Formulas · Mathematics

Measures of dispersion

4 formulas, each with worked examples. Revision sheet · Practise these formulas

Class 11

Mean deviation about the mean

M . D . (x ˉ) = \frac{\sum ∣ x _{i} - x ˉ ∣}{n}

What each letter means
$M . D . (x ˉ)$	the mean deviation about the mean: the average distance of the values from their mean
$\sum ∣ x_{i} - x ˉ ∣$	the sum of the distances of the values from the mean
$n$	how many values there are

The mean deviation is the average distance of the values from their mean: find each value's distance from the mean (ignoring the sign), add them, and divide by how many there are.

Why it works

The plain differences xᵢ − x̄ always add up to 0, the positives cancelling the negatives, so they say nothing about spread. Taking their sizes |xᵢ − x̄| stops the cancelling, and averaging them gives a typical distance from the centre.

When to use it

Measuring how spread out a set of values is around its mean, in the same units as the data.

How to use it

Find the mean x̄.
Find each distance |xᵢ − x̄|, always positive.
Add the distances and divide by n. For a frequency table, multiply each distance by its frequency and divide by the total frequency.

To remember: The average distance from the average.

Other forms

Rearranged $\sum ∣ x_{i} - x ˉ ∣ = n \times M . D . (x ˉ)$ The total of the distances.
To remember itFor a frequency table: M.D. = Σfᵢ|xᵢ − x̄| ÷ Σfᵢ.

Worked examples

Story

A shop sold 3, 5, 7, 9 and 11 kg of rice on five days. Find the mean deviation about the mean.

The mean is 35 ÷ 5 = 7; the distances from it are 4, 2, 0, 2 and 4.
$4 + 2 + 0 + 2 + 4 = 12$
$\frac{12}{5} = 2.4$

Answer: 2.4

Picture

Four points on a number line are at 1, 3, 5 and 11. On average, how far are they from their balance point, the mean?

The mean is 20 ÷ 4 = 5; the distances from it are 4, 2, 0 and 6.
$4 + 2 + 0 + 6 = 12$
$\frac{12}{4} = 3$

Answer: 3

Direct

Find the mean deviation about the mean of 6, 7, 10, 12, 13, 4, 8, 12.

The mean is 72 ÷ 8 = 9, and the distances from 9 are 3, 2, 1, 3, 4, 5, 1 and 3.
$3 + 2 + 1 + 3 + 4 + 5 + 1 + 3 = 22$
$\frac{22}{8} = 2.75$

Answer: 2.75

Reverse

The mean deviation of 10 values about their mean is 1.5. What is the total of their distances from the mean?

Try it first, then show the working

$1.5 \times 10 = 15$

Answer: 15

Exam

Find the mean deviation about the mean for x = 2, 5, 6, 8, 10, 12 with frequencies 2, 8, 10, 7, 8, 5.

Try it first, then show the working

The total frequency is 40 and Σfᵢxᵢ = 300, so the mean is 300 ÷ 40 = 7.5.
Each frequency times the distance from 7.5, added: 11 + 20 + 15 + 3.5 + 20 + 22.5 = 92.
$\frac{92}{40} = 2.3$

Answer: 2.3

Common mistake: Keeping the signs of xᵢ − x̄: then the sum is always 0.

Class 11

Standard deviation

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

What each letter means
$σ$	the standard deviation; its square σ² is the variance
$\sum (x_{i} - x ˉ)^{2}$	the sum of the squared distances of the values from the mean
$n$	how many values there are

The variance is the average of the squared distances from the mean; the standard deviation σ is its square root, which brings it back to the units of the data.

Why it works

Squaring the distances, like taking their sizes, stops positives and negatives cancelling, and it gives large distances more weight. The square root at the end undoes the squaring of the units (cm² back to cm).

When to use it

The standard measure of spread: comparing how consistent two sets of data are, and in all later statistics.

How to use it

Find the mean x̄.
Square each distance xᵢ − x̄ and add the squares.
Divide by n for the variance, then take the square root for σ.

To remember: Square, average, root.

Other forms

Same formula, another form $σ^{2} = \frac{\sum ( x _{i} - x ˉ ) ^{2}}{n}$ The variance.

Worked examples

Story

Five plants are 4, 6, 8, 10 and 12 cm tall. Find the standard deviation of their heights.

The mean is 8; the squared distances from it are 16, 4, 0, 4 and 16.
$16 + 4 + 0 + 4 + 16 = 40$
$\frac{40}{5} = 22$

Answer: 2.8284271247461903

Picture

Four points on a number line are at 1, 3, 5 and 7. Find the standard deviation of their positions.

The mean is 4; the squared distances from it are 9, 1, 1 and 9.
$\frac{20}{4} = 5$

Answer: 2.23606797749979

Direct

Find the variance and the standard deviation of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

The mean is 150 ÷ 10 = 15.
The squared distances from 15 add up to 81 + 49 + 25 + 9 + 1 + 1 + 9 + 25 + 49 + 81 = 330.
$\frac{330}{10} = 33$
So the variance is 33, and the standard deviation is √33, about 5.74.

Answer: 5.744562646538029

Reverse

Twenty values have standard deviation 3. What is the sum of their squared distances from the mean?

Try it first, then show the working

$3^{2} \times 20 = 180$

Answer: 180

Exam

Which set is more spread out: A = 2, 4, 6 or B = 1, 4, 7?

