Unit 6 Chapter 2 Measures of Central Tendency: Class 10 Economics Notes | Unit 6 Chapter 2 SEE Guide Nepal

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Unit 6: Quantitative Methods in Economics

Class 10 Economics

Unit 6 Chapter 2 Measures of Central Tendency

For SEE board exam preparation: Complete theoretical notes and fully solved numerical exercises on Mean, Median, and Mode

Welcome to the complete study guide on Unit 6 Chapter 2 Measures of Central Tendency under Quantitative Methods in Economics. This is Chapter 2 of Unit 6 for Class 10 Economics students in Nepal preparing for their SEE board exams.

Here you will find structured theoretical notes on Mean (Arithmetic Mean), Median, and Mode for Individual, Discrete, and Continuous Series — along with all fully solved textbook numerical problems.

Explore our complete study list here: Class 10 Economics Notes.

1. Theoretical Concepts

1. Introduction: Unit 6 Chapter 2 Measures of Central Tendency

A value that lies between the minimum and maximum values of a dataset and represents all values in a statistical series is called a Measure of Central Tendency. These measures include Mean, Median, and Mode. In this chapter, we study the calculation of the Arithmetic Mean, Median, and Mode under Measures of Central Tendency.

2.1 Mean (Arithmetic Mean)

The Mean is calculated by dividing the sum of all values in a dataset by the total number of values. It represents the average value of the data. Among the measures of central tendency, the Mean is the most widely used and easiest to calculate. It is also called the Arithmetic Mean. In economics, it is used to find the average value of income, consumption, expenditure, production, and similar variables. The methods of calculating the Arithmetic Mean for Individual, Discrete, and Continuous Series are presented below.

(a) Individual Series

A series in which the value of each variable or item is given separately is called an Individual Series. Individual series data does not have frequencies. The following formula is used to calculate the simple Arithmetic Mean from an individual series:

$$\overline{X} = \frac{\Sigma X}{n}$$

where,

$\overline{X}$ = Arithmetic Mean

$\Sigma X$ = Sum of all observations

$n$ = Total number of observations

(b) Discrete Series

If a series gives both the values of the variable and their corresponding frequencies, it is called a Discrete Series. The following formula is used to find the Arithmetic Mean from a discrete series:

$$\overline{X} = \frac{\Sigma fx}{N}$$

where,

$\overline{X}$ = Arithmetic Mean

$f$ = Frequency

$x$ = Value of the variable

$\Sigma fx$ = Sum of products of each value and its corresponding frequency

$N = \Sigma f$ = Sum of all frequencies

(c) Continuous Series

In a continuous series, the values of variables are given in the form of class intervals, each with a corresponding frequency. To find the simple Arithmetic Mean of a continuous series, the mid-value of each class interval must first be calculated. The following formula is used with the direct method:

$$\overline{X} = \frac{\Sigma fm}{N}$$

where,

$\overline{X}$ = Arithmetic Mean

$m$ = Mid-value of the class interval

$f$ = Frequency

$\Sigma fm$ = Sum of products of mid-values and their corresponding frequencies

$N = \Sigma f$ = Sum of all frequencies

2.2 Median

The value that lies exactly in the middle of a series or dataset is called the Median. When calculating the median, the given data must first be arranged in ascending or descending order. It divides any series into two equal halves. Since the median identifies the average based on the positional location of values in an ordered series rather than their numerical size, it is also called the Positional Average. When there are large gaps between values in a dataset, the mean may not give a reliable average — in such cases, the median is used. In economics, it is used to study the distribution of income and wealth, economic inequality, and similar topics.

(a) Individual Series

To calculate the median from an individual series, first arrange the given data in ascending or descending order, then apply the following formula:

$$\text{Position of Median } (M_d) = \left(\frac{n+1}{2}\right)^{\text{th}} \text{ item}$$

where $n$ = Total number of observations.

Two situations arise when calculating the median from an individual series:

i. When the total number of items is odd: When the total number of items is odd, the middle item of the arranged series directly gives the Median.

ii. When the total number of items is even: When the total number of items is even, the two middle items of the series are added and divided by 2 to find the Median.

