Dispersion in Statistics 

In statistics, we often use measures of central tendency such as mean, median, or mode to describe the typical value of a dataset. However, two different datasets can have the same central tendency but vary greatly in their spread. To understand how widely the values are scattered around the central value, we use dispersion in statistics. 

1.0What is Dispersion?

Dispersion is the measure of the degree to which data values are spread out around a central value. In simple terms, it tells us how consistent or variable the data is.

• If dispersion is small, the data points are closely packed around the central value.
• If dispersion is large, the data values are more spread out.

Thus, both central tendency and dispersion together give a complete description of the dataset.

Also Read: Probability and Statistics

2.0Importance of Dispersion

• It helps compare the consistency of two or more datasets.
• It gives reliability to averages and measures of central tendency.
• It is useful in economics, business, and scientific research for analyzing variability.

3.0Measure of Central Tendency and Dispersion

While central tendency provides a single representative value (mean, median, mode), measure of dispersion explains how much the data differs from this central value. Both must be studied together for accurate statistical analysis.

4.0Measures of Dispersion in Statistics

There are several ways of measuring the dispersion of data. These are broadly classified into:

1. Absolute Measures of Dispersion

These are expressed in the same units as the original data. Common methods include:

• Range = Largest value – Smallest value
• Quartile Deviation (Semi-Interquartile Range)
• Mean Deviation
• Standard Deviation

2. Relative Measures of Dispersion

These are ratios or percentages that allow comparison between datasets of different units. Examples include:

• Coefficient of Range
• Coefficient of Quartile Deviation
• Coefficient of Variation

5.0Methods of Dispersion in Statistics

(i) Range

The simplest measure, showing the difference between the maximum and minimum values.

(ii) Quartile Deviation

Measures the spread of the middle 50% of data.

(iii) Mean Deviation

Mean deviation is average of the deviations of all values from a central tendency (mean, median, or mode).

(iv) Standard Deviation

Most commonly used measure, showing how much data deviates from the mean.

Also Practice: Mean Median Mode Questions

6.0Shortcut formula for grouped variance

With step deviations $u=\frac{x-a}{h}$ ,

$\bar{x}=a+h\frac{\sum fu}{n},{\sigma }^2=h^2\left[\frac{\sum f{u}^2}{n}-{\left(\frac{\sum fu}{n}\right)}^2\right].$

7.0Example of Dispersion

Consider the marks obtained by two students in five subjects:

• Student A: 50, 52, 48, 49, 51
• Student B: 30, 70, 20, 90, 40

Both may have the same mean (around 50), but the dispersion in statistics for Student B is much larger, showing inconsistency compared to Student A.

8.0Solved Examples on Dispersion in Statistics

Example 1: The marks obtained by 10 students in a test are: 12, 15, 18, 20, 22, 25, 28, 30, 35, 40. Find the range of marks.

Solution:

Range = Maximum value − Minimum value = 40 - 12 = 28 

Answer: Range = 28

Example 2: The weights (in kg) of 5 students are: 40, 42, 43, 45, 50. Find the mean deviation about the mean.

Solution:

1. Mean: $\bar{x}=\frac{40+42+43+45+50}{5}=\frac{220}{5}=44$
2. Deviations from mean $\left|x_i-\bar{x}\right|$: 

|40 - 44| = 4, |42 - 44| = 2, |43 - 44| = 1, |45 - 44| = 1, |50 - 44| = 6 

3. Mean Deviation: $MD=\frac{\sum |x_i-\bar{x}|}{n}=\frac{4+2+1+1+6}{5}=\frac{14}{5}=2.8$

Mean Deviation = 2.8

Example 3: The data is: 2, 4, 6, 8, 10. Find the variance and standard deviation.

Solution:

1. Mean :

$\bar{x}=\frac{2+4+6+8+10}{5}=\frac{30}{5}=6$

2. Deviations squared ${\left(x_i-\bar{x}\right)}^2$:

$(2-6{)}^2=16,(4-6{)}^2=4,(6-6{)}^2=0,(8-6{)}^2=4,(10-6{)}^2=16$

3. Variance: 

${\sigma }^2=\frac{\sum {\left(x_i-\bar{x}\right)}^2}{n}=\frac{16+4+0+4+16}{5}=\frac{40}{5}=8$

4. Standard Deviation: $\sigma =\sqrt{8}=2.828$

Answer: Variance = 8, Standard Deviation ≈ 2.83

Example 4: The marks of 8 students are: 15, 18, 20, 22, 25, 30, 35, 40. Find the Quartile Deviation.

