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6b... missed the 6...Alice123 question number please
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6b... missed the 6...Alice123 question number please
hmm well i will give some listRyan123 can you explain this please
okay so if order doesnt matter its 3876Same year bi n ii..... HELP!!! n also Jiyad Ahsan if u arent busy
thanx alot man !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
okay part (ii) since there are 4 groups each with 3 possessions so you have6b... missed the 6...
upper quartile possibility is 75% so P(Z<63-51/s.d)=.75.....12=.6745sd therefore sd=17,8http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_62.pdf
need help on question 1(ii) ! please
part 3, sorry it took so much time
am sure about this...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!!
thankyou!upper quartile possibility is 75% so P(Z<63-51/s.d)=.75.....12=.6745sd therefore sd=17,8
you mean to say i will have to do all this for so little marks :OThis 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
I wont take that long to solve it, if you understand well.you mean to say i will have to do all this for so little marks :O
Yes you are.. for your first q, let me give you the main idea. If you still have trouble solving it, ill post the whole workinghttp://papers.xtremepapers.com/CIE/Cambridge International A and AS 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 .
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