IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: cs on October 22, 2010, 03:10:49 am
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June 2010 Q1,
how do you know if that question is asking for binomial or P and C..?
A bottle of sweets contains 13 red sweets, 13 blue sweets, 13 green sweets and 13 yellow sweets. 7 sweets are selected at random . Find the probability that exactly 3 of them are red.
My first instinct is to do X~B (7, 13/52)
and then P(x=3) then continue,
but the answer is found using 13C3 X 39C4 over 52C7
anyone can help me?
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June 2010 Q1,
how do you know if that question is asking for binomial or P and C..?
A bottle of sweets contains 13 red sweets, 13 blue sweets, 13 green sweets and 13 yellow sweets. 7 sweets are selected at random . Find the probability that exactly 3 of them are red.
My first instinct is to do X~B (7, 13/52)
and then P(x=3) then continue,
but the answer is found using 13C3 X 39C4 over 52C7
anyone can help me?
Nope.......you cannot use binomial here. This is because you don't even now what are the colours of the selected sweets. So you cannot take if fro granted.
Moreover the word exactly gives you a hint that you should be using combinations since the selected sweets must be 3 red and the rest a ny other colour.
So firstly, the probability of 3 red ---> 13C3
Since you have been told that there is exactly 3 red and not more, this implies that the rest(4) should be of other colours out of only 39 sweets ---> 39C4
Since you are finding probability, you should divide the above product by the probability of taking out 7 sweets = 52C7
Hence the solution is (13C3 x 39C4)/52C7
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Thank you Deadly King, you are always here to solve my doubt.. =)
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I would like to add more, the probability of success in " choosing a red sweet " is NOT constant, it keeps on changing everytime you make a selection since the sweets are not replaced. Furthermore, the trials aren't independent. Both of those conditions are essential for binomial distribution.
hope that helped, by the way, your doing your exam in this nov session??
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Thank you Deadly King, you are always here to solve my doubt.. =)
Anytime dear :)
By the way when are you having your stats exams ??? 3 Nov ???
Thanks for your info mate :D
Very much appreciated :)
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Thanks both of you, yes i am sitting for the exam on 3 Nov, good luck!
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Thanks both of you, yes i am sitting for the exam on 3 Nov, good luck!
Same day for me as well ;)
Though I'll be sitting for 62 :-\
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Same day for me as well ;)
Though I'll be sitting for 62 :-\
hahaha meeee 2 !!! i will be sitting 62 , where are you living??
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hahaha meeee 2 !!! i will be sitting 62 , where are you living??
Hehe......that's nice :D
Am from Mauritius..............and you ???
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Hehe......that's nice :D
Am from Mauritius..............and you ???
Wowww nice, muritaniaaaa loool, thats tooo far from here :P I am Jordanian living in the UAE, so you are Muratious ya3ny??
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Wowww nice, muritaniaaaa loool, thats tooo far from here :P I am Jordanian living in the UAE, so you are Muratious ya3ny??
Hehe........Not Mauritania but Mauritius, tiny island in the Indian Ocean :D
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I would like to add more, the probability of success in " choosing a red sweet " is NOT constant, it keeps on changing everytime you make a selection since the sweets are not replaced. Furthermore, the trials aren't independent. Both of those conditions are essential for binomial distribution.
hope that helped, by the way, your doing your exam in this nov session??
When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.
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When you say "the trials aren't independent", what does that exactly mean? I get the part that they arent constant.
The trials are NOT independent ----> each trial depends on the previous trial since the previous trial decreased the total number of sweets in the bag, so the probability will depend on how many sweets have been removed and have they been replaced or not. ;)
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Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2
why 6! X 7C3?
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Deadly King and others, i have another doubt, CIE Nov 06 Q 6 part 2
why 6! X 7C3?
I can't really explain the workings from the marking scheme but I did it another way which was approved by my teacher.
First we find out the number of arrangements in which all the 9 persons can be arranged. This will be (9! +(7! x 3!))
9! => There are nine persons which can be arranged in any way.
(7! x 3!) => There are 6 men and 3 women. We'll be taking all the women to be standing beside each other. Hence they are taken as one possiblity only which makes 7!. The 3 women can then be arranged in any way 3!.
Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
8! = > One woman is taken as one possiblity and the other two as another possiblity. In all there will be 8 possiblities, after adding the six men.
Number of arrangements : (9! +(7! x 3!)) - 3(8! x 2!) = 151200
Hope you understand :)
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I can't really explain the workings from the marking scheme but I did it another way which was approved by my teacher.
