• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Mathematics: Post your doubts here!

Messages
3
Reaction score
0
Points
11
Hye guys,i have a question,I have came across some histograms and i have no idea which one is a correct one. Why some histograms their class boundary are in decimals(0.5) while some are in round numbers? I mean when you draw it,some draw from the 0.5 while some draw on round numbers?
 
Messages
176
Reaction score
104
Points
53
Hye guys,i have a question,I have came across some histograms and i have no idea which one is a correct one. Why some histograms their class boundary are in decimals(0.5) while some are in round numbers? I mean when you draw it,some draw from the 0.5 while some draw on round numbers?
Well. It depends on the intervals in the group data. If an Upper Boundary of the Interval is the same value as the Lower Boundary of the next interval, then there's no need to add/subtract .5 to the corresponding boundaries. I.e,
0-100, 100-200

But if it is 1-99, 100-149. Then subtract .5 from Lower boundary and add .5 to the upper.
.5-99.5, 99.5-149.5.
There Simply cannot be empty space WITHIN a histogram unless an interval has frequency of 0
 
Messages
81
Reaction score
5
Points
18
I love these questions. They are the easiest, that's why. :p

Within 12 g implies the least is 20-12 = 8 and highest = 20+12 =32
so P(8<X<32)
P( 8-20 / σ <Z< 32-20 / σ)
P(-12/σ <Z < 12/σ)
Seems familiar? :D
then its probability is 0.94? o_O
 
Messages
90
Reaction score
24
Points
8
Hye guys,i have a question,I have came across some histograms and i have no idea which one is a correct one. Why some histograms their class boundary are in decimals(0.5) while some are in round numbers? I mean when you draw it,some draw from the 0.5 while some draw on round numbers?
its like where you have continuous boundaries like 0<t<20,20<t<40 and so on then u dont have subtract .5 but if the boundaries are like 10<t<30 , 31<t<40 then ul have to subtract .5 and make it like 9.5<t<30.5 ,30.5<t<40.5 and so on
 
Messages
90
Reaction score
24
Points
8
im not able to solve Q2...help !
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
and also i hv seen similar questions in many years recently,can some one tell me how to go about it ?
alright here you go
i)When ever you have to find the original mean=E(x-a)/n +a
so here the mean is given as 28.325. Find the value of n from the equation by substituting the value of E(x-a). 28.325-a=E(x-a)/n where a=25 n=40
For the s.d you should remember that s.d isnt affected by constant shifts in the data such as -a.
assume that x-a =y so E(x-a)=E(y) and hence E(x-a)^2=E(y)^2
s.d=underoot(E(y)^2/n-(E(y)/n)^2)
s.d=9.11
ii)check your data booklet one of the formula for s.d is underoot(E(x)^2/n - (E(x)/n)^2). Since you know that the s.d is 9.11 and that the original mean is 28.325 substitute the given values and find the value of (E(x)/n )^2.
 
Top