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Year 8 Maths Investigating Data

Analysing Frequency Tables

20 practice questions 1 video lesson Theory + worked examples
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Theory

A frequency table lists each value \(x\) once beside its frequency \(f\). Add an \(fx\) column and a cumulative frequency column, then \(\text{mean} = \dfrac{\sum fx}{\sum f}\) and the median is read from the running total. This Year 8 maths page shows how.

The columns. The value is \(x\) and its frequency is \(f\). The \(fx\) column is value \(\times\) frequency, and the cumulative frequency is a running total of \(f\). The symbol \(\sum\) means “add up”.

How many values: \(\sum f\), the sum of the frequency column; never add the value column. Range: largest \(x\) minus smallest \(x\). Mode: the value \(x\) in the row with the largest \(f\).

Mean: each row contributes \(fx\) to the total, so \(\text{mean} = \dfrac{\sum fx}{\sum f}\).

Median: find the middle position(s) from \(\sum f\), then read down the cumulative frequency column to the first row that reaches each position.

Relative frequency of a row \(= \dfrac{f}{\sum f}\); multiply by \(100\) for a percentage. The relative frequencies add to \(1\).

A missing frequency can be found from the total: the frequencies add to \(\sum f\).

Grouped data. The class centre is halfway between the ends of a class. The modal class has the largest \(f\). Using the class centres as \(x\) gives only an estimate of the mean.

Nights of sport A column graph of 30 students: 0 nights 4, 1 night 8, 2 nights 10, 3 nights 6, 4 nights 2. 0 2 4 6 8 10 12 0 4 1 8 2 10 3 6 4 2 Nights of sport (x) Frequency (f)
Thirty students: \(\sum fx = 0 + 8 + 20 + 18 + 8 = 54\), so the mean is \(54 \div 30 = 1.8\) nights. The two middle values are both \(2\) (cumulative \(4, 12, 22\)), so the median is \(2\); the mode is \(2\) and the range is \(4\).

The four columns for the nights of sport:

xffxCumulative
0404
18812
2102022
361828
42830
Total3054
Divide by \(\sum f\), not by the number of rows. The mean is \(\sum fx \div \sum f\); averaging the values in the first column ignores how often each happened.
The mode is a value. If the largest frequency is \(10\), in the row \(x = 2\), the mode is \(2\), not \(10\).

Statistics from a frequency table

  1. Add the \(f\) column for the number of values.
  2. Build the \(fx\) column and add it for the total; mean \(= \sum fx \div \sum f\).
  3. Build the cumulative frequency column; read down to the middle position(s) for the median.
  4. Read the mode (largest \(f\)) and the range (largest \(x\) \(-\) smallest \(x\)).

Grouped data

  1. Find each class centre.
  2. Multiply centre \(\times\) frequency, add, and divide by \(\sum f\) for an estimated mean.
Example 1 — Range, mode, median, mean
Scores \(4\) to \(10\) have frequencies \(1, 2, 3, 4, 3, 2, 1\). Find
(i) the range
(ii) the mode
(iii) the median
(iv) the mean.
Solution

There are \(16\) scores, so the median is halfway between the 8th and 9th. The cumulative frequencies are \(1, 3, 6, 10, \ldots\), so both are \(7\).

\(\text{Range}\)\(=\)\(10 - 4 = 6\)
\(\text{Median}\)\(=\)\(\dfrac{7 + 7}{2} = 7\)
\(\sum fx\)\(=\)\(4 + 10 + 18 + 28\)
\(\)\(\)\(+\ 24 + 18 + 10\)
\(\)\(=\)\(112\)
\(\text{Mean}\)\(=\)\(\dfrac{112}{16} = 7\)
range 6, mode 7, median 7, mean 7

(i) \(6\)
(ii) \(7\) (frequency \(4\))
(iii) \(7\)
(iv) \(7\)

Example 2 — From a list to a table
Test scores: \(5, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9\). Set up a frequency table and find the mean.
Solution
Score xffx
5210
6424
7321
8540
9218
Total16113
\(\bar{x}\)\(=\)\(\dfrac{113}{16}\)
\(\)\(\approx\)\(7.1\)
113÷16≈7.1

The mean is about \(7.1\).