First we find out the number of arrangements in which all the 9 persons can be arranged. This will be (9! +(7! x 3!))
9! => There are nine persons which can be arranged in any way.
(7! x 3!) => There are 6 men and 3 women. We'll be taking all the women to be standing beside each other. Hence they are taken as one possiblity only which makes 7!. The 3 women can then be arranged in any way 3!.
Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
8! = > One woman is taken as one possiblity and the other two as another possiblity. In all there will be 8 possiblities, after adding the six men.
Number of arrangements : (9! +(7! x 3!)) - 3(8! x 2!) = 151200
Hope you understand :)
Good job!
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Good job!
Glad to have been of help :)
I just hope cs understands it as well ;)
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yea. there's working behind my pass year book, and they gave 2 types of working, i did the way you did, and manage to get 9! and 7!X3!, i don't get this part(when two women are together):
Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
I got 8! X 2!, why is there a 3?
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yea. there's working behind my pass year book, and they gave 2 types of working, i did the way you did, and manage to get 9! and 7!X3!, i don't get this part(when two women are together):
Then we need to find the number of arrangements in which two women are next to each other => 3(8! x 2!)
I got 8! X 2!, why is there a 3?
The three is because there are three women ;)
Let there be 3 women : W1, W2 and W3
Possible arrangement are :
i) W1W2 and W3
ii) W1W3 and W2
iii) W2W3 and W1
So there are 3 ways in which the women could be arranged :D
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DK son 8)............u r awesome 8).................+rep
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DK son 8)............u r awesome 8).................+rep
Thanks daddy ;D
Son resembles his dad ;)
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Ohh i always thought its 3!, there's only 3 ways, ok i understand now.. Thank you!
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Ohh i always thought its 3!, there's only 3 ways, ok i understand now.. Thank you!
Glad you understood ;)
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Question guys
There are 3 married couples and 4 single people
What are the ways of choosing 5 people from these if:
i) 2 Married couples are chosen
ii) 1 Married couple is chosen
My answers i) 3C2 * 6C1 = 18
ii) 3C1 * 8C3 = 168
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Question guys
There are 3 married couples and 4 single people
What are the ways of choosing 5 people from these if:
i) 2 Married couples are chosen
ii) 1 Married couple is chosen
My answers i) 3C2 * 6C1 = 18
ii) 3C1 * 8C3 = 168
Do you mind saying which year it is?
i think you missed something in the question here :-[
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even i think i did miss something. this is a question my friend asked me that he got from a handout worksheet so its not from a paper. anyways dont worry about t bro :)
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even i think i did miss something. this is a question my friend asked me that he got from a handout worksheet so its not from a paper. anyways dont worry about t bro :)
Ok........if you are saying so.
But I've done a number similar to this, however i can't remember which year :-[
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so which exams do you have next buddy?
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so which exams do you have next buddy?
Statistics (S1) and two chem papers on wednesday. I believe it's the same for you, nan?
Then physics practicals on thursday ;)
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yes thats right :) thought i have computing tomorrow :/ haha
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yes thats right :) thought i have computing tomorrow :/ haha
Hehe.......Good luck then br0 :)
I don't do computing :D
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question! ;D
The following question is from 9709_w09_qp_62:
(i need the solution for 4 a(ii) and b)
4 (a) (i) Find how many different four-digit numbers can be made using only the digits 1, 3, 5 and 6
with no digit being repeated.
(ii) Find how many different odd numbers greater than 500 can be made using some or all of
the digits 1, 3, 5 and 6 with no digit being repeated.
(b) Six cards numbered 1, 2, 3, 4, 5, 6 are arranged randomly in a line. Find the probability that the
cards numbered 4 and 5 are not next to each other.
Thanks in advance ;)
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question! ;D
The following question is from 9709_w09_qp_62:
(i need the solution for 4 a(ii) and b)
4 (a) (i) Find how many different four-digit numbers can be made using only the digits 1, 3, 5 and 6
with no digit being repeated.
(ii) Find how many different odd numbers greater than 500 can be made using some or all of
the digits 1, 3, 5 and 6 with no digit being repeated.
(b) Six cards numbered 1, 2, 3, 4, 5, 6 are arranged randomly in a line. Find the probability that the
cards numbered 4 and 5 are not next to each other.
Thanks in advance ;)
b) cards numbered 4 and 5 next to each other : 5!× 2!
6 cards can be arranged in 6!
6! - (5!×2!) = 480
probability : 480/6! = 2/3
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I have a question !! how can I know which measure of central tendency is the most appropriate to represent a set of data ??!!
thank youu :)
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I have a question !! how can I know which measure of central tendency is the most appropriate to represent a set of data ??!!
thank youu :)
Am not sure myself.
