(i) Median Class:

Cumulative frequencies: 8, 17, 27, 39, 48.
n = 48, n/2 = 24. The cumulative frequency just exceeding 24 is 27 (class 30–45).
Median class = 30–45

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(ii) Probability of calling an even number (first ball):

Even numbers from 1 to 75: 2, 4, 6, … 74 → 37 even numbers
Total balls = 75

$$P(\text{even}) = \frac{37}{75}$$

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(iii) Median:

Median class = 30–45, l = 30, f = 10, cf = 17, h = 15, n/2 = 24

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 30 + \frac{24 - 17}{10} \times 15$$

$$= 30 + \frac{7 \times 15}{10} = 30 + 10.5 = \mathbf{40.5}$$

Source: Statistics, Median of Grouped Data

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• Part (i): Build the cumulative frequency table; the class where cumulative frequency first reaches or exceeds n/2 = 24 is the median class.
• Part (ii): Balls are 1–75; even numbers are 2, 4, …, 74 = 37 numbers. Total = 75. Many students mistakenly write 38 — remember 75 is odd so evens = 37.
• Part (iii): Apply the standard median formula. Carry cf of the class before the median class (17, not 27). This is a common error to avoid.

(b) median

The middle-most observation of data arranged in order is called the median.

Source: Statistics, Section 13.4 (Median of Grouped Data), Chapter 13

The textbook explicitly defines median as "a measure of central tendency which gives the value of the middle-most observation in the data." Mode is the most frequent value, mean is the average, and deviation is not a measure of central tendency. For a 1-mark MCQ, just writing the option letter and the term is sufficient.

(d) 0

Since $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$, we have $\Sigma f_i x_i = \bar{x} \cdot \Sigma f_i$. Therefore, $\sum f_i(x_i - \bar{x}) = \Sigma f_i x_i - \bar{x}\Sigma f_i = \bar{x}\Sigma f_i - \bar{x}\Sigma f_i = 0$.

Source: Chapter 13, Section 13.2 Mean of Grouped Data

This is a standard property of the arithmetic mean: the sum of deviations of observations from their mean is always zero. Expanding $\sum f_i(x_i - \bar{x}) = \sum f_i x_i - \bar{x}\sum f_i$, and substituting the formula for $\bar{x}$, both terms cancel. Examiners expect you to identify option (d) and optionally show the one-step algebraic justification.

Frequency Distribution Table with Cumulative Frequency:

| Class | Frequency (f) | cf |
|-------|---------------|----|
| 0–15 | 15 | 15 |
| 15–30 | 10 | 25 |
| 30–45 | 12 | 37 |
| 45–60 | 9 | 46 |
| 60–75 | 8 | 54 |
| 75–90 | 10 | 64 |
| 90–105 | 6 | 70 |

Total n = 70

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Median:

$\dfrac{n}{2} = 35$

Cumulative frequency just greater than 35 is 37 (class 30–45).

∴ Median class = 30–45, $l = 30$, $cf = 25$, $f = 12$, $h = 15$

$$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 30 + \left(\frac{35 - 25}{12}\right) \times 15$$

$$= 30 + \frac{150}{12} = 30 + 12.5 = \mathbf{42.5}$$

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Mode:

Maximum frequency = 15 → Modal class = 0–15

$l = 0$, $f_1 = 15$, $f_0 = 0$ (no preceding class), $f_2 = 10$, $h = 15$

$$\text{Mode} = 0 + \left(\frac{15 - 0}{2(15) - 0 - 10}\right) \times 15 = \frac{15}{20} \times 15 = \mathbf{11.25}$$

Source: Chapter 13, Sections 13.3 and 13.4

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• Median: Find n/2, locate the class whose cumulative frequency first exceeds n/2 — that is the median class. Then apply the formula. Here cf = 25 (class before 30–45) and f = 12.
• Mode: The modal class has the highest frequency (15 for 0–15). Since it is the first class, $f_0 = 0$. Substitute carefully in the mode formula.
• Common mistakes: using wrong cf (must be of the class preceding median class), and forgetting $f_0 = 0$ when modal class is the first interval.

Given: Mean = 28, $\Sigma f_i$ = 100

Step 1: Find p and q using $\Sigma f_i = 100$

$$12 + p + 27 + 20 + q + 6 = 100$$
$$65 + p + q = 100$$
$$p + q = 35 \quad \text{...(1)}$$

Step 2: Calculate $\Sigma f_i x_i$ using class marks

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 5 | 12 | 60 |
| 10–20 | 15 | p | 15p |
| 20–30 | 25 | 27 | 675 |
| 30–40 | 35 | 20 | 700 |
| 40–50 | 45 | q | 45q |
| 50–60 | 55 | 6 | 330 |

$$\Sigma f_i x_i = 1765 + 15p + 45q$$

Step 3: Apply mean formula

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \Rightarrow 28 = \frac{1765 + 15p + 45q}{100}$$

$$2800 = 1765 + 15p + 45q$$
$$15p + 45q = 1035$$
$$p + 3q = 69 \quad \text{...(2)}$$

Step 4: Solve equations (1) and (2)

Subtract (1) from (2):
$$2q = 34 \Rightarrow q = 17$$
$$p = 35 - 17 = 18$$

$$\boxed{p = 18, \quad q = 17}$$

Source: Chapter 13, Section 13.2 (Mean of Grouped Data)

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• Two conditions are always needed for two unknowns: (i) sum of frequencies = total, (ii) mean formula.
• Use class marks (midpoints) for $x_i$ in the direct method.
• Set up simultaneous equations and solve — examiners award marks at each step, so show all working clearly.
• Don't skip the table; it earns step marks and keeps calculation organised.

(A) 0.3

Using the step-deviation formula: $\bar{x} = a + h\bar{u}$

$$64 = 62.5 + 5 \times \bar{u} \implies 5\bar{u} = 1.5 \implies \bar{u} = 0.3$$

The key formula is $\bar{x} = a + h\bar{u}$. Substitute the known values ($\bar{x} = 64$, $a = 62.5$, $h = 5$) and solve for $\bar{u}$. Many students mistakenly choose 0.5 (dividing 1.5 by 3) or 1.5 (forgetting to divide by $h$). Always isolate $\bar{u}$ by dividing the difference $(\bar{x} - a)$ by $h$.

(a) Median Age:

| Age Group | Frequency | Cumulative Frequency |
|-----------|-----------|----------------------|
| 15–25 | 8 | 8 |
| 25–35 | 10 | 18 |
| 35–45 | 15 | 33 |
| 45–55 | 25 | 58 |
| 55–65 | 40 | 98 |
| 65–75 | 24 | 122 |
| 75–85 | 18 | 140 |

Total N = 140, so N/2 = 70.

Median class = 55–65 (cf = 58, f = 40, l = 55, h = 10)

$$\text{Median} = 55 + \frac{70 - 58}{40} \times 10 = 55 + 3 = \mathbf{58}$$

(b) Finding x:

Original Σfi·xi (midpoints: 20, 30, 40, 50, 60, 70, 80):

Σfi·xi = 8(20)+10(30)+15(40)+25(50)+40(60)+24(70)+18(80)
= 160+300+600+1250+2400+1680+1440 = 7830

New total frequency = 140 + x; New Σfi·xi = 7830 + 70x

$$58 = \frac{7830 + 70x}{140 + x}$$

$$8120 + 58x = 7830 + 70x \implies 290 = 12x \implies \boxed{x = \frac{290}{12} \approx 24.17}$$

Since x must be a whole number, x ≈ 24 (or exact value $x = \dfrac{145}{6}$, accept as per calculation).

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• Part (a): Always build cumulative frequency, identify median class where cf first exceeds N/2, then apply the formula correctly. Here median = 58.
• Part (b): Use midpoints for mean calculation. Set up the equation with (140 + x) in denominator and solve. If the answer is non-integer, present the exact fraction — CBSE accepts it. Show all steps clearly for full marks.

Given: Mode = 75, modal class = 65–80, $f_0$ = 6, $f_2$ = 8, $l$ = 65, $h$ = 15.