Example 3 — An fx column
The \(20\) members of a debating club took part in \(1\) to \(5\) debates, with frequencies \(2, 5, 6, 4, 3\). Find the mean.
Solution
\(\sum fx\)\(=\)\(2 + 10 + 18 + 16 + 15\)
\(\)\(=\)\(61\)
\(\text{Mean}\)\(=\)\(\dfrac{61}{20} = 3.05\)
61÷20=3.05

\(3.05\) debates per member.

Example 4 — Median from cumulative frequency
\(24\) students packed their own lunch on \(0\) to \(5\) days last week, with frequencies \(3, 4, 5, 6, 4, 2\).
(i) Find the median.
(ii) What percentage packed lunch on more than \(3\) days?
Solution

Cumulative frequencies: \(3, 7, 12, 18, 22, 24\). The 12th value is \(2\) and the 13th is \(3\).

\(\text{Median}\)\(=\)\(\dfrac{2 + 3}{2} = 2.5\)
\(\dfrac{4 + 2}{24}\)\(=\)\(0.25 = 25\%\)
2.5 days 25

(i) \(2.5\) days
(ii) \(25\%\)

Example 5 — A missing frequency
A canteen served \(25\) customers. Items bought \(1\) to \(5\) had frequencies \(5, k, 6, 3, 1\).
(i) Find \(k\).
(ii) Find the mean.
(iii) Find the median.
Solution
\(5 + k + 6 + 3 + 1\)\(=\)\(25\)
\(k\)\(=\)\(10\)
\(5 + 20 + 18\)\(=\)\(43\)
\(43 + 12 + 5\)\(=\)\(60\)
\(\text{Mean}\)\(=\)\(\dfrac{60}{25} = 2.4\)
k =102.42

(i) \(k = 10\)
(ii) \(2.4\)
(iii) The 13th value: running totals \(5, 15\) put the 6th to 15th at \(2\), so the median is \(2\).

Example 6 — Grouped data
Fun-run times of \(30\) runners: \(16\)–\(20\) min (\(4\)), \(21\)–\(25\) (\(11\)), \(26\)–\(30\) (\(10\)), \(31\)–\(35\) (\(5\)).
(i) Find the class centre of \(26\)–\(30\).
(ii) Name the modal class.
(iii) Estimate the mean.
Solution
\(\dfrac{26 + 30}{2}\)\(=\)\(28\)
\(\sum fx\)\(=\)\(72 + 253 + 280 + 165\)
\(\)\(=\)\(770\)
\(\text{Mean}\)\(\approx\)\(770 \div 30\)
\(\)\(\approx\)\(25.7\)
282125 about 25.7 min

(i) \(28\)
(ii) \(21\)–\(25\) minutes
(iii) About \(25.7\) minutes; only an estimate, since every runner in a class counts as its centre.

Common pitfalls

Adding the value column. The number of values is \(\sum f\), not the sum of the \(x\) column.
Giving the frequency as the mode. The mode is the value with the largest frequency, not the frequency itself.
Range from the frequencies. The range uses the values: \(2\) to \(11\) gives \(9\), however large the frequencies.
Calling a grouped mean exact. Class centres give an estimate of the mean only.

Frequently asked questions

How do you find the mean from a frequency table?

Multiply each value by its frequency (\(fx\)), add these, then divide by the total frequency.

What is cumulative frequency?

A running total of the frequencies, row by row.

How do you find the median from a frequency table?

Find the middle position from the total frequency, then go down the cumulative frequency column to the first row that reaches it.

What is relative frequency?

A row’s frequency divided by the total frequency: \(8\) out of \(40\) is \(0.2\).

What is a class centre?

The number halfway between the ends of a class: for \(10\)–\(14\) it is \(12\).