Solution:

1. Arrange data (already in ascending order).

Number of terms n = 8.

2. $Q_1=\frac{n+1}{4}\text{ th term }=\frac{9}{4}\text{ th term }=2.25\text{ th term. }$

So, $Q_1$= 18 + 0.25(20 - 18) = 18.5.

3. $Q_3=\frac{3(n+1)}{4}\text{ th term }=\frac{27}{4}\text{ th term }=6.75\text{ th term }.$

So, $Q_3$ = 30 + 0.75(35-30) = 33.75.

4. Quartile Deviation

$QD=\frac{Q_3-Q_1}{2}=\frac{33.75-18.5}{2}=\frac{15.25}{2}=7.625$

Answer: Quartile Deviation = 7.63 (approx.)

Example 5: Find the variance, standard deviation, and mean deviation about the median for the data 6, 8, 9, 11, 13.

Solution: 

Mean $\bar{x}=\frac{6+8+9+11+13}{5}=9.4$.

Variance ${\sigma }^2=\frac{1}{5}\sum (x-\bar{x}{)}^2$

$(6-9.4{)}^2=11.56,(8-9.4{)}^2=1.96,(9-9.4{)}^2=0.16,(11-9.4{)}^2=2.56,(13-9.4{)}^2=12.96$

Sum = 29.2 = 29.2. 

Hence ${\sigma }^2=29.2/5=5.84,\sigma =\sqrt{5.84}\approx \mathrm{2.415.}$

Median = 9

Mean deviation about median:

${\mathrm{MD}}_{\text{median }}=\frac{|6-9|+|8-9|+|9-9|+|11-9|+|13-9|}{5}=\frac{10}{5}=2.$

Example 6: For the distribution x: 2, 4, 6, 8 with frequencies f: 3, 5, 7, 5, find the mean and standard deviation.

Solution

n = 3 + 5 + 7 + 5 = 20.

$\sum fx=2\cdot 3+4\cdot 5+6\cdot 7+8\cdot 5=108$

Mean $\bar{x}=108/20=5.4$ _.

Compute $\sum f(x-\bar{x}{)}^2$

(2−5.4)2⋅3=11.56⋅3=34.68,(4−5.4)2⋅5=1.96⋅5=9.80,(6−5.4)2⋅7=0.36⋅7=2.52,(8−5.4)2⋅5=6.76⋅5=33.80.

$\begin{array}{rl}& (2-5.4{)}^2\cdot 3=11.56\cdot 3=34.68\\ & (4-5.4{)}^2\cdot 5=1.96\cdot 5=9.80\\ & (6-5.4{)}^2\cdot 7=0.36\cdot 7=2.52\\ & (8-5.4{)}^2\cdot 5=6.76\cdot 5=33.80\end{array}$

Total = 80.80. Hence ${\sigma }^2=80.80/20=4.04,\sigma \approx \mathrm{2.010.}$

Example 7: For grouped data:

Find the mean and standard deviation using step deviation. Take class width h = 10.

Solution

Midpoints $x_i$: 5, 15, 25, 35. Choose a = 25, $u_i=(x_i-a)/h$ gives -2, -1, 0, 1.

∑fu=5(−2)+9(−1)+12(0)+4(1)=−15

$\sum fu=5(-2)+9(-1)+12(0)+4(1)=-15,n=30$.

Mean $\bar{x}=a+h\frac{\sum fu}{n}=25+10\left(-\frac{15}{30}\right)=20.$

For variance, use ${\sigma }^2=h^2\left[\frac{\sum f{u}^2}{n}-{\left(\frac{\sum fu}{n}\right)}^2\right].$

$\sum f{u}^2=5\cdot 4+9\cdot 1+12\cdot 0+4\cdot 1=33.$

Hence ${\sigma }^2=10^2\left[\frac{33}{30}-{\left(\frac{-15}{30}\right)}^2\right]=100(1.1-0.25)=85.$

So $\sigma =\sqrt{85}\approx \mathrm{9.220.}$

Example 8: Two data sets are A: 40, 50, 60, 70, 80 and B: 55, 60, 65, 70, 75. Which is more consistent? Use the coefficient of variation.

