Measure of Central Tendency

This can be computed from individual value/data as well as grouped data. By mathematically the formula is

$\bar{x} = \frac{\sum x}{n}$

Example 1

Compute means from the following data

5, 8, 3, 2, 7, 9, 8, 2, 2, 4

Solution

Mean = (5 + 8 + 3 + 2 + 7 + 9 + 8 + 2 + 2 + 4) / 10

= 50 / 10

Mean = 5

Example 2

Given the following data calculate the mean

24, 24, 23, 20, 12, 0

Solution

Mean = (24 + 24 + 23 + 20 + 12 + 0) / 6

= 103 / 6 = 17.2

Mean = 17.2

Example 3

Given the following data compute mean

Class interval	Frequency
50-59	2
60-69	4
70-79	6
80-89	3
90-99	2

Solution

Mean = ∑fx / ∑f

$\text{Mean} = \frac{\sum (f \times x)}{\text{Total Frequency}}$

CI	F	CM (X)	FX
50-59	2	54.5	109
60-69	4	64.5	258
70-79	6	74.5	447
80-89	3	84.5	253.5
90-99	2	94.5	189
	17		1256.5

Mean = 1256.5 / 17 = 73.9

Mean = 73.9

Mean deviation is the average amount by which individual values differ from the mean, irrespective of signs.

Example Calculation

Compute the mean deviation from the following data:

5, 8, 3, 2, 7, 9, 8, 2, 2, 4

Step 1: Calculate the Mean (x̄)

$\bar{x} = \frac{5 + 8 + 3 + 2 + 7 + 9 + 8 + 2 + 2 + 4}{10} = \frac{50}{10} = 5$

Step 2: Calculate |Xi - x̄| (Absolute Deviations)

| Xi | Xi - x̄ | |Xi - x̄| | | --- | --- | --- | | 5 | 0 | 0 | | 8 | 3 | 3 | | 3 | -2 | 2 | | 2 | -3 | 3 | | 7 | 2 | 2 | | 9 | 4 | 4 | | 8 | 3 | 3 | | 2 | -3 | 3 | | 2 | -3 | 3 | | 4 | -1 | 1 |

Step 3: Sum of Absolute Deviations

$\sum |X_i - \bar{x}| = 0 + 3 + 2 + 3 + 2 + 4 + 3 + 3 + 3 + 1 = 24$

Step 4: Calculate the Mean Deviation

$\text{Mean Deviation} = \frac{\sum |X_i - \bar{x}|}{n} = \frac{24}{10} = 2.4$

The mean deviation of the given dataset is 2.4.

Mean square deviation is the average of the square deviation or is the mean deviation square.

Mathematically:

$\text{Mean Square Deviation} = \frac{\sum (X_i - \bar{X})^2}{n}$

Where:

• $X_i$: Individual data points
• $\bar{X}$: Mean of the data set
• $n$: Number of data points

Example. Compute mean square deviation from the following data

5, 8, 3, 2, 7, 9, 8, 2, 2, 4

Solution

x̄ = 5

X	X - x̄	(X - x̄)²
5	0	0
8	3	9
3	-2	4
2	-3	9
7	2	4
9	4	16
8	3	9
2	-3	9
2	-3	9
4	-1	1
Total		70

MSD = 70 / 10 = 7

Example 2. Given the following data compute mean deviation from the formula

$\bar{x} = \frac{\sum (x - \bar{x})}{\sum f}$

Class interval	Frequency	Class mark	Frequency × class mark	(X - x̄)	(X - x̄)²	F(X - x̄)²
50-59	2	54.5	109	-19.4	376.36	753.72
60-69	4	64.5	258	-9.4	88.36	353.44
70-79	6	74.5	447	0.6	0.36	2.16
80-89	3	84.5	253.5	10.6	112.36	337.08
90-99	2	94.5	189	20.6	424.36	848.72

x̄ = 1256 / 17 = 73.9

Is the root of variation OR is the root of the mean square deviation.

Mathematically:

$\sigma = \sqrt{\frac{\sum (X_i - \bar{X})^2}{n}}$

For grouped data.

