Chapter 13 — STATISTICS

Concept Used

For individual observations, mean and variance are defined as:

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

\[ \sigma^2 = \frac{1}{N} \sum (x_i - \overline{x})^2 \]

Mean gives the central tendency of data, while variance measures spread around the mean.

Solution Roadmap

• Step 1: Count total observations \(N\)
• Step 2: Compute \(\sum x_i\)
• Step 3: Calculate mean \(\overline{x}\)
• Step 4: Find deviation \((x_i - \overline{x})\)
• Step 5: Square deviations
• Step 6: Compute variance

Solution

Step 1: Number of observations \[ N = 8 \]

Step 2: Sum of observations \[ \sum x_i = 6+7+10+12+13+4+8+12 = 72 \]

Step 3: Mean \[ \overline{x} = \frac{72}{8} = 9 \]

Step 4 & 5: Compute deviations and squares

\(x_i\)	\(x_i - 9\)	\((x_i - 9)^2\)
6	-3	9
7	-2	4
10	1	1
12	3	9
13	4	16
4	-5	25
8	-1	1
12	3	9
\(\sum x_i = 72\)	0	\(\sum (x_i - \overline{x})^2 = 74\)

Step 6: Variance \[ \sigma^2 = \frac{1}{N} \sum (x_i - \overline{x})^2 = \frac{74}{8} = 9.25 \]

Final Answer:
Mean = \(9\)
Variance = \(9.25\)

Visual Insight (Spread Around Mean)

The points are scattered around the mean value. Larger spread indicates higher variance.

Exam Significance

• Very frequent in CBSE Board exams (direct formula-based question)
• Foundation concept for JEE (Statistics + Probability)
• Used in data interpretation and variance minimization problems
• Helps in understanding dispersion (important for advanced stats)

Concept Used

For individual discrete data:

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

\[ \sigma^2 = \frac{1}{N}\sum x_i^2 - \overline{x}^2 \]

Standard results:

\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]

\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]

These are fundamental summation formulas frequently used in algebra, sequences, and statistics.

Solution Roadmap

• Step 1: Define dataset \(1,2,3,\dots,n\)
• Step 2: Compute mean using sum formula
• Step 3: Use variance identity \(\sigma^2 = \frac{1}{n}\sum x_i^2 - \overline{x}^2\)
• Step 4: Substitute summation formulas
• Step 5: Simplify algebra carefully (no skipping steps)

Solution

Step 1: Data set \[ x_i = 1,2,3,\dots,n \quad \text{and} \quad N=n \]

Step 2: Mean \[ \overline{x} = \frac{1}{n}\sum_{i=1}^{n} i \]

Substitute sum: \[ = \frac{1}{n} \cdot \frac{n(n+1)}{2} \]

Cancel \(n\): \[ \overline{x} = \frac{n+1}{2} \]

Step 3: Variance formula \[ \sigma^2 = \frac{1}{n}\sum_{i=1}^{n} i^2 - \overline{x}^2 \]

Substitute values: \[ = \frac{1}{n} \cdot \frac{n(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 \]

Cancel \(n\): \[ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \]

Step 4: Take common factor \((n+1)\) \[ = (n+1)\left(\frac{2n+1}{6} - \frac{n+1}{4}\right) \]

Step 5: Take LCM of 6 and 4 → 12

\[ = (n+1)\left(\frac{2(2n+1) - 3(n+1)}{12}\right) \]

Expand numerator: \[ = (n+1)\left(\frac{4n+2 - 3n - 3}{12}\right) \]

Simplify: \[ = (n+1)\left(\frac{n - 1}{12}\right) \]

Final multiplication: \[ \sigma^2 = \frac{(n+1)(n-1)}{12} \]

\[ \sigma^2 = \frac{n^2 - 1}{12} \]

Final Answer:
Mean = \(\frac{n+1}{2}\)
Variance = \(\frac{n^2 - 1}{12}\)

Visual Insight

Natural numbers are symmetrically distributed about the mean. As \(n\) increases, spread increases, hence variance grows proportional to \(n^2\).

