nov 08 paper.urgent!

[vids]

pls can any1 explain me histograms?
nd d q 6 (b)ii. of nov 08 p4

[I Jimmy I]

Attach the paper?
So that I'll explain to u

[vids]

here u go

[vids]

n hw do u go abt histograms...
d formaula which is fd=f/range
d ans is nt cuming right 4 d nov 08 q

[Priceless]

LOL ya sure np. hwevr wat du u wanna bout histograms. dey r just like bar charts but only der bars touch. u draw dem wen u hv continuous groups of data like in da ques in O/N 08 ppr Q6. however dey told not 2 draw a histogram so 2 work dat out. du this.
in this case u hv 2 work out da frequency density which u find out boy doing FD= Frequency/Group width. usually FD is equal 2 da height of da column so 11.5=115/(2x) itz 2x becuz if u c da width of da column 20-22 is 2 n since 115/2 duz not equal 11.5 der has 2 be sum constatnt. so u can work out da constant now. 2x=115/11.5=10 so x=10/2=5. so now u can find out da heights=FD of da other columns now. FD=frequency/width x 2. so 4 10<m<20 u wud 2 FD=35/(10x5)=0.7 n if u check in da mark scheme itz da rite answer. u can work out da heights like this 4 da other columns aswell.

Hope i helped n i hope u understud- itz a bit hard 2 explain. neway best of luck wid ur xams!

[Ascer Zayan]

pls can any1 explain me histograms?
nd d q 6 (b)ii. of nov 08 p4

to Draw to histogram first find the F.d (Frequency density = Frequncy divided by class width)
then Plot F.d on the y axis and intervals on x-axis
for Nov2008
Find the F.d for 20<m<22 which equals to 57.5 and find all F.d for the rest (for first one it is 3.5,then57.5,then12.5 then 4)
then  use ratio
57.5-11.5
3.5-X
cross multiply X=0.7
then repeat with others  with ratios
AND VOLA

[2009scared]

Thanks a lot!

[I Jimmy I]

Ok listen,
For 20< x < 22 it says that the height is 11.5 which can be done by dividing 115 by 2, and then by 5
Cuz for these type of questions u need to divide the frequency by the range and then divide or multiply or add or subtract to get a similar answer to the given one...
This questions is quite easy however I still have a  doubt in finding the possible values of the median but the rest is fine...