IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: nitcomp on May 10, 2009, 12:16:23 pm

Title: histogram and medians doubt!
Post by: nitcomp on May 10, 2009, 12:16:23 pm

hello everyone!
i just wanted to know if anyone here can help me in finding
A MEDIAN FROM A HISTOGRAM... ???

thanks in advance! :)

Title: Re: histogram and medians doubt!
Post by: nid404 on May 10, 2009, 01:00:57 pm

I'll try but u need to give a question or I won't be able to explain.....I'm extremely bad at that. Just give the class interval and frequency.
I might be able to give u the median.

Title: Re: histogram and medians doubt!
Post by: nitcomp on May 10, 2009, 04:32:56 pm

i have uploaded sample questions of gcse boards... :)
no doubt in specific, but just generally,
how can a histogram alone be used to find a median??  ??? ???

i think this type of question is a possible candidate for the IGCSE finals coming up..
so, please help!

thanks ;D

Title: Re: histogram and medians doubt!
Post by: SGVaibhav on May 10, 2009, 04:55:37 pm

solutions plz
Mr. Paul, can u please show us how to find median from a histogram.

Title: Re: histogram and medians doubt!
Post by: nitcomp on May 10, 2009, 05:08:06 pm

sorry, no mark scheme.....

i am sure astarmathsandphysics can help?

Title: Re: histogram and medians doubt!
Post by: nid404 on May 10, 2009, 05:26:21 pm

Generally to find median in any grouped frequency one uses this formula - (n+1)/2 where n stands for the sum of frequencies. Take q3 for example where u need to find the median score. There are 320 students, so median is 160.5(using the formula). S0 u check which range the 160 and 161 student lies. This is between 90-110(by adding up the frequencies u get this(range * frequency density)) The hundredth student got a 90. So the 160th student got how much?
5(fre density)* x=60(frequency, 160-100)
x=12, therefore 160th student got 102.
and
5*x=61
x=12.2
161th student got 102.2
therefore median marks equal= (102+102.2)/2
=102.1

Correct me if i'm wrong. This is how i've been doin it. Mike Folland's textbook says this. Astar can explain in a better way i suppose but i thought i'll try.

Title: Re: histogram and medians doubt!
Post by: astarmathsandphysics on May 10, 2009, 10:10:55 pm

This looks ok to me but some books use n/2

Title: Re: histogram and medians doubt!
Post by: nid404 on May 11, 2009, 04:52:59 am

I was wondering if one can make a cumulative frequency graph using the data and then find out the median. It becomes easier.

Title: Re: histogram and medians doubt!
Post by: SGVaibhav on May 11, 2009, 06:55:05 am

i think for finding median in a cumulative frequency graph, we have to have n/2.
for normal cases (n+1)/2.
am i right?

Title: Re: histogram and medians doubt!
Post by: Christy on May 11, 2009, 06:59:21 am

ya .. ur right :D

Title: Re: histogram and medians doubt!
Post by: anusha500 on May 11, 2009, 09:09:59 am

you can also find the area. and divide by 2.
the value of the x-axis gives u the median.

there is also another formula. but tht is VERY complicated.

Median = L +I * N/2 - F
                      ______
                          f

where:
L = lower limit of the interval containing the median
I = width of the interval containing the median
N = total number of respondents
F = cumulative frequency corresponding to the lower limit
f = number of cases in the interval containing the median