Exam Significance

• Direct formula-based question in CBSE Board exams
• Frequently used in JEE Main numerical simplification problems
• Forms base for sequences, series, and summation problems
• Important in variance minimization and algebraic manipulation questions
• Shortcut memory result: variance of first \(n\) natural numbers = \(\frac{n^2-1}{12}\)

Concept Used

Multiples of a number form an Arithmetic Progression (AP).

Here: \[ 3,6,9,12,\dots,30 \]

Mean of AP: \[ \overline{x} = \frac{\text{first term} + \text{last term}}{2} \]

Variance: \[ \sigma^2 = \frac{1}{N}\sum (x_i - \overline{x})^2 \]

Important shortcut: If a dataset is multiplied by a constant \(a\), then variance becomes \(a^2\) times.

Solution Roadmap

• Step 1: Identify AP and number of terms
• Step 2: Compute mean using AP formula
• Step 3: Find deviations from mean
• Step 4: Square deviations and sum
• Step 5: Compute variance

Solution

Step 1: Data \[ 3,6,9,12,15,18,21,24,27,30 \quad \Rightarrow \quad N=10 \]

Step 2: Mean (AP shortcut) \[ \overline{x} = \frac{3 + 30}{2} = \frac{33}{2} = 16.5 \]

Step 3 & 4: Deviations and squares

\(x_i\)	\(x_i - 16.5\)	\((x_i - 16.5)^2\)
3	-13.5	182.25
6	-10.5	110.25
9	-7.5	56.25
12	-4.5	20.25
15	-1.5	2.25
18	1.5	2.25
21	4.5	20.25
24	7.5	56.25
27	10.5	110.25
30	13.5	182.25
		\(\sum (x_i-\overline{x})^2 = 742.5\)

Step 5: Variance \[ \sigma^2 = \frac{742.5}{10} = 74.25 \]

Final Answer:
Mean = \(16.5\)
Variance = \(74.25\)

Shortcut Insight (High Value)

Multiples of 3 = \(3 \times (1,2,3,\dots,10)\)

Variance of \(1\) to \(10\): \[ = \frac{10^2 - 1}{12} = \frac{99}{12} = 8.25 \]

Multiply by \(3^2 = 9\): \[ \sigma^2 = 9 \times 8.25 = 74.25 \]

This avoids long calculations in competitive exams.

Visual Insight

The data is symmetrically distributed about the mean, leading to balanced deviations on both sides.

Exam Significance

• Very common CBSE Board question (AP-based statistics)
• Direct shortcut application in JEE Main
• Important concept: scaling effect on variance
• Useful in data transformation problems
• Time-saving trick can reduce 2–3 minutes in exams

Concept Used

For discrete frequency distribution:

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

\[ \sigma^2 = \frac{1}{N} \sum f_i (x_i - \overline{x})^2 \]

Here, frequency \(f_i\) represents how many times each observation occurs.

Solution Roadmap

• Step 1: Compute total frequency \(N\)
• Step 2: Compute \(\sum f_i x_i\)
• Step 3: Calculate mean
• Step 4: Find deviations \((x_i - \overline{x})\)
• Step 5: Compute \(f_i(x_i-\overline{x})^2\)
• Step 6: Compute variance

Solution

Step 1: Total frequency \[ N = 2+4+7+12+8+4+3 = 40 \]

Step 2: Compute \(\sum f_i x_i\)

\[ = (2\times6) + (4\times10) + (7\times14) + (12\times18) + (8\times24) + (4\times28) + (3\times30) \]

\[ = 12 + 40 + 98 + 216 + 192 + 112 + 90 = 760 \]

Step 3: Mean \[ \overline{x} = \frac{760}{40} = 19 \]

Step 4 & 5: Deviations and weighted squares

\(x_i\)	\(f_i\)	\(x_i - 19\)	\((x_i - 19)^2\)	\(f_i(x_i - 19)^2\)
6	2	-13	169	338
10	4	-9	81	324
14	7	-5	25	175
18	12	-1	1	12
24	8	5	25	200
28	4	9	81	324
30	3	11	121	363
	\(N=40\)			\(\sum f_i(x_i-\overline{x})^2 = 1736\)

Step 6: Variance \[ \sigma^2 = \frac{1736}{40} = 43.4 \]

Final Answer:
Mean = \(19\)
Variance = \(43.4\)

Visual Insight

Higher frequencies near \(x=18\) and \(x=24\) pull the mean towards the center. Spread around mean determines variance.