Using the formula:
$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h$$

$$75 = 65 + \left(\frac{f_1 - 6}{2f_1 - 6 - 8}\right) \times 15$$

$$10 = \frac{15(f_1 - 6)}{2f_1 - 14}$$

$$10(2f_1 - 14) = 15(f_1 - 6)$$

$$20f_1 - 140 = 15f_1 - 90$$

$$5f_1 = 50 \implies f_1 = 10$$

The frequency of the modal class is 10.

Source: Chapter 13, Section 13.3 Mode of Grouped Data

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• Examiners expect you to write the mode formula, substitute all known values clearly, and solve the linear equation step by step.
• Key variables: $l$ = lower limit of modal class, $h$ = class width, $f_1$ = modal class frequency (unknown), $f_0$ = preceding frequency, $f_2$ = succeeding frequency.
• Don't forget: $h = 80 - 65 = 15$.
• Both marks are awarded for correct substitution and correct final answer; show working to secure full credit.

Cumulative Frequency Table:

| Height (cm) | Frequency (f) | Cumulative Frequency (cf) |
|---|---|---|
| 130–135 | 4 | 4 |
| 135–140 | 11 | 15 |
| 140–145 | 12 | 27 |
| 145–150 | 7 | 34 |
| 150–155 | 10 | 44 |
| 155–160 | 6 | 50 |

Here, $n = 50$, so $\dfrac{n}{2} = 25$.

The cumulative frequency just greater than 25 is 27, which belongs to class 140–145.

∴ Median class = 140–145

$l = 140,\quad cf = 15,\quad f = 12,\quad h = 5$

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h = 140 + \left(\frac{25 - 15}{12}\right) \times 5$$

$$= 140 + \frac{50}{12} = 140 + 4.17 = \mathbf{144.17 \text{ cm}}$$

Source: Chapter 13, Section 13.4

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• Always build the cumulative frequency (cf) column first.
• Locate the median class: the class whose cf first exceeds $n/2$.
• Substitute carefully into the formula; examiners award marks for the table (1 mark), identifying the median class with correct values of $l, cf, f, h$ (1 mark), and the final calculation (1 mark).
• Round to 2 decimal places unless instructed otherwise.

Let assumed mean $a = 12.5$

| Class | $f_i$ | $x_i$ | $d_i = x_i - 12.5$ | $f_i d_i$ |
|-------|--------|--------|----------------------|------------|
| 0–5 | 8 | 2.5 | –10 | –80 |
| 5–10 | 7 | 7.5 | –5 | –35 |
| 10–15 | 10 | 12.5 | 0 | 0 |
| 15–20 | 13 | 17.5 | 5 | 65 |
| 20–25 | 12 | 22.5 | 10 | 120 |
| Total | 50 | | | 70 |

Using the formula:

$$\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 12.5 + \frac{70}{50} = 12.5 + 1.4 = \mathbf{13.9}$$

Source: Chapter 13, Section 13.2 (Assumed Mean Method)

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• Choose $a$ as the class mark of the middle class (here 10–15, so $a = 12.5$).
• $d_i = x_i - a$: find deviation of each class mark from assumed mean.
• Formula: $\bar{x} = a + \dfrac{\Sigma f_i d_i}{\Sigma f_i}$
• Examiners award marks for: correct table (class marks, $d_i$, $f_i d_i$), correct $\Sigma f_i = 50$, $\Sigma f_i d_i = 70$, and final answer 13.9. Show all table columns clearly.

Finding class marks (midpoints):

| Class | $f_i$ | $x_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 5 | 5 | 25 |
| 10–20 | 18 | 15 | 270 |
| 20–30 | 15 | 25 | 375 |
| 30–40 | f | 35 | 35f |
| 40–50 | 6 | 45 | 270 |
| Total | $44 + f$ | | $940 + 35f$ |

Using the formula: $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$

$$25 = \frac{940 + 35f}{44 + f}$$

$$25(44 + f) = 940 + 35f$$

$$1100 + 25f = 940 + 35f$$

$$1100 - 940 = 35f - 25f$$

$$160 = 10f$$

$$\boxed{f = 16}$$

Source: Chapter 13 – Statistics, Section 13.2 Mean of Grouped Data

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• Examiner expects: A complete table with $x_i$ and $f_i x_i$, correct substitution into the mean formula, and clear algebraic steps to isolate $f$.
• Key formula: $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$ (Direct Method).
• Common mistake: Forgetting to include $f$ in $\Sigma f_i$ (denominator). Both numerator and denominator must contain $f$.
• For 3 marks: 1 mark for the table, 1 mark for setting up the equation, 1 mark for correct value of $f$.

Since mode = 55 lies in class 45–60, this is the modal class.

Here, $l = 45$, $h = 15$, $f_1 = 15$, $f_0 = x$, $f_2 = 10$

Using Mode formula:

$$55 = 45 + \left(\frac{15 - x}{2(15) - x - 10}\right) \times 15$$

$$10 = \frac{(15 - x) \times 15}{20 - x}$$

$$10(20 - x) = 15(15 - x)$$

$$200 - 10x = 225 - 15x$$

$$5x = 25$$

$$\boxed{x = 5}$$

Source: Chapter 13, Section 13.3 – Mode of Grouped Data

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• Since mode = 55 falls in 45–60, that is the modal class. The class before it (30–45) has frequency $x$, so $f_0 = x$.
• Substitute directly into the mode formula and solve the linear equation.
• Examiners award 1 mark for correct identification of modal class/formula setup and 1 mark for the correct value of $x = 5$.

(i) Finding x and the mean mass:

Total apples = 250

$$20 + 60 + 70 + x + 60 = 250$$
$$210 + x = 250 \implies x = 40$$

Mean by Direct Method:

| Mass (g) | $f_i$ | $x_i$ (mid-point) | $f_i x_i$ |
|---|---|---|---|
| 80–100 | 20 | 90 | 1800 |
| 100–120 | 60 | 110 | 6600 |
| 120–140 | 70 | 130 | 9100 |
| 140–160 | 40 | 150 | 6000 |
| 160–180 | 60 | 170 | 10200 |
| Total | 250 | | 33700 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{33700}{250} = \textbf{134.8 g}$$

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(ii) Finding the modal mass:

The class 120–140 has the highest frequency (70), so it is the modal class.

Here: $l = 120$, $f_1 = 70$, $f_0 = 60$, $f_2 = 40$, $h = 20$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 120 + \frac{70-60}{2(70)-60-40} \times 20$$

$$= 120 + \frac{10}{40} \times 20 = 120 + 5 = \textbf{125 g}$$

Source: Statistics, Chapter 13, Sections 13.2 and 13.4

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• For (i): Always find the unknown frequency first using $\Sigma f_i = $ total. Then use mid-points for the direct method. Show the table clearly — examiners award marks step-by-step.
• For (ii): Identify the modal class as the one with the highest frequency. Apply the formula correctly: $l$ = lower limit, $f_1$ = modal class frequency, $f_0$ = preceding class frequency, $f_2$ = succeeding class frequency, $h$ = class width. Don't confuse $f_0$ and $f_2$.

(I) Modal Class:
The class with the highest frequency is 600–800 (frequency = 7).
∴ Modal class = 600–800

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(II) Median:

| Class | Frequency | Cumulative Frequency |
|-------|-----------|----------------------|
| 200–400 | 2 | 2 |
| 400–600 | 4 | 6 |
| 600–800 | 7 | 13 |
| 800–1000 | 4 | 17 |
| 1000–1200 | 2 | 19 |
| 1200–1400 | 3 | 22 |
| 1400–1600 | 1 | 23 |
| 1600–1800 | 1 | 24 |

Total $n = 24$, so $\dfrac{n}{2} = 12$

Median class = 600–800 (cf just before = 6, f = 7, l = 600, h = 200)

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 600 + \frac{12 - 6}{7} \times 200 = 600 + \frac{1200}{7} \approx \textbf{771.43 mm}$$

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(III) Good Rainfall Sub-divisions:
Sub-divisions with rainfall ≥ 1000 mm: 2 + 3 + 1 + 1 = 7 sub-divisions

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• (I) Modal class is simply the class with the highest frequency — no calculation needed.
• (II) Always build the cf table first; identify median class where cf first exceeds n/2. Apply the formula correctly showing all steps.
• (III) Add frequencies of all classes from 1000 mm onwards (1000–1200, 1200–1400, 1400–1600, 1600–1800). Examiners look for correct addition.

