Grouped data and the assumed mean · Statistics: data & averages

When there are many different values (marks out of 100, ages, weights), a table with one column per value would be enormous. So we group the values into classes such as 1–10, 11–20, 21–30, and count how many fall in each.

The words you need

For the class 21–30:

Word	Meaning	For 21–30
class limits	the numbers written	21 and 30
class boundaries	halfway to the next class	20.5 and 30.5
class width (size)	upper boundary − lower boundary	$30.5 - 20.5 = 10$
class mark (mid-point)	the middle of the class	$\frac{21 + 30}{2} = 25.5$

The mean of grouped data

We don’t know the exact values inside a class, so we let the class mark stand for every value in it. Then it’s the same as a frequency table:

\bar x = \frac{\sum fx}{\sum f}, \quad \text{where } x \text{ is the class mark}

The answer is an estimate, and that’s fine: it’s what the question expects.

The mean from a frequency tableFill in the table

class	f	class mark x	f × x
1 – 10	3
11 – 20	7
21 – 30	12
31 – 40	10
41 – 50	6
51 – 60	2
total	40

The class mark is the middle of the class: (lower + upper) ÷ 2, e.g. (1 + 10) ÷ 2 = 5.5. Then multiply by the frequency for each row and add up the column.

Fill in the class marks and $f \times x$ for each row. A cell turns green when it’s right. Then try the assumed mean method on the same table.

The assumed mean method

Pick a class mark near the middle, call it $A$ (the assumed mean), and work with $d = x - A$ instead of $x$ . The numbers are smaller, many $d$ ‘s are negative and cancel, and

\bar x = A + \frac{\sum fd}{\sum f}

It always gives the same answer as $\frac{\sum fx}{\sum f}$ . Use it when the question asks for it, or when the class marks are large.

Worked example · WAEC 2023

WAEC 2023 · Paper 2 · Q11 (a)

The table shows the frequency distribution of the ages of fifty members of a family.

Ages (years)	1–5	6–10	11–15	16–20	21–25	26–30
Frequency	7	12	4	6	11	10

Calculate, correct to two decimal places, the:

mean;

Class marks
$\frac{1 + 5}{2} = 3$ , then add the class width 5 each time: $3, 8, 13, 18, 23, 28$ .
Think first. What is the class mark of 1–5? Of 6–10?
Choose A and find d
$d = -15, -10, -5, 0, 5, 10$ .
Think first. Take $A = 18$ . What are the values of $d = x - 18$ ?
Multiply by f and add

Ages $f$ $x$ $d$ $fd$
1–5 7 3 −15 −105
6–10 12 8 −10 −120
11–15 4 13 −5 −20
16–20 6 18 0 0
21–25 11 23 5 55
26–30 10 28 10 100
$\sum f = 50$ $\sum fd = -90$
The mean
$\bar x = 18 + \frac{-90}{50} = 18 - 1.8 = 16.20 \text{ years}$
Directly: $\sum fx = 810$ and $810 \div 50 = 16.20$ , the same.

Ages	$f$	$x$	$d$	$fd$
1–5	7	3	−15	−105
6–10	12	8	−10	−120
11–15	4	13	−5	−20
16–20	6	18	0	0
21–25	11	23	5	55
26–30	10	28	10	100
	$\sum f = 50$			$\sum fd = -90$

The median and mode of grouped data

Find the class first, then estimate inside it.

Median. With $N = \sum f$ values, the median is the $\frac N2$ th. Keep a running total of the frequencies to find the class it’s in, then

\text{median} = L + \left(\frac{\frac N2 - F}{f_m}\right) \times c

where $L$ is the lower boundary of the median class, $F$ the running total before it, $f_m$ its frequency and $c$ the class width. (The cumulative frequency curve in the next topic does the same job by drawing.)

Mode. The modal class has the highest frequency. With $\Delta_1$ = (its frequency − the one before) and $\Delta_2$ = (its frequency − the one after):

\text{mode} = L + \left(\frac{\Delta_1}{\Delta_1 + \Delta_2}\right) \times c

This is exactly what the crossed lines on a histogram measure; you’ll see them in the charts lesson.

Worked example · NECO 2024

NECO 2024 · Paper 2 · Q12 (a)

The table shows the scores of candidates.

Marks (%)	1–10	11–20	21–30	31–40	41–50	51–60
Frequency	26	36	38	30	15	5

Calculate, correct to one decimal place, the (i) mean; (ii) median; (iii) modal score.

The mean
$\sum f = 150$ , adding the frequencies.
$\begin{aligned} \sum fx &= 26(5.5) + 36(15.5) \\ &\quad + 38(25.5) + 30(35.5) \\ &\quad + 15(45.5) + 5(55.5) \\ &= 3695 \end{aligned}$
$\bar x = \frac{3695}{150} \approx 24.6$ , dividing $\sum fx$ by $\sum f$ .
Think first. Class marks are 5.5, 15.5, …. What are $\sum f$ and $\sum fx$ ?
The median class
Running totals: $26, 62, 100, \ldots$ . The 75th is after the 62nd, so it’s in 21–30, with $L = 20.5$ , $F = 62$ , $f_m = 38$ , $c = 10$ .
Think first. The median is the 75th score. Which class is it in?
The median
$20.5 + \frac{75 - 62}{38} \times 10 = 20.5 + 3.42 \approx 23.9$
The mode
$\Delta_1 = 38 - 36 = 2$ and $\Delta_2 = 38 - 30 = 8$ :
$20.5 + \frac{2}{2 + 8} \times 10 = 22.5$
Think first. The modal class is 21–30. Find $\Delta_1$ and $\Delta_2$ .

Your turn

WAEC 2012 · Paper 2 · Q10 (a)

Marks	60–64	65–69	70–74	75–79	80–84	85–89	90–94	95–99
Frequency	2	3	6	11	8	7	2	1

The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the:

(a)
mean;

Worked solution (try it first)

(a)

Mean $= A + \frac{\sum fd}{\sum f}$
$= 77 + \frac{70}{40}$
$= 77 + 1.75$
$= 78.75$ .

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