Exam Significance

• Very common CBSE Board question (discrete series)
• Tests understanding of weighted mean
• Foundation for grouped data (next level)
• Frequently appears in JEE Main data interpretation
• Table-based solving improves speed and accuracy

Smart Shortcut (Advanced)

Use shortcut: \[ \sigma^2 = \frac{\sum f_i x_i^2}{N} - \overline{x}^2 \]

Faster for MCQs and competitive exams.

Concept Used

For discrete frequency distribution:

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

\[ \sigma^2 = \frac{1}{N} \sum f_i (x_i - \overline{x})^2 \]

Deviation must always be taken as \(x_i - \overline{x}\) (with sign), not absolute value.

Solution Roadmap

• Step 1: Compute total frequency \(N\)
• Step 2: Compute \(\sum f_i x_i\)
• Step 3: Calculate mean
• Step 4: Find deviations \((x_i - \overline{x})\)
• Step 5: Compute \(f_i(x_i-\overline{x})^2\)
• Step 6: Compute variance

Solution

Step 1: Total frequency \[ N = 3+2+3+2+6+3+3 = 22 \]

Step 2: Compute \(\sum f_i x_i\)

\[ = (3\times92) + (2\times93) + (3\times97) + (2\times98) + (6\times102) + (3\times104) + (3\times109) \]

\[ = 276 + 186 + 291 + 196 + 612 + 312 + 327 = 2200 \]

Step 3: Mean \[ \overline{x} = \frac{2200}{22} = 100 \]

Step 4 & 5: Deviations and weighted squares

\(x_i\)	\(f_i\)	\(x_i - 100\)	\((x_i - 100)^2\)	\(f_i(x_i - 100)^2\)
92	3	-8	64	192
93	2	-7	49	98
97	3	-3	9	27
98	2	-2	4	8
102	6	2	4	24
104	3	4	16	48
109	3	9	81	243
	\(N=22\)			\(\sum f_i(x_i-\overline{x})^2 = 640\)

Step 6: Variance

\[ \sigma^2 = \frac{640}{22} \]

\[ = 29.0909\ldots \approx 29.09 \]

Final Answer:
Mean = \(100\)
Variance = \(29.09\)

Visual Insight

Data is tightly clustered around 100, hence variance is relatively small.

Exam Significance

• Classic CBSE Board question (discrete series)
• Tests accuracy in deviation handling
• Very common in JEE Main MCQs
• Mean = 100 is a key observation → simplifies calculations
• Helps in mastering weighted distributions

Smart Shortcut (Competitive Exams)

Use identity: \[ \sigma^2 = \frac{\sum f_i x_i^2}{N} - \overline{x}^2 \]

Faster and avoids deviation table in MCQs.

Concept Used (Short-Cut / Assumed Mean Method)

Let assumed mean \(a = 64\)

Define: \[ y_i = x_i - a \]

Mean: \[ \overline{x} = a + \frac{\sum f_i y_i}{N} \]

Variance: \[ \sigma^2 = \frac{1}{N} \sum f_i y_i^2 - \left(\frac{\sum f_i y_i}{N}\right)^2 \]

Since values are consecutive, \(h = 1\)

Solution Roadmap

• Step 1: Choose assumed mean \(a\)
• Step 2: Compute \(y_i = x_i - a\)
• Step 3: Compute \(f_i y_i\) and \(f_i y_i^2\)
• Step 4: Find totals
• Step 5: Compute mean
• Step 6: Compute variance and standard deviation

Solution

Step 1: Take assumed mean \[ a = 64 \]

Step 2 & 3: Construct table

\(x_i\)	\(f_i\)	\(y_i = x_i - 64\)	\(y_i^2\)	\(f_i y_i\)	\(f_i y_i^2\)
60	2	-4	16	-8	32
61	1	-3	9	-3	9
62	12	-2	4	-24	48
63	29	-1	1	-29	29
64	25	0	0	0	0
65	12	1	1	12	12
66	10	2	4	20	40
67	4	3	9	12	36
68	5	4	16	20	80
	\(\sum f_i = 100\)			\(\sum f_i y_i = 0\)	\(\sum f_i y_i^2 = 286\)

Step 4: Mean

\[ \overline{x} = a + \frac{\sum f_i y_i}{N} = 64 + \frac{0}{100} = 64 \]

Step 5: Variance

\[ \sigma^2 = \frac{1}{100}(286) - 0^2 = 2.86 \]

Step 6: Standard Deviation

\[ \sigma = \sqrt{2.86} \approx 1.69 \]

Final Answer:
Mean = \(64\)
Standard Deviation ≈ \(1.69\)

Visual Insight

Data is highly concentrated around 64, hence very small standard deviation.

