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Averages from tables

Free Printable Averages from tables PDF Worksheets, Revision Guide & GCSE Past Paper Questions (Grade 4)

Download free averages from tables worksheets (PDF), GCSE maths past paper questions and worked solutions for Averages from tables (4W). This is a Grade 4 statistics topic from the GCSE questions by topic collection for Foundation and Higher tier crossover GCSE Maths (Edexcel, AQA, OCR and IGCSE).

The free PDF worksheets table below brings together printable question booklets (Qs) and worked answers or mark schemes (Ans) from Maths Genie, Maths4Everyone, Corbett Maths, Maths Made Easy, Dr Frost, Naiker Maths, Dr Austin, 1st Class Maths, Metatutor, GCSE Maths Rocket, E Classroom and PMT, along with past paper question booklets.

Below the worksheet links you can read our averages from tables revision notes and worked examples with diagrams, try real GCSE past paper averages from tables questions with step-by-step solutions, or download a PDF of all averages from tables worksheet links.

Free Averages from tables PDF Worksheets & Answers

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Guide to Averages from tables: Explanation, Diagrams & Worked Examples

Download Printable One Page Guide (PDF)

Explanation:

For an ungrouped frequency table, find the mean by multiplying each value x by its frequency f, then divide the total of the fx column by the total frequency Σf (never divide by the number of rows!).

For a grouped frequency table (with intervals like 0 < t ≤ 10), you do not know the exact values, so find the midpoint of each interval first and calculate Estimated Mean = Σ(f × midpoint)Σf. The modal class is the interval with the highest frequency.

Example 1: Find the mean number of goals per match for 20 matches.

Goals (x)0123Total
Freq (f)476320
f × x0712928
Mean = ΣfxΣf = 2820 = 1.4 goals

Example 2: Estimate the mean time from the grouped table.

Time t0<t≤1010<t≤2020<t≤30Total
Freq (f)610420
Midpoint51525—
f × Mid30150100280
Estimated Mean = 28020 = 14 mins (Modal class: 10 < t ≤ 20)
Scanned GCSE Exam Questions & Worked Solutions

GCSE Past Paper Averages from tables Questions & Answers

Past Paper Example 1 IGCSE Nov 2023 2F Q17 / 2H Q1 4 marks
The table shows information about the lengths, in minutes, of 50 telephone calls.
Length of telephone call(m minutes)Frequency0 < m ≤ 585 < m ≤ 10210 < m ≤ 15615 < m ≤ 20420 < m ≤ 251225 < m ≤ 3018
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the total length, in minutes, of these telephone calls. [3 marks]
Worked Solution & Mark Scheme (4 marks)
Part (a) Solution:
Highest frequency is 18 for 25 < m ≤ 30 B1
Answer (a): 25 < m ≤ 30
Part (b) Solution:
Σ(f × midpoint) = 8×2.5 + 2×7.5 + 6×12.5 + 4×17.5 + 12×22.5 + 18×27.5 = 20 + 15 + 75 + 70 + 270 + 495 = 945 minutes M2 A1
Answer (b): 945 minutes
Past Paper Example 2 IGCSE Nov 2024 1F Q18 / 1H Q1 5 marks
The table shows some information about the hourly rates of pay of 60 workers.
Hourly rate of pay(p dollars)Frequency10 < p ≤ 151815 < p ≤ 201620 < p ≤ 251425 < p ≤ 30830 < p ≤ 354
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean hourly rate of pay of the 60 workers. [4 marks]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
Highest frequency is 18 in 10 < p ≤ 15 B1
Answer (a): 10 < p ≤ 15
Part (b) Solution:
(12.5×18 + 17.5×16 + 22.5×14 + 27.5×8 + 32.5×4) ÷ 60 = 1170 ÷ 60 = 19.5 M3 A1
Answer (b): 19.5 dollars
Past Paper Example 3 IGCSE Jan 2021 1F Q18 / 1H Q2 5 marks
The table gives information about the speeds, in kilometres per hour, of 80 motorbikes as each pass under a bridge.
Speed(s kilometres per hour)Frequency40 < s ≤ 501050 < s ≤ 601660 < s ≤ 701970 < s ≤ 802380 < s ≤ 9012
(a) Write down the modal class. [1 mark]
(b) Work out an estimate for the mean speed of the motorbikes as they pass under the bridge.
Give your answer correct to 3 significant figures.
[4 marks]
Worked Solution & Mark Scheme (5 marks)
Part (a) Solution:
Highest frequency is 23 in 70 < s ≤ 80 B1
Answer (a): 70 < s ≤ 80
Part (b) Solution:
(10×45 + 16×55 + 19×65 + 23×75 + 12×85) ÷ 80 = (450 + 880 + 1235 + 1725 + 1020) ÷ 80 = 5310 ÷ 80 = 66.375 ≈ 66.4 M3 A1
Answer (b): 66.4 kilometres per hour

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