\[ \frac{N}{2} = \frac{26}{2} = 13 \]

Median lies between 13th and 14th observations

From cumulative frequency:

• 13th observation lies in \(x = 7\)
• 14th observation also lies in \(x = 7\)

\[ M = 7 \]

Step 3: Compute deviations and weighted deviations

\(x_i\)	\(f_i\)	\(|x_i - 7|\)	\(f_i|x_i - 7|\)
5	8	2	16
7	6	0	0
9	2	2	4
10	2	3	6
12	2	5	10
15	6	8	48
			\(\sum f_i|x_i - M| = 84\)

Step 4: Apply formula

\[ \text{Mean Deviation} = \frac{84}{26} \] \[ = 3.23 \]

Final Answer:

\[ \boxed{3.23} \]

Concept Significance

• Important for CBSE board exams (case-study + table questions)
• Helps in solving cumulative frequency based problems
• Frequently appears in MCQs (CUET, JEE Main)
• Strengthens understanding of median-based dispersion

Theory

For discrete frequency distribution, Mean Deviation about Median uses the median as the central value and considers frequencies as weights.

Formula:

\[ \text{Mean Deviation about Median} = \frac{1}{\sum f_i} \sum f_i |x_i - M| \]

Concept insight:

• Median divides data into two equal parts
• Less affected by extreme values
• Useful for skewed distributions

Solution Roadmap

• Compute cumulative frequency
• Find median position
• Determine median value \(M\)
• Compute \(|x_i - M|\)
• Multiply by frequencies
• Sum and divide by total frequency

Visual Intuition

Solution

Step 1: Compute cumulative frequency

\(x_i\)	\(f_i\)	\(cf\)
15	3	3
21	5	8
27	6	14
30	7	21
35	8	29
	\(N = 29\)

Step 2: Find median position

\[ \frac{N+1}{2} = \frac{30}{2} = 15 \]

Median is the 15th observation

From cumulative frequency:

• 14th observation → 27
• 15th observation → 30

\[ M = 30 \]

Step 3: Compute deviations and weighted deviations

\(x_i\)	\(f_i\)	\(|x_i - 30|\)	\(f_i|x_i - 30|\)
15	3	15	45
21	5	9	45
27	6	3	18
30	7	0	0
35	8	5	40
			\(\sum f_i|x_i - M| = 148\)

Step 4: Apply formula

\[ \text{Mean Deviation} = \frac{148}{29} \] \[ \approx 5.1 \]

Final Answer:

\[ \boxed{5.1} \]

Concept Significance

• Important for CBSE board exams (median + cf based questions)
• Common in MCQs and case-based questions (CUET, JEE Main)
• Strengthens understanding of cumulative frequency
• Useful in real-life skewed data analysis

Theory

For grouped data, Mean Deviation about Mean is calculated using class marks (midpoints) and frequencies.

Key formulas:

\[ x_i = \text{class mark} = \frac{\text{lower limit + upper limit}}{2} \] \[ \overline{x} = a + \left(\frac{\sum f_i d_i}{\sum f_i}\right)\times h \] \[ \text{Mean Deviation} = \frac{1}{N}\sum f_i |x_i - \overline{x}| \]

Concept insight:

• Grouped data requires approximation using class marks
• Step-deviation simplifies calculations
• Mean acts as balancing point of distribution

Solution Roadmap

• Find class marks \(x_i\)
• Choose assumed mean \(a\)
• Compute \(d_i = \frac{x_i - a}{h}\)
• Calculate \(\sum f_i d_i\) and mean
• Compute \(|x_i - \overline{x}|\)
• Multiply by frequencies
• Divide by total frequency

Visual Intuition

Solution

Step 1: Compute class marks \(x_i\)

\[ 50,\;150,\;250,\;350,\;450,\;550,\;650,\;750 \]

Step 2: Choose assumed mean

\[ a = 350,\quad h = 100 \]

Step 3: Compute \(d_i = \frac{x_i - 350}{100}\)

Class	\(f_i\)	\(x_i\)	\(d_i\)	\(f_i d_i\)
0–100	4	50	-3	-12
100–200	8	150	-2	-16
200–300	9	250	-1	-9
300–400	10	350	0	0
400–500	7	450	1	7
500–600	5	550	2	10
600–700	4	650	3	12
700–800	3	750	4	12
	\(N = 50\)			\(\sum f_i d_i = 4\)

