maths help very very urgent!

[godfather93]

someone please explain question Q6) b)ii)?
I'm getting the wrong answer.

[astarmathsandphysics]

Am not at home. Can someone else try

[godfather93]

someone else please!!

[godfather93]

i solved it. thread closed.

[Ghost Of Highbury]

okk..heres how u do it..
---------------RULE---------------------------------------------------
if class width = n * standard width, then, height = frequency/n
--------------------------------------------------------------------------
first lets find the standard width...the  height given = 11.5, its frequency is 115
therefore 11.5 = 115/n...=> n = 10
class width = 10 * standard width
2              =  10 * standard width
therefore standard width = 0.2
-----------------------------------------------------
now use this formula with standard width 0.2 to calculate the heights..
height = fequency * (standard width/classwidth)
1st - h = 35 * 0.2/10 = 0.7
2nd - 11.5 (given)
3rd - h = 26 * 0.2/2 = 2.6
4th - h = 24 * 0.2/6 = 0.8
----------
chk the markscheme for verification
good luck
byee

[Ghost Of Highbury]

okk...
good luck
byee

[godfather93]

thanks but i already solved it!
But thank you for your response!!!

[godfather93]

thank you!!!
Hopefully the exam will be good.

[nitcomp]

ok let me explain...

first things first... u should know this FREQUENCY = HEIGHT X CLASS WIDTH, ok?
lets take the given example...

freq=115
class width=2

with the help of the formula, height= 115/2
                                                        =57.5

now the height (freq. density) SHOULD equal 57.5, but it doesnt..
it equals 11.5 cm..

now u will have to find the "FACTOR'' by which the height is reduced..

11/57.5 which equals 0.2, ok?

now find the supposed height of the other column..

(example..3rd column...)
height = 26/2 = 13...

now multiply 13 into the factor...

13*0.2 = 2.6!
there you go...
same thing with other columns....


if u dont get me, please notify,, i will try again

[nitcomp]

aww.. u just got it!
aneways..
i tried!!