SPI Mathematics · Statistics

Mean of Grouped Data

For a frequency table, each class is stood in for by its class mark, and the mean is the sum of frequency times class mark divided by the total frequency.

All Statistics notes

Why this matters

Ten questions in the bank, seven of them EASY. Most hand you the totals and ask for the mean or a missing number in one line, by the direct, assumed mean or step deviation formula. One HARD question asks for two missing frequencies from the mean and the total.

Concept 1 of 3: Class marks and the direct method

In a table you do not know the exact values inside a class, so you treat every value in it as sitting at the middle of the class. Then frequency times class mark is that class's share of the total.

Definition

Class mark $= \dfrac{\text{lower limit} + \text{upper limit}}{2}$ , the mid-point of the class.
Class size (width) $=$ upper limit $-$ lower limit. It is not the class mark.
Mean $= \dfrac{\sum f_i x_i}{\sum f_i}$ , with $x_i$ the class marks.
Divide by the total frequency $\sum f_i$ , never by the number of classes.
Two missing frequencies need two equations: one from the total frequency, one from $\sum f_i x_i = \bar{x} \times \sum f_i$ .

Direct method

\text{class mark} = \frac{\text{lower} + \text{upper}}{2} \qquad \bar{x} = \frac{\sum f_i x_i}{\sum f_i}

$f_i$ frequency of a class
$x_i$ class mark of that class

Worked example

The classes 0-10, 10-20, 20-30 and 30-40 have frequencies 3, 5, 8 and 4. Find the mean.

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 1 · Statistics · Mean of Grouped Data

The mean of the following frequency distribution is

Class Interval	0-6	6-12	12-18	18-24	24-30
Frequency	7	5	10	12	6

Multiplying by a class limit

Each frequency is multiplied by the class mark, the mid-point. Using the lower or upper limit shifts the mean by half a class.

Dividing by the number of classes

The mean of a table is

\sum f_i x_i

over the total frequency

\sum f_i

. A table with 6 classes and 50 students divides by 50, not 6.

Using only one equation for two unknown frequencies

When two frequencies are missing, the total frequency gives one equation and the mean gives the other. Solve them together, then check that the frequencies add to the total.

Drill 4 more on class marks and the direct method

Concept 2 of 3: Assumed mean method

Large class marks make big products. Pick one class mark near the middle as a guess a, measure every class mark from it, and the small deviations tell you how far the true mean is from the guess.

Definition

Choose an assumed mean a, usually a class mark near the middle.
Deviation $d_i = x_i - a$ , measured from the class mark $x_i$ of each class.
Mean $= a + \dfrac{\sum f_i d_i}{\sum f_i}$ .
Classes below a have negative deviations; keep the signs.

Assumed mean method

d_i = x_i - a \qquad \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

aassumed mean
$d_i$ deviation of a class mark from a

Worked example

Class marks 15, 25, 35, 45 have frequencies 4, 6, 7, 3. Find the mean taking a = 25.

Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Statistics · Mean of Grouped Data

The mean of the data when

\sum f_i d_i = 435

\sum f_i = 30

and a = 47.5 is

Dropping the minus signs

Class marks below the assumed mean give negative deviations. Adding them as positive numbers makes

\sum f_i d_i

too large and the mean too high.

Drill 2 more on assumed mean method

Concept 3 of 3: Step deviation method

When the classes all have the same width h, every deviation is a whole number of steps of h. Divide by h to work with small numbers, then multiply back by h at the end.

Definition

$u_i = \dfrac{x_i - a}{h}$ , where h is the class size.
Mean $= a + h \times \dfrac{\sum f_i u_i}{\sum f_i}$ .
If $u_i = \frac{x_i - 40}{5}$ is given, read off a = 40 and h = 5.

Step deviation method

u_i = \frac{x_i - a}{h} \qquad \bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}

hclass size, the same for every class

Worked example

Class marks 15, 25, 35, 45, 55 have frequencies 2, 3, 5, 6, 4. Find the mean by the step deviation method with a = 35 and h = 10.

Practice this conceptself-check · 2 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Statistics · Mean of Grouped Data

u_i = \frac{x_i - 25}{10}

\Sigma f_iu_i = 20

\Sigma f_i = 100

, then

\overline{x}

Forgetting to multiply by h

In the step deviation method,

\frac{\sum f_i u_i}{\sum f_i}

counts steps, not marks. Multiply it by the class size h before adding it to a.

Drill 1 more on step deviation method

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Class marks and the direct method
Direct method
$\text{class mark} = \frac{\text{lower} + \text{upper}}{2} \qquad \bar{x} = \frac{\sum f_i x_i}{\sum f_i}$
Assumed mean method
Assumed mean method
$d_i = x_i - a \qquad \bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$
Step deviation method
Step deviation method
$u_i = \frac{x_i - a}{h} \qquad \bar{x} = a + h \times \frac{\sum f_i u_i}{\sum f_i}$

Watch out for (5)

Multiplying by a class limit
→ Class marks and the direct method
Dividing by the number of classes
→ Class marks and the direct method
Using only one equation for two unknown frequencies
→ Class marks and the direct method
Dropping the minus signs
→ Assumed mean method
Forgetting to multiply by h
→ Step deviation method

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