Exam Significance

• Direct Board question: “Use short-cut method”
• Highly important for JEE Main (time-saving)
• Assumed mean reduces heavy calculations
• When \(\sum f_i y_i = 0\), mean becomes instant
• Strong conceptual base for grouped data (next level)

Smart Insight

Choosing assumed mean at center (64) made: \[ \sum f_i y_i = 0 \]

This is the ideal case → fastest possible solution in exams.

Concept Used (Step-Deviation Method)

For grouped data:

\[ x_i = \text{class mark} = \frac{\text{lower} + \text{upper}}{2} \]

Let assumed mean \(a = 105\), class width \(h = 30\)

\[ y_i = \frac{x_i - a}{h} \]

Mean: \[ \overline{x} = a + \left(\frac{\sum f_i y_i}{N}\right)h \]

Variance: \[ \sigma^2 = h^2 \left[\frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2 \right] \]

Solution Roadmap

• Step 1: Compute class marks \(x_i\)
• Step 2: Choose assumed mean \(a\)
• Step 3: Compute coded variable \(y_i\)
• Step 4: Compute \(f_i y_i\) and \(f_i y_i^2\)
• Step 5: Calculate mean
• Step 6: Compute variance carefully

Solution

Step 1: Class marks \[ 15, 45, 75, 105, 135, 165, 195 \]

Step 2: Take \[ a = 105, \quad h = 30 \]

Step 3 & 4: Construct table

Class	\(f_i\)	\(x_i\)	\(y_i\)	\(f_i y_i\)	\(y_i^2\)	\(f_i y_i^2\)
0–30	2	15	-3	-6	9	18
30–60	3	45	-2	-6	4	12
60–90	5	75	-1	-5	1	5
90–120	10	105	0	0	0	0
120–150	3	135	1	3	1	3
150–180	5	165	2	10	4	20
180–210	2	195	3	6	9	18
	\(N=30\)			\(\sum f_i y_i = 2\)		\(\sum f_i y_i^2 = 76\)

Step 5: Mean

\[ \overline{x} = 105 + \left(\frac{2}{30}\right)\times 30 \]

\[ = 105 + 2 = 107 \]

Step 6: Variance

\[ \sigma^2 = 30^2 \left[\frac{76}{30} - \left(\frac{2}{30}\right)^2 \right] \]

First compute each term: \[ \frac{76}{30} = 2.5333 \]

\[ \left(\frac{2}{30}\right)^2 = \frac{4}{900} = 0.00444 \]

Subtract: \[ 2.5333 - 0.00444 = 2.5289 \]

Multiply by \(900\): \[ \sigma^2 = 900 \times 2.5289 = 2276 \]

Final Answer:
Mean = \(107\)
Variance = \(2276\)

Visual Insight

Data is concentrated around the central class (90–120), creating a near symmetric distribution.

Exam Significance

• Very important CBSE Board question (grouped data)
• Step-deviation method is fastest for large intervals
• Common in JEE Main numerical problems
• Tests understanding of coding transformation
• Avoids large-number calculations → high accuracy

Smart Insight

Choosing central class (105) as assumed mean minimizes deviations: \[ y_i \in \{-3,-2,-1,0,1,2,3\} \]

This drastically reduces calculation effort.

