Measures of Central Tendency | MGT 202 Business Statistics

TU BBS MGT 202 Unit 3: arithmetic, geometric and harmonic means, median, partition values, mode, formulas and exam numericals.

Oct 28, 2026

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This unit covers the geometric mean, the harmonic mean (simple and weighted) and the mode, with worked numericals, followed by board exam questions on measures of central tendency listed by exam year.

Geometric Mean

Formulas

Individual data:

$$\text{G.M.} = \operatorname{Antilog}\left(\frac{\sum \log x}{n}\right)$$

Discrete and continuous data:

$$\text{G.M.} = \operatorname{Antilog}\left(\frac{\sum f\log x}{N}\right)$$

where $N = \sum f$, $f$ = frequency, and $x$ is the mid-value of the class for continuous data.

Combined Geometric Mean

$$\log G = \frac{N_{1}\log G_{1} + N_{2}\log G_{2}}{N_{1} + N_{2}}$$

Worked Examples

Example 1: Discrete series

Calculate the G.M. from the following data.

Value (x)	Frequency (f)	log x	f log x
10	20	1.0000	20.0000
12	28	1.0792	30.2176
15	30	1.1761	35.2830
18	27	1.2553	33.8931
20	25	1.3010	32.5250
	N = 130		$\sum f\log x = 151.9187$

$$\begin{aligned} \text{G.M.} &= \operatorname{Antilog}\left(\frac{\sum f\log x}{N}\right) = \operatorname{Antilog}\left(\frac{151.9187}{130}\right) \\ &= \operatorname{Antilog}(1.1686) \\ &= 14.74 \end{aligned}$$

Example 2: Continuous series

Find the geometric mean of the following distribution.

Percent increase	No. of companies (f)	Mid-value (x)	log x	f log x
10–20	20	15	1.1761	23.5220
20–30	22	25	1.3979	30.7538
30–40	18	35	1.5441	27.7938
40–50	10	45	1.6532	16.5320
	N = 70			$\sum f\log x = 98.6016$

$$\begin{aligned} \text{G.M.} &= \operatorname{Antilog}\left(\frac{98.6016}{70}\right) = \operatorname{Antilog}(1.4086) \\ &= 25.62 \end{aligned}$$

Harmonic Mean

Formulas

Individual data:

$$\text{H.M.} = \frac{n}{\sum \frac{1}{x}}$$

where $n$ = number of observations.

Discrete and continuous data:

$$\text{H.M.} = \frac{N}{\sum f\cdot\frac{1}{x}} = \frac{N}{\sum \frac{f}{x}}$$

where $N = \sum f$, $f$ = frequency, and $x$ is the mid-value of the class for continuous data.

Weighted harmonic mean:

$$\text{H.M.}_{w} = \frac{\sum w}{\sum \frac{w}{x}}$$

Worked Examples

Example 1: Individual data

Find the H.M. of the rate of production (per hour): 20, 25, 28, 30, 35.

x	1/x
20	0.0500
25	0.0400
28	0.0357
30	0.0333
35	0.0286
n = 5	$\sum \frac{1}{x} = 0.1876$

$$\text{H.M.} = \frac{n}{\sum \frac{1}{x}} = \frac{5}{0.1876} = 26.65$$

Example 2: Continuous series (average speed)

Find the average speed of the cars.

Speed (km/hr)	Mid-value (x)	No. of cars (f)	f/x
40–60	50	20	0.4000
60–80	70	30	0.4286
80–100	90	20	0.2222
100–120	110	10	0.0909
		N = 80	$\sum \frac{f}{x} = 1.1417$

$$\text{H.M.} = \frac{N}{\sum \frac{f}{x}} = \frac{80}{1.1417} = 70.07\ \text{km/hr}$$

Example 3: Weighted harmonic mean

A cyclist covers his first three miles at an average speed of 8 mph, another two miles at 3 mph and the last two miles at 2 mph. Find the average speed for the entire journey.

Speed (x)	Distance (w)	1/x	w/x
8	3	0.125	0.3750
3	2	0.333	0.6667
2	2	0.500	1.0000
	$\sum w = 7$		$\sum \frac{w}{x} = 2.0417$

$$\text{H.M.}_{w} = \frac{\sum w}{\sum \frac{w}{x}} = \frac{7}{2.0417} = 3.43\ \text{mph}$$

Mode

A variate value which is repeated the maximum number of times in the given data is called the mode ($M_0$). It is mainly used in the field of business and fashion.

A set of data having two mode values is called bimodal.
A set of data having more than two mode values is called multi-modal.

Formula for Continuous Series

$$\begin{aligned} M_{0} &= L + \frac{\Delta_{1}}{\Delta_{1}+\Delta_{2}} \times h \\ \Delta_{1} &= f_{1} - f_{0}, \qquad \Delta_{2} = f_{1} - f_{2} \end{aligned}$$

where $L$ = lower limit of the modal class, $f_1$ = frequency of the modal class, $f_0$ = frequency of the class preceding it, $f_2$ = frequency of the class succeeding it, and $h$ = class width. The modal class is the class corresponding to the maximum frequency.

Relation between Mean, Median and Mode

$$M_{0} = 3M_{d} - 2\bar{x}$$

Worked Examples

Example 1: Mode after regrouping (unequal classes)

Calculate the most repeated value after regrouping the frequency distribution given below.

