Measures of dispersion from aprk

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I now have all the content needed. Here is the complete topic from Park's Textbook of Preventive and Social Medicine (APRK), p. 952-953:

Measures of Dispersion

Park's Textbook of Preventive and Social Medicine, p. 952

Measures of central tendency (like the mean) tell us where data is centered, but they don't tell us how spread out or variable the values are. Measures of dispersion describe the scatter or variability of a dataset.
The three main measures are:
  1. (a) The Range
  2. (b) The Mean Deviation
  3. (c) The Standard Deviation

(a) The Range

The range is the simplest measure of dispersion.
Range = Highest value - Lowest value
Example: Diastolic BP of 10 individuals: 83, 75, 81, 79, 71, 90, 75, 95, 77, 94
  • Highest = 95, Lowest = 71
  • Range = 71 to 95, or expressed as the difference = 24
For grouped data, the range = difference between the mid-points of the extreme categories.
Limitation: The range only considers extreme values and tells nothing about the spread of values in between. It is therefore of limited practical importance.

(b) The Mean Deviation (M.D.)

The mean deviation is the average of the deviations from the arithmetic mean (ignoring + and - signs).
Formula:
$$\text{M.D.} = \frac{\sum |x - \bar{x}|}{n}$$
Example: Diastolic BP of 10 individuals: 83, 75, 81, 79, 71, 95, 75, 77, 84, 90
BP (x)Arithmetic Mean (x̄)Deviation (x - x̄)
8381+2
7581-6
81810
7981-2
7181-10
9581+14
7581-6
7781-4
8481+3
9081+9
Total = 810Total = 56 (ignoring ± sign)
Mean = 810 / 10 = 81
M.D. = 56 / 10 = 5.6

(c) The Standard Deviation (S.D.)

The standard deviation is the most frequently used measure of dispersion. It is defined as the Root-Mean-Square Deviation, denoted by the Greek letter σ (sigma).
Formula:
$$\text{S.D.} = \sqrt{\frac{\sum (x - \bar{x})^2}{n}}$$
  • For sample size > 30: use denominator n
  • For sample size < 30: use denominator (n - 1) to correct for underestimation
$$\text{S.D.} = \sqrt{\frac{\sum (x - \bar{x})^2}{n - 1}}$$

Steps to Calculate S.D.:

  1. Find the deviation of each value from the mean: (x - x̄)
  2. Square each deviation: (x - x̄)²
  3. Add all the squared deviations: Σ(x - x̄)²
  4. Divide by n (or n-1 for small samples)
  5. Take the square root
Example: Same diastolic BP values: 83, 75, 81, 79, 71, 95, 75, 77, 84, 90
x(x - x̄)(x - x̄)²
83+24
75-636
810-
79-24
71-10100
95+14196
75-636
77-416
84+39
90+981
x̄ = 81, n = 10Total = 482
$$\text{S.D.} = \sqrt{\frac{482}{10-1}} = \sqrt{\frac{482}{9}} = \sqrt{53.55} = \textbf{7.31}$$

Interpretation of S.D.:

  • It is an abstract number giving an idea of the "spread" of the data
  • The larger the S.D., the greater the dispersion of values around the mean
  • Its full significance is best appreciated in the context of the Normal Distribution curve

Quick Comparison

MeasureFormulaBest Used When
RangeMax - MinQuick rough estimate
Mean DeviationΣ|x-x̄| / nSimple average spread
Standard Deviation√(Σ(x-x̄)²/n)Most precise; used in further statistics
The Standard Deviation is always preferred because it is mathematically tractable and forms the basis for tests of significance, normal distribution, and confidence intervals.
- Park's Textbook of Preventive and Social Medicine, p. 952-953

What does difference between the mid point of extreme categories mean ?

This is a straightforward conceptual clarification - no tools needed.

"Difference between mid-points of extreme categories" - Explained

This phrase applies specifically when data is presented as a frequency distribution table with class intervals (grouped data), not as raw individual values.

What is a "category" here?

A category = a class interval, e.g., 60-70, 70-80, 80-90, etc.
The extreme categories = the first (lowest) class interval and the last (highest) class interval.

What is the "mid-point" of a class interval?

$$\text{Mid-point} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$$

Example

Suppose diastolic BP is recorded as a frequency distribution:
Class IntervalFrequency
60 - 703
70 - 8012
80 - 9018
90 - 1009
100 - 1102
  • Lowest (first) category = 60-70, its mid-point = (60+70)/2 = 65
  • Highest (last) category = 100-110, its mid-point = (100+110)/2 = 105
$$\text{Range} = 105 - 65 = \textbf{40}$$

Why use mid-points instead of actual extreme values?

In grouped data, you don't know the exact individual values - you only know they fall somewhere within their class interval. You can't say the actual lowest value is 60 or the actual highest is 110. The mid-point (65 and 105) is the best representative value for each class, so the range is calculated using those.

In contrast - for raw (ungrouped) data:

You use the actual highest and lowest values directly (e.g., 95 - 71 = 24), as shown in Park's example.
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