Option B: increases by 3

If each observation $x_i$ is increased by 3, the new mean $= \dfrac{\Sigma(x_i+3)}{n} = \dfrac{\Sigma x_i + 3n}{n} = \bar{x} + 3$.

The mean is the sum of all observations divided by $n$. Adding 3 to every observation adds $3n$ to the total sum; dividing by $n$ gives an increase of exactly 3 in the mean. Options A, C, D are wrong because the mean does change, it changes by 3 (not 6 or 3n). This is a standard property of mean tested frequently in MCQs.

Converting to a frequency table: 0–10: 3, 10–20: 9, 20–30: 15, 30–40: 30, 40–50: 18, 50–60: 5.

The class with the highest frequency is 30–40.

Answer: (C) 30 – 40

Convert the cumulative frequency table to a regular frequency table by subtracting successive values (e.g., 57 − 27 = 30 for 30–40). The class with the maximum frequency is the modal class. Here, 30–40 has frequency 30, which is the highest — so it is the modal class.

Step 1: Set up the cumulative frequency table

Total students = 230

| Marks | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 10–20 | 12 | 12 |
| 20–30 | 30 | 42 |
| 30–40 | x | 42 + x |
| 40–50 | 65 | 107 + x |
| 50–60 | y | 107 + x + y |
| 60–70 | 25 | 132 + x + y |
| 70–80 | 18 | 150 + x + y |

Step 2: Form equations

Since total = 230:
$$150 + x + y = 230 \implies x + y = 80 \quad \cdots (1)$$

Step 3: Find median class

$\dfrac{n}{2} = \dfrac{230}{2} = 115$

The cf just greater than 115 is (107 + x), which falls in class 40–50.

So median class = 40–50; $l = 40$, $f = 65$, $cf = 42 + x$, $h = 10$

Step 4: Apply median formula

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h$$

$$46 = 40 + \left(\frac{115 - (42 + x)}{65}\right) \times 10$$

$$6 = \frac{(73 - x) \times 10}{65}$$

$$6 \times 65 = 10(73 - x)$$

$$390 = 730 - 10x$$

$$10x = 340 \implies \boxed{x = 34}$$

From (1): $y = 80 - 34 = \boxed{46}$

∴ x = 34 and y = 46

Source: Chapter 13, Section 13.4 Median of Grouped Data

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• Two conditions are always used: (i) sum of all frequencies = total, giving one equation; (ii) median formula, giving the second equation.
• Identify the median class first by finding $n/2 = 115$ and locating which class's cumulative frequency first exceeds 115. Since cf up to 30–40 is $42+x$ and up to 40–50 is $107+x$, for the median class to be 40–50 we need $42+x < 115 \leq 107+x$, which is satisfied for $x = 34$.
• Show all substitution steps clearly — CBSE awards marks for each step.

Answer: C — 52 is the mode of the data.

Since the maximum number of students obtained 52 marks, 52 is the observation that occurs most frequently, which is the mode.

Mode is defined as the value that appears most often in a data set. "Maximum number of students obtained 52 marks" directly means 52 has the highest frequency — that is the definition of mode. Mean requires calculation of average; median is the middle value; range is the difference between highest and lowest values.

Using the empirical relation: Mode = 3 Median – 2 Mean

Given: Mode = 10, Mean + Median = 25

Let Mean = $x$, Median = $25 - x$

$10 = 3(25 - x) - 2x$
$10 = 75 - 3x - 2x$
$5x = 65$
$x = 13$

∴ Mean = 13 and Median = 12 → Option B

The key formula from the textbook (Chapter 13, Remarks after Section 13.4) is the empirical relationship: 3 Median = Mode + 2 Mean. Set up two equations using this and Mean + Median = 25, then solve simultaneously. Examiners expect you to recall and apply this formula directly.

Step 1: Calculate Mean (Direct Method)

| Weight (kg) | f | x (mid-value) | fx |
|---|---|---|---|
| 38–40 | 3 | 39 | 117 |
| 40–42 | 2 | 41 | 82 |
| 42–44 | 4 | 43 | 172 |
| 44–46 | 5 | 45 | 225 |
| 46–48 | 14 | 47 | 658 |
| 48–50 | 4 | 49 | 196 |
| 50–52 | 3 | 51 | 153 |
| Total | 35 | | 1603 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{1603}{35} = 45.8 \text{ kg}$$

Step 2: Calculate Median

Cumulative frequencies: 3, 5, 9, 14, 28, 32, 35.

$n = 35$, so $\dfrac{n}{2} = 17.5$

The cumulative frequency just greater than 17.5 is 28 → Median class = 46–48

Here: $l = 46$, $cf = 14$, $f = 14$, $h = 2$

$$\text{Median} = l + \left(\frac{\frac{n}{2} - cf}{f}\right) \times h = 46 + \left(\frac{17.5 - 14}{14}\right) \times 2$$

$$= 46 + \frac{3.5}{14} \times 2 = 46 + 0.5 = \textbf{46.5 kg}$$

Step 3: Difference

$$\text{Difference} = \text{Median} - \text{Mean} = 46.5 - 45.8 = \boxed{0.7 \text{ kg}}$$

Source: Statistics, Chapter 13 (Mean of Grouped Data §13.2; Median of Grouped Data §13.4)

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• Mean uses the direct formula $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$ with midpoints as $x_i$.
• Median requires building cumulative frequency, finding $n/2 = 17.5$, identifying the median class (where cf first exceeds 17.5), then applying the formula.
• The answer is Median − Mean = 0.7 kg (positive, since the distribution is skewed left by the dominant 46–48 class).
• Examiners award marks at each step: the table (1–2 marks), mean calculation (1 mark), median calculation with correct cf and formula (2 marks), and final difference (stated clearly).

Option C: 28 and 29

Since n = 100 (even), median = average of 50th and 51st observations. From cumulative frequency table, cumulative frequency up to marks 28 is 50, and up to 29 is 78. So the 50th observation is 28 and 51st is 29.

Source: Chapter 13, Section 13.4 Median of Grouped Data

• Arrange data in ascending order and build a cumulative frequency table.
• For even n, median = average of (n/2)th and (n/2 + 1)th = 50th and 51st observations.
• CF up to 28 = 6+20+24 = 50, so the 50th observation is 28; CF up to 29 = 78, so the 51st observation is 29.
• Examiners expect you to identify the correct pair using the cumulative frequency table — not just guess from the options.

The empirical relationship is 3 Median = Mode + 2 Mean, so $x = 3$, $y = 2$, $z = 1$.

Therefore, $x + y + z = 3 + 2 + 1 = \mathbf{6}$.

Answer: A — 6

Source: Chapter 13, Section 13.4 (Remarks)

The NCERT textbook states the empirical relation as: 3 Median = Mode + 2 Mean (i.e., 3 Median = 1 Mode + 2 Mean). Matching with $x$ Median + $y$ Mean = $z$ Mode gives $x=3$, $y=2$, $z=1$, so $x+y+z=6$. Examiners expect you to recall this standard formula exactly as given in the textbook.

