Measure of Dispersion

Q: What is the coefficient of variation used for?

The coefficient of variation, standard deviation divided by mean times 100, expresses the spread as a percentage of the mean. Because it has no unit, it compares the variability of distributions with different means or units. The distribution with the smaller coefficient of variation is more consistent, for example the steadier of two batsmen.

Q: What is the variance of the first n natural numbers?

The variance of 1, 2, ..., n is (n squared - 1)/12, so the standard deviation is the square root of (n squared - 1)/12. The same value holds for any n consecutive integers, since shifting does not change variance. For an arithmetic progression with common difference d, multiply the variance by d squared.

MathsStatisticsFor JEE aspirants

Measures of dispersion tell how widely data is scattered around its average. Two data sets can have the same mean but very different spread, so the mean alone is not enough. The main measures of dispersion are the range, the mean deviation, the variance and the standard deviation, while the coefficient of variation compares the variability of different distributions. This page gives their formulas for ungrouped and grouped data, their properties and solved examples for JEE Main and JEE Advanced.

Key Formulas - Quick Reference

Range $=$ largest value $-$ smallest value
Mean deviation about $a$ : $MD (a) = \frac{1}{n} \sum ∣ x_{i} - a ∣$ (ungrouped), where $a = \overset{x}{ˉ}$ or the median $M$
Mean deviation for a frequency distribution: $MD (a) = \frac{1}{N} \sum f_{i} ∣ x_{i} - a ∣$
Variance: $σ^{2} = \frac{1}{n} \sum (x_{i} - \overset{x}{ˉ})^{2}$ $= \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2}$ , and $σ = σ^{2}$
Frequency distribution: $σ^{2} = \frac{\sum f _{i} x _{i}^{2}}{N} - \overset{x}{ˉ}^{2}$
Step deviation: $σ_{x}^{2} = h^{2} σ_{u}^{2}$ , where $u_{i} = \frac{x _{i} - a}{h}$
First $n$ natural numbers: $σ^{2} = \frac{n ^{2} - 1}{12}$
$Var (a x_{i} + b) = a^{2} Var (x_{i})$ and $σ_{a x + b} = ∣ a ∣ σ$
Combined variance: $σ^{2} = \frac{n _{1} ( σ _{1}^{2} + d _{1}^{2} ) + n _{2} ( σ _{2}^{2} + d _{2}^{2} )}{n _{1} + n _{2}}$ , where $d_{i} = \overset{x}{ˉ}_{i} - \overset{x}{ˉ}$
Coefficient of variation: $CV = \frac{σ}{x ˉ} \times 100$

1. What Is Dispersion?

An average gives the centre of the data but says nothing about how the values are spread around it. Dispersion (scatter) is the degree to which the observations spread about a central value. A small dispersion means the values are close to the average; a large dispersion means they are scattered.

Figure 1: Both batsmen average 50 runs, yet B's scores are far more scattered. The mean alone cannot show this, so we need a measure of dispersion such as the range or

σ

The measures of dispersion in the syllabus are:

Range: the gap between the largest and the smallest value.
Mean deviation: the average distance of the observations from the mean or the median.
Variance and standard deviation: based on the squared distances from the mean.
Coefficient of variation: standard deviation as a percentage of the mean, used to compare distributions.

2. Range

Range $= L - S$ , where $L$ is the largest value and $S$ is the smallest value. For grouped data, range $=$ upper limit of the last class $-$ lower limit of the first class.

The range is quick to find, but it uses only the two extreme values. One unusual observation changes it completely, and it says nothing about how the values in between are spread. That is why the measures below use every observation.

3. Mean Deviation

The mean deviation about a point $a$ is the arithmetic mean of the absolute deviations $∣ x_{i} - a ∣$ of the observations from $a$ . It is usually taken about the mean $\overset{x}{ˉ}$ or the median $M$ .

3.1 Mean Deviation for Ungrouped Data

MD (a) = \frac{1}{n} i = 1 \sum n ∣ x_{i} - a ∣

Find the central value $a$ (mean or median, as asked).
Write the absolute deviations $∣ x_{i} - a ∣$ . The modulus makes every distance positive.
Add them and divide by $n$ .

3.2 Mean Deviation for Discrete and Grouped Data

MD (a) = \frac{1}{N} i = 1 \sum n f_{i} ∣ x_{i} - a ∣, N = \sum f_{i}

For grouped data, $x_{i}$ is the class mark. The mean comes from the frequency formula; the median comes from the grouped median formula $l + (\frac{N /2 - c f}{f}) \times h$ .

3.3 Key Facts about Mean Deviation

Least about the median: $\sum ∣ x_{i} - a ∣$ is smallest when $a = M$ , so the mean deviation about the median is never more than about any other point (Figure 3).
Never more than the standard deviation: the mean deviation about the mean is at most $σ$ .
Drawback: the modulus is awkward in algebra, so mean deviation is hard to use for combined groups or further theory. Squaring the deviations avoids this, which leads to the variance.