(b) Discrete Series

To calculate the median from a discrete series, first arrange the data values in ascending or descending order (with corresponding frequencies arranged accordingly), then calculate the cumulative frequency ($c.f.$) and apply the following formula:

$$\text{Position of Median } (M_d) = \left(\frac{N+1}{2}\right)^{\text{th}} \text{ item’s value}$$

After finding the position, look at the cumulative frequency column for the value equal to or just greater than the median position — the value of the variable corresponding to that cumulative frequency is the Median.

(c) Continuous Series

To find the median from a continuous series, cumulative frequencies are calculated as in the discrete series. Then the median class is identified using:

$$\text{Position of } M_d = \left(\frac{N}{2}\right)^{\text{th}} \text{ item}$$

After identifying the median class, the median is calculated using:

$$M_d = L + \frac{\frac{N}{2} – c.f.}{f} \times i$$

where,

$L$ = Lower limit of the median class interval

$\frac{N}{2}$ = Position of the median

$c.f.$ = Cumulative frequency of the class interval preceding the median class

$f$ = Frequency of the median class interval

$i$ = Size (width) of the class interval

2.3 Mode

The value that appears most frequently in any dataset or series is called the Mode. In other words, the value with the highest frequency in a dataset is the Mode. However, it is not always true that the most frequently repeated item is necessarily the mode — sometimes two or more values may appear equally often, in which case a different method is used to calculate the Mode. In economics, it is used in areas such as identifying consumer behaviour and market analysis.

(a) Individual Series

A series presented without corresponding frequencies is called an individual series. The Mode from an individual series is calculated as follows:

i. Unimodal (single mode): Among the values in the data, whichever value appears most often is the Mode.

ii. Bimodal or Multimodal: When two or more values appear equally often, the following empirical formula is used to calculate the Mode:

$$\text{Mode} = 3 \times \text{Median} – 2 \times \text{Mean}$$

(b) Discrete Series

A series presented with the frequencies of each value is called a discrete series. In a discrete series, the value with the highest frequency is the Mode.

(c) Continuous Series

In a continuous series, the Mode lies within the class interval that has the highest frequency. The following formula is used to calculate the Mode from a continuous series:

$$M_o = L + \left(\frac{f_1 – f_0}{2f_1 – f_0 – f_2}\right) \times i$$

where,

$L$ = Lower limit of the modal class

$f_0$ = Frequency of the class preceding the modal class

$f_1$ = Frequency of the modal class

$f_2$ = Frequency of the class succeeding the modal class

$i$ = Class interval (width)

Complete Solved Exercise

Question 1: Answer the following questions:

(a) Write the formula for calculating the Arithmetic Mean from an Individual Series.
Answer: The formula for calculating the Arithmetic Mean ($\overline{X}$) from an individual series is: $$\overline{X} = \frac{\Sigma X}{n}$$ where,

$\overline{X}$ = Arithmetic Mean

$\Sigma X$ = Sum of all observations

$n$ = Total number of observations

(b) Calculate the Arithmetic Mean from the given data:

i. Weight: 25, 27, 23, 20, 10
Solution: The weight values ($X$): $25, 27, 23, 20, 10$. Total number of observations ($n$) $= 5$
Sum of weights ($\Sigma X$) $= 25 + 27 + 23 + 20 + 10 = 105$
Applying the formula: $$\overline{X} = \frac{\Sigma X}{n} = \frac{105}{5} = 21$$ Therefore, the Arithmetic Mean weight $= \mathbf{21}$.

ii. Age: 15, 17, 19, 16, 18
Solution: The age values ($X$): $15, 17, 19, 16, 18$. Total observations ($n$) $= 5$
Sum of ages ($\Sigma X$) $= 15 + 17 + 19 + 16 + 18 = 85$
$$\overline{X} = \frac{85}{5} = 17$$ Therefore, the Arithmetic Mean age $= \mathbf{17}$ years.

iii. Calculate the Arithmetic Mean for the weights of 10 female students in Class 10.
Weight: 45, 38, 38, 42, 32, 30, 33, 39, 42, 45
Solution: Values ($X$): $45, 38, 38, 42, 32, 30, 33, 39, 42, 45$. Total ($n$) $= 10$
Sum ($\Sigma X$) $= 45 + 38 + 38 + 42 + 32 + 30 + 33 + 39 + 42 + 45 = 384$
$$\overline{X} = \frac{384}{10} = 38.4$$ Therefore, the average weight of the female students $= \mathbf{38.4}$ kg.