Mean:

| Age | $f_i$ | $x_i$ (mid-value) | $f_i x_i$ |
|---|---|---|---|
| 5–15 | 6 | 10 | 60 |
| 15–25 | 11 | 20 | 220 |
| 25–35 | 21 | 30 | 630 |
| 35–45 | 23 | 40 | 920 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 5 | 60 | 300 |
| Total | 80 | | 2830 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2830}{80} = \textbf{35.375 years}$$

Mode:

The class with the highest frequency is 35–45 (frequency = 23).
So, Modal class = 35–45, $l = 35$, $f_1 = 23$, $f_0 = 21$, $f_2 = 14$, $h = 10$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h$$

$$= 35 + \frac{23 - 21}{2(23) - 21 - 14} \times 10 = 35 + \frac{2}{11} \times 10 = 35 + 1.82 = \textbf{36.82 years}$$

Source: Exercise 13.2, Q.1; Chapter 13

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• Mean: Use Direct Method — find mid-value ($x_i$) of each class, multiply by frequency, sum up, then divide by total frequency (80).
• Mode: Identify the modal class (highest frequency = 23, class 35–45). Apply the formula with $l$ = lower limit, $f_1$ = modal class frequency, $f_0$ = preceding class frequency, $f_2$ = succeeding class frequency, $h$ = class width.
• Examiners expect the full table for mean (awarded ~2 marks) and correct substitution in mode formula (~2 marks) + final answers (~1 mark). Never skip the formula step.

(c) 13

Using the empirical relationship: Mode = 3 Median − 2 Mean = 3(11) − 2(10) = 33 − 20 = 13.

Source: Chapter 13, Section 13.4 (Remarks)

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The key formula to remember is the empirical relationship between the three measures of central tendency:

Mode = 3 Median − 2 Mean

This is stated in the Remarks under Section 13.4. Simply substitute Mean = 10 and Median = 11 to get Mode = 33 − 20 = 13. Examiners expect you to recall and apply this formula directly for such 1-mark questions.

(c) classes preceding the median class

The cumulative frequency of a class is obtained by adding the frequencies of all the classes preceding the given class.

Source: Chapter 13, Section 13.5 Summary (Point 3)

The textbook's Summary explicitly states: "The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class." In the median formula, cf refers to the cumulative frequency of the class just before the median class. Students often confuse this with "up to the median class" — but the median class's own frequency is not included in cf.

Step 1: Cumulative Frequency Table

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 65–85 | 4 | 4 |
| 85–105 | 5 | 9 |
| 105–125 | x | 9 + x |
| 125–145 | 20 | 29 + x |
| 145–165 | 14 | 43 + x |
| 165–185 | y | 43 + x + y |
| 185–205 | 4 | 47 + x + y |

Step 2: Using total frequency = 68

$$47 + x + y = 68 \implies x + y = 21 \quad \text{...(1)}$$

Step 3: Finding median class

$\dfrac{n}{2} = \dfrac{68}{2} = 34$

cf just before 125–145 is $9 + x$, and cf of 125–145 is $29 + x$.
Since median = 137 lies in class 125–145, this is the median class.

Here: $l = 125,\ f = 20,\ cf = 9 + x,\ h = 20$

Step 4: Applying median formula

$$137 = 125 + \left(\frac{34 - (9+x)}{20}\right) \times 20$$

$$12 = 25 - x$$

$$x = 13$$

Step 5: From (1):

$$y = 21 - 13 = 8$$

$$\boxed{x = 13, \quad y = 8}$$

Source: Chapter 13, Section 13.4 Median of Grouped Data

---

• First form the equation from total frequency (sum = 68) to get x + y = 21.
• Identify the median class by locating where n/2 = 34 falls in cumulative frequencies — it must be 125–145 since median 137 lies there.
• Apply the median formula with correct l, cf, f, h values to get x = 13.
• Substitute back into equation (1) to get y = 8.
• Examiner checks: correct cf column, right median class, correct substitution, and both values stated clearly.

Using the empirical relationship: 3 Median = Mode + 2 Mean

3(25.2) = 26.1 + 2 Mean → 75.6 – 26.1 = 2 Mean → Mean = 49.5/2 = 24.75

Answer: (A) 24·75

The key formula is the empirical relation: 3 Median = Mode + 2 Mean (stated in Chapter 13, Remarks section). Rearranging gives Mean = (3 Median – Mode) / 2. Substitute Median = 25.2, Mode = 26.1 to get (75.6 – 26.1)/2 = 24.75. Examiners expect you to recall and apply this formula directly.

Modal Age:

The highest frequency is 33, corresponding to class 35–40.
So, modal class = 35–40, $l = 35$, $h = 5$, $f_1 = 33$, $f_0 = 21$, $f_2 = 11$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 35 + \frac{33-21}{66-21-11} \times 5 = 35 + \frac{12}{34} \times 5 = 35 + 1.76 = \textbf{36.76 years}$$

Median Age:

Cumulative frequency table:

| Age | f | cf |
|---|---|---|
| 15–20 | 2 | 2 |
| 20–25 | 4 | 6 |
| 25–30 | 18 | 24 |
| 30–35 | 21 | 45 |
| 35–40 | 33 | 78 |
| 40–45 | 11 | 89 |
| 45–50 | 3 | 92 |
| 50–55 | 6 | 98 |
| 55–60 | 2 | 100 |

$n = 100$, so $\frac{n}{2} = 50$.
cf just below 50 is 45 (class 30–35); hence median class = 35–40.

$l = 35$, $cf = 45$, $f = 33$, $h = 5$

$$\text{Median} = l + \frac{\frac{n}{2} - cf}{f} \times h = 35 + \frac{50 - 45}{33} \times 5 = 35 + \frac{25}{33} = 35 + 0.76 = \textbf{35.76 years}$$

Source: Chapter 13, Sections 13.3 (Mode) and 13.4 (Median of Grouped Data)

---

• Modal class = class with highest frequency (33 → 35–40). Apply the standard mode formula.
• Median class = class where cumulative frequency first crosses $n/2 = 50$; cf reaches 45 after 30–35, so 35–40 is the median class.
• Examiners award marks for: correct identification of modal/median class, correct substitution in formula, and correct final values. Always show the cf table for median.

(D) Assertion (A) is false, but Reason (R) is true.

The sum of first n natural numbers is $\frac{n(n+1)}{2}$ (R is true), so Mean $= \frac{n(n+1)/2}{n} = \frac{n+1}{2}$, not $\frac{n-1}{2}$. Hence A is false.

• R is true: The standard formula $1+2+\cdots+n = \frac{n(n+1)}{2}$ is correct.
• A is false: Dividing the sum by n gives mean $= \frac{n+1}{2}$, not $\frac{n-1}{2}$ (e.g., mean of 1, 2, 3 is 2 = $\frac{3+1}{2}$, not $\frac{3-1}{2} = 1$).
• In Assertion–Reason questions, always verify A independently before checking if R explains it.

Using the empirical relationship: Mode = 3 Median − 2 Mean

Mode = 3 × 43.4 − 2 × 43 = 130.2 − 86 = 44.2

Answer: (C) 44.2

CBSE expects students to know the empirical formula Mode = 3 Median − 2 Mean, which relates the three measures of central tendency for a moderately skewed distribution. Substitute the given values directly: Mode = 3(43.4) − 2(43) = 130.2 − 86 = 44.2. This formula is a standard result covered in Chapter 13 (Statistics).

Given: Mean = 35, $\Sigma f_i = 25$

Step 1: From the sum of frequencies:

$$1 + x + 5 + 7 + y + 3 + 1 = 25$$
$$17 + x + y = 25$$
$$\boxed{x + y = 8} \quad \text{...(i)}$$

Step 2: Find class marks and compute $\Sigma f_i x_i$:

| Class | $f_i$ | $x_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 1 | 5 | 5 |
| 10–20 | $x$ | 15 | $15x$ |
| 20–30 | 5 | 25 | 125 |
| 30–40 | 7 | 35 | 245 |
| 40–50 | $y$ | 45 | $45y$ |
| 50–60 | 3 | 55 | 165 |
| 60–70 | 1 | 65 | 65 |

$$\Sigma f_i x_i = 605 + 15x + 45y$$

Step 3: Apply mean formula:

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} \Rightarrow 35 = \frac{605 + 15x + 45y}{25}$$

$$875 = 605 + 15x + 45y$$
$$15x + 45y = 270$$
$$\boxed{x + 3y = 18} \quad \text{...(ii)}$$

Step 4: Solve equations (i) and (ii):

Subtracting (i) from (ii):
$$2y = 10 \Rightarrow y = 5$$
$$x = 8 - 5 = 3$$

$$\therefore x = 3, \quad y = 5$$

Source: Statistics, Section 13.2 Mean of Grouped Data, Chapter 13

---

• Two equations, two unknowns — examiners expect you to set up both equations clearly: one from $\Sigma f_i = 25$ and one from the mean formula.
• Always use class marks (midpoints) as $x_i$ for grouped data mean.
• Show the table of $f_i x_i$ neatly — it earns method marks even if arithmetic slips.
• Label your equations (i) and (ii) and show the subtraction step clearly.
• Final answer must state both values of $x$ and $y$ explicitly.