Try it first, then show the working

Both have mean 4. For A the squared distances add up to 4 + 0 + 4 = 8; for B they add up to 9 + 0 + 9 = 18.
$\frac{18}{3} = 6$
The variances are 8/3 and 6, so B, with the larger standard deviation, is more spread out.

Answer: B

Common mistake: Forgetting the square root (giving the variance), or squaring the sum of the distances instead of adding the squares.

Class 11

Standard deviation, the shortcut form

σ = \frac{\sum x _{i}^{2}}{n} - x ˉ^{2}

What each letter means
$σ$	the standard deviation
$\sum x_{i}^{2}$	the sum of the squares of the values
$n$	how many values there are
$x ˉ$	the mean of the values

The variance is the mean of the squares minus the square of the mean; this saves working out every distance from the mean.

Why it works

Multiply out Σ(xᵢ − x̄)² = Σxᵢ² − 2x̄Σxᵢ + n x̄². Since Σxᵢ = n x̄, this is Σxᵢ² − n x̄². Dividing by n gives Σxᵢ²/n − x̄².

When to use it

When the mean is not a whole number (so the distances are awkward), and when only the totals Σxᵢ and Σxᵢ² are known.

How to use it

Add the values and add their squares.
Divide each total by n: that gives x̄ and the mean of the squares.
Take x̄² from the mean of the squares, and take the square root.

To remember: The mean of the squares minus the square of the mean.

Other forms

Same formula, another form $σ^{2} = \frac{\sum x _{i}^{2}}{n} - x ˉ^{2}$ The variance.

Worked examples

Story

Eight test marks are 2, 4, 4, 4, 5, 5, 7 and 9. Find their standard deviation.

The mean is 40 ÷ 8 = 5, and the squares add up to 4 + 16 + 16 + 16 + 25 + 25 + 49 + 81 = 232.
$\frac{232}{8} - 5^{2} = 4$
$4 = 2$

Answer: 2

Picture

Five points on a number line are at 1, 2, 3, 4 and 5. Find the standard deviation of their positions.

The squares add up to 55 and the mean is 3.
$\frac{55}{5} - 3^{2} = 2$

Answer: 1.4142135623730951

Direct

For 6, 8, 10, …, 24 (ten values, mean 15), the squares add up to 2580. Find the variance with the shortcut.

$\frac{2580}{10} - 1 5^{2} = 33$
The same variance, 33, as the long way; σ = √33.

Answer: 5.744562646538029

Reverse

Ten values have mean 6 and standard deviation 2. Find the sum of their squares.

Try it first, then show the working

$10 \times (4 + 36) = 400$

Answer: 400

Exam

The mean and variance of 7 observations are 8 and 16. Five of them are 2, 4, 10, 12 and 14. Find the other two.

Try it first, then show the working

The total is 7 × 8 = 56, and the five add up to 42, so the other two add up to 14.
The sum of squares is 7 × (16 + 64) = 560, and the five give 460, so the other two give 100.
$6^{2} + 8^{2} = 100$
The two observations are 6 and 8.

Answer: 6 and 8

Common mistake: Taking (Σxᵢ)² for Σxᵢ²: square each value first, then add.

Class 11

Coefficient of variation

C V = \frac{σ}{x ˉ} \times 100

What each letter means
$C V$	the coefficient of variation: the standard deviation as a percentage of the mean
$σ$	the standard deviation
$x ˉ$	the mean (not 0)

The coefficient of variation is the standard deviation as a percentage of the mean. The set with the smaller CV is the more consistent (less variable) one.

Why it works

A spread of 5 means a lot around a mean of 10 but little around a mean of 1000. Dividing by the mean measures the spread relative to the size of the values, and with no units, so different sets can be compared fairly.

When to use it

Comparing the variability or consistency of two sets with different means or different units: two batsmen's scores, two factories' wages.

How to use it

Find the mean and the standard deviation of each set.
Divide σ by x̄ and multiply by 100.
The smaller CV is the more consistent set.

To remember: Spread as a percentage of the size.

Other forms

Rearranged $σ = \frac{C V \times x ˉ}{100}$ The standard deviation from the CV and the mean.

Worked examples

Story

Batsman A averages 50 runs with standard deviation 15; batsman B averages 40 with standard deviation 8. Who is more consistent?

$\frac{15}{50} \times 100 = 30$
$\frac{8}{40} \times 100 = 20$
B has the smaller CV, so B is more consistent.

Answer: B, with CV 20 against 30

Picture

Rods cut by a machine have mean length 120 cm and standard deviation 6 cm. Find the coefficient of variation of their lengths.

$\frac{6}{120} \times 100 = 5$

Answer: 5

Direct

A set has standard deviation 5 and mean 50. Find its coefficient of variation.

$\frac{5}{50} \times 100 = 10$

Answer: 10

Reverse

A set has a coefficient of variation of 25 and a mean of 40. Find its standard deviation.

Try it first, then show the working

$\frac{10}{40} \times 100 = 25$

Answer: 10

Exam

Two plants of a factory pay the same average monthly wage, ₹2500, with variances 81 and 100. Which plant's wages vary more?

Try it first, then show the working

The standard deviations are √81 = 9 and √100 = 10.
$\frac{9}{2500} \times 100 = 0.36$
$\frac{10}{2500} \times 100 = 0.4$
The second plant has the larger CV, so its wages vary more.

Answer: The second plant (CV 0.4 against 0.36)

Common mistake: Comparing standard deviations alone when the means differ a lot.