iv. Calculate the average temperature from the table below:

Day	Sunday	Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
Temperature (°C)	15°	16°	16°	17°	17°	14°	12°

Solution: Values ($X$): $15, 16, 16, 17, 17, 14, 12$. Total days ($n$) $= 7$
Sum ($\Sigma X$) $= 15 + 16 + 16 + 17 + 17 + 14 + 12 = 107$
$$\overline{X} = \frac{107}{7} \approx 15.29$$ Therefore, the average temperature $\approx \mathbf{15.29°C}$.

v. If the Mean of the series 12, 18, 25, 15, x, 30 is 20, find the value of x.
Solution: Series: $12, 18, 25, 15, x, 30$. Total items ($n$) $= 6$; Mean ($\overline{X}$) $= 20$
$\Sigma X = 12 + 18 + 25 + 15 + x + 30 = 100 + x$
$$20 = \frac{100 + x}{6}$$ $$120 = 100 + x$$ $$\boxed{x = 20}$$

(c) Why is the Median called the Positional Average?
Answer: When any dataset or series is arranged in ascending or descending order, the value of the item at the exact middle (centre) of the series is the Median. Rather than basing the average on the numerical size of the values, the Median determines the average based on the position (location) of the value in the ordered series. Since the Median divides the series into two equal halves and represents a specific positional value in the ordered series, it is called the Positional Average.

(d) Calculate the Median from the given data:

i. 10, 18, 19, 20, 22
Solution: Data in ascending order: $10, 18, 19, 20, 22$. Total items ($n$) $= 5$ (odd)
$$M_d = \left(\frac{5+1}{2}\right)^{\text{th}} \text{ item} = 3^{\text{rd}} \text{ item} = 19$$ Therefore, Median $(M_d) = \mathbf{19}$.

ii. 20, 18, 15, 12, 10, 7
Solution: Data in ascending order: $7, 10, 12, 15, 18, 20$. Total items ($n$) $= 6$ (even)
$$M_d = \left(\frac{6+1}{2}\right)^{\text{th}} \text{ item} = 3.5^{\text{th}} \text{ item}$$ $$M_d = \frac{3^{\text{rd}} \text{ item} + 4^{\text{th}} \text{ item}}{2} = \frac{12 + 15}{2} = 13.5$$ Therefore, Median $(M_d) = \mathbf{13.5}$.

iii. 30, 40, 50, 45, 60, 55, 65, 70, 75
Solution: Data in ascending order: $30, 40, 45, 50, 55, 60, 65, 70, 75$. Total items ($n$) $= 9$ (odd)
$$M_d = \left(\frac{9+1}{2}\right)^{\text{th}} \text{ item} = 5^{\text{th}} \text{ item} = 55$$ Therefore, Median $(M_d) = \mathbf{55}$.

(e) Calculate the Mode from the given data:

i. 10, 12, 19, 19, 22, 15
Solution: The value $19$ appears most often (twice).
Therefore, Mode $(M_o) = \mathbf{19}$.

ii. 20, 10, 18, 15, 10, 12, 10, 7, 7
Solution: The value $10$ appears most often (three times).
Therefore, Mode $(M_o) = \mathbf{10}$.

iii.

Age of Student ($X$)	22	23	24	25	29	31
No. of Students ($f$)	10	20	28	40	22	16

Solution: The highest frequency is $f = 40$, corresponding to age $X = 25$ years.
Therefore, Mode $(M_o) = \mathbf{25}$ years.

iv.

Marks ($X$)	20	22	25	30	34	40
No. of Students ($f$)	2	3	4	7	5	3

Solution: The highest frequency is $f = 7$, corresponding to marks $X = 30$.
Therefore, Mode $(M_o) = \mathbf{30}$.

Income Rs. (thousands) ($X$)	200	220	205	300	340	400
No. of Students ($f$)	20	23	44	35	25	13

Solution: The highest frequency is $f = 44$, corresponding to income $X = 205$ thousand.
Therefore, Mode $(M_o) = \mathbf{Rs.\ 205\ thousand}$ (i.e., Rs. 2,05,000).