Concept Used (Step-Deviation Method)

For grouped data:

\[ x_i = \text{class mark} = \frac{\text{lower} + \text{upper}}{2} \]

Take assumed mean \(a = 25\), class width \(h = 10\)

\[ y_i = \frac{x_i - a}{h} \]

Mean: \[ \overline{x} = a + \left(\frac{\sum f_i y_i}{N}\right)h \]

Variance (step-deviation form): \[ \sigma^2 = h^2 \left[\frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2 \right] \]

Solution Roadmap

• Step 1: Compute class marks \(x_i\)
• Step 2: Choose assumed mean \(a\)
• Step 3: Compute coded variable \(y_i\)
• Step 4: Compute \(f_i y_i\), \(f_i y_i^2\)
• Step 5: Calculate mean
• Step 6: Compute variance step-by-step

Solution

Step 1: Class marks \[ 5, 15, 25, 35, 45 \]

Step 2: \[ a = 25, \quad h = 10 \]

Step 3 & 4: Table

Class	\(f_i\)	\(x_i\)	\(y_i\)	\(y_i^2\)	\(f_i y_i\)	\(f_i y_i^2\)
0–10	5	5	-2	4	-10	20
10–20	8	15	-1	1	-8	8
20–30	15	25	0	0	0	0
30–40	16	35	1	1	16	16
40–50	6	45	2	4	12	24
	\(N=50\)				\(\sum f_i y_i = 10\)	\(\sum f_i y_i^2 = 68\)

Step 5: Mean

\[ \overline{x} = 25 + \left(\frac{10}{50}\right)\times 10 \]

\[ = 25 + 2 = 27 \]

Step 6: Variance (fully expanded)

\[ \sigma^2 = 10^2 \left[\frac{68}{50} - \left(\frac{10}{50}\right)^2 \right] \]

Compute terms:

\[ \frac{68}{50} = 1.36 \]

\[ \left(\frac{10}{50}\right)^2 = \frac{100}{2500} = 0.04 \]

Subtract: \[ 1.36 - 0.04 = 1.32 \]

Multiply: \[ \sigma^2 = 100 \times 1.32 = 132 \]

Final Answer:
Mean = \(27\)
Variance = \(132\)

Visual Insight

Highest frequency lies near 30–40 class, slightly shifting mean above 25.

Exam Significance

• Very important CBSE Board grouped-data question
• Step-deviation method saves time in large intervals
• Common in JEE Main data interpretation
• Tests coding + variance accuracy
• Essential for higher statistics (Class 12)

Smart Insight

Choosing central class (25) simplifies: \[ y_i = -2,-1,0,1,2 \]

This is the most efficient computation strategy in exams.

Concept Used (Step-Deviation Method)

Grouped data → use class marks:

\[ x_i = \frac{\text{lower} + \text{upper}}{2} \]

Choose assumed mean \(a = 92.5\), class width \(h = 5\)

\[ y_i = \frac{x_i - a}{h} \]

Mean: \[ \overline{x} = a + \left(\frac{\sum f_i y_i}{N}\right)h \]

Variance: \[ \sigma^2 = h^2 \left[\frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2 \right] \]

Solution Roadmap

• Step 1: Compute class marks
• Step 2: Choose assumed mean
• Step 3: Compute coded values
• Step 4: Compute totals
• Step 5: Mean
• Step 6: Variance
• Step 7: Standard deviation

Solution

Step 1: Class marks \[ 72.5, 77.5, 82.5, 87.5, 92.5, 97.5, 102.5, 107.5, 112.5 \]

Step 2: \[ a = 92.5, \quad h = 5 \]

Step 3 & 4: Table

Class	\(f_i\)	\(x_i\)	\(y_i\)	\(y_i^2\)	\(f_i y_i\)	\(f_i y_i^2\)
70–75	3	72.5	-4	16	-12	48
75–80	4	77.5	-3	9	-12	36
80–85	7	82.5	-2	4	-14	28
85–90	7	87.5	-1	1	-7	7
90–95	15	92.5	0	0	0	0
95–100	9	97.5	1	1	9	9
100–105	6	102.5	2	4	12	24
105–110	6	107.5	3	9	18	54
110–115	3	112.5	4	16	12	48
	\(N=60\)				\(\sum f_i y_i = 6\)	\(\sum f_i y_i^2 = 254\)

Step 5: Mean

\[ \overline{x} = 92.5 + \left(\frac{6}{60}\right)\times 5 \]

\[ = 92.5 + 0.5 = 93 \]

