Mathematics: Post your doubts here!

[Anika Raisa]

not as good as expected... i screwed up Qs 9 n 10

I screwed up the subsitution integretion n 10! N also some simple few!!

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q6 (ii) anyone ??
Jiyad Ahsan

(i) okay umm in the first part you found that n was 10 by solving o.75^n < 0.06

(ii) since mean is np (n is the nymber of trials and p is the probability of success)
you get the mean which is 10.5 (by 14 x o.75)
we find the mean because the question said that the two nearest values to that mean would give the highest possibility so we will use 10 and 11
using 10 we use binomial distribution and we get 0.22 and by using 11 we get o.24
so the answer is 11

(iii) you have to use binomial distribution 2 times
first you find the probabilities for 12, 13 and 14 and add them.. use n =14, p= 0.75 and q = 0.25
the answer will be 0.2811
now use this value as p, and find q by 1 - p which will be 0.7189 now
=5C3 x (o.2811^3) x (0.7189^2)
=0.115
we do the second distribution because we need to find the probability that it happens in 3 out of 5 months

 

[Anika Raisa]

GOOD LUCK!!

Thnk u! Same to u!

 

[aniketjain]

hey can anyone tell me tht whn do we have to use 0.5 in normal distribution in the value of X..... its costing a lot of marks .. asap plzz !!

 

[Anika Raisa]

Dear people sitting for S1 post your doubts here!
https://www.xtremepapers.com/commun...st-your-doubts-here.25911/page-15#post-564306

Because S1 n M2 are on the same date n so its likely your post may remain unanswered if posted here!!

 

[aniketjain]

Dear people sitting for S1 post your doubts here!
https://www.xtremepapers.com/commun...st-your-doubts-here.25911/page-15#post-564306

Because S1 n M2 are on the same date n so its likely your post may remain unanswered if posted here!!

thnk u

 

[Anika Raisa]

hey can anyone tell me tht whn do we have to use 0.5 in normal distribution in the value of X..... its costing a lot of marks .. asap plzz !!

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!!

 

[Anika Raisa]

Thank you so much for remembering me among all these amazing people and tagging me. Feels good
Insha'Allah the paper will be good and easy for everyone.
Pray for all of us sis

hw was P3??
Mine ws ok ! Screwed up many!! Had a an unfavourable condition on tym b4 xam!! Most Ppl here all actuali went through lot of trouble when they had to go to the venue!
Hoping gt is low n S1 wud help boost d grade... Pray please!

 

[Jiyad Ahsan]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_61.pdf
Q-2 (ii)
Well can someone please explain how basically to do these types of questions basically ?
Jiyad Ahsan

the mean is always (Σ(x-a)/n ) + a,
133/n + 25 = 28.325 since assumed mean,a is given 25 and Σ(x-a) is given 133 and mean is given 28.325
so we find n=40, then just use the s.d formula that i gave before
and u'll have variance = 82.99 and s.d = 9.11
for part (ii)
the normal formula for variance is
variance = (Σx^2/n) - (mean^2) so
82.99 = Σx^2/40 - (28.325)^2
Σx^2 = [ 82.99 + (28.325)^2 ] x 40
Σx^2 = 35412

 

[Anika Raisa]

Assalamu 'Alaykum May Allah Help you...ur welcome-although i was so useless

Nope u werent a bit useless in fact useful with dua post .... May Allah bless u! N pray 4 me please!!! Last xam didnt go as epected ! Hoping others wud go well!!!

 

[Jiyad Ahsan]

"it is found that the weights of 94% of the letters are within 12 g of the mean" , what does this mean ?
94% is the mean but then what is X ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
Q- 7 anyone ?
Jiyad Ahsan

that satement is in Q5
they say that 94% is within 12g of the mean (which is 20g)
which means that if you take 20+12 = 32 (which is 12 g away form the mean) or 20-12=8 the probability of (8<x<32) is 94% (94/100 = 0.94)
so the mean would be 20, x would be 8 or 32 and the probability will be o.94

 

[Ryan123]

that satement is in Q5
they say that 94% is within 12g of the mean (which is 20g)
which means that if you take 20+12 = 32 (which is 12 g away form the mean) the probability is 94% (94/100 = 0.94)
so the mean would be 20, x would be 32 and the probability will be o.94

No no you got it wrong...you have to consider it to be within 12g of mean so it can 12g more/less than mean so its from 8<X<32 with prob 0.94

 

[aniketjain]

y

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!!

a i m talking about normal approximation .. i know how to calculate mean and all but in many markschemes and all thy take the value of X 0.5 less thn the exact value of X or 0.5 more thn the exact value of X .

 

[Jiyad Ahsan]

No no you got it wrong...you have to consider it to be within 12g of mean so it can 12g more/less than mean so its from 8<X<32 with prob 0.94

lol i meant just that, but i missed it out thnx for pointing it out
though i think you can use just one 8 or 32

 

[Alice123]

No no you got it wrong...you have to consider it to be within 12g of mean so it can 12g more/less than mean so its from 8<X<32 with prob 0.94

isnt the probability 0.97???

 

[Jiyad Ahsan]

y
a i m talking about normal approximation .. i know how to calculate mean and all but in many markschemes and all thy take the value of X 0.5 less thn the exact value of X or 0.5 more thn the exact value of X .

listen man, if in binomial distribution the value of np and nq exceeds 5 you have to use normal approximation
if it says
less than ( < ) you minus 0.5 from the x
greater than ( > ) you add o.5 to x
if its less than or equals to ( ≤ ) you add 0.5
if its greater than or equals to ( ≥ ) you minus 0.5

Ryan123 this^ right?

 

[Jiyad Ahsan]

isnt the probability 0.97???

yeah i know alice i read the table as well and the value in the ms is on 0.97 while the in the question its given quite clearly '94%'
probably the value in the ms is wrong... maybe

 

[Ryan123]

isnt the probability 0.97???

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

 

[Jiyad Ahsan]

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

awesome ! i didnt even remember that formula
Zenia ZZ the solution you were looking for. Q5 from november 11 / paper 61

 

[Jiyad Ahsan]

by the way if someone could please explain (hopefully before our paper ) why and when (in what conditions) in normal distribution do we subtract from 1,
from what i've learned till now..
Φ(-a) = 1 - Φ(a) here to remove the negative sign
P(Z > a) = 1 - Φ(a) here (apparently) to remove the greater than ( > ) sign

also there was something about the fact that we cant read values less than o.5 off the table, so you had to 1 minus that too.. can someone explain please..

Ryan123 can you explain this please