4. Variance and Standard Deviation

4.1 Definitions

The variance $σ^{2}$ is the mean of the squared deviations from the arithmetic mean. The standard deviation $σ$ is the positive square root of the variance, so it has the same unit as the data.

σ^{2} = \frac{1}{n} i = 1 \sum n (x_{i} - \overset{x}{ˉ})^{2}, σ^{2} = \frac{1}{N} i = 1 \sum n f_{i} (x_{i} - \overset{x}{ˉ})^{2}, σ = σ^{2}

Figure 2: For 2, 3, 7, 8, 10 (mean 6), averaging

∣ x_{i} - \overset{x}{ˉ} ∣

gives the mean deviation 2.8 and averaging

(x_{i} - \overset{x}{ˉ})^{2}

gives the variance 9.2, so

σ \approx 3.03

. Squaring gives the far-away values 2 and 10 much more weight.

4.2 Formulas That Avoid Computing Every Deviation

Expanding $(x_{i} - \overset{x}{ˉ})^{2}$ and using $\sum x_{i} = n \overset{x}{ˉ}$ gives a faster form: variance $=$ mean of the squares $-$ square of the mean.

σ^{2} = \frac{\sum x _{i}^{2}}{n} - (\frac{\sum x _{i}}{n})^{2}, σ^{2} = \frac{\sum f _{i} x _{i}^{2}}{N} - (\frac{\sum f _{i} x _{i}}{N})^{2}

4.3 Shortcut and Step-Deviation Methods

With deviations $d_{i} = x_{i} - a$ from an assumed mean $a$ :

σ^{2} = \frac{\sum f _{i} d _{i}^{2}}{N} - (\frac{\sum f _{i} d _{i}}{N})^{2}

With step deviations $u_{i} = \frac{x _{i} - a}{h}$ (use $h$ = class width for grouped data):

σ^{2} = h^{2} [\frac{\sum f _{i} u _{i}^{2}}{N} - (\frac{\sum f _{i} u _{i}}{N})^{2}]

These work because subtracting $a$ does not change the variance, while dividing by $h$ divides the variance by $h^{2}$ . So we find the variance of the small numbers $u_{i}$ and multiply by $h^{2}$ . For the standard deviation, multiply $σ_{u}$ by $h$ .

4.4 Variance of the First $n$ Natural Numbers

Using $\sum k = \frac{n ( n + 1 )}{2}$ and $\sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$ :

σ^{2} = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{4} = \frac{n ^{2} - 1}{12}, σ = \frac{n ^{2} - 1}{12}

Any $n$ consecutive integers, or any A.P. with common difference 1, has this same variance, because shifting does not change it. An A.P. with common difference $d$ has variance $\frac{d ^{2} ( n ^{2} - 1 )}{12}$ .

Exam Trick

Recognise the pattern first. The variance of $2, 4, 6, 8, 10$ is $2^{2} \times \frac{5 ^{2} - 1}{12} = 8$ , and the variance of $101, 102, \dots, 200$ equals that of $151, 152, \dots, 250$ . No table is needed.

5. Properties of Variance and Standard Deviation

5.1 Change of Origin and Scale

Adding or subtracting a constant $λ$ to every value leaves the spread unchanged: $Var (x_{i} + λ) = Var (x_{i})$ .
Multiplying every value by $k$ multiplies the variance by $k^{2}$ and the standard deviation by $∣ k ∣$ .
Together: $Var (a x_{i} + b) = a^{2} Var (x_{i})$ and $σ_{a x + b} = ∣ a ∣ σ_{x}$ . So variance is independent of change of origin but not of change of scale.

Figure 4: Adding a constant slides the data without changing its spread, so

σ

is unchanged. Multiplying by 2 doubles every gap, so

σ

doubles and the variance becomes 4 times.

5.2 Mean Square Deviation about Any Point

The mean of the squared deviations about any point $a$ splits into the variance plus a non-negative term:

\frac{1}{n} \sum (x_{i} - a)^{2} = σ^{2} + (\overset{x}{ˉ} - a)^{2}

So the mean square deviation is least, and equal to $σ^{2}$ , exactly when $a = \overset{x}{ˉ}$ (Figure 3). Equivalently, $\frac{\sum x _{i}^{2}}{n} \geq (\frac{\sum x _{i}}{n})^{2}$ .

Figure 3: For 2, 3, 7, 8, 10, the mean deviation about

A

is smallest at the median (2.6 at

A = 7

). The mean square deviation

σ^{2} + (\overset{x}{ˉ} - A)^{2}

is smallest at the mean, where it equals the variance 9.2.