(c) 15

Using the empirical relationship: $3 \text{ Median} = \text{Mode} + 2\text{ Mean}$
$= 21 + 2(12) = 21 + 24 = 45$
$\therefore \text{Median} = 45 \div 3 = \mathbf{15}$

Source: Chapter 13, Section 13.4 (Remarks)

---

The key formula to remember is the empirical relationship: $3 \text{ Median} = \text{Mode} + 2\text{ Mean}$. This is stated in the Remarks of Section 13.4. Substituting directly gives Median = 15. Examiners expect students to recall and apply this formula correctly — no derivation needed for an MCQ.

Cumulative Frequency Table:

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 1400–1550 | 6 | 6 |
| 1550–1700 | 13 | 19 |
| 1700–1850 | 25 | 44 |
| 1850–2000 | 10 | 54 |

Here, $n = 54$, so $\dfrac{n}{2} = 27$.

The cumulative frequency just greater than 27 is 44, which belongs to class 1700–1850.

∴ Median class = 1700–1850

Here, $l = 1700$, $cf = 19$, $f = 25$, $h = 150$

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h = 1700 + \left(\frac{27 - 19}{25}\right) \times 150$$

$$= 1700 + \frac{8 \times 150}{25} = 1700 + 48 = \mathbf{1748}$$

Source: Chapter 13, Section 13.4 Median of Grouped Data

---

• Always build the cf column first, then find $n/2$.
• The median class is the one whose cf first exceeds $n/2$ — a common error is picking the class where cf equals $n/2$ exactly.
• Correctly identify $l$ (lower limit of median class), $cf$ (cf of the preceding class), $f$ (frequency of median class), and $h$ (class width = 150 here).
• Examiners award marks for: cf table (1 mark), identifying median class with correct values (1 mark), and correct formula application giving 1748 (1 mark).

Let assumed mean a = 125, class size h = 10.

| Weight (kg) | $f_i$ | $x_i$ | $d_i = x_i - 125$ | $f_i d_i$ |
|---|---|---|---|---|
| 100–110 | 4 | 105 | –20 | –80 |
| 110–120 | 12 | 115 | –10 | –120 |
| 120–130 | 23 | 125 | 0 | 0 |
| 130–140 | 8 | 135 | 10 | 80 |
| 140–150 | 3 | 145 | 20 | 60 |
| Total | 50 | | | –60 |

$$\bar{x} = a + \frac{\Sigma f_i d_i}{\Sigma f_i} = 125 + \frac{-60}{50} = 125 - 1.2 = \mathbf{123.8 \text{ kg}}$$

Mean weight of animals = 123.8 kg

Source: Chapter 13, Section 13.2 (Assumed Mean Method)

---

• Choose assumed mean a = 125 (midpoint of the middle class 120–130, which has the highest frequency — a natural central value).
• Calculate deviations $d_i = x_i - a$ for each class mark.
• Apply the formula: $\bar{x} = a + \dfrac{\Sigma f_i d_i}{\Sigma f_i}$.
• Examiners award marks for: correct table with $f_i$, $x_i$, $d_i$, $f_id_i$; correct $\Sigma f_i = 50$ and $\Sigma f_id_i = -60$; and correct final calculation.
• Note the negative sum of $f_id_i$ — don't drop the minus sign.

The highest frequency is 23, so the modal class = 45–55.

Here, $l = 45$, $f_1 = 23$, $f_0 = 22$, $f_2 = 14$, $h = 10$

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

$$= 45 + \left(\frac{23 - 22}{2(23) - 22 - 14}\right) \times 10 = 45 + \frac{1}{10} \times 10 = 45 + 1 = \mathbf{46}$$

Source: Chapter 13, Section 13.3

---

• Modal class is the class with the highest frequency (23 → 45–55).
• Identify $f_0$ (preceding class frequency = 22) and $f_2$ (succeeding class frequency = 14) carefully — a common error is swapping these.
• Examiner awards 1 mark for correct identification of modal class/values and 1 mark for correct substitution and final answer.

Using the Direct Method:

| Daily Expenditure (₹) | $f_i$ | $x_i$ (Class Mark) | $f_i x_i$ |
|---|---|---|---|
| 100 – 120 | 8 | 110 | 880 |
| 120 – 140 | 3 | 130 | 390 |
| 140 – 160 | 8 | 150 | 1200 |
| 160 – 180 | 6 | 170 | 1020 |
| 180 – 200 | 5 | 190 | 950 |
| Total | 30 | | 4440 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4440}{30} = \textbf{₹148}$$

The mean daily expenditure of the employees is ₹148.

Source: Chapter 13, Section 13.2 — Mean of Grouped Data (Direct Method)

---

• Class mark $x_i = \dfrac{\text{Upper limit} + \text{Lower limit}}{2}$ is used to represent each class interval.
• The Direct Method formula is $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$.
• Always show the table with all columns — examiners award marks for the working, not just the final answer.
• Verify: $\Sigma f_i = 30$ ✓ (matches total employees given in the question).

Cumulative Frequency Table:

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 0–10 | 3 | 3 |
| 10–20 | 5 | 8 |
| 20–30 | 11 | 19 |
| 30–40 | 10 | 29 |
| 40–50 | x | 29 + x |
| 50–60 | 3 | 32 + x |
| 60–70 | 2 | 34 + x |

Total $n = 34 + x$, so $\dfrac{n}{2} = \dfrac{34+x}{2}$

Median = 34.5 lies in class 30–40.
So: $l = 30,\ f = 10,\ cf = 19,\ h = 10$

Using the formula:
$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h$$

$$34.5 = 30 + \left(\frac{\dfrac{34+x}{2} - 19}{10}\right) \times 10$$

$$4.5 = \frac{34+x}{2} - 19$$

$$4.5 + 19 = \frac{34+x}{2}$$

$$23.5 \times 2 = 34 + x$$

$$47 = 34 + x$$

$$\boxed{x = 13}$$

Source: Chapter 13, Section 13.4 Median of Grouped Data

---

• First identify the median class: since median = 34.5, it lies in 30–40.
• Express $n$ in terms of $x$ and substitute into the median formula.
• The key step students miss: $cf$ before the median class is 19 (cumulative frequency up to 20–30), and $f = 10$ for the median class 30–40.
• Solve the linear equation carefully — $x = 13$ makes $n = 47$, so $n/2 = 23.5$, which is indeed greater than $cf = 19$ and less than $29$, confirming 30–40 is the correct median class.

The class with the highest frequency is 40–50 (frequency = 17), so the modal class = 40–50.

Here, $l = 40,\ f_1 = 17,\ f_0 = 12,\ f_2 = 4,\ h = 10$

$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 40 + \left(\frac{17-12}{34-12-4}\right) \times 10$$

$$= 40 + \frac{5}{18} \times 10 = 40 + 2.78 = \mathbf{42.78}$$

Source: Chapter 13, Section 13.3

---

• Examiners award 1 mark for correctly identifying the modal class and writing the values of $l, f_1, f_0, f_2, h$, and 1 mark for correct substitution and final answer.
• The modal class is always the one with the highest frequency (here 40–50, f = 17), not the first class.
• $f_0$ is the frequency of the class before the modal class (30–40 = 12) and $f_2$ is the class after (50–60 = 4). Getting these mixed up is a common error.