Example 2 from the following data below calculate the standard deviation

CI	F	CL MK	FX	X - x̄	(X - x̄)²	F(X - x̄)²
5-9	2	7	14	-14.4	207.36	414.72
10-14	11	12	132	-9.4	88.36	971.96
15-19	38	17	646	-4.4	19.36	735.68
20-24	54	22	1188	0.6	0.36	19.44
25-29	28	27	756	5.6	31.36	878.08
30-34	9	32	288	10.6	112.36	1011.24
35-39	1	37	37	15.6	243.36	243.36

$\text{s.d} = \sqrt{\frac{4261.7142}{n}}$

s.d = √30.011

s.d = 5.48

Is the central or middle value in a series of ordered value OR is the mid-point between the two figures.

NB For odd distribution median is the central value. For the even distribution median is obtained by picking middle value and add them then dividing by two.

Example. Find the median of the following data

46, 50, 70, 69, 60, 51, 55, 84, 72

Solution

1st arrange in ascending order

46, 50, 51, 55, 60, 69, 70, 72, 84

The median = 60

Example 2

Find the median of the following distribution

46, 50, 72, 70, 69, 60, 51, 55, 84, 71, 69, 35

Median = (60 + 69) / 2 = 129 / 2 = 64.5

Thus median = 64.5

Example 3

Find the median of the following distribution

48, 61, 75, 78, 100, 23, 56, 16

Solution

1st arrange in ascending order

16, 23, 56, 61, 75, 78, 100

Median = (56 + 61) / 2 = 117 / 2 = 58.5

Thus median = 58.5

Mathematically

$M = L_1 + \frac{\frac{N}{2} - n_b}{n_w} \times c$

Example. Calculate median for the following data

CI	F
20-24	3
25-29	7
29-34	8
35-39	12
40-44	9
45-49	6
50-54	4
55-59	1

Solution

M = L1 + (N/2 - nb) / nw × c

M = 34.5 + (50/2 - 18) / 12 × 5

Median = 34.5 + 2.92 = 37.42

Median = 37.42

Example 2 calculate median from the following data

Class interval	Frequency
40-49	4
50-59	8
60-69	10
70-79	7
80-89	7
90-99	4

Solution

M = L1 + (N/2 - nb) / nw × c

Data given:

• N = 40
• L1 = 59.5
• nw = 10
• nb = 12
• c = 10

Thus median = 59.5 + (40/2 - 12) / 10 × 10

= 59.5 + 8 = 67.5

Median = 67.5

Is the frequently occurring value.

Mode class is the class with the highest frequency.

Types of Mode

i. Uni mode is the value with single mode.

Example of uni mode: 51, 50, 60, 51, 60, 61, 62, 40, 41, 70, 51, 80, 7

Hence mode = 51

ii. Bi mode is the value with double mode.

Example: 61, 50, 60, 52, 60, 61, 62, 40, 41, 70, 68, 38, 80, 70, 100

Hence mode = 60 and 70

Mathematically

$M_o = L_1 + \frac{t_1}{t_1 + t_2} \times c$

Where:

• $L_1$ = Lower class boundary
• $t_1$ = Frequency of the modal class minus frequency below the modal class
• $t_2$ = Frequency of the modal class minus frequency above the modal class
• $c$ = Class interval

Example. Compute the mode in the table below

Class interval	Frequency
0-4	5
5-9	2
10-14	1
15-19	6
20-24	5
25-29	1
30-34	4
35-39	3
40-44	2
45-49	1

Data given:

• L1 = 14.5
• t1 = 5
• t2 = 1
• c = 5

From formula:

Mode = L1 + (t1 / (t1 + t2)) × c

Mode = 14.5 + (5 / (5 + 1)) × 5

Mode = 14.5 + 4.17 = 18.67 ≈ 18.5

The mode = 18.5

Range is the difference between highest value and the lowest value.

Mathematically

Range = Highest value - Lowest value

R = H - L

Example. Calculate the range of the following data

8, 22, 18, 20, 21, 13, 16, 30, 46, 10, 2, 22

Solution

R = h - l

= 46 - 2 = 44

Thus range = 44