Mathematics: Post your doubts here!

[Esme]

Aww ... I was close to messing up that part ... thanks to the tension that Q5 built up .. but thank god i got it in the end

I sort of panicked in the end when I realised how little time is left :|

 

[MustafaMotani]

I sort of panicked in the end when I realised how little time is left :|

same at my end ..

 

[Ryan123]

I sort of panicked in the end when I realised how little time is left :|

if there was time i could get some around 70 because in checking i would have figured out the vectors and another mistake but now i am looking below 60 it sucks. i mean such a long exam definitely the lengthiest from 2002-2012

 

[Esme]

if there was time i could get some around 70 because in checking i would have figured out the vectors and another mistake but now i am looking below 60 it sucks. i mean such a long exam definitely the lengthiest from 2002-2012

I just needed 5 more minutes..just 5!

 

[Ryan123]

I just needed 5 more minutes..just 5!

i had to walk run and ride a three wheel driven by a human(rikshaw) for over 40mins in the rain and then entered the exam hall. half the energy was gone there oh well got s1 and a good as mark to look forward too

 

[Esme]

i had to walk run and ride a three wheel driven by a human(rikshaw) for over 40mins in the rain and then entered the exam hall. half the energy was gone there oh well got s1 and a good as mark to look forward too

Sorry to hear about that mate
Best of luck for stats. Insha'Allah it'll go well

 

[Zenia ZZ]

it is from 8 to 20 so P(8<X<20)=.96 so therefore P(-12/r<Z<12/r)=.96 which is 12/r=1.96/2 therefore r = 5.84

no i still don't get it i am sorry ! once again with a little more detail please .

 

[Ryan123]

no i still don't get it i am sorry ! once again with a little more detail please .

sorry did something wrong in the first time took prob as .96 when it actually is .94 really sorry bout that
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

 

[Alice123]

Let me know how did you do that? did you get x=2a?
Well for my school many got stuck after one point

yea...

 

[MyraMylo]

can someone tell me if when we are suposed to use 6.5 incase of lesser than 7 in case of normal approximation/normal distribution?
Because sometimes they use 7 as whole and i don't seem to get the difference between these type of questions!

 

[Ryan123]

can someone tell me if when we are suposed to use 6.5 incase of lesser than 7 in case of normal approximation/normal distribution?
Because sometimes they use 7 as whole and i don't seem to get the difference between these type of questions!

when they ask you for approximation you are supposed to use 6.5. Also in many sums they dont ask you those and still you got to use 6.5. As far as i know if you have to get the mean/standard deviation from the binomial formula(np and Rootnpq) then you need to use 6.5 incase of less than 7.
And when just for normal when you got the mean/s.d or you have to find out using the normal then you shall use 7

 

[MyraMylo]

when they ask you for approximation you are supposed to use 6.5. Also in many sums they dont ask you those and still you got to use 6.5. As far as i know if you have to get the mean/standard deviation from the binomial formula(np and Rootnpq) then you need to use 6.5 incase of less than 7.
And when just for normal when you got the mean/s.d or you have to find out using the normal then you shall use 7

Thanks a lot but are you sure about this...as i don't expect anything good from p3 so i really need to ace my s1...

 

[Alice123]

PART i)
The question said that only Rhesus + is taken, so if we note down the probabilities for all the Rhesus+=
A+= 0.35
B+= 0.08
AB+= 0.03
O+= 0.37

Now, what you do is, you draw up a probability distribution table for the Rhesus + values. So first of all calculate the total probability of the Rhesus + values= 0.35+0.08+0.03+037= 0.83

So what will the probability of selecting a person with the blood group of A+? It will be 0.35/0.83. For B+ 0.08/0.83. For AB+ 0.03/0.83 and for O+ 0.37/.83.

The question asks to find the probability that fewer than 3 are group O+. Now, you have think. It can either be O+, or it can't be O+. So the immediate thought that should hit you is that you have to use a binomial approximation. So the probability of a success (The probabolity of selecting a person who is O+) is 0.37/.83. The probability of a failure (Not selecting a person who's O+) is (0.35+0.08+0.03)/0.83= .46/.83
So P(X<3) means that you have to use the binomial expansion for 0, 1 and 2.
9C0*(.37/.83)^0*(0.46/.83)^9 + 9C1*(.37/.83)^1*(0.46/.83)^8 + 9C2*(.37/.83)^2*(0.46/.83)^7
=0.1555= 0.156

PART ii)
Now the question asks you to find a probability that involves a huge number. Using a binomial approximation can be very time consuming. You could use a normal approximation (np>5)
You have to find the probabily that more than 60 people are group A+. For using normal approximation, firstly, you have standardize 60.5. For which you need the mean and the standard deviation.
To find the mean we use np (total number* probability), and to find the variance we use npq (total number*probability*(1-probability))
So mean= np= 150*.35=52.5
And variance= npq= 150*.35*(1-.35)= 34.125
So standard deviation= square root of variance= 5.84

So now you have to standardize 60 i.e P(X>60). However, the questions asks you to find the probability of MORE than 60 people. so you have to take P(X>60.5).
Now first of all standardize 60.5 using (60.5-mean)/(standard deviation)
(60.5-52.5)/5.84
This give yous the Z value which is 1.369
Now you are to find P(X>1.369) using the normal distribution table. The value corresponding to 1.369 is 0.3145. Since you have to find the probability greater than 0.329, you have to subtract 1 from the probability
So 1-03145= 0.0855


Hope this helps!

in part 2, why arent we arent we taking p as o.35/0.83???

 

[Jiyad Ahsan]

Oh thank you so much !! but one more question please . which mean is used then for the standard deviation calculation ?

oh umm right i'll edit it in the original post as well
they give you two things right? the and the
so you have to do Σ(x-a)^2 _ ( Σ(x-a) )^2 just put the two numbers they give you in the question in formula (obviously the Σ(x-a)^2 given would be larger
n n oh and the mean that i stated before is basically the Σ(x-a)/n i.e. the smaller number divided by n..

i'll just edit the original post so u can understand better

 

[Anika Raisa]

Thanks for your help n support throughout..I pray for all of u
littlecloud11
PhyZac
Esme
iKhaled
Anika Raisa
Dug
and everyone who helped me n whose name i forgot to mention
Goodluck to all of u!!!

Best of luck 4 awl xams! Hw ws P3?

 

[Anika Raisa]

Thanks for your help n support throughout..I pray for all of u
littlecloud11
PhyZac
Esme
iKhaled
Anika Raisa
Dug
and everyone who helped me n whose name i forgot to mention
Goodluck to all of u!!!

Best of luck 4 awl xams! Hw ws P3?

 

[Alice123]

Best of luck 4 awl xams! Hw ws P3?

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

 

[Anika Raisa]

it went terrible... the 1st 7 questions made me feel like i wud get full marks n then ....... felt like murdering the a***hole who made the question

Same here!!! Fingers crossed hoping 4 best!! Pray every1! pLease!

 

[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