Using the Direct Method: $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$

| Classes | $f_i$ | $x_i$ (class mark) | $f_i x_i$ |
|---------|-------|---------------------|-----------|
| 25–30 | 14 | 27.5 | 385.0 |
| 30–35 | 22 | 32.5 | 715.0 |
| 35–40 | 16 | 37.5 | 600.0 |
| 40–45 | 6 | 42.5 | 255.0 |
| 45–50 | 5 | 47.5 | 237.5 |
| 50–55 | 3 | 52.5 | 157.5 |
| 55–60 | 4 | 57.5 | 230.0 |
| Total | 70 | | 2580.0 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2580}{70} = \mathbf{36.86}$$

The mean of the distribution is 36.86 (approx).

Source: Chapter 13, Section 13.2 Mean of Grouped Data

---

• Class mark $x_i$ = (lower limit + upper limit) / 2 for each class. Always write this column clearly.
• Use the Direct Method here since the values are small and manageable.
• $\Sigma f_i = 70$ is the total frequency — verify this first to avoid errors.
• Examiner awards marks for: correct class marks, correct $f_i x_i$ products, correct $\Sigma f_i x_i$, and correct final division. Show each step.

(a) decreases by 2

When every term is decreased by 2, the new mean = old mean − 2, so the mean also decreases by 2.

If original mean is $\bar{x} = \dfrac{\sum x_i}{n}$, and each term becomes $(x_i - 2)$, new mean $= \dfrac{\sum(x_i-2)}{n} = \bar{x} - 2$. The mean shifts by exactly the same constant subtracted from every term. Examiners expect you to state the correct option and give a one-line reason.

(b) 21

Using the empirical relation: Mode = 3 Median − 2 Mean = 3(15) − 2(12) = 45 − 24 = 21.

The key formula is the empirical relationship: 3 Median = Mode + 2 Mean, given in the Remarks of Section 13.4. Rearranging: Mode = 3 Median − 2 Mean. Always remember this formula for MCQs and short-answer questions.

Step 1: Find x

Total families = 200
∴ 24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200
172 + x = 200 → x = 28

Step 2: Median (using step-deviation / cumulative frequency table, $h = 500$)

| Class | $f$ | $cf$ |
|-------|-----|------|
| 1000–1500 | 24 | 24 |
| 1500–2000 | 40 | 64 |
| 2000–2500 | 33 | 97 |
| 2500–3000 | 28 | 125 |
| 3000–3500 | 30 | 155 |
| 3500–4000 | 22 | 177 |
| 4000–4500 | 16 | 193 |
| 4500–5000 | 7 | 200 |

$\dfrac{n}{2} = 100$. The cf just exceeding 100 is 125 (class 2500–3000).

$l = 2500,\ cf = 97,\ f = 28,\ h = 500$

$$\text{Median} = 2500 + \frac{100 - 97}{28} \times 500 = 2500 + \frac{1500}{28} = 2500 + 53.57 \approx \textbf{₹2553.57}$$

Step 3: Mean (step-deviation method, $a = 2750,\ h = 500$)

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i-2750}{500}$ | $f_i u_i$ |
|-------|--------|--------|-------------------------------|-----------|
| 1000–1500 | 24 | 1250 | –3 | –72 |
| 1500–2000 | 40 | 1750 | –2 | –80 |
| 2000–2500 | 33 | 2250 | –1 | –33 |
| 2500–3000 | 28 | 2750 | 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 | 200 | | | –35 |

$$\bar{x} = 2750 + \frac{-35}{200} \times 500 = 2750 - 87.5 = \textbf{₹2662.50}$$

Source: Chapter 13, Sections 13.2 & 13.4

---

• Finding x: always start by using $\Sigma f = n$ (total). This fetches 1 mark.
• Median: identify median class where cumulative frequency first exceeds $n/2 = 100$; cf up to 2000–2500 is 97, so the next class 2500–3000 is the median class. Apply the standard formula carefully.
• Mean: step-deviation method is easiest with large $x_i$; choose $a$ as the middle class mark. Keep signs of $u_i$ correct — here $\Sigma f_i u_i$ is negative, so mean < assumed mean.
• Show the full table; examiners award marks for the working, not just the final answer.

(a) Mode = 3 Median − 2 Mean

The empirical relationship given in the NCERT textbook (Chapter 13, Remarks after Section 13.4) is: 3 Median = Mode + 2 Mean, which rearranges to Mode = 3 Median − 2 Mean. Remember this formula exactly — it is a standard 1-mark question in board exams.

Mean (Direct Method):

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 5 | 7 | 35 |
| 10–20 | 15 | 14 | 210 |
| 20–30 | 25 | 13 | 325 |
| 30–40 | 35 | 12 | 420 |
| 40–50 | 45 | 20 | 900 |
| 50–60 | 55 | 11 | 605 |
| 60–70 | 65 | 15 | 975 |
| 70–80 | 75 | 8 | 600 |
| Total | | 100 | 4070 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{4070}{100} = \textbf{40.7}$$

---

Median:

| Class | $f_i$ | $cf$ |
|-------|--------|------|
| 0–10 | 7 | 7 |
| 10–20 | 14 | 21 |
| 20–30 | 13 | 34 |
| 30–40 | 12 | 46 |
| 40–50 | 20 | 66 |
| 50–60 | 11 | 77 |
| 60–70 | 15 | 92 |
| 70–80 | 8 | 100 |

$n = 100$, so $\dfrac{n}{2} = 50$.

The cumulative frequency just greater than 50 is 66 → Median class = 40–50.

$l = 40,\ cf = 46,\ f = 20,\ h = 10$

$$\text{Median} = l + \left(\frac{\dfrac{n}{2} - cf}{f}\right) \times h = 40 + \left(\frac{50 - 46}{20}\right) \times 10 = 40 + 2 = \textbf{42}$$

Source: Chapter 13, Sections 13.2 and 13.4

---

• Mean: Use the Direct Method — multiply each class mark $x_i$ by its frequency $f_i$, sum the products, divide by $\Sigma f_i = 100$.
• Median: Build the cumulative frequency column, find $n/2 = 50$, identify the class whose cf first exceeds 50 (that's 40–50 with cf = 66), then apply the formula. Examiners expect the full table, identification of median class, and substitution into the formula with correct values of $l$, $cf$, $f$, $h$.

(i) Median Class:

Total students = 8 + 10 + 13 + 6 + 3 = 40; N/2 = 20

Cumulative frequencies: 8, 18, 31, 37, 40

Since 20th observation falls in the class 40–60, the median class is 40–60.

---

(ii) Mean:

| Class | $x_i$ (mid-value) | $f_i$ | $f_i x_i$ |
|-------|-------------------|--------|------------|
| 0–20 | 10 | 8 | 80 |
| 20–40 | 30 | 10 | 300 |
| 40–60 | 50 | 13 | 650 |
| 60–80 | 70 | 6 | 420 |
| 80–100 | 90 | 3 | 270 |
| Total | | 40 | 1720 |

$$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1720}{40} = \textbf{43 seconds}$$

OR Mode:

Modal class = 40–60 (highest frequency = 13)

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 40 + \frac{13-10}{26-10-6} \times 20 = 40 + \frac{3}{10} \times 20 = \textbf{46 seconds}$$

---

(iii) Students who took less than 60 seconds = 8 + 10 + 13 = 31 students

Source: Statistics, Data Analysis

---

• (i) Always find cumulative frequency first; whichever class contains the N/2-th value is the median class.
• (ii) For mean, use midpoints. For mode, identify modal class (highest frequency) then apply the formula carefully — examiners check correct substitution of $f_0$, $f_1$, $f_2$.
• (iii) "Less than 60 seconds" means add frequencies of 0–20, 20–40, and 40–60 only.

Option (D) 3

Mean = $\dfrac{2 + 9 + (x+6) + (2x+3) + 5 + 10 + 5}{7} = 7$

$\Rightarrow 2x + 40 = 49 \Rightarrow 2x = 9$...