Value	0–10	10–20	20–40	40–50	50–60	60–80	80–100
Frequency	10	15	50	20	10	20	5

Since the given distribution has unequal class sizes, adjustment is required. Regrouping into classes of width 20:

Value	Frequency (f)
0–20	10 + 15 = 25
20–40	50
40–60	20 + 10 = 30
60–80	20
80–100	5

Modal class = class corresponding to the maximum frequency (50) = 20–40, so $L = 20$, $f_1 = 50$, $f_0 = 25$, $f_2 = 30$, $h = 20$.

$$\begin{aligned} \Delta_{1} &= 50 - 25 = 25, \qquad \Delta_{2} = 50 - 30 = 20 \\ M_{0} &= 20 + \frac{25}{25+20} \times 20 \\ &= 20 + 11.11 = 31.11 \end{aligned}$$

Example 2: Missing frequencies from a given mode

If the modal marks for a group of 94 students are 54, 10 students got marks between 0–20, 30 students between 40–60 and 14 students between 80–100, find the number of students getting marks between 20–40 and 60–80, if the maximum marks of the test were 100.

Let $a$ and $b$ be the missing frequencies of the 20–40 and 60–80 classes.

Marks	No. of students (f)
0–20	10
20–40	a ($f_0$)
40–60	30 ($f_1$)
60–80	b ($f_2$)
80–100	14
	$N = 54 + a + b$

$$\begin{aligned} 94 &= 54 + a + b \;\Rightarrow\; a + b = 40 \;\Rightarrow\; b = 40 - a \quad \text{(i)} \\ M_{0} &= L + \frac{f_{1}-f_{0}}{(f_{1}-f_{0})+(f_{1}-f_{2})} \times h \quad (\text{modal class } 40\text{–}60) \\ 54 &= 40 + \frac{30-a}{60-(a+b)} \times 20 \\ 14 &= \frac{30-a}{60-40} \times 20 = 30 - a \\ a &= 16, \qquad b = 40 - 16 = 24 \end{aligned}$$

The missing frequencies are 16 and 24 respectively.

Important Questions

2078 Exam

Find the combined mean of the following data: [2 marks]

Group A B C

Number 200 250 300

Mean 25 10 15

Group	A	B	C
Number	200	250	300
Mean	25	10	15

2079 Exam

Before and after the implementation of an economic program to uplift the economic condition of a community, the following information was found. [15 marks]

Monthly income (Rs. '00) Prior to the plan: no. of families After the plan: no. of families

4 – 6 10 65

6 – 8 70 –

8 – 10 35 37

10 – 12 20 15

12 – 14 10 15

14 – 16 3 5

16 – 18 2 5
1. Find the highest income of the poorest 40% of the population before and after the plan.
2. Find the lowest income of the richest 40% of the population before and after the plan.
3. Obtain the limits of income of the middle 50% of families before and after the plan.
The following table shows the monthly income of workers of two manufacturing companies A and B: [2 marks]

Company A Company B

Mean monthly income Rs. 2550 Rs. 2800

Number of workers 100 110
An enquiry into the budget of middle class families in a certain locality of a city gave the following information: [5 marks]

Expenses on Food Rent Clothing Fuel Misc.

Weight 30% 15% 20% 10% 25%

Price in 2021 (Rs.) 160 60 110 30 80

Price in 2022 (Rs.) 184 70 135 55 100

What is the cost of living index number of 2022 as compared to 2021? If an employee's monthly salary is Rs. 40,500 in the base period, what should be his/her salary in the current period?

Monthly income (Rs. '00)	Prior to the plan: no. of families	After the plan: no. of families
4 – 6	10	65
6 – 8	70	–
8 – 10	35	37
10 – 12	20	15
12 – 14	10	15
14 – 16	3	5
16 – 18	2	5

	Company A	Company B
Mean monthly income	Rs. 2550	Rs. 2800
Number of workers	100	110

Expenses on	Food	Rent	Clothing	Fuel	Misc.
Weight	30%	15%	20%	10%	25%
Price in 2021 (Rs.)	160	60	110	30	80
Price in 2022 (Rs.)	184	70	135	55	100

2081 Exam

The following table shows the average daily wages of workers in cities A and B. Find the combined average wage of the workers. [2 marks]

City A B

Average daily wage (Rs.) 1000 1200

No. of workers 250 200
The following table shows the marks distribution of students of a college: [10 marks]

Marks 0–10 10–20 20–30 30–40 40–50 50–60 60–70

No. of students 8 15 20 27 22 18 10

Calculate the limits of marks obtained by the middle 80% of the students.

City	A	B
Average daily wage (Rs.)	1000	1200
No. of workers	250	200

Marks	0–10	10–20	20–30	30–40	40–50	50–60	60–70
No. of students	8	15	20	27	22	18	10

2082 Exam

Find the lowest marks of the top 10 percent of the students from the following marks distribution: [10 marks]

Marks 0–10 10–20 20–30 30–40 40–50 50–60 60–70 70–80

Students 4 5 14 17 25 18 10 7

Marks	0–10	10–20	20–30	30–40	40–50	50–60	60–70	70–80
Students	4	5	14	17	25	18	10	7

2083 Exam

The mean of 100 items was 50. Later on it was found that two items were misread as 92 and 8 instead of 192 and 88. Find the correct mean. [2 marks]

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