5.3 Combined Variance of Two Groups

Let two groups have sizes $n_{1}, n_{2}$ , means $\overset{x}{ˉ}_{1}, \overset{x}{ˉ}_{2}$ and variances $σ_{1}^{2}, σ_{2}^{2}$ . With combined mean $\overset{x}{ˉ} = \frac{n _{1} x ˉ _{1} + n _{2} x ˉ _{2}}{n _{1} + n _{2}}$ and $d_{1} = \overset{x}{ˉ}_{1} - \overset{x}{ˉ}$ , $d_{2} = \overset{x}{ˉ}_{2} - \overset{x}{ˉ}$ :

σ^{2} = \frac{n _{1} ( σ _{1}^{2} + d _{1}^{2} ) + n _{2} ( σ _{2}^{2} + d _{2}^{2} )}{n _{1} + n _{2}} = \frac{n _{1} σ _{1}^{2} + n _{2} σ _{2}^{2}}{n _{1} + n _{2}} + \frac{n _{1} n _{2} ( x ˉ _{1} - x ˉ _{2} ) ^{2}}{( n _{1} + n _{2} ) ^{2}}

The second form shows that the combined variance is the weighted average of the two variances plus an extra amount due to the gap between the two means. If the means are equal, the extra term is zero.

Exam Trick

Corrected variance: when a wrong observation $w$ is replaced by the correct value $c$ , correct both totals first: new $\sum x = \sum x - w + c$ and new $\sum x^{2} = \sum x^{2} - w^{2} + c^{2}$ . Then use $σ^{2} = \frac{\sum x ^{2}}{n} - \overset{x}{ˉ}^{2}$ with the corrected mean.

JEE Advanced

Bounds on the variance. If every observation lies in $[a, b]$ , then $σ^{2} \leq \frac{( b - a ) ^{2}}{4}$ , with equality when half the values equal $a$ and half equal $b$ . Also $σ = 0$ only when all observations are equal, and the mean deviation about the mean never exceeds $σ$ .

6. Coefficient of Variation: Comparing Distributions

The coefficient of variation is the standard deviation expressed as a percentage of the mean:

CV = \frac{σ}{x ˉ} \times 100

CV has no unit, so it can compare data with different means or units. The distribution with the smaller CV is less variable, that is, more consistent. If two distributions have equal means, comparing their standard deviations is enough.

Batsman (Figure 1)	Mean	$σ$	CV	Conclusion
A	50	$6 \approx 2.45$	$\approx 4.9%$	more consistent
B	50	$\approx 13.63$	$\approx 27.3%$	less consistent

7. Solved Examples

Solved Example 1

The range of the observations 2, 3, 5, 9, 8, 7, 6, 5, 7, 4, 3 is
(A) 11
(B) 7
(C) 5.5
(D) 6

Solution:

Largest value $L = 9$ , smallest value $S = 2$ .
Range $= 9 - 2 = 7$ .

Answer: (B) 7

Solved Example 2

The standard deviation of the 7 scores 1, 2, 3, 4, 5, 6, 7 is

(A) 4

(B) 2

(D) none of these

Solution:

These are the first 7 natural numbers, so $σ^{2} = \frac{7 ^{2} - 1}{12} = 4$ .
$σ = 2$ .

Answer: (B) 2

Solved Example 3

The variance of 2, 4, 6, 8, 10 is

(A) 8

(B) $8$

(D) none of these

Solution:

The data is $2 \times (1, 2, 3, 4, 5)$ .
Variance of 1 to 5 $= \frac{5 ^{2} - 1}{12} = 2$ .
Multiplying by 2 multiplies the variance by $2^{2} = 4$ , so the variance $= 8$ .

Answer: (A) 8

Solved Example 4

The coefficients of variation of two series are 75% and 90%, and their standard deviations are 15 and 18. Their means are
(A) 10
(B) 20
(C) 30
(D) 40

Solution:

$\overset{x}{ˉ} = \frac{σ}{CV} \times 100$ .
First series: $\overset{x}{ˉ} = \frac{15}{75} \times 100 = 20$ . Second series: $\overset{x}{ˉ} = \frac{18}{90} \times 100 = 20$ .
Both series have mean 20.

Answer: (B) 20

Solved Example 5

A variable

x

takes values

x_{i}

with

a \leq x_{i} \leq b

for

i = 1, 2, \dots, n

. Then

(A) $a \leq Var (x) \leq b$

(B) $a^{2} \leq Var (x) \leq b^{2}$

(D) $(b - a)^{2} \geq Var (x)$

Solution:

Both $x_{i}$ and $\overset{x}{ˉ}$ lie in $[a, b]$ , so $∣ x_{i} - \overset{x}{ˉ} ∣ \leq b - a$ for every $i$ .
Each squared deviation is at most $(b - a)^{2}$ , so their mean $Var (x) \leq (b - a)^{2}$ .
(A), (B) and (C) fail in general: if all $x_{i}$ are equal to a positive $a$ , the variance is 0.