Hope this helps :)
http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html
http://www.quickmba.com/stats/centralten/
By the way thanks for the answers.:)
+rep
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Am not sure myself.
Hope this helps :)
http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html
http://www.quickmba.com/stats/centralten/
By the way thanks for the answers.:)
+rep
thank youu !!
and ur so welcome !! =]
thanks for the +rep
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Am not sure myself.
Hope this helps :)
http://simon.cs.vt.edu/sosci/Site/MMM/mmm.html
http://www.quickmba.com/stats/centralten/
By the way thanks for the answers.:)
+rep
Excellent info!
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another question !! ( sorry !! :P)
(just iv and v)
7 Nine cards, each of a different colour, are to be arranged in a line.
(i) How many different arrangements of the 9 cards are possible?
The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card?
Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
(iii) How many different arrangements in total of 3 cards are possible?
(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card?
(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green
card?
thanks =]
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b) cards numbered 4 and 5 next to each other : 5!× 2!
6 cards can be arranged in 6!
6! - (5!×2!) = 480
probability : 480/6! = 2/3
Thanks alot! +rep ;)
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Thanks alot! +rep ;)
I'm just glad to help =]
thanks for +rep
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another question !! ( sorry !! :P)
(just iv and v)
7 Nine cards, each of a different colour, are to be arranged in a line.
(i) How many different arrangements of the 9 cards are possible?
The 9 cards include a pink card and a green card.
(ii) How many different arrangements do not have the pink card next to the green card?
Consider all possible choices of 3 cards from the 9 cards with the 3 cards being arranged in a line.
(iii) How many different arrangements in total of 3 cards are possible?
(iv) How many of the arrangements of 3 cards in part (iii) contain the pink card?
(v) How many of the arrangements of 3 cards in part (iii) do not have the pink card next to the green
card?
thanks =]
We have to start from part iii to continue with part iv and v
a permutation of a set of values is an arrangement of those values into a particular order.
so 9p3=504
iv)We have to assume that the pink card has been chosen in a group of 3 cards so now that group has 2 places remaining
8P2=56
There will be three groups in total which can be made from a group of 9 cards and 56 times 8 cards other th&n the pink can fit in two places of a group in which the pink card has already been chosen.
so we multiply 56 * 3 =168
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thank youu soo soo much that was very helpful
+rep =]
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__ __ 7C1
7C1 __ __
There will be two ways the pink and Green card can shift between themselves as 2!=2
and there are two ways they can stay together
2*2*7=28
504-28=476
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Explained it in a easier way
~Sf staff at your service
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Explained it in a easier way
~Sf staff at your service
aaawww you guys are just amazing thank youu =]
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sorry guys i have another question.. :-[
Eggs produced at a farm are packaged in boxes of six. Assume that, for any egg, the probability that it is broken when it reaches the retail outlet is 0.1 , independent of all other eggs. A box is said to be bad if it contains at least two broken eggs. Calculate the probability that a randomly selected box is bad.
Ten boxes are chosen at random. Find the probability that just two of these boxes are bad.
It is known that, in fact, breakages are more likely to occur after the eggs have been packed into boxes, and while they are being transported to the retail outlet. Explain why this fact is likely to invalidate the calculation.
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the answer is as follows:
0.114; 0.2223
leave the last part of the question... i understand the answer now
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Give me 10 minutes, i will explain.
alrite thank you! ;D
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sorry guys i have another question.. :-[
Eggs produced at a farm are packaged in boxes of six. Assume that, for any egg, the probability that it is broken when it reaches the retail outlet is 0.1 , independent of all other eggs. A box is said to be bad if it contains at least two broken eggs. Calculate the probability that a randomly selected box is bad.
Ten boxes are chosen at random. Find the probability that just two of these boxes are bad.
It is known that, in fact, breakages are more likely to occur after the eggs have been packed into boxes, and while they are being transported to the retail outlet. Explain why this fact is likely to invalidate the calculation.