(f) If the Mean and Median of a series are 22 and 25 respectively, find the value of the Mode.
Solution: Mean $(\overline{X}) = 22$; Median $(M_d) = 25$
Using the empirical formula: $$M_o = 3 \times M_d – 2 \times \overline{X}$$ $$M_o = 3(25) – 2(22) = 75 – 44 = 31$$ Therefore, Mode $= \mathbf{31}$.

Question 2: Solve the following problems:

(a) Calculate the Mean from the given data:

i.

$X$	5	10	15	20
$f$	2	4	3	1

Solution:

$X$	$f$	$fx$
5	2	10
10	4	40
15	3	45
20	1	20
Total	$N=10$	$\Sigma fx=115$

$$\overline{X} = \frac{\Sigma fx}{N} = \frac{115}{10} = \mathbf{11.5}$$

ii.

Age ($X$)	20	30	40	50	60
Number ($f$)	3	5	4	2	1

Solution:

Age ($X$)	Number ($f$)	$fx$
20	3	60
30	5	150
40	4	160
50	2	100
60	1	60
Total	$N=15$	$\Sigma fx=530$

$$\overline{X} = \frac{530}{15} \approx \mathbf{35.33} \text{ years}$$

iii.

Marks ($X$)	20	19	25	31	18	15
No. of Students ($f$)	3	7	4	12	10	5

Solution:

Marks ($X$)	Students ($f$)	$fx$
20	3	60
19	7	133
25	4	100
31	12	372
18	10	180
15	5	75
Total	$N=41$	$\Sigma fx=920$

$$\overline{X} = \frac{920}{41} \approx \mathbf{22.44}$$

iv.

Income Rs. (thousands) ($X$)	20	25	35	50	60	75
No. of Families ($f$)	3	12	8	5	10	7

Solution:

Income ($X$)	Families ($f$)	$fx$
20	3	60
25	12	300
35	8	280
50	5	250
60	10	600
75	7	525
Total	$N=45$	$\Sigma fx=2015$

$$\overline{X} = \frac{2015}{45} \approx \mathbf{Rs.\ 44.78\ thousand}$$

(b) The Mean of the following data is 18. Find the value of f:

Expenditure ($X$)	10	15	20	25
Frequency ($f$)	4	$f$	6	2

Solution:

$X$	$f$	$fx$
10	4	40
15	$f$	$15f$
20	6	120
25	2	50
Total	$N=12+f$	$\Sigma fx=210+15f$

Given: Mean $(\overline{X}) = 18$
$$18 = \frac{210 + 15f}{12 + f}$$ $$18(12 + f) = 210 + 15f$$ $$216 + 18f = 210 + 15f$$ $$3f = 6 \implies \boxed{f = 2}$$

(c) Calculate the Mean from the given data (monthly saving in Rs. thousands):

Monthly Saving Rs. (thousands) ($X$)	5	10	15	20	25	30	35
No. of Families ($f$)	6	12	18	10	6	4	2

Solution:

Saving ($X$)	Families ($f$)	$fx$
5	6	30
10	12	120
15	18	270
20	10	200
25	6	150
30	4	120
35	2	70
Total	$N=58$	$\Sigma fx=960$

$$\overline{X} = \frac{960}{58} \approx \mathbf{Rs.\ 16.55\ thousand}$$

(d) The table shows the monthly income of 50 households. Calculate the average household income.

Income Rs. (thousands) ($X$)	15	18	20	12	30	16	35
No. of Households ($f$)	7	10	5	8	6	10	4

Solution:

Income ($X$)	Households ($f$)	$fx$
15	7	105
18	10	180
20	5	100
12	8	96
30	6	180
16	10	160
35	4	140
Total	$N=50$	$\Sigma fx=961$

$$\overline{X} = \frac{961}{50} = \mathbf{Rs.\ 19.22\ thousand}$$

(e) The table shows the number of students and marks obtained. Calculate the Arithmetic Mean.

Marks Obtained ($X$)	65	38	40	52	50	46	35
No. of Students ($f$)	4	8	5	3	2	7	6

Solution:

Marks ($X$)	Students ($f$)	$fx$
65	4	260
38	8	304
40	5	200
52	3	156
50	2	100
46	7	322
35	6	210
Total	$N=35$	$\Sigma fx=1552$

$$\overline{X} = \frac{1552}{35} \approx \mathbf{44.34}$$

(f) The Mean of the following data is 25. Find the value of k.

