Measures of Dispersion – 1 (Statistics for Psychologists)

In statistics, the measures of dispersion help us find how the data is distributed. For example, 1, 0.5, 0.25, 0.125… are ‘very’ close to each other; while 1, 10, 40, 600… are ‘too far’ away. By developing tools for dispersion, we try to identify how ‘spread’ the observations are.

Requisites of Ideal Measures of Dispersion

We noticed how none of the central tendencies we discussed earlier were ‘ideal’. Before we discuss Measures of Dispersion, let’s look at the requisites of an ideal measure. According to Yule (whom we familiarised ourselves with in the last unit) and Kendall, an ideal measure of dispersion should possess following properties.

  1. It should be easy to calculate and simple to follow.
  2. It should be rigidly defined: For the same data, all the methods should produce the same result.
  3. It should be based on all the observations.
  4. It should be useful for further mathematical treatment.
  5. It should be affected as little as possible by fluctuations of sampling.

Measures of Dispersion


Range is the simplest measure of dispersion – and is given by the difference between two extreme observations. If A and B are the smallest and greatest observations, then range is B – A

Example 1: Find the range of the distribution: 10, 50, 22, 44, 10, 16, 45
Solution: Maximum, B = 50 and Minimum, A = 10. So, range = B – A = 50 – 10 = 40.

Example 2: Find the range of the following distribution:


Solution: Range = 9 – (- 1) = 10

Though it is rigidly defined and easy to calculate, it is not based on all observations.

Quartiles and Quartile Deviation

Remember Median? How we picked median is finding the N/2th observation and spotting the cumulative frequency ‘just’ greater than that value.

Similarly, we can find two other ‘quarter’ measurements. N/4th observation, called ‘first quartile’ (denoted by Q­1) and 3N/4th observation, called ‘third quartile’ (denoted by Q3).

Quartile Deviation (also called semi-interquartile range) is given by

Q = ½ (Q1 + Q3), where Q1 is first quartile and Q3 is third quartile

Example 4: Find the median, first and third quartiles of the distribution and therefore find the quartile deviation.


Solution: Here, the value of x’s are already sorted.

XfCumulative Frequency
2108+10 = 18
31118+11= 29
41629+16= 45
52045+20= 65
62565+25= 90
71590+15= 105
89105+9= 114
96114+6= 120
 N = 120 

N = 120.

Now, to find median, N/2 = 60.
The cumulative frequency just above 60 is 65. The value corresponding to c.f. 65 is the median.
Hence, Median = 5.

Now, to find 1st quartile, Q1, N/4 = 30.
The cumulative frequency just above 30 is 45. The value corresponding to c.f. 45 is the first quartile.
Hence, Q1 = 4.

Now, to find 3rd quartile, Q3, 3N/4 = 90.
The cumulative frequency just above 90 is 90 itself. The value corresponding to itis the third quartile.
Hence, Q3 = 6.

Quartile Deviation, Q = Q = ½ (Q1 + Q3) = Q = ½ (4+ 6) = 5

Example 5: Find the quartile deviation of the following distribution:


(Try it yourself!)

Quartile deviation might be better than Range as it makes use of 50% of data. Range only considers the smallest and the highest. But it also ignores the rest of the 50% of data – hence is not a reliable measure.


  1. There’s one more quartile, Q2. But you already know what it is! Can you guess?
  2. Can you guess what will be Q0 and Q4?
  3. Can you guess why Q1, Q2, Q3 are called ‘quartiles’?

Mean Deviation

If xi | fi, i = 1, 2, …, n is the frequency distribution, then mean deviation from the average A (can be mean, median or mode) is given by

\text{Mean Deviation}=\frac1N\sum_i f_i |x_i-A|,\text{  }N=\sum f_i

where  represents the modulus or the absolute value of the deviation x_i - A  – negative sign, if it is there, is ignored. Mean deviation is based on all observations; it is a better measure of dispersion than range of quartile deviation. But the step of ignoring sign using  makes it unfit for further mathematical treatment.

Example 6: Find the Mean Deviation of the Following Distribution about Median: 3, 3, 3, 6, 9, 12, 12

Solution: A = Median = 6

∑f7∑f |x-A|=24

N= ∑f = 7

Mean Deviation about Median =\frac1N\sum_i f_i |x_i-A|=\frac{24}7 = 3.429

Example 7,8: Find the Mean Deviation (i) about Mean and (ii) Mode for the same distribution above. (Try it yourself!)


  • The Mean Deviation about Median will always be the least compared to Mean Deviation about other measures of central tendencies!
  • Mean deviation about any average will always be positive or zero!

Mean deviation is based on all observations; it is a better measure of dispersion than range or quartile deviation. But the step of ignoring sign using  makes it unfit for further mathematical treatment.

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