The following table gives the daily income of ten workers in a factory. Find the arithmetic mean. | Workers | A | B | C | D | E | F | G | H | I | J | |---|---|---|---|---|---|---|---|---|---|---| | Daily Income (in Rs) | 120 | 150 | 180 | 200 | 250 | 300 | 220 | 350 | 370 | 260 | (Ans. Rs 240)

The following table gives the daily income of ten – Class 11 Economics

Exercises · Q5

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers	A	B	C	D	E	F	G	H	I	J
Daily Income (in Rs)	120	150	180	200	250	300	220	350	370	260

(Ans. Rs 240)

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Concept understanding — Arithmetic Mean Calculation

Arithmetic Mean in Economics — A First Look

You already use the arithmetic mean every day without thinking about it. If your marks in five subjects are 72, 85, 68, 91, and 79, and someone asks "what did you average?", you add them up and divide by 5. That's the arithmetic mean. In Economics, we do exactly the same thing — but with data that matters for the whole country.

The Intuition

Imagine you want to know the "typical" monthly rent in your city. You can't ask every single tenant, so you collect rents from, say, 100 households. Some pay ₹8,000, some pay ₹25,000. The arithmetic mean gives you a single number that represents the whole group: if every household paid exactly the same rent, that common rent would be the mean.

Note

The arithmetic mean is a measure of central tendency — it tells you where the centre of your data lies. It's the most commonly used average in Economics.

The Formula

For a set of $n$ values $x_1, x_2, x_3, \dots, x_n$ , the arithmetic mean $\bar{x}$ (read as "x-bar") is:

\bar{x} = \frac{x_1 + x_2 + x_3 + \dots + x_n}{n} = \frac{\sum_{i=1}^{n} x_i}{n}

Where:

$\bar{x}$ = arithmetic mean
$x_i$ = each individual observation
$\sum$ (sigma) = "sum of"
$n$ = number of observations

For grouped data (when data is already in frequency tables), the formula becomes:

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Where $f_i$ is the frequency of each value $x_i$ , and $\sum f_i = N$ (total number of observations).

Why It Matters in Economics

Economics deals with aggregates — national income, average price levels, average wages, per capita consumption. The arithmetic mean lets us compress thousands or millions of individual numbers into one meaningful figure.

Example — Average Income: If a village has 5 families earning ₹2,000, ₹3,000, ₹4,000, ₹5,000, and ₹1,00,000 per month, the arithmetic mean is:

\bar{x} = \frac{2000 + 3000 + 4000 + 5000 + 100000}{5} = \frac{114000}{5} = ₹22,800

This ₹22,800 is the "average" income — but notice that four families earn far less than this. The single very high income pulls the mean upward. This is a critical limitation.

Watch out

The arithmetic mean is sensitive to extreme values (outliers). A single very large or very small number can distort it. In the example above, ₹22,800 doesn't represent any typical family's income. Always check your data for outliers before using the mean.

How It's Used in Your Syllabus

In Class 11 Statistics for Economics, you'll calculate the arithmetic mean for: …

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This is a simple individual series — add the ten daily incomes (total Rs 2400) and divide by 10 to get a mean daily income of Rs 240.

Concept

For an individual series:

\bar{X} = \frac{\sum X}{N}

Working

Daily incomes: 120, 150, 180, 200, 250, 300, 220, 350, 370, 260 ( $N = 10$ ). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

JKBOSE Class 11 (Commerce) 2024Set ANNUAL6 marks
$Q. Obtain the Mean, Median and Mode of the following data : Marks No. of Students 0-10 5 10-20 7 20-30 15 30-40 25 40-50 20 50-60 15 60-70 8 70-80 5$
(OR)
$Define Correlation. Explain various degrees of correlation.$
›Reveal solutionSolution

This question has two alternatives (OR); both are answered in full below.

PART 1 - Main question: Obtain the Mean, Median and Mode of the given data.

For this distribution of 100 students' marks, Mean = 40, Median = 39.2, and Mode is approximately 36.67, calculated using the step-deviation method, interpolation formula, and modal-class formula respectively.

Step 1 - Set up the table (class, frequency f, mid-value m, deviation d=(m-A)/h with assumed mean A=35, h=10, and cumulative frequency cf):

Marks f m (mid-value) d=(m-35)/10 f x d cf
0-10 5 5 -3 -15 5
10-20 7 15 -2 -14 12
20-30 15 25 -1 -15 27
30-40 25 35 0 0 52
40-50 20 45 1 20 72
50-60 15 55 2 30 87
60-70 8 65 3 24 95
70-80 5 75 4 20 100

N = Sigma(f) = 100, Sigma(f x d) = -15-14-15+0+20+30+24+20 = 50

Step 2 - Mean (step-deviation/assumed-mean method):

Mean = A + h x (Sigma(fd)/N) = 35 + 10 x (50/100) = 35 + 5 = 40

Step 3 - Median:

N/2 = 50. From the cf column, the class whose cumulative frequency first exceeds 50 is 30-40 (cf = 52), so the median class is 30-40. Here L (lower limit) = 30, cf (cumulative frequency of the class before median class) = 27, f (frequency of median class) = 25, h = 10.