$X$	20	30	40	$k$	60	70
$f$	4	6	7	8	10	4

Solution:

$X$	$f$	$fx$
20	4	80
30	6	180
40	7	280
$k$	8	$8k$
60	10	600
70	4	280
Total	$N=39$	$1420+8k$

Note: Using Mean $\approx 46.67$ (the textbook answer gives $k = 50$):
$$\frac{1420 + 8k}{39} = 46.667$$ $$1420 + 8k = 1820$$ $$8k = 400 \implies \boxed{k = 50}$$

(g) Calculate the average daily wages of families from the table (wages in Rs. hundreds):

Daily Wages Rs. (hundreds)	100–200	200–300	300–400	400–500
No. of Families ($f$)	5	9	12	4

Solution: This is a Continuous Series. Mid-values ($m$) must be calculated.

Daily Wages	Mid-value ($m$)	Families ($f$)	$fm$
100–200	150	5	750
200–300	250	9	2250
300–400	350	12	4200
400–500	450	4	1800
Total		$N=30$	$\Sigma fm=9000$

$$\overline{X} = \frac{9000}{30} = \mathbf{300} \text{ (hundreds)}$$ Therefore, average daily wages $= \mathbf{Rs.\ 300}$ (hundred) i.e., Rs. 30,000.

(h) Calculate the average monthly income of families from the table (income in Rs. thousands):

Monthly Income Rs. (thousands)	10–20	20–30	30–40	40–50	50–60
No. of Families ($f$)	6	14	20	8	2

Solution:

Income	Mid-value ($m$)	Families ($f$)	$fm$
10–20	15	6	90
20–30	25	14	350
30–40	35	20	700
40–50	45	8	360
50–60	55	2	110
Total		$N=50$	$\Sigma fm=1610$

$$\overline{X} = \frac{1610}{50} = \mathbf{Rs.\ 32.2\ thousand}$$

(i) Calculate the average monthly expenditure of families from the less-than cumulative frequency table:

Monthly Expenditure (Rs. thousands)	Families ($c.f.$)
Less than 5	6
Less than 7	20
Less than 9	30
Less than 11	40
Less than 13	45

Solution: Converting to ordinary frequency table:

Class	Mid-value ($m$)	Freq. ($f$)	$fm$
3–5	4	6	24
5–7	6	14	84
7–9	8	10	80
9–11	10	10	100
11–13	12	5	60
Total		$N=45$	$\Sigma fm=348$

$$\overline{X} = \frac{348}{45} \approx \mathbf{Rs.\ 7.73\ thousand}$$

(j) Calculate the average monthly expenditure from the more-than cumulative frequency table:

Monthly Expenditure (Rs. thousands)	Families ($c.f.$)
More than 20	50
More than 30	30
More than 40	22
More than 50	16
More than 60	6

Solution: Converting to ordinary frequency table:

Class	Mid-value ($m$)	Freq. ($f$)	$fm$
20–30	25	20	500
30–40	35	8	280
40–50	45	6	270
50–60	55	10	550
60–70	65	6	390
Total		$N=50$	$\Sigma fm=1990$

$$\overline{X} = \frac{1990}{50} = \mathbf{Rs.\ 39.8\ thousand}$$

(k) Calculate the Median from the given data:

i. Number of children per family:

No. of Children ($X$)	0	1	2	3	4	5
No. of Families ($f$)	3	7	10	5	2	1

Solution:

Children ($X$)	Families ($f$)	$c.f.$
0	3	3
1	7	10
2	10	20
3	5	25
4	2	27
5	1	28
Total	$N=28$	—

$$M_d \text{ position} = \frac{N+1}{2} = \frac{29}{2} = 14.5^{\text{th}} \text{ item}$$ The $c.f.$ equal to or just greater than $14.5$ is $20$, corresponding to $X = 2$.
Therefore, Median $(M_d) = \mathbf{2}$.

ii. Daily expenditure of students:

Daily Expenditure Rs. ($X$)	100	120	150	180	200
No. of Students ($f$)	5	8	12	7	3

Solution:

Expenditure ($X$)	Students ($f$)	$c.f.$
100	5	5
120	8	13
150	12	25
180	7	32
200	3	35
Total	$N=35$	—

$$M_d \text{ position} = \frac{36}{2} = 18^{\text{th}} \text{ item}$$ The $c.f.$ just greater than $18$ is $25$, corresponding to $X = 150$.
Therefore, Median daily expenditure $= \mathbf{Rs.\ 150}$.

iii. Marks of students:

Marks ($X$)	10	12	15	18	20	22
No. of Students ($f$)	4	6	8	5	3	2

Solution: Building the $c.f.$ table: $4, 10, 18, 23, 26, 28$. Total $N = 28$.
$$M_d \text{ position} = 14.5^{\text{th}} \text{ item}$$ $c.f.$ just greater than $14.5$ is $18$, corresponding to $X = 15$.
Therefore, Median marks $= \mathbf{15}$.

iv. Daily income of workers:

Daily Income Rs. ($X$)	500	1000	1200	1880	2300	3000
No. of Workers ($f$)	3	5	12	8	6	1

Solution: $c.f.$: $3, 8, 20, 28, 34, 35$. Total $N = 35$.
$$M_d \text{ position} = 18^{\text{th}} \text{ item}$$ $c.f.$ just greater than $18$ is $20$, corresponding to $X = 1200$.
Therefore, Median daily income $= \mathbf{Rs.\ 1200}$.

v. Marks (Continuous Series):

Marks	0–10	10–20	20–30	30–40	40–50
Frequency ($f$)	5	10	15	8	2

Solution: $c.f.$: $5, 15, 30, 38, 40$. $N = 40$.
Median position $= \frac{40}{2} = 20^{\text{th}}$ item → Median class: $20–30$ (since $c.f. = 30$ is the first $c.f. \geq 20$).
$L = 20$, $c.f. = 15$, $f = 15$, $i = 10$
$$M_d = 20 + \frac{20 – 15}{15} \times 10 = 20 + 3.33 = \mathbf{23.33}$$

vi. Annual income of farmers (Rs. lakhs):

Annual Income (Rs. lakhs)	1–2	2–3	3–4	4–5	5–6
No. of Farmers ($f$)	7	11	15	5	2

Solution: $c.f.$: $7, 18, 33, 38, 40$. $N = 40$.
Median position $= 20^{\text{th}}$ item → Median class: $3–4$.
$L = 3$, $c.f. = 18$, $f = 15$, $i = 1$
$$M_d = 3 + \frac{20 – 18}{15} \times 1 = 3 + 0.133 \approx \mathbf{3.13} \text{ lakhs}$$

vii. Marks — Inclusive Class Intervals:

Marks	5–9	10–14	15–19	20–24	25–29	30–34
Frequency ($f$)	5	7	10	15	5	4

Solution: Converting inclusive to exclusive (subtract 0.5 from lower, add 0.5 to upper limits):
Exclusive classes: $4.5–9.5$, $9.5–14.5$, $14.5–19.5$, $19.5–24.5$, $24.5–29.5$, $29.5–34.5$
$c.f.$: $5, 12, 22, 37, 42, 46$. $N = 46$.
Median position $= 23^{\text{rd}}$ item → Median class: $19.5–24.5$.
$L = 19.5$, $c.f. = 22$, $f = 15$, $i = 5$
$$M_d = 19.5 + \frac{23 – 22}{15} \times 5 = 19.5 + 0.333 \approx \mathbf{19.83}$$

viii. Weight of students — more-than cumulative frequency:

Weight	No. of Students
More than 30	50
More than 35	35
More than 40	20
More than 45	15
More than 50	5

Solution: Converting to ordinary frequency and $c.f.$:

Class	$f$	$c.f.$
30–35	15	15
35–40	15	30
40–45	5	35
45–50	10	45
50–55	5	50
Total	$N=50$	—

Median position $= 25^{\text{th}}$ item → Median class: $35–40$.
$L = 35$, $c.f. = 15$, $f = 15$, $i = 5$
$$M_d = 35 + \frac{25 – 15}{15} \times 5 = 35 + 3.33 = \mathbf{38.33}$$

(l) Age distribution of 100 people — more-than cumulative frequency. Calculate the Median.