Recalculating: $2 + 9 + x + 6 + 2x + 3 + 5 + 10 + 5 = 3x + 40$

$\dfrac{3x + 40}{7} = 7 \Rightarrow 3x + 40 = 49 \Rightarrow 3x = 9 \Rightarrow x = 3$

Answer: D) 3

Sum all terms: $2 + 9 + (x+6) + (2x+3) + 5 + 10 + 5 = 3x + 40$. Set $\frac{3x+40}{7} = 7$, solve to get $x = 3$. Watch out for carefully collecting like terms ($x + 2x = 3x$) — a common slip.

Mode:

The class with the highest frequency is 35–45 (f = 23).
So, Modal class = 35–45, l = 35, f₁ = 23, f₀ = 21, f₂ = 14, h = 10.

$$\text{Mode} = l + \left(\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right) \times h = 35 + \left(\frac{23 - 21}{2(23) - 21 - 14}\right) \times 10$$

$$= 35 + \frac{2}{11} \times 10 = 35 + 1.8 = \textbf{36.8 years}$$

---

Mean (Direct Method): Taking class marks $x_i$:

| Age | $f_i$ | $x_i$ | $f_i x_i$ |
|---|---|---|---|
| 5–15 | 6 | 10 | 60 |
| 15–25 | 11 | 20 | 220 |
| 25–35 | 21 | 30 | 630 |
| 35–45 | 23 | 40 | 920 |
| 45–55 | 14 | 50 | 700 |
| 55–65 | 5 | 60 | 300 |
| Total | 80 | | 2830 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2830}{80} = \textbf{35.375 years}$$

Interpretation: The mode (36.8 years) is the age group most common among patients. The mean (35.375 years) is the average age of all patients admitted.

Source: Chapter 13, Exercise 13.2, Q.1

---

• Modal class = class with highest frequency (23), not just the highest frequency value.
• Apply the standard mode formula; examiners check correct substitution of f₀ (preceding class freq.) and f₂ (succeeding class freq.).
• For mean, the direct method is simplest here since x_i values are manageable. Use class marks (midpoints) as representatives.
• The interpretation line is required by the question ("Compare and interpret") — don't skip it even in a 5-mark answer.
• Σf must equal total patients = 80; verify this before computing mean.

(A) 12

Using the empirical relation: $3 \text{ Median} = \text{Mode} + 2\text{ Mean}$

Given: Mode − Median = 24, so Mode = Median + 24.

$$3\text{ Median} = (\text{Median} + 24) + 2\text{ Mean}$$
$$\text{Median} - \text{Mean} = 12$$

The key formula is the empirical relationship 3 Median = Mode + 2 Mean (from Chapter 13, Remarks). Rearranging gives Mode − Mean = 3(Median − Mean), and Mode − Median = 2(Median − Mean). So Median − Mean = (Mode − Median)/2 = 24/2 = 12. Remember this shortcut for MCQs.

Given data:

| Age (years) | Frequency | Cumulative Frequency |
|-------------|-----------|----------------------|
| 15–19 | 62 | 62 |
| 20–24 | 132 | 194 |
| 25–29 | 96 | 290 |
| 30–34 | 37 | 327 |
| 35–39 | 13 | 340 |
| 40–44 | 11 | 351 |
| 45–49 | 10 | 361 |
| 50–54 | 4 | 365 |

Total N = 365

(i) The highest frequency is 132, corresponding to class 20–24.
∴ Lower limit of modal class = 20

(ii) N/2 = 365/2 = 182.5
The cumulative frequency just greater than 182.5 is 194 (class 20–24).
∴ Median class = 20–24

(iii) Empirical relationship:
$$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$$

---

• Modal class is the class with the highest frequency (132); its lower limit is simply the starting value of that class.
• Median class is found by locating N/2 = 182.5 in the cumulative frequency column — the class whose cumulative frequency first exceeds this value.
• The empirical formula (Mode = 3 Median − 2 Mean) is a standard result from the Statistics chapter; students must reproduce it exactly as given in the NCERT textbook.

Option (A) 20

Sum of 5 observations = 15 × 5 = 75; Sum of first 3 = 14 × 3 = 42; Sum of last 3 = 17 × 3 = 51.
Third observation = 42 + 51 − 75 = 20.

The third observation is counted in both "first three" and "last three" groups. So: 3rd obs = (Sum of first 3) + (Sum of last 3) − (Total sum) = 42 + 51 − 75 = 20. This is a standard trick question on means.

Finding class marks and $f_i x_i$:

| Marks ($x_i$) | No. of students ($f_i$) | $f_i x_i$ |
|---|---|---|
| 5 | 12 | 60 |
| 15 | 23 | 345 |
| 25 | 34 | 850 |
| 35 | 25 | 875 |
| 45 | 6 | 270 |
| Total | 100 | 2400 |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2400}{100} = \mathbf{24}$$

The mean marks of the students is 24.

Source: Chapter 13, Section 13.2 – Mean of Grouped Data (Direct Method)

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• Class mark = (lower limit + upper limit) ÷ 2. For 0–10, class mark = 5; for 10–20, class mark = 15, etc.
• Multiply each class mark by its frequency to get $f_i x_i$, then sum all values.
• Apply the direct method formula: $\bar{x} = \dfrac{\Sigma f_i x_i}{\Sigma f_i}$.
• Since class marks and frequencies are small, the Direct Method is the appropriate and simplest choice here.
• Always show the table clearly — examiners award marks for the working, not just the final answer.

(B) mode

The mode gives the value that occurs most frequently. To know the marks obtained by the maximum number of students, the teacher needs the mode.

Mode = value with the highest frequency. The textbook states: "In situations which require establishing the most frequent value or most popular item, the mode is the best choice." Here, "marks obtained by maximum number of students" directly means the most frequently occurring marks, so mode is the correct measure — not mean (average of all) or median (middle value).

(A) increases by 2

When each observation is increased by 2, the middle value also shifts up by 2, so the median increases by 2.

Median depends on the positional value of the middle observation. If every observation increases by a constant k, the entire data shifts by k, so the median also increases by k. Here k = 2, so the answer is (A). Options (B) and (D) are incorrect; (C) would be true only if the data were unchanged.

(i) Modal Class:
The class with the highest frequency is 600 – 800 (frequency = 7).
∴ Modal class = 600 – 800

(ii) Mean of the given data:

| Class | $x_i$ (mid-value) | $f_i$ | $f_i x_i$ |
|-------|-------------------|--------|------------|
| 200–400 | 300 | 3 | 900 |
| 400–600 | 500 | 4 | 2000 |
| 600–800 | 700 | 7 | 4900 |
| 800–1000 | 900 | 4 | 3600 |
| 1000–1200 | 1100 | 3 | 3300 |
| 1200–1400 | 1300 | 3 | 3900 |
| | | $\Sigma f_i = 24$ | $\Sigma f_i x_i = 18600$ |

$$\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{18600}{24} = \textbf{775 mm}$$

(iii) Sub-divisions with at least 800 mm rainfall = 4 + 3 + 3 = 10 sub-divisions

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• (i) Modal class is simply the class with the highest frequency — just identify it, no calculation needed.
• (ii) Use Direct Method for mean: calculate mid-values, multiply by frequency, sum up, then divide by total frequency. Median would require cumulative frequency and locating the median class; mean is simpler here.
• (iii) "At least 800 mm" means classes 800–1000, 1000–1200, and 1200–1400, so add their frequencies: 4+3+3 = 10. Examiners look for correct addition of relevant classes.

Option (D) 3

Mean = $\dfrac{2+9+(x+6)+(2x+3)+5+10+5}{7} = 7$

$\Rightarrow 2+9+x+6+2x+3+5+10+5 = 49$

$\Rightarrow 3x + 40 = 49 \Rightarrow 3x = 9 \Rightarrow x = 3$

Apply the mean formula: sum of all observations ÷ total number of observations = given mean. Set up the equation, simplify by collecting like terms, and solve for x. There are 7 observations, so multiply mean (7) by 7 to get 49. Watch out for correctly expanding $(x+6)$ and $(2x+3)$ before adding.

