Median from Histograms - LONG

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Hi all
I hope you can help me understand how I can estimate the median from a
histogram

On the histogram I have x- axis = time (in months)
y-axis = frequency density

There are 5 bars in the histogram

1st bar has area = 2 and the classwidth = 2
2nd bar has area = 14 and the classwidth = 2
3rd bar has area = 40 and the class width = 4
4th bar has area = 32 and the classwidth = 4
5th bar has area = 72 and the classwidth = 12

The total area of the histogram is 160, so half that area is 80
So the median lies somewhere around the 80th position

What I have done so far:

The median lies somewhere in the 4th bar and I have labelled some of the
area in the 4th bar, A and the rest of the area in the 4th bar I have
labelled B

On the left hand side of the 80th position,(which lies somewhere between 8
to 12 months on the x-axis) I have:

2+14+40+A (Where A is the unknown area)

On the RHS

B+72 (Where B is the unknown area)

Equating the two I get 56+A =B+72

Total area
A+B = 32 (classwidth x freq density = 4x32)

So substituting I get A = 24

But now I am stuck and do not understand how to find the median from this

I know the median lies from 8 months onwards so the median is: 8+ ????

Can someone help? I hope I have explained the problem clearly enough

TIA
sheri

--
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well dance

Archived from group: uk>education>maths

Brian Reay · Joined: 05 Aug 2007 Posts: 215

"~Bitzchick~" wrote in message $c_1.10037@text.news.blueyonder.co.uk...
> Hi all
> I hope you can help me understand how I can estimate the median from a
> histogram
>
>
> On the histogram I have x- axis = time (in months)
> y-axis = frequency density
>
> There are 5 bars in the histogram
>
> 1st bar has area = 2 and the classwidth = 2
> 2nd bar has area = 14 and the classwidth = 2
> 3rd bar has area = 40 and the class width = 4
> 4th bar has area = 32 and the classwidth = 4
> 5th bar has area = 72 and the classwidth = 12
>
> The total area of the histogram is 160, so half that area is 80
> So the median lies somewhere around the 80th position

Think of it in terms of area needed.

You need an area equivalent to 80. The 1st , 2nd, 3rd bar give you 2+14+40
= 56, so you need 24.

Your 4th bar has area 32 and so you need 3/4 of its area. Your median is
therefore 3/4 into the class width of the 4th bar.

(You can also think in terms of FD. The FD of the 4th bar is 8 and so you
need 3 units of the 4th bar's class interval,
but the area approach is more straight forward.)

--
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www.g8osn.org.uk
See my items for sale on my webpage

Now your amateur licence is free, why not send at least £15 per year to
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Barb Knox · Joined: 05 Aug 2007 Posts: 118

In article ,
"Brian Reay" wrote:

> "~Bitzchick~" wrote in message
> $c_1.10037@text.news.blueyonder.co.uk...
> > Hi all
> > I hope you can help me understand how I can estimate the median from a
> > histogram
> >
> >
> > On the histogram I have x- axis = time (in months)
> > y-axis = frequency density
> >
> > There are 5 bars in the histogram
> >
> > 1st bar has area = 2 and the classwidth = 2
> > 2nd bar has area = 14 and the classwidth = 2
> > 3rd bar has area = 40 and the class width = 4
> > 4th bar has area = 32 and the classwidth = 4
> > 5th bar has area = 72 and the classwidth = 12
> >
> > The total area of the histogram is 160, so half that area is 80
> > So the median lies somewhere around the 80th position
>
> Think of it in terms of area needed.
>
> You need an area equivalent to 80. The 1st , 2nd, 3rd bar give you 2+14+40
> = 56, so you need 24.
>
> Your 4th bar has area 32 and so you need 3/4 of its area. Your median is
> therefore 3/4 into the class width of the 4th bar.
>
> (You can also think in terms of FD. The FD of the 4th bar is 8 and so you
> need 3 units of the 4th bar's class interval,
> but the area approach is more straight forward.)

Doesn't "estimate the median" here mean "estimate the median VALUE"? In
which case it would be the height of the 4th bar, which is 32/4 = 8.

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| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Brian Reay wrote:
> "~Bitzchick~" wrote in

> Think of it in terms of area needed.

Yes that is what I've been trying to do; in terms of a fraction of an area

>
> You need an area equivalent to 80. The 1st , 2nd, 3rd bar give you
> 2+14+40 = 56, so you need 24.

yes I've got that

>
> Your 4th bar has area 32
>and so you need 3/4 of its area.

Ah, so 24 = 0.75 of 32 (0.75x( 4x8) ) I think I understand now

>Your median
> is therefore 3/4 into the class width of the 4th bar.

so 0.75x4 = 3

So you mean my median will be 8+3 = 11

>
> (You can also think in terms of FD. The FD of the 4th bar is 8 and
> so you need 3 units of the 4th bar's class interval,
> but the area approach is more straight forward.)

Yes. yes, yes I understand now

Thank you very much for all your help Brian

sheri

~Bitzchick~ · Joined: 05 Aug 2007 Posts: 18

Barb Knox wrote:
> Doesn't "estimate the median" here mean "estimate the median VALUE"?
> In which case it would be the height of the 4th bar, which is 32/4 =
> 8.
Thanks Barb, but the median time lies a little over the 8, so it's 8+ (a
fraction of the classwidth=3) = 11

Thanks

sheri

Brian Reay · Joined: 05 Aug 2007 Posts: 215

"Barb Knox" wrote in message @lust.ihug.co.nz...
>>
>> Think of it in terms of area needed.
>>
>> You need an area equivalent to 80. The 1st , 2nd, 3rd bar give you
>> 2+14+40
>> = 56, so you need 24.
>>
>> Your 4th bar has area 32 and so you need 3/4 of its area. Your median is
>> therefore 3/4 into the class width of the 4th bar.
>>
>> (You can also think in terms of FD. The FD of the 4th bar is 8 and so
>> you
>> need 3 units of the 4th bar's class interval,
>> but the area approach is more straight forward.)
>
> Doesn't "estimate the median" here mean "estimate the median VALUE"? In
> which case it would be the height of the 4th bar, which is 32/4 = 8.

The vertical axis on a histogram is frequency density and is not directly
related to the data, only the class width (frequency / class width).
Thus the 8 (in this case) is related to FD, not the data value.

Thing of a cumulative frequency curve, you don't read the median off the
vertical (frequency) axis, do you?

Brian

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