Age	Number
More than 10	100
More than 20	80
More than 30	65
More than 40	43
More than 50	20
More than 60	12
More than 70	2

Solution: Converting and building $c.f.$:

Class	$f$	$c.f.$
10–20	20	20
20–30	15	35
30–40	22	57
40–50	23	80
50–60	8	88
60–70	10	98
70–80	2	100
Total	$N=100$	—

Median position $= 50^{\text{th}}$ item → Median class: $30–40$.
$L = 30$, $c.f. = 35$, $f = 22$, $i = 10$
$$M_d = 30 + \frac{50 – 35}{22} \times 10 = 30 + 6.82 \approx \mathbf{36.82}$$

(m) Marks — less-than cumulative frequency. Calculate the Median.

Marks	No. of Students
Less than 20	5
Less than 30	7
Less than 40	22
Less than 50	30
Less than 60	35
Less than 70	40

Solution: Converting to ordinary frequency and $c.f.$: $f$: $5, 2, 15, 8, 5, 5$; $c.f.$: $5, 7, 22, 30, 35, 40$. $N = 40$.
Median position $= 20^{\text{th}}$ item → Median class: $30–40$.
$L = 30$, $c.f. = 7$, $f = 15$, $i = 10$
$$M_d = 30 + \frac{20 – 7}{15} \times 10 = 30 + 8.67 = \mathbf{38.67}$$

(n) Calculate the Mode from the given data:

i. Present the following data in a frequency table and find the Mode:
12, 10, 13, 14, 12, 11, 15, 13, 12, 10, 14, 13, 12, 11, 15, 12, 13, 14, 11, 12
Solution:

Value ($X$)	Frequency ($f$)
10	2
11	3
12	6
13	4
14	3
15	2

The highest frequency is $f = 6$, corresponding to $X = 12$.
Therefore, Mode $(M_o) = \mathbf{12}$.

ii. Calculate the Mode from the continuous series:

Marks	0–10	10–20	20–30	30–40	40–50
Frequency ($f$)	5	10	15	8	2

Solution: Modal class $= 20–30$ (highest $f = 15$).
$L = 20$, $f_1 = 15$, $f_0 = 10$, $f_2 = 8$, $i = 10$
$$M_o = 20 + \frac{15-10}{2(15)-10-8} \times 10 = 20 + \frac{5}{12} \times 10 = 20 + 4.17 = \mathbf{24.17}$$

iii. Daily wages (Rs. hundreds):

Daily Wages (Rs. hundreds)	100–200	200–300	300–400	400–500	500–600
No. of Families ($f$)	5	9	12	15	13

Solution: Modal class $= 400–500$ (highest $f = 15$).
$L = 400$, $f_1 = 15$, $f_0 = 12$, $f_2 = 13$, $i = 100$
$$M_o = 400 + \frac{15-12}{2(15)-12-13} \times 100 = 400 + \frac{3}{5} \times 100 = 400 + 60 = \mathbf{460}$$ Therefore, Mode $= \mathbf{Rs.\ 460}$ (hundred).

iv. Marks series:

Marks	0–5	5–10	10–15	15–20	20–25	25–30
Frequency ($f$)	8	10	10	6	12	5

Solution: Modal class $= 20–25$ (highest $f = 12$).
$L = 20$, $f_1 = 12$, $f_0 = 6$, $f_2 = 5$, $i = 5$
$$M_o = 20 + \frac{12-6}{2(12)-6-5} \times 5 = 20 + \frac{6}{13} \times 5 = 20 + 2.31 = \mathbf{22.31}$$

v. Wages (Rs. hundreds):

Wages (Rs. hundreds)	100–200	200–300	300–400	400–500	500–600
No. of Families ($f$)	5	6	12	7	2

Solution: Modal class $= 300–400$ (highest $f = 12$).
$L = 300$, $f_1 = 12$, $f_0 = 6$, $f_2 = 7$, $i = 100$
$$M_o = 300 + \frac{12-6}{2(12)-6-7} \times 100 = 300 + \frac{6}{11} \times 100 = 300 + 54.55 = \mathbf{354.55}$$ Therefore, Mode wages $= \mathbf{Rs.\ 354.55}$ (hundred).