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Mathematics: Post your doubts here!

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Messed up vectors.. I took too long to understand the last part of that question... Might loose a mark or two elsewhere too.
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 :)
 
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the questions werent very hard i did really bad though but it was lengthy really lengthy and so i made silly mistakes in differential eqn and vectors 2nd part was really quite easy if i had a good read of it. i had like 10mins for both vector questions and i messed up in the 2nd one badly

It happens bro ... I didnt substitute y^2 in Q.5 silliest mistake anyone could do -.-
 
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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
 
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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! :(
 
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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
 
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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
 
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no i still don't get it o_O 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
 
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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!
 
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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
 
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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...
 
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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???
 
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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
 
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