Solution
For A: mean = 60. Variance

$\frac{400+100+0+100+400}{5}=200\Rightarrow {\sigma }_A=\sqrt{200}=\mathrm{14.142.}$

${\mathrm{CV}}_A=\frac{14.142}{60}\times 100\approx 23.57\%.$

For B: mean = 65. Variance $\frac{100+25+0+25+100}{5}=50\Rightarrow {\sigma }_B=\sqrt{50}=\mathrm{7.071.}$

${\mathrm{CV}}_B=\frac{7.071}{65}\times 100\approx 10.88\%.$
Lower CV implies higher consistency. Set B is more consistent.

Example 9: Two groups have $n_1=40,{\bar{x}}_1=50,{\sigma }_1=6$ and $n_2=60,{\bar{x}}_2=56,{\sigma }_2=8$ _. Find the combined mean and combined standard deviation.

Solution

Combined mean $\bar{x}=\frac{n_1{\bar{x}}_1+n_2{\bar{x}}_2}{n_1+n_2}=\frac{40\cdot 50+60\cdot 56}{100}=53.6$.

Combined variance

$\begin{array}{rl}& {\sigma }^2=\frac{n_1\left[{\sigma }_1^2+{\left({\bar{x}}_1-\bar{x}\right)}^2\right]+n_2\left[{\sigma }_2^2+{\left({\bar{x}}_2-\bar{x}\right)}^2\right]}{n_1+n_2}.\\ & {\left({\bar{x}}_1-\bar{x}\right)}^2=(-3.6{)}^2=12.96,{\left({\bar{x}}_2-\bar{x}\right)}^2=(2.4{)}^2=\mathrm{5.76.}\\ & n_1[\cdot ]=40(36+12.96)=1958.4,n_2[\cdot ]=60(64+5.76)=\mathrm{4185.6.}\end{array}$

Sum = 6144.0. Hence ${\sigma }^2=61.44,\sigma \approx 7.84$ _.

Example 10: If Y = 3X - 4 and ${\sigma }_Y=15$, find ${\sigma }_X$. Comment on the effect of shift on dispersion.

Solution
For a linear change $Y=aX+b,{\sigma }_Y=|a|{\sigma }_X$. Hence ${\sigma }_X=15/3=5$ _.

The shift b = -4 does not affect variance or standard deviation.

9.0Practice Questions on Dispersion in Statistics

1. Data: 7, 8, 12, 15, 15, 18, 21. Find range, variance, and standard deviation.
2. For x: 1, 3, 5, 7 with f: 2, 4, 6, 8, find the mean and standard deviation.
3. Grouped data:

Use step deviation to find the mean and standard deviation. Take h = 10.

4. Two samples:

Sample A has mean 48 and standard deviation 5 for n = 25.

Sample B has mean 52 and standard deviation 7 for n = 35.

Find the combined mean and standard deviation.

5. Compare consistency using CV.

Set P: mean = 80, standard deviation = 12.

Set Q: mean = 72, standard deviation = 9.

6. If each observation in a data set is multiplied by k, show how variance and standard deviation change. Apply to k = 0.5 when original variance is 20.
7. A distribution has mean 30 and variance 49. Find variance and standard deviation after the transformation Z = 2X + 10.
8. In a discrete distribution with mean 1010, the deviations from 10 are −4, −2, −1, 0, 1, 3, 3 with frequencies 1, 2, 3, 4, 3, 2, 1. Compute variance using the shortcut ${\sigma }^2=\frac{\sum f{d}^2}{n}-{\left(\frac{\sum fd}{n}\right)}^2$ where d = x - 10.
9. Grouped data:

Find the coefficient of variation.

10. For the first n natural numbers, prove that variance $=\frac{n^2-1}{12}$ . Verify for n = 5 by direct calculation.

Also Check: Probability and Statistics previous year questions with solutions