Answer: (D) $(b - a)^{2} \geq Var (x)$ . The sharper bound is $\frac{( b - a ) ^{2}}{4}$ .

Solved Example 6

The runs scored by a batsman in ten innings are 34, 38, 42, 44, 46, 48, 54, 55, 63, 70. The mean deviation about the median is
(A) 8.6
(B) 7.6
(C) 8
(D) none of these

Solution:

The data is already in order and $n = 10$ , so median $= \frac{46 + 48}{2} = 47$ .
$\sum ∣ x_{i} - 47∣ = 13 + 9 + 5 + 3 + 1 + 1 + 7 + 8 + 16 + 23 = 86$ .
Mean deviation $= \frac{86}{10} = 8.6$ . (Shortcut: sum of values above $M$ minus sum below $M$ , $290 - 204 = 86$ .)

Answer: (A) 8.6

Solved Example 7

For a set of 100 observations, taking the assumed mean as 4, the sum of the deviations is

- 11

and the sum of the squares of these deviations is 275. The coefficient of variation is
(A) 41.13%
(B) 40.13%
(C) 42.13%
(D) none of these

Solution:

$\overset{x}{ˉ} = 4 + \frac{- 11}{100} = 3.89$ .
$σ^{2} = \frac{275}{100} - (\frac{- 11}{100})^{2} = 2.75 - 0.0121 = 2.7379$ , so $σ \approx 1.6547$ .
$CV = \frac{1.6547}{3.89} \times 100 \approx 42.54%$ , which matches none of (A), (B), (C).

Answer: (D) none of these (CV ≈ 42.54%)

Solved Example 8

The mean of five observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, the other two are
(A) 9, 4
(B) 2, 11
(C) 7, 6
(D) 5, 8

Solution:

Let the other two be $a$ and $b$ . Sum: $a + b + 9 = 5 \times 4.4 = 22$ , so $a + b = 13$ .
Variance: $\frac{a ^{2} + b ^{2} + 1 + 4 + 36}{5} - 4. 4^{2} = 8.24$ , so $\frac{a ^{2} + b ^{2} + 41}{5} = 8.24 + 19.36 = 27.6$ .
$a^{2} + b^{2} + 41 = 138$ , so $a^{2} + b^{2} = 97$ .
$2 ab = 1 3^{2} - 97 = 72$ , so $ab = 36$ . The numbers are the roots of $t^{2} - 13 t + 36 = 0$ : $t = 9, 4$ .

Answer: (A) 9, 4

Solved Example 9

For ten pairs of values,

\sum x = 235

\sum y = 250

\sum x^{2} = 6750

and

\sum y^{2} = 6840

. The standard deviations of

x

and

y

are
(A) 11.08, 7.68
(B) 11.02, 7.58
(C) 11.48, 7.48
(D) none of these

Solution:

$σ_{x}^{2} = \frac{6750}{10} - 23. 5^{2} = 675 - 552.25 = 122.75$ , so $σ_{x} \approx 11.08$ .
$σ_{y}^{2} = \frac{6840}{10} - 2 5^{2} = 684 - 625 = 59$ , so $σ_{y} \approx 7.68$ .

Answer: (A) 11.08, 7.68

Solved Example 10

The mean square deviations of

n

observations about

- 2

and about

2

are 18 and 10 respectively. The standard deviation is
(A) 1
(B) 2
(C) 3
(D) 4

Solution:

Use $\frac{1}{n} \sum (x_{i} - a)^{2} = σ^{2} + (\overset{x}{ˉ} - a)^{2}$ .
About $- 2$ : $σ^{2} + (\overset{x}{ˉ} + 2)^{2} = 18$ . About $2$ : $σ^{2} + (\overset{x}{ˉ} - 2)^{2} = 10$ .
Subtracting: $8 \overset{x}{ˉ} = 8$ , so $\overset{x}{ˉ} = 1$ . Then $σ^{2} = 18 - 9 = 9$ and $σ = 3$ .

Answer: (C) 3

Solved Example 11

Find the mean deviation about the mean of 3, 4, 5, 6, 7.

Solution:

$\overset{x}{ˉ} = \frac{25}{5} = 5$ .
$\sum ∣ x_{i} - 5∣ = 2 + 1 + 0 + 1 + 2 = 6$ .
Mean deviation $= \frac{6}{5} = 1.2$ .

Answer: 1.2

Solved Example 12

Find the mean deviation about the mean for the following data.

$x_{i}$	3	9	17	23	27
$f_{i}$	8	10	12	9	5

Solution:

$x_{i}$	$f_{i}$	$f_{i} x_{i}$	$∣ x_{i} - 15∣$	$f_{i} ∣ x_{i} - 15∣$
3	8	24	12	96
9	10	90	6	60
17	12	204	2	24
23	9	207	8	72
27	5	135	12	60
Total	44	660		312

$\overset{x}{ˉ} = \frac{660}{44} = 15$ .
Mean deviation $= \frac{312}{44} \approx 7.09$ .

