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How do you find the median of a histogram?

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    I'm doing AQA non-modular maths paper 2 tomorrow...... revising at the mo, but I can't find in any of my books how to find the median of a histogram.

    I'm sure we're expected to know this but i've wasted about an hours revision time looking through all the books to find some notes on how to do this.... without success!

    Sorry for posting a boring thread but I'm getting in a bit of a flap!! :confused:
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    Any mathematicians around!?
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    http://www.analytictech.com/mb313/avghist.htm
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    Surely they'll give you data, from which you can interpret the median more easily. If not, you had better get good at counting squares!

    The mean of data like that is easy to calculate, but you specified median.. Erk! I should know - I am doing F. Maths next year. I'll make it my mission to see if there is an easier way than counting sqaures....

    Won't be two ticks!
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    The point with half the area to the left, half to the right.
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    (Original post by Rosalily)
    Any mathematicians around!?
    It is not possible to find the mean or median of the data which provided a histogram, but a ``best'' estimate for the mean or median can be calculated, and bounds on where the mean or median can be are obtainable. The ``best'' estimates are obtained by assuming that the data is uniformly spread within each class.

    \noindent {\sl Example}: Consider a histogram which has 10 data in the class with class mark 100, 12 data in the class with class mark 125, 20 data in the class with class mark 150, 8 data in the class with class mark 175, and 5 data in the class with class mark 200. What can you say about the mean and median of the data?

    The ``best'' estimate for the mean is obtained by assuming the data is uniformly spread within each interval; for purposes of calculating the mean, this is equivalent to assuming that all the data lie on the class marks. In this example the ``best'' estimate for the mean is $$ \mu = \frac{10 \times 100 + 12 \times 125 + 20 \times 150 + 8 \times 175 + 5 \times 200}{55} = 143\frac{7}{11}. $$

    In order to get bounds on the mean, it is necessary to know the class boundaries, which are halfway between the class marks. Adding or subtracting $\frac{25}{2}$ from the class marks provides the class boundaries 87.5, 112.5, 137.5, etc. The least possible mean would occur if all of the data in each class were at the lower class boundary. In this example the least possible mean is $$ \mu = \frac{10 \times 87.5 + 12 \times 112.5 + 20 \times 137.5 + 8 \times 162.5 + 5 \times 187.5}{55} = 131\frac{3}{22}. $$ Similarly, the greatest possible value for the mean is $156\frac{3}{22}$. \vspace*{2

    basically u cant, bu it is near the peak, or some *****, figure it out..

    do u have geography decision making tomorrow and do u know what is on it?
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    I've got it now


    You find the tallest bar, and draw diagonally across to the corner of the bar next to it, on both sides

    The point where the lines intersect is the median
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    My A-level book has virtually every other kind of average you can calculate (Arithmetic mean, Weighted mean, Geometric mean, Harmonic mean!), but suggests that you get counting for the median. Surely what yours is talking about is the Modal class?

    For instance, I could have 1 very tall bar at the beginning, but a higher collective frequency density of bars past that bar.... Yes - the way you suggest MUST be the mode.
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    Plus that would fit with the equation we learnt in class:

    Mode = L + c( ∂1 / (∂1 + ∂2)):

    Where ∂1 is the height from the tallest on the lower class side, and ∂2 is the height from the tallest on the higher class side, and where L is the lower boundary of the modal class, and c is the width of the modal class!

    The intersection method results in the same thing!
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    i dont thiink this is gcse level. I think it is higher like A levels. For mode we only need to know is that it is the number which repeats the most, has the highest frequency.
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    (Original post by IZZY!)
    i dont thiink this is gcse level. I think it is higher like A levels. For mode we only need to know is that it is the number which repeats the most, has the highest frequency.
    no, it's gcse i reckon. you just havent met this idea in this way. you do edexcel anyway...
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    (Original post by henryt)
    Plus that would fit with the equation we learnt in class:

    Mode = L + c( ∂1 / (∂1 + ∂2)):

    Where ∂1 is the height from the tallest on the lower class side, and ∂2 is the height from the tallest on the higher class side, and where L is the lower boundary of the modal class, and c is the width of the modal class!

    The intersection method results in the same thing!

    Therefore I'll stick with the intersection method... much simpler!
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    But remember, when using the intersection method, you're tying to find the MODE. You said you wanted the MEDIAN?

    The median is simply the point where half the frequency density is on one side, and (naturally!) the other half is on the other side.
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    I thought that the mode was just the tallest bar?
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    I'm so confused!!
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    I thought that the mode is the group with the highest frequency and the median is the line that you draw. :confused: anyone agree?
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    Well - It depends how you define the mode. I am looking at the VALUE of the mode, whereas you are looking at determining the Modal CLASS.

    They'll specify which they want.

    However, the median is the middle piece of data. This doesn't neccessarily have to lie in the modal class (see stem/leaf diag below).

    0 | 1,2,3,4,5,6,7,8,9
    1 | 1,2,3,4,5,6,7,8
    2 |
    3 | 3
    4 | 4
    5 |

    The modal class is the first class ("0"), but the median is the first value in class "1" (of value 11).
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    Awesome, it came up and I managed to do it! I LOVE TSR!
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    I wasnt on TSR last night and THERE WAS A DAMN HISTOGRAM!!!
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    Ooooommmmggggg it came up I knew it would!

    I got a bit confused but hopefully did ok! :confused:
Updated: June 15, 2005
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