Mean Calculation (Direct Method):

| Class | $f_i$ | $x_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 0–10 | 8 | 5 | 40 |
| 10–20 | 7 | 15 | 105 |
| 20–30 | 15 | 25 | 375 |
| 30–40 | 20 | 35 | 700 |
| 40–50 | 12 | 45 | 540 |
| 50–60 | 8 | 55 | 440 |
| 60–70 | 10 | 65 | 650 |
| Total | $\Sigma f_i = 80$ | | $\Sigma f_i x_i = 2850$ |

$$\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i} = \frac{2850}{80} = \textbf{35.625}$$

Mode Calculation:

The highest frequency is 20, corresponding to class 30–40 → Modal class = 30–40.

Here, $l = 30,\ f_1 = 20,\ f_0 = 15,\ f_2 = 12,\ h = 10$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 30 + \frac{20-15}{40-15-12} \times 10$$

$$= 30 + \frac{5}{13} \times 10 = 30 + 3.846 \approx \textbf{33.85}$$

Source: Chapter 13, Sections 13.2 and 13.3

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• Mean: Use mid-points ($x_i$) of each class; apply $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$.
• Modal class: The class with the highest frequency (20 here, i.e., 30–40).
• In the mode formula, $f_0$ = frequency of class before modal class, $f_2$ = frequency of class after modal class. Don't mix these up.
• Show the full table for Mean — examiners award step marks for correct $x_i$ and $f_i x_i$ columns.
• Round mode to 2 decimal places unless told otherwise.

Finding the missing frequency f:

| Class | $x_i$ | $f_i$ | $f_i x_i$ |
|-------|--------|--------|------------|
| 11–13 | 12 | 7 | 84 |
| 13–15 | 14 | 6 | 84 |
| 15–17 | 16 | 9 | 144 |
| 17–19 | 18 | 13 | 234 |
| 19–21 | 20 | f | 20f |
| 21–23 | 22 | 5 | 110 |
| 23–25 | 24 | 4 | 96 |
| Total | | $44 + f$ | $752 + 20f$ |

Given $\bar{x} = 18$:

$$18 = \frac{752 + 20f}{44 + f}$$

$$18(44 + f) = 752 + 20f$$

$$792 + 18f = 752 + 20f$$

$$40 = 2f \implies \boxed{f = 20}$$

Finding the Mode:

The highest frequency is 20, so the modal class is 19–21.

Here, $l = 19$, $h = 2$, $f_1 = 20$, $f_0 = 13$, $f_2 = 5$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 19 + \frac{20-13}{40-13-5} \times 2$$

$$= 19 + \frac{7}{22} \times 2 = 19 + \frac{7}{11} = 19 + 0.636 \approx \boxed{19.64}$$

Source: Chapter 13, Sections 13.2 and 13.3

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• Finding f: Use the direct method formula $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$. Set it equal to 18 and solve the linear equation for f.
• Modal class: The class with the highest frequency (here, 19–21 with f = 20) is the modal class.
• Mode formula: Examiners expect you to correctly identify $f_0$ (preceding class frequency = 13) and $f_2$ (succeeding class frequency = 5). Show substitution clearly for full marks.
• Write the table neatly — it earns method marks even if arithmetic slips occur.

Using the empirical relationship: 3 Median = Mode + 2 Mean

$3 \text{ Median} = 15x + 2(18x) = 15x + 36x = 51x$

$\text{Median} = 17x$

Answer: (C) $17x$

The key formula (given as a Remark in Section 13.4) is 3 Median = Mode + 2 Mean. Substituting Mode = 15x and Mean = 18x gives Median = 17x. Examiners expect you to recall and apply this empirical relationship directly — no derivation needed for an MCQ.

Mean (using assumed mean method, a = 52.5, h = 15):

| Class | $f_i$ | $x_i$ | $u_i = \frac{x_i - 52.5}{15}$ | $f_i u_i$ |
|-------|--------|--------|-------------------------------|-----------|
| 0–15 | 4 | 7.5 | –3 | –12 |
| 15–30 | 8 | 22.5 | –2 | –16 |
| 30–45 | 11 | 37.5 | –1 | –11 |
| 45–60 | 14 | 52.5 | 0 | 0 |
| 60–75 | 10 | 67.5 | 1 | 10 |
| 75–90 | 7 | 82.5 | 2 | 14 |
| 90–105| 6 | 97.5 | 3 | 18 |
| Total | 60 | | | 3 |

$$\bar{x} = a + \frac{\sum f_i u_i}{\sum f_i} \times h = 52.5 + \frac{3}{60} \times 15 = 52.5 + 0.75 = \textbf{53.25}$$

Mode:

The highest frequency is 14, so the modal class = 45–60.

$l = 45,\ f_1 = 14,\ f_0 = 11,\ f_2 = 10,\ h = 15$

$$\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h = 45 + \frac{14-11}{28-11-10} \times 15 = 45 + \frac{3}{7} \times 15 = 45 + 6.43 = \textbf{51.43}$$

Mean = 53.25, Mode = 51.43

Source: Chapter 13, Section 13.2 (Mean) and Section 13.3 (Mode)

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• Mean: The step-deviation (assumed mean) method is most efficient here since all class widths are equal (h = 15). Take a = 52.5 (midpoint of middle class). Calculate $u_i$ and $f_i u_i$, sum them, and apply the formula.
• Mode: Identify the modal class as the one with the highest frequency (14 → class 45–60). Then substitute $l, f_1, f_0, f_2, h$ into the mode formula. Examiners award marks for correctly identifying the modal class, writing the formula with correct substitution, and the final answer.
• Show the table neatly — 1–2 marks are often for working, not just the final answer.

Step 1: Form equations using total frequency = 90

| Class | Frequency | Cumulative Frequency (cf) |
|-------|-----------|--------------------------|
| 20–30 | p | p |
| 30–40 | 15 | p + 15 |
| 40–50 | 25 | p + 40 |
| 50–60 | 20 | p + 60 |
| 60–70 | q | p + 60 + q |
| 70–80 | 8 | p + 68 + q |
| 80–90 | 10 | p + 78 + q |

Sum of frequencies: $p + 15 + 25 + 20 + q + 8 + 10 = 90$

$$p + q + 78 = 90 \implies p + q = 12 \quad \cdots (1)$$

Step 2: Identify median class

$\dfrac{n}{2} = \dfrac{90}{2} = 45$

Median = 50, so the median class is 40–50 (cf just before it is $p + 15$, and cf after it is $p + 40$, which must be ≥ 45).

Here: $l = 40,\ f = 25,\ cf = p + 15,\ h = 10$

Step 3: Apply median formula

$$50 = 40 + \left(\frac{45 - (p+15)}{25}\right) \times 10$$

$$10 = \frac{(30 - p)}{25} \times 10$$

$$25 = 30 - p \implies p = 5$$

Step 4: Find q

From (1): $5 + q = 12 \implies q = 7$

$$\boxed{p = 5, \quad q = 7}$$

Source: Statistics, Section 13.4 Median of Grouped Data, Chapter 13

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• Two equations are needed for two unknowns: one from total frequency, one from the median formula.
• The median class is the class containing the 45th observation. Since cf up to 40–50 is $p+40$ and the median lies in this interval, apply the formula with $cf = p+15$ (cf of the preceding class).
• Examiners award marks for: setting up the cf table correctly (1 mark), forming equation (1) from total frequency (1 mark), identifying median class (1 mark), applying formula and solving for p (1 mark), finding q (1 mark).
• Always write the final boxed values clearly.

Using the empirical relationship: Mode = 3 Median − 2 Mean

Mode = 3 × 40 − 2 × 43 = 120 − 86 = 34

The correct option is A) 34.

The empirical relationship between the three measures of central tendency is:
Mode = 3 Median − 2 Mean

This formula is stated in Chapter 13. Always substitute Median first, then Mean. Here: 3(40) − 2(43) = 120 − 86 = 34. This is a standard 1-mark application question in board exams.