Answer: 7.09 (approx.)

Solved Example 13

Find the standard deviation of the first

n

natural numbers.

Solution:

$σ^{2} = \frac{1}{n} \sum k^{2} - (\frac{1}{n} \sum k)^{2}$ .
$\frac{1}{n} \sum k^{2} = \frac{( n + 1 ) ( 2 n + 1 )}{6}$ and $\frac{1}{n} \sum k = \frac{n + 1}{2}$ .
$σ^{2} = \frac{( n + 1 ) ( 2 n + 1 )}{6} - \frac{( n + 1 ) ^{2}}{4} = \frac{( n + 1 ) ( 4 n + 2 - 3 n - 3 )}{12} = \frac{n ^{2} - 1}{12}$ .

Answer: $σ = \frac{n ^{2} - 1}{12}$

Solved Example 14

\sum_{i = 1}^{18} (x_{i} - 8) = 9

and

\sum_{i = 1}^{18} (x_{i} - 8)^{2} = 45

, find the standard deviation of

x_{1}, x_{2}, \dots, x_{18}

Solution:

Let $d_{i} = x_{i} - 8$ . Shifting does not change the variance, so $σ_{x}^{2} = σ_{d}^{2}$ .
$σ_{d}^{2} = \frac{45}{18} - (\frac{9}{18})^{2} = \frac{5}{2} - \frac{1}{4} = \frac{9}{4}$ .
$σ_{x} = \frac{3}{2}$ .

Answer: $\frac{3}{2}$

Solved Example 15

Find the coefficient of variation of the first

n

natural numbers.

Solution:

$σ = \frac{n ^{2} - 1}{12}$ and $\overset{x}{ˉ} = \frac{n + 1}{2}$ .
$CV = \frac{( n - 1 ) ( n + 1 )}{12} \times \frac{2}{n + 1} \times 100 = 100 \frac{4 ( n - 1 )}{12 ( n + 1 )}$ .
$CV = 100 \frac{n - 1}{3 ( n + 1 )}$ .

Answer: $100 \frac{n - 1}{3 ( n + 1 )}$

Solved Example 16

Determine the variance of the following distribution.

Class	0-2	2-4	4-6	6-8	8-10
Frequency	2	7	12	19	9

Solution:

Class	Class mark $x_{i}$	$f_{i}$	$u_{i} = \frac{x _{i} - 7}{2}$	$f_{i} u_{i}$	$f_{i} u_{i}^{2}$
0-2	1	2	$- 3$	$- 6$	18
2-4	3	7	$- 2$	$- 14$	28
4-6	5	12	$- 1$	$- 12$	12
6-8	7	19	0	0	0
8-10	9	9	1	9	9
Total		49		$- 23$	67

$N = 49$ , $h = 2$ .
$σ^{2} = 2^{2} [\frac{67}{49} - (\frac{- 23}{49})^{2}] = 4 \times \frac{67 \times 49 - 529}{2401} = \frac{11016}{2401}$ .
$σ^{2} \approx 4.59$ .

Answer: about 4.59

Solved Example 17

The means of two samples of sizes 200 and 300 are 25 and 10, and their standard deviations are 3 and 4. The variance of the combined sample of size 500 is
(A) 64
(B) 65.2
(C) 67.2
(D) 64.2

Solution:

Combined mean $= \frac{200 \times 25 + 300 \times 10}{500} = 16$ .
$d_{1} = 25 - 16 = 9$ and $d_{2} = 10 - 16 = - 6$ .
$σ^{2} = \frac{200 ( 9 + 81 ) + 300 ( 16 + 36 )}{500} = \frac{18000 + 15600}{500} = 67.2$ .

Answer: (C) 67.2

Solved Example 18

The variance of the first 20 natural numbers is

(A) $\frac{133}{4}$

(B) $\frac{379}{12}$

(D) $\frac{399}{4}$

Solution:

$σ^{2} = \frac{1}{20} \times \frac{20 \times 21 \times 41}{6} - (\frac{21}{2})^{2} = \frac{287}{2} - \frac{441}{4}$ .
$σ^{2} = \frac{133}{4}$ , which agrees with $\frac{2 0 ^{2} - 1}{12} = \frac{399}{12} = \frac{133}{4}$ .

Answer: (A) $\frac{133}{4}$

Solved Example 19

The mean and variance of a series of 5 numbers are 8 and 24. The mean and variance of another series of 3 numbers are also 8 and 24. The variance of the combined series is
(A) 20
(B) 24
(C) 25
(D) 42

Solution:

$σ^{2} = \frac{5 ( 24 ) + 3 ( 24 )}{8} + \frac{5 \times 3}{( 5 + 3 ) ^{2}} (8 - 8)^{2}$ .
The means are equal, so the second term is 0 and $σ^{2} = \frac{192}{8} = 24$ .