(i) at least two eggs broken: so we find the probability that less than two eggs and then subtract it from one :
(6C1) (0.1)^1 (0.9)^5 + (6C0)(0.1)^0 (0.9)^6 = 0.8857
1- 0.8857=0.144
(ii) from the ten boxes they want the probability of two boxes :
(10C2) ( 0.144)^2 (0.8857)^8 = 0.223
(iii) the event that eggs are broken is not independent of the event that any other egg is broken ( not sure )
i hope it's right because I'm not really sure !!! hope it helps =]
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+rep
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(i) at least two eggs broken: so we find the probability that less than two eggs and then subtract it from one :
(6C1) (0.1)^1 (0.9)^5 + (6C0)(0.1)^0 (0.9)^6 = 0.8857
1- 0.8857=0.144
(ii) from the ten boxes they want the probability of two boxes :
(10C2) ( 0.144)^2 (0.8857)^8 = 0.223
(iii) the event that eggs are broken is not independent of the event that any other egg is broken ( not sure )
i hope it's right because I'm not really sure !!! hope it helps =]
Thanks a lot again!! +rep again :D
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your sooo welcome I'm just happy to help ;)
thank youu guys for +rep :D
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Explained it in a easier way
~Sf staff at your service
I like that ;)
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Can some1 help!?
may/june 2010 paper 61 q5 (ii)
Please explain, the mark scheme is useless on this one
thanx in advance
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Can some1 help!?
may/june 2010 paper 61 q5 (ii)
Please explain, the mark scheme is useless on this one
thanx in advance
Read the attached Examiners Report for help.
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soo how did you guys do ?? I sat for 62 and I thought it was very tricky :(
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soo how did you guys do ?? I sat for 62 and I thought it was very tricky :(
Yeah.......i sat for the same paper as well.
I made some monstrous mistakes that I can't help imagine how i managed to do that >:( :(
It sure was tricky :-\
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Yeah.......i sat for the same paper as well.
I made some monstrous mistakes that I can't help imagine how i managed to do that >:( :(
It sure was tricky :-\
I DID THE SAME THING >:(
can you believe that anyone could get the approximation question wrong i did !!! i forgot to put .5 thing :( :(
what did you get for the second question the standard deviation, mean and the ?(x-50) and
?(x-50)^2 ??
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I DID THE SAME THING >:(
can you believe that anyone could get the approximation question wrong i did !!! i forgot to put .5 thing :( :(
what did you get for the second question the standard deviation, mean and the ?(x-50) and
?(x-50)^2 ??
Am not sure about the approximation too >:(
I can't remember the mean but I think i got the summation to be 712.
Tell me your values, i'll see if i remember :-[
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for the summation of (x-50) i got 24 and for the summation of (x-50)^2 i got the same thing 712 or 715 something like that but i cant remember the exact number !! for the s.d i got 6. something or 7.something and for the new mean i got 52.7 !!
and how about the last question (iii) ?? i got 9030 and I'm pretty sure it's wrong !! :-[
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for the summation of (x-50) i got 24 and for the summation of (x-50)^2 i got the same thing 712 or 715 something like that but i cant remember the exact number !! for the s.d i got 6. something or 7.something and for the new mean i got 52.7 !!
and how about the last question (iii) ?? i got 9030 and I'm pretty sure it's wrong !! :-[
thats gr8! my values for question 2 were exactly the same as yours :) last question, hmm i really cant remember.. though they were HUGE numbers lol
and what did u get for the 6 mark q'n? i got 8/11
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yaaay !! ;)
i got 8/11 too :D
and I almost died solving the last question reallyy tricky :P
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haha hi5 :) yes last question required a lot of thinking! lol
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thats gr8! my values for question 2 were exactly the same as yours :) last question, hmm i really cant remember.. though they were HUGE numbers lol
and what did u get for the 6 mark q'n? i got 8/11
8/11 is the correct answer. ;)
Last number, huge numbers as well except for the two probabilities.
for the summation of (x-50) i got 24 and for the summation of (x-50)^2 i got the same thing 712 or 715 something like that but i cant remember the exact number !! for the s.d i got 6. something or 7.something and for the new mean i got 52.7 !!
and how about the last question (iii) ?? i got 9030 and I'm pretty sure it's wrong !! :-[
Apart from the s.d, I got more or less same answers. :D
I think I messed up the s.d :-[
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yaaay !! ;)
i got 8/11 too :D
and I almost died solving the last question reallyy tricky :P
niceeeeee i got 8/11 aswell lool the lasst question was helllllllll!! I remember getting a 2/3 or something like that in the very last part of the last question :-\
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niceeeeee i got 8/11 aswell lool the lasst question was helllllllll!! I remember getting a 2/3 or something like that in the very last part of the last question :-\
Yeah...........some of my friends got 2/3 as well.
I guess it is the right answer...........but I could not link it :(
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niceeeeee i got 8/11 aswell lool the lasst question was helllllllll!! I remember getting a 2/3 or something like that in the very last part of the last question :-\
I got 2/3 too !! =] how about (iii) what did you get ??
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I got 2/3 too !! =] how