Step 4: Calculate mean

\[ \overline{x} = 350 + \left(\frac{4}{50}\right)\times 100 = 350 + 8 = 358 \]

Step 5: Compute \(|x_i - \overline{x}|\) and \(f_i|x_i - \overline{x}|\)

\(x_i\)	\(|x_i - 358|\)	\(f_i|x_i - 358|\)
50	308	1232
150	208	1664
250	108	972
350	8	80
450	92	644
550	192	960
650	292	1168
750	392	1176
		\(\sum f_i|x_i-\overline{x}| = 7896\)

Step 6: Apply formula

\[ \text{Mean Deviation} = \frac{7896}{50} \] \[ = 157.92 \]

Final Answer:

\[ \boxed{157.92} \]

Concept Significance

• Highly important for CBSE board exams (case-study questions)
• Step-deviation method saves time in competitive exams
• Frequently asked in JEE Main data interpretation
• Real-life application: income inequality analysis

Theory

For grouped data, Mean Deviation about Mean is computed using class marks (midpoints) and frequencies.

Key formulas:

\[ x_i = \frac{\text{lower limit + upper limit}}{2} \] \[ \overline{x} = a + \left(\frac{\sum f_i d_i}{\sum f_i}\right)\times h \] \[ \text{Mean Deviation} = \frac{1}{N}\sum f_i |x_i - \overline{x}| \]

Concept insight:

• Grouped data uses class midpoints as representatives
• Step-deviation simplifies arithmetic
• Mean represents central tendency of grouped distribution

Solution Roadmap

• Find class marks \(x_i\)
• Choose assumed mean \(a\)
• Compute \(d_i = \frac{x_i-a}{h}\)
• Find \(\sum f_i d_i\)
• Compute mean \(\overline{x}\)
• Calculate \(|x_i-\overline{x}|\)
• Multiply by frequencies and divide by \(N\)

Visual Intuition

Solution

Step 1: Compute class marks \(x_i\)

\[ 100,\;110,\;120,\;130,\;140,\;150 \]

Step 2: Choose assumed mean

\[ a = 120,\quad h = 10 \]

Step 3: Compute \(d_i = \frac{x_i - 120}{10}\)

Class	\(f_i\)	\(x_i\)	\(d_i\)	\(f_i d_i\)
95–105	9	100	-2	-18
105–115	13	110	-1	-13
115–125	26	120	0	0
125–135	30	130	1	30
135–145	12	140	2	24
145–155	10	150	3	30
	\(N = 100\)			\(\sum f_i d_i = 53\)

Step 4: Calculate mean

\[ \overline{x} = 120 + \left(\frac{53}{100}\right)\times 10 \] \[ = 120 + 5.3 = 125.3 \]

Step 5: Compute \(|x_i - \overline{x}|\) and \(f_i|x_i - \overline{x}|\)

\(x_i\)	\(|x_i - 125.3|\)	\(f_i|x_i - 125.3|\)
100	25.3	227.7
110	15.3	198.9
120	5.3	137.8
130	4.7	141.0
140	14.7	176.4
150	24.7	247.0
		\(\sum f_i|x_i-\overline{x}| = 1128.8\)

Step 6: Apply formula

\[ \text{Mean Deviation} = \frac{1128.8}{100} \] \[ = 11.28 \]

Final Answer:

\[ \boxed{11.28} \]

Concept Significance

• Very important for CBSE board exams (grouped data case-study)
• Step-deviation method reduces calculation time in exams
• Frequently asked in JEE Main data-based MCQs
• Used in real-life analysis like height/health statistics

Theory

For grouped data, Mean Deviation about Median is calculated using class midpoints and frequencies, where median is obtained using interpolation formula.