Answer: (B) 24

Solved Example 20

The mean deviation about the median of 340, 150, 210, 240, 300, 310, 320 is
(A) 52.4
(B) 52.5
(C) 52.86
(D) none of these

Solution:

Arrange: 150, 210, 240, 300, 310, 320, 340. With $n = 7$ , the median is the 4th value, 300.
$\sum ∣ x_{i} - 300∣ = 150 + 90 + 60 + 0 + 10 + 20 + 40 = 370$ .
Mean deviation $= \frac{370}{7} \approx 52.86$ .

Answer: (C) 52.86

Solved Example 21

The variance of the data below is

Size of item	3.5	4.5	5.5	6.5	7.5	8.5	9.5
Frequency	3	7	22	60	85	32	8

(A) 1.29
(B) 2.19
(C) 1.32
(D) none of these

Solution:

$x_{i}$	$f_{i}$	$d_{i} = x_{i} - 6.5$	$f_{i} d_{i}$	$f_{i} d_{i}^{2}$
3.5	3	$- 3$	$- 9$	27
4.5	7	$- 2$	$- 14$	28
5.5	22	$- 1$	$- 22$	22
6.5	60	0	0	0
7.5	85	1	85	85
8.5	32	2	64	128
9.5	8	3	24	72
Total	217		128	362

$σ^{2} = \frac{362}{217} - (\frac{128}{217})^{2} \approx 1.668 - 0.348 = 1.320$ .

Answer: (C) 1.32

Solved Example 22

The mean and variance of 5 observations are 4 and 5.2. If three of them are 1, 2 and 6, the remaining two are
(A) 2, 9
(B) 5, 6
(C) 4, 7
(D) 3, 8

Solution:

Let them be $x_{1}, x_{2}$ . Sum: $x_{1} + x_{2} + 9 = 20$ , so $x_{1} + x_{2} = 11$ .
$\sum (x_{i} - 4)^{2} = 5 \times 5.2 = 26$ : $(x_{1} - 4)^{2} + (x_{2} - 4)^{2} + 9 + 4 + 4 = 26$ , so $(x_{1} - 4)^{2} + (x_{2} - 4)^{2} = 9$ .
With $x_{2} = 11 - x_{1}$ : $(x_{1} - 4)^{2} + (7 - x_{1})^{2} = 9$ gives $x_{1}^{2} - 11 x_{1} + 28 = 0$ , so $x_{1} = 4$ or 7.

Answer: (C) 4, 7

Solved Example 23

The mean deviation about the mean of the series

a, a + d, a + 2 d, \dots, a + 2 n d (d > 0)

(A) $\frac{n + 1}{2 n + 1} d$

(B) $\frac{n ( n + 1 )}{2 n + 1} d$

(D) none of these

Solution:

There are $2 n + 1$ terms, symmetric about the middle term, so $\overset{x}{ˉ} = a + n d$ .
$\sum ∣ x_{i} - \overset{x}{ˉ} ∣ = 2 (n d + (n - 1) d + \dots + d) = 2 d \times \frac{n ( n + 1 )}{2} = n (n + 1) d$ .
Mean deviation $= \frac{n ( n + 1 ) d}{2 n + 1}$ .

Answer: (B) $\frac{n ( n + 1 )}{2 n + 1} d$

Solved Example 24

Batsman P scores a mean of 50 runs with standard deviation 15. Batsman Q scores a mean of 40 runs with standard deviation 10. Which batsman is more consistent?

Solution:

$CV_{P} = \frac{15}{50} \times 100 = 30%$ .
$CV_{Q} = \frac{10}{40} \times 100 = 25%$ .
The smaller CV shows less variability, so Q is more consistent, although P scores more on average.

Answer: Batsman Q

Solved Example 25

Find the mean deviation about the mean for the marks of 100 students: 0-10: 8, 10-20: 30, 20-30: 40, 30-40: 12, 40-50: 10.

Solution:

Class	Class mark $x_{i}$	$f_{i}$	$∣ x_{i} - 23.6∣$	$f_{i} ∣ x_{i} - 23.6∣$
0-10	5	8	18.6	148.8
10-20	15	30	8.6	258
20-30	25	40	1.4	56
30-40	35	12	11.4	136.8
40-50	45	10	21.4	214
Total		100		813.6

$\overset{x}{ˉ} = \frac{\sum f _{i} x _{i}}{N} = \frac{40 + 450 + 1000 + 420 + 450}{100} = 23.6$ .
Mean deviation $= \frac{813.6}{100} \approx 8.14$ marks.

Answer: about 8.14 marks

Solved Example 26

Find the variance and standard deviation of the marks of the 100 students in Solved Example 25.

