Frequency Tables and Averages

The three averages — mean, median and mode — along with the ability to calculate them from frequency tables and grouped frequency tables, are among the most commonly tested statistics topics on all three Edexcel papers.

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Key Vocabulary

Term	Definition
Mean	Sum of all values ÷ number of values
Median	The middle value when data is arranged in order
Mode	The most frequently occurring value
Modal class	The class interval with the highest frequency (grouped data)
Frequency	How many times a value or class occurs
Estimated mean	Mean calculated from grouped data using midpoints (an estimate because we don't know exact values)

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Averages from a List of Data

Worked Example 1

Find the mean, median and mode of: 3, 5, 7, 7, 8, 10, 12.

Solution:

• Mean = (3 + 5 + 7 + 7 + 8 + 10 + 12) ÷ 7 = 52 ÷ 7 = 7.43 (2 d.p.)
• Median: 7 values → median position = (7 + 1) ÷ 2 = 4th value → median = 7
• Mode = 7 (appears twice; all others appear once)

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Averages from a Frequency Table

When data is presented in a frequency table, you use the fx column to find the mean.

Worked Example 2

The table shows the number of pets owned by 30 students. Find the mean, median and mode.

Pets (x)	Frequency (f)	f × x
0	4	0
1	9	9
2	8	16
3	5	15
4	3	12
5	1	5
Total	30	57

Solution:

• Mean = $\frac{\sum fx}{\sum f}$∑f∑fx​ = 57 ÷ 30 = 1.9 pets
• Median: 30 values → median is the average of the 15th and 16th values. Cumulative frequencies are 4, 13, 21, 26, 29, 30. The 15th and 16th both lie in the "2 pets" row (cumulative frequency reaches 21 by x = 2). Median = 2 pets
• Mode = 1 pet (highest frequency = 9)

Worked Example 3

The table shows the number of siblings for 50 students. Find the median number of siblings.

Siblings	0	1	2	3	4	5
Frequency	6	14	18	8	3	1

Solution:

• n = 50, so median position = (50 + 1) ÷ 2 = 25.5th value (average of 25th and 26th).
• Cumulative frequencies: 6, 20, 38, 46, 49, 50.
• The 25th value lies in the "2 siblings" row (CF reaches 38 at x = 2). So does the 26th.
• Median = 2 siblings

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Estimating the Mean from a Grouped Frequency Table

When data is grouped, we do not know exact values — we use the midpoint of each class interval as our estimate.

Worked Example 4

The table shows the time (t minutes) taken by 40 students to complete a puzzle.

Time (t minutes)	Frequency (f)	Midpoint (x)	f × x
$0 < t \leq 5$0<t≤5	3	2.5	7.5
$5 < t \leq 10$5<t≤10	8	7.5	60
$10 < t \leq 15$10<t≤15	14	12.5	175
$15 < t \leq 20$15<t≤20	10	17.5	175
$20 < t \leq 25$20<t≤25	5	22.5	112.5
Total	40		530

Solution:

• Estimated mean = 530 ÷ 40 = 13.25 minutes
• Modal class = $10 < t \leq 15$10<t≤15 (highest frequency = 14)
• Median class: Median position = (40 + 1) ÷ 2 = 20.5th value. Cumulative frequencies: 3, 11, 25, 35, 40. The 20.5th value lies in the $10 < t \leq 15$10<t≤15 class.

Step-by-step method for the estimated mean:

1. Find the midpoint of each class: (lower bound + upper bound) ÷ 2.
2. Multiply each midpoint by its frequency to get the fx column.
3. Add the fx column to get Σfx.
4. Divide by Σf.

Worked Example 5

Estimate the mean height of 60 children from the table below.

Height (h cm)	Frequency
$120 \leq h < 130$120≤h<130	5
$130 \leq h < 140$130≤h<140	14
$140 \leq h < 150$140≤h<150	22
$150 \leq h < 160$150≤h<160	15
$160 \leq h < 170$160≤h<170	4

Solution:

Class	f	Midpoint	fx
120–130	5	125	625
130–140	14	135	1,890
140–150	22	145	3,190
150–160	15	155	2,325
160–170	4	165	660
Total	60		8,690

Estimated mean = 8,690 ÷ 60 = 144.83 cm (2 d.p.)

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Averages from a Frequency Table — Exam Style

Worked Example 6

The table shows the marks scored by 25 students in a test.

Mark	3	4	5	6	7	8
Frequency	2	5	6	7	3	2

(a) Calculate the mean mark. (b) Find the median mark. (c) Write down the mode. (d) Which average best represents the data? Give a reason.

Solution:

(a) Σfx = (3×2) + (4×5) + (5×6) + (6×7) + (7×3) + (8×2) = 6 + 20 + 30 + 42 + 21 + 16 = 135 Mean = 135 ÷ 25 = 5.4

(b) 25 values → median is the 13th value. Cumulative frequencies: 2, 7, 13, 20, 23, 25 The 13th value is in the "5 marks" row → Median = 5

(c) Mode = 6 (highest frequency = 7)

(d) The mean (5.4) is best because it uses all the data and there are no extreme outliers to distort it. Alternatively, the median (5) is appropriate because it is not affected by the slightly higher or lower values.

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Finding the Mean When One Value is Unknown

Worked Example 7

Five numbers have a mean of 8. Four of the numbers are 5, 7, 9 and 11. Find the fifth number.

Solution:

• Mean = sum ÷ count → 8 = sum ÷ 5 → sum = 40
• Sum of the four known values = 5 + 7 + 9 + 11 = 32
• Fifth number = 40 − 32 = 8

Worked Example 8

The mean of 10 numbers is 17. When another number is added, the mean rises to 18. What is the added number?

Solution:

• Original sum = 10 × 17 = 170
• New sum = 11 × 18 = 198
• Added number = 198 − 170 = 28

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Combining Means

Worked Example 9

Class A has 20 students with a mean test score of 64. Class B has 30 students with a mean of 72. Find the overall mean.

Solution:

• Total for A = 20 × 64 = 1,280
• Total for B = 30 × 72 = 2,160
• Overall mean = (1,280 + 2,160) ÷ (20 + 30) = 3,440 ÷ 50 = 68.8

Common mistake: Do NOT just average 64 and 72 to get 68. The classes have different sizes so you must weight by the number of students in each.

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Choosing the Appropriate Average