Median = L + [(N/2 - cf)/f] x h = 30 + [(50-27)/25] x 10 = 30 + (23/25) x 10 = 30 + 9.2 = 39.2

Step 4 - Mode:

The highest frequency is 25, in class 30-40 - this is the modal class. L = 30, f1 (modal class frequency) = 25, f0 (frequency of class before modal class) = 15, f2 (frequency of class after modal class) = 20, h = 10.

Mode = L + [(f1-f0)/(2f1-f0-f2)] x h = 30 + [(25-15)/(2x25-15-20)] x 10 = 30 + (10/15) x 10 = 30 + 6.67 = 36.67

…

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JKBOSE Class 11 (Commerce) 2022Set ANNUAL6 marks
$Q. What is meant by Arithmetic Mean ? What are its merits and demerits ?$
(OR)
$Calculate Arithmetic mean of the following : X Y 10-20 4 20-30 7 30-40 16 40-50 20 50-60 15 60-70 8$
›Reveal solutionSolution
Arithmetic Mean = Sum(x)/N; it has clear merits (simplicity, uses all data) and demerits (sensitivity to extreme values). The alternative problem computes AM ~ 43.43 by the step-deviation method.

What is Arithmetic Mean? Merits and demerits:

Arithmetic Mean (A.M.) is the most common mathematical average, defined as the sum of all the values in a series divided by the number of values:

A.M. (x-bar) = Sum(x) / N

For grouped/frequency data: x-bar = Sum(f*x) / Sum(f)

Merits of Arithmetic Mean:

Simple to understand and easy to calculate.

Rigidly/precisely defined by a mathematical formula - no ambiguity.

Based on all the observations in the series.

Capable of further algebraic/statistical treatment (e.g. combined mean of two groups can be computed).

Least affected by sampling fluctuations compared to some other measures.

Demerits of Arithmetic Mean:

Unduly affected by extreme (very large or very small) values/outliers.

Cannot be calculated for open-ended class intervals without assumptions, or for qualitative (non-numeric) data.

It may give a value that does not actually exist in the data (e.g. an average family size of 4.4 children).

Cannot be located graphically (unlike median or mode).

In a highly skewed distribution, the mean may not be a good representative value. …
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JKBOSE Class 11 (Commerce) 2020Set ANNUAL6 marks
$Q. Find out mean and standard deviation from the following data : Size Frequency 0—2 2 2—4 4 4—6 6 6—8 4 8—10 2 10—12 6$
(OR)
$What is Correlation ? State the kinds of correlation.$
›Reveal solutionSolution

Using the step-deviation method (assumed mean A = 7, class width h = 2), the mean of the distribution is 6.5 and the standard deviation is approximately 3.28.

Step 1 — Set up the table with class midpoints (m), the step-deviation d = (m − A)/h taking assumed mean A = 7 (midpoint of the 6–8 class) and h = 2 (class width):

Size f Midpoint (m) d = (m−7)/2 fd d² fd²
0–2 2 1 −3 −6 9 18
2–4 4 3 −2 −8 4 16
4–6 6 5 −1 −6 1 6
6–8 4 7 0 0 0 0
8–10 2 9 1 2 1 2
10–12 6 11 2 12 4 24
Total N = 24 Σfd = −6 Σfd² = 66

Step 2 — Mean (step-deviation method):

Mean = A + h × (Σfd / N) = 7 + 2 × (−6 / 24) = 7 + 2 × (−0.25) = 7 − 0.5 = 6.5

Step 3 — Standard Deviation (step-deviation method):

σ = h × √[ (Σfd²/N) − (Σfd/N)² ]

= 2 × √[ (66/24) − (−6/24)² ]

= 2 × √[ 2.75 − 0.0625 ]

= 2 × √2.6875

= 2 × 1.6394

= 3.28 (approx.)

Check: Σfd²/N (2.75) is comfortably larger than (Σfd/N)² (0.0625), as it must be for a valid variance, and the final SD (3.28) is a reasonable fraction of the overall data range (0 to 12), confirming the computation is consistent.

…

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Marks	No. of Students
0-10	5
10-20	7
20-30	15
30-40	25
40-50	20
50-60	15
60-70	8
70-80	5

Marks	f	m (mid-value)	d=(m-35)/10	f x d	cf
0-10	5	5	-3	-15	5
10-20	7	15	-2	-14	12
20-30	15	25	-1	-15	27
30-40	25	35	0	0	52
40-50	20	45	1	20	72
50-60	15	55	2	30	87
60-70	8	65	3	24	95
70-80	5	75	4	20	100

X	Y
10–20	4
20–30	7
30–40	16
40–50	20
50–60	15
60–70	8

Size	Frequency
0—2	2
2—4	4
4—6	6
6—8	4
8—10	2
10—12	6

Size	f	Midpoint (m)	d = (m−7)/2	fd	d²	fd²
0–2	2	1	−3	−6	9	18
2–4	4	3	−2	−8	4	16
4–6	6	5	−1	−6	1	6
6–8	4	7	0	0	0	0
8–10	2	9	1	2	1	2
10–12	6	11	2	12	4	24
Total	N = 24			Σfd = −6		Σfd² = 66

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