Solution:

Class mark $x_{i}$	$f_{i}$	$u_{i} = \frac{x _{i} - 25}{10}$	$f_{i} u_{i}$	$f_{i} u_{i}^{2}$
5	8	$- 2$	$- 16$	32
15	30	$- 1$	$- 30$	30
25	40	0	0	0
35	12	1	12	12
45	10	2	20	40
Total	100		$- 14$	114

$σ^{2} = 1 0^{2} [\frac{114}{100} - (\frac{- 14}{100})^{2}] = 100 (1.14 - 0.0196) = 112.04$ .
$σ = 112.04 \approx 10.58$ marks.

Answer: $σ^{2} = 112.04$ , $σ \approx 10.58$ marks

Practice Questions

The standard deviation of some data is 6. If each observation is increased by 1 (or decreased by 1), the new standard deviation is:Answer: 6
The variance of the first $n$ natural numbers is:Answer: $\frac{n ^{2} - 1}{12}$ , so the standard deviation is $\frac{n ^{2} - 1}{12}$
The coefficient of variation of a series is 50 and its standard deviation is 21.2. Its arithmetic mean is:Answer: 42.4
Variance is independent of change of: (A) origin only (B) scale only (C) origin and scale (D) none of theseAnswer: (A) origin only
Which of the following is not a measure of dispersion? (A) mean (B) variance (C) mean deviation (D) rangeAnswer: (A) mean
If the standard deviation of $a_{1}, a_{2}, \dots, a_{n}$ is $σ$ , the standard deviation of $λ a_{1}, λ a_{2}, \dots, λ a_{n}$ is:Answer: $∣ λ ∣ σ$
The coefficient of variation and standard deviation of a distribution are 50 and 20. Its mean is:Answer: 40
The standard deviation of 25 numbers is 40. If each number is increased by 5, the new standard deviation is:Answer: 40
For $x_{1} < x_{2} < \dots < x_{101}$ , the mean deviation about a number $k$ is least when $k$ equals:Answer: $x_{51}$ , the median
The mean deviation from the mean of $- 1, 0, 4$ is:Answer: 2
If the standard deviation of $1, 2, 3, \dots, 10$ is $σ$ , the standard deviation of $11, 12, 13, \dots, 20$ is:Answer: $σ$
If each observation of data with variance $σ^{2}$ is increased by $λ$ , the new variance is:Answer: $σ^{2}$
If each of $n$ observations with standard deviation $σ$ is multiplied by a constant $c$ , the new standard deviation is:Answer: $∣ c ∣ σ$ (equal to $c σ$ when $c > 0$ )
The mean deviation about the median is: (A) greater than about any other number (B) less than about any other number (C) equal to that about any other number (D) maximum if all values are positiveAnswer: (B) less than about any other number
If the standard deviation of a variate $x$ is 10, the standard deviation of $50 + 5 x$ is:Answer: 50
The median and standard deviation of a distribution are 20 and 4. If each item is increased by 2:Answer: the median becomes 22 and the standard deviation stays 4
If each observation of data with variance $σ^{2}$ is multiplied by $λ$ , the new variance is:Answer: $λ^{2} σ^{2}$
The standard deviation of a variate $x$ is $σ$ . If $a$ , $b$ , $c$ are constants, the standard deviation of $\frac{a x + b}{c}$ is:Answer: $\frac{a}{c} σ$
In an experiment with 15 observations, $\sum x^{2} = 2830$ and $\sum x = 170$ . One observation 20 was wrong and is replaced by 30. The corrected variance is: [AIEEE 2003]Answer: 78
Consider the statements: (a) the mode can be computed from a histogram; (b) the median is not independent of change of scale; (c) the variance is independent of change of origin and scale. Which are correct? [AIEEE 2004]Answer: only (a) and (b)
In a series of $2 n$ observations, half equal $a$ and the rest equal $- a$ . If the standard deviation is 2, then $∣ a ∣$ is: [AIEEE 2004]Answer: 2
Let $x_{1}, \dots, x_{n}$ satisfy $\sum x_{i}^{2} = 400$ and $\sum x_{i} = 80$ . A possible value of $n$ among 12, 9, 18, 15 is: [AIEEE 2005]Answer: 18 (since $n \geq 16$ )
Population A has observations $101, 102, \dots, 200$ and population B has $151, 152, \dots, 250$ . If $V_{A}$ and $V_{B}$ are their variances, $\frac{V _{A}}{V _{B}}$ is: [AIEEE 2006]Answer: 1

