GCSEMathsWJECMaths & Numeracy (Double Award)HigherRevision NotesStatistics & ProbabilityHistogramsMedian from a Histogram

Median from a Histogram (WJEC GCSE Maths & Numeracy (Double Award)): Revision Note

Exam code: 3320

Roger B

Written by: Roger B

Reviewed by: Jamie Wood

Updated on

WJEC Maths & Numeracy (Double Award): HigherEdexcel Statistics: Higher

Median from a Histogram

How do I find the median for data on a histogram?

  • A histogram is a way of representing grouped data as a diagram

  • The connection between the frequency density shown on the histogram and the frequency that would be shown in a grouped data table is given by the formula

    • frequency space density equals fraction numerator frequency over denominator class space width end fraction

  • If you are asked to estimate a median for data in a histogram there are two options:

    • You can recreate the grouped data table using the frequency density formula, and then use linear interpolation

      • Linear interpolation is explained in Median from Grouped Data

    • Or you can work out the estimated median directly from the histogram

      • See the following Worked Example for how to do this

Worked Example

The histogram shows the weight, in kg, of 60 new-born bottlenose dolphins.

A histogram for the weights of newborn bottlenose dolphins

Find an estimate for the median weight of the dolphins in the sample, giving your answer correct to two decimal places.

Answer:

60 over 2 equals 30, so to estimate the median we need to find the weight of the 30th dolphin

To find the frequencies represented by the different bars, rearrange  frequency space density equals fraction numerator frequency over denominator class space width end fraction  as  frequency equals frequency space density cross times class space width

 
4 minus 8 space kg colon space space frequency equals 1 cross times 4 equals 4 8 minus 10 space kg colon space space frequency equals 8 cross times 2 equals 16 10 minus 12 space kg colon space space frequency equals 9.5 cross times 2 equals 19
 

The first two classes have a cumulative frequency of 4 plus 16 equals 20
So the median is going to be '10 dolphins into' the 10-12 kg class

The height (frequency density) of the 10-12 kg bar is 9.5
We need to find what width would give a frequency of 10
Use  frequency space density equals fraction numerator frequency over denominator class space width end fraction   and solve for width
 

9.5 equals 10 over width 9.5 cross times width equals 10 width equals fraction numerator 10 over denominator 9.5 end fraction equals 20 over 19 equals 1.0526... equals 1.05 space open parentheses 2 space straight d. straight p. close parentheses

 
That means that the median lies 1.05 kg into the 10-12 kg class interval

10 plus 1.05 equals 11.05

A histogram showing the part of the 10-12 kg class interval calculated in the answer

 
Estimated median = 11.05 kg (2 d.p.)

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    Author
    Reviewer
    Roger B

    Author: Roger B

    Expertise: Maths Content Creator

    Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

    Jamie Wood

    Reviewer: Jamie Wood

    Expertise: Maths Content Creator

    Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.