Key formulas:

\[ M = l + \left(\frac{\frac{N}{2} - c}{f}\right)\times h \] \[ \text{Mean Deviation} = \frac{1}{N}\sum f_i |x_i - M| \]

Concept insight:

• Median lies in the class where cumulative frequency exceeds \(N/2\)
• Grouped data requires interpolation for accuracy
• Deviation is measured from class midpoints

Solution Roadmap

• Compute cumulative frequency
• Find median class
• Apply median formula
• Compute class midpoints
• Calculate deviations \(|x_i - M|\)
• Multiply by frequencies
• Divide by \(N\)

Visual Intuition

Solution

Step 1: Compute cumulative frequency

Class	\(f_i\)	\(cf\)
0–10	6	6
10–20	8	14
20–30	14	28
30–40	16	44
40–50	4	48
50–60	2	50
	\(N = 50\)

Step 2: Find median class

\[ \frac{N}{2} = 25 \]

Median class is \(20–30\) (first cf ≥ 25)

Step 3: Apply median formula

\[ l = 20,\quad c = 14,\quad f = 14,\quad h = 10 \] \[ M = 20 + \left(\frac{25 - 14}{14}\right)\times 10 \] \[ = 20 + \frac{11}{14}\times 10 = 20 + 7.86 = 27.86 \]

Step 4: Compute midpoints

\[ 5,\;15,\;25,\;35,\;45,\;55 \]

Step 5: Compute deviations and weighted deviations

\(x_i\)	\(|x_i - 27.86|\)	\(f_i|x_i - M|\)
5	22.86	137.16
15	12.86	102.88
25	2.86	40.04
35	7.14	114.24
45	17.14	68.56
55	27.14	54.28
		\(\sum f_i|x_i-M| = 517.16\)

Step 6: Apply formula

\[ \text{Mean Deviation} = \frac{517.16}{50} \] \[ = 10.34 \]

Final Answer:

\[ \boxed{10.34} \]

Concept Significance

• Highly important for CBSE board exams (median class + interpolation)
• Strengthens understanding of grouped data statistics
• Common in JEE Main case-study and MCQs
• Useful in real-life data like marks distribution analysis

Theory

For grouped data, Mean Deviation about Median is calculated using interpolated median and class midpoints.

Key formulas:

\[ M = l + \left(\frac{\frac{N}{2} - c}{f}\right)\times h \] \[ \text{Mean Deviation} = \frac{1}{N}\sum f_i |x_i - M| \]

Concept insight:

• Median class contains the \( \frac{N}{2} \)th observation
• Class boundaries are used for continuous data
• Midpoints represent each class

Solution Roadmap

• Convert classes to continuous form
• Compute cumulative frequency
• Locate median class
• Apply median formula
• Compute midpoints
• Calculate \(|x_i - M|\)
• Multiply by frequencies and divide by \(N\)

Visual Intuition

Solution

Step 1: Convert to continuous classes

\[ 15.5–20.5,\;20.5–25.5,\;25.5–30.5,\;30.5–35.5,\;35.5–40.5,\;40.5–45.5,\;45.5–50.5,\;50.5–55.5 \]

Step 2: Compute cumulative frequency

Class	\(f_i\)	\(cf\)
15.5–20.5	5	5
20.5–25.5	6	11
25.5–30.5	12	23
30.5–35.5	14	37
35.5–40.5	26	63
40.5–45.5	12	75
45.5–50.5	16	91
50.5–55.5	9	100
	\(N = 100\)

Step 3: Find median class

\[ \frac{N}{2} = 50 \]

Median class is \(35.5–40.5\)

Step 4: Apply median formula

\[ l = 35.5,\; c = 37,\; f = 26,\; h = 5 \] \[ M = 35.5 + \left(\frac{50 - 37}{26}\right)\times 5 \] \[ = 35.5 + \frac{13}{26}\times 5 = 35.5 + 2.5 = 38 \]

Step 5: Compute midpoints

\[ 18,\;23,\;28,\;33,\;38,\;43,\;48,\;53 \]

Step 6: Compute deviations and weighted deviations

\(x_i\)	\(|x_i - 38|\)	\(f_i|x_i - M|\)
18	20	100
23	15	90
28	10	120
33	5	70
38	0	0
43	5	60
48	10	160
53	15	135
		\(\sum f_i|x_i-M| = 735\)

Step 7: Apply formula

\[ \text{Mean Deviation} = \frac{735}{100} \] \[ = 7.35 \]

Final Answer:

\[ \boxed{7.35} \]

Concept Significance

• Very important CBSE board question (median + grouped data)
• Tests interpolation concept clearly
• Frequently appears in JEE Main data interpretation
• Applicable in demographic and age-distribution studies