Common Mistakes to Avoid

Watch out

Forgetting the square root: the standard deviation is $σ^{2}$ , and the coefficient of variation uses $σ$ , not $σ^{2}$ .
Thinking that adding a constant to every value changes the variance or standard deviation. Only the mean, median and mode shift.
Writing $σ_{k x} = k σ$ for negative $k$ . The standard deviation is $∣ k ∣ σ$ and the variance is $k^{2} σ^{2}$ .
In the step-deviation method, multiplying the variance by $h$ instead of $h^{2}$ .
Dropping the modulus in mean deviation. Without $∣ ∣$ , the deviations from the mean add up to 0.
Taking the mean deviation about the mean when the question asks about the median, or the other way round.
Averaging two variances to get the combined variance. The gap between the group means adds the extra term $\frac{n _{1} n _{2} ( x ˉ _{1} - x ˉ _{2} ) ^{2}}{( n _{1} + n _{2} ) ^{2}}$ .
For a corrected variance, fixing only $\sum x$ . Both $\sum x$ and $\sum x^{2}$ must be corrected before recomputing.

Frequently Asked Questions

What are measures of dispersion?

Measures of dispersion show how spread out the observations are around a central value. The common ones are the range, the mean deviation, the variance and the standard deviation. Two data sets can share a mean yet have very different spread, so an average is incomplete without a measure of dispersion.

What is the difference between variance and standard deviation?

Variance is the mean of the squared deviations from the mean, so its unit is the square of the data unit. Standard deviation is the positive square root of the variance and has the same unit as the data. For marks in a test, the variance is in marks squared while the standard deviation is in marks.

Why is standard deviation preferred over mean deviation?

Mean deviation uses absolute values, which are hard to handle algebraically. Standard deviation squares the deviations instead, so it gives simple rules for combining groups, correcting data and changing origin or scale. It also gives more weight to values far from the mean, which makes it more sensitive to real spread.

How does adding or multiplying by a constant affect variance and standard deviation?

Adding or subtracting a constant does not change the variance or the standard deviation, because every value moves by the same amount. Multiplying every value by $k$ multiplies the variance by $k^{2}$ and the standard deviation by $∣ k ∣$ . So the variance of $a x + b$ is $a^{2}$ times the variance of $x$ .

What is the coefficient of variation used for?

The coefficient of variation, $\frac{σ}{x ˉ} \times 100$ , expresses the standard deviation as a percentage of the mean. Because it has no unit, it compares the variability of distributions with different means or units. The distribution with the smaller coefficient of variation is more consistent, for example the steadier of two batsmen.

How do you find the combined variance of two groups?

First find the combined mean. Then add each group's variance to the square of the distance of its mean from the combined mean, weight by group size and divide by the total size: $σ^{2} = \frac{n _{1} ( σ _{1}^{2} + d _{1}^{2} ) + n _{2} ( σ _{2}^{2} + d _{2}^{2} )}{n _{1} + n _{2}}$ . Simply averaging the variances is wrong unless the means are equal.

What is the variance of the first n natural numbers?

The variance of $1, 2, \dots, n$ is $\frac{n ^{2} - 1}{12}$ , so the standard deviation is $\frac{n ^{2} - 1}{12}$ . The same value holds for any $n$ consecutive integers, since shifting does not change variance. For an arithmetic progression with common difference $d$ , multiply the variance by $d^{2}$ .

How are variance and standard deviation asked in JEE Main and JEE Advanced?

JEE Main often asks for corrected variance, missing observations from a given mean and variance, combined variance, and the effect of shifting or scaling data. JEE Advanced adds algebraic data sets, such as arithmetic progressions or binomial coefficients, and inequalities like $\frac{\sum x _{i}^{2}}{n} \geq \overset{x}{ˉ}^{2}$ that restrict possible values of $n$ .

Previous year questions on Measure of Dispersion

21 questions from past papers, each with a step-by-step solution.

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1. What Is Dispersion?

2. Range

3. Mean Deviation

3.1 Mean Deviation for Ungrouped Data

3.2 Mean Deviation for Discrete and Grouped Data

3.3 Key Facts about Mean Deviation

4. Variance and Standard Deviation

4.1 Definitions

4.2 Formulas That Avoid Computing Every Deviation

4.3 Shortcut and Step-Deviation Methods

4.4 Variance of the First n Natural Numbers

5. Properties of Variance and Standard Deviation

5.1 Change of Origin and Scale

5.2 Mean Square Deviation about Any Point

5.3 Combined Variance of Two Groups

6. Coefficient of Variation: Comparing Distributions

7. Solved Examples

Common Mistakes to Avoid

Frequently Asked Questions

What are measures of dispersion?

What is the difference between variance and standard deviation?

Why is standard deviation preferred over mean deviation?

How does adding or multiplying by a constant affect variance and standard deviation?

What is the coefficient of variation used for?

How do you find the combined variance of two groups?

What is the variance of the first n natural numbers?

How are variance and standard deviation asked in JEE Main and JEE Advanced?

Previous year questions on Measure of Dispersion

Ready to master Statistics?

4.4 Variance of the First $n$ Natural Numbers