Step 6: Variance (fully expanded)

\[ \sigma^2 = 25 \left[\frac{254}{60} - \left(\frac{6}{60}\right)^2 \right] \]

\[ \frac{254}{60} = 4.2333 \]

\[ \left(\frac{6}{60}\right)^2 = \frac{36}{3600} = 0.01 \]

\[ 4.2333 - 0.01 = 4.2233 \]

\[ \sigma^2 = 25 \times 4.2233 = 105.58 \]

Step 7: Standard Deviation

\[ \sigma = \sqrt{105.58} \approx 10.27 \]

Final Answer:
Mean = \(93\)
Variance ≈ \(105.58\)
Standard Deviation ≈ \(10.27\)

Visual Insight

Maximum frequency occurs near 90–95, pulling the mean slightly above 92.5.

Exam Significance

• Very important CBSE Board long question
• Tests full step-deviation mastery
• Common in JEE Main (numerical-based)
• Combines mean + variance + SD in one problem
• Accuracy + speed both required

Smart Insight

Choosing \(a = 92.5\) (central class) makes: \[ y_i = -4 \text{ to } 4 \]

This minimizes calculations and errors — ideal exam strategy.

Calculate the standard deviation and mean diameter of the circles.

Concept Used

Classes are inclusive, so convert into continuous form using correction factor: \[ \pm 0.5 \]

Grouped data formulas (step-deviation):

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

\[ \sigma^2 = h^2 \left[\frac{\sum f_i y_i^2}{N} - \left(\frac{\sum f_i y_i}{N}\right)^2 \right] \]

Solution Roadmap

• Step 1: Convert to continuous classes
• Step 2: Find class marks \(x_i\)
• Step 3: Choose assumed mean \(a\)
• Step 4: Compute \(y_i\), \(f_i y_i\), \(f_i y_i^2\)
• Step 5: Compute mean
• Step 6: Compute variance
• Step 7: Standard deviation

Solution

Step 1: Continuous classes

\[ 32.5–36.5,\ 36.5–40.5,\ 40.5–44.5,\ 44.5–48.5,\ 48.5–52.5 \]

Step 2: Class marks \[ 34.5,\ 38.5,\ 42.5,\ 46.5,\ 50.5 \]

Step 3: \[ a = 42.5,\quad h = 4 \]

Step 4: Table

Class	\(f_i\)	\(x_i\)	\(y_i\)	\(y_i^2\)	\(f_i y_i\)	\(f_i y_i^2\)
32.5–36.5	15	34.5	-2	4	-30	60
36.5–40.5	17	38.5	-1	1	-17	17
40.5–44.5	21	42.5	0	0	0	0
44.5–48.5	22	46.5	1	1	22	22
48.5–52.5	25	50.5	2	4	50	100
	\(N=100\)				\(\sum f_i y_i = 25\)	\(\sum f_i y_i^2 = 199\)

Step 5: Mean

\[ \overline{x} = 42.5 + \left(\frac{25}{100}\right)\times 4 \]

\[ = 42.5 + 1 = 43.5 \]

Step 6: Variance (fully expanded)

\[ \sigma^2 = 16 \left[\frac{199}{100} - \left(\frac{25}{100}\right)^2 \right] \]

\[ \frac{199}{100} = 1.99 \]

\[ \left(\frac{25}{100}\right)^2 = \frac{625}{10000} = 0.0625 \]

\[ 1.99 - 0.0625 = 1.9275 \]

\[ \sigma^2 = 16 \times 1.9275 = 30.84 \]

Step 7: Standard Deviation

\[ \sigma = \sqrt{30.84} \approx 5.55 \]

Final Answer:
Mean Diameter = \(43.5\) mm
Standard Deviation ≈ \(5.55\) mm

Visual Insight

Higher frequencies in larger diameter classes shift the mean towards the right.

Exam Significance

• Classic CBSE Board case: inclusive → continuous conversion
• Tests full pipeline: correction → coding → variance → SD
• Very common in JEE Main numerical problems
• High chance of mistakes if steps are skipped
• Must-master for Class 12 Statistics

Smart Insight

Choosing central class (42.5) makes: \[ y_i = -2,-1,0,1,2 \]

This minimizes computation → fastest exam approach.