Mathematics: Post your doubts here!

[Alice123]

and Jiyad Ahsan

basically it says its within 12 from mean right? So that basically means its mean+12 and mean-12 which are 8 and 32.​

So P(8<X<32)=.94...P(8-20/r<Z<32-20/r)=.94....P(-12/r<Z<12/r)=.94 so if you remember that P(-a<z<b)=P(Z<b)+P(z<a)-1 apply that here so it becomes 2.P(Z<12/r)-1=.94 so P(Z<12/r)=(.94+1)/2....P(Z<12/r)=.97 find at what point prob is .97....so you get 12/r=1.881 ....therefore r=12/1.881=6.38​

and abt the normal think that you asked i am not sure abt the np and nq>5 thing but i think you are right because in a theory ans they said you cant use normal since np<5 so i am guessing you are right i havent done all the papers but from what i saw the easiest way to do is use binomial when its small or they ask you too use normal approximation when its will take too long to do the binomial like its 200+ items or something or when they ask you to i am not sure whether i am entirely right but i think thats it

Same year bi n ii..... HELP!!! n also Jiyad Ahsan if u arent busy

 

[Jiyad Ahsan]

Same year bi n ii..... HELP!!! n also Jiyad Ahsan if u arent busy

Alice123 question number please

 

[Alice123]

Alice123 question number please

6b... missed the 6...

 

[Ryan123]

Ryan123 can you explain this please

hmm well i will give some list
P(Z<a)=Φa then P(Z<-a)=1-Φa then P(Z>a)=1-Φa then P(Z>-a)=Φa then P(a<Z<b)=Φb-Φa then P(-a<Z<b)=Φb+Φa-1 then P(-a<Z<-b)=Φa-Φb hmm thats all i remmber atm

 

[Jiyad Ahsan]

Same year bi n ii..... HELP!!! n also Jiyad Ahsan if u arent busy

okay so if order doesnt matter its 3876
and since 4 are selected you multiply by 4!
=93024
another way to explain it is that
nCr = n! / r! (n-r)! - - - and nPr = n!/ (n-r)!
so if you cancel it out
nPr = nCr x r!

and i'm solving b(ii) right now,will post it in a bit

 

[Zarif009]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_62.pdf

Need help in question 1 (ii) please

 

[Zenia ZZ]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w12_qp_62.pdf
Q5 (b ii and iii) please !
Jiyad Ahsan

 

[Jiyad Ahsan]

hmm well i will give some list
P(Z<a)=Φa then P(Z<-a)=1-Φa then P(Z>a)=1-Φa then P(Z>-a)=Φa then P(a<Z<b)=Φb-Φa then P(-a<Z<b)=Φb+Φa-1 then P(-a<Z<-b)=Φa-Φb hmm thats all i remmber atm

thanx alot man !

 

[Jiyad Ahsan]

6b... missed the 6...

okay part (ii) since there are 4 groups each with 3 possessions so you have
3! x 3! x 3! x 3! (or 3!^4 do that the 3 possessions can be arranges within their group) multiplied by 4! (so the 4 groups can be arranged as well)
3!^4 x 4!
=31104

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_62.pdf
Q5 (b ii and iii) please !
Jiyad Ahsan

part (i)
12P8
=19958400

part (ii)
so frances and mary do not sit together,
just find out the number of ways in which they DO sit together and subtract from the total
if they sit together then 11P7 x 2P2 - - - - (11P7 because since they sit together you count them as one unit, 2P2 because they can arrange themselves in their group, sit left or right)
11P7 x 2P2
=3326400
this is when they do sit together, so subtracting from total ( part (i))
19958400 - 3326400
=16632000

part (iii)
i'll just upload the working for it in a bit

 

[Zarif009]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_62.pdf

need help on question 1(ii) ! please

 

[Ryan123]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_62.pdf

need help on question 1(ii) ! please

upper quartile possibility is 75% so P(Z<63-51/s.d)=.75.....12=.6745sd therefore sd=17,8

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_62.pdf
Q5 (b ii and iii) please !
Jiyad Ahsan

part 3, sorry it took so much time
oh and i forgot to mention ( though its obvious)
the denominators cancel out a part of the numerators so only the 8P8 's are left

 

[Jiyad Ahsan]

ok everyone, im outta here, i have physics as well tomorrow
Goodluck everyone, insha'Allah you'll do great

 

[hassanhijazi1995]

Um i m nt clear abt ur question bt i m assuming dat u r talking about Normal approximation...

So when u u use it when u have calculated probability using bionomal or ur mean n variance using np n npq respectively!!

am sure about this...
when they give you in the question the mean and the S.D no need to add 0.5 or subtract 0.5

but when they give you the probabilty of success and failure and u need to calculat the mean and S.D. here you will +- 0.5

 

[Zarif009]

upper quartile possibility is 75% so P(Z<63-51/s.d)=.75.....12=.6745sd therefore sd=17,8

thankyou!

 

[Yousif Mukkhtar]

Can anyone explain why they gave Z= 1.882 and p=0.97? Q5) I)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf

 

[Zenia ZZ]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_61.pdf
please help me with(question 6 ii )
and for question 7 (a ii ) 7! x 7! (is for arranging the friends and the partners ) but we are multiplying by 2 because either the friends could be first or the partners ? am i right ?? i want to confirm .

 

[A star]

This is very tricky question. Took me time to understand, anyway,


Alice123
you are right about (20+12<x<20-12)

now you have to imagine the graph in your head, and you have to come into a conclusion that it is symmetrical.

Now imagine the z of 20+12 that is 32 is x
So the z of 20-12 that is 8 will be a negative number -x

Both of them are x but one is negative, because they are the same distance from the mean

Let the probability of x is p
so the probability of -x is 1-p

we know that difference of the two probability is 0.94

so p - (1-p) = 0.94
p - 1 + p = 0.94
2p = 0.95 + 1 = 1.94
p = 0.97

Now is p is 0.97 z is from the table, 1.882 (or 1.881)

now continue normally

1.882 = 32-20 / sd
sd = 12/1.882
=6.38

you mean to say i will have to do all this for so little marks :O

 

[PhyZac]

you mean to say i will have to do all this for so little marks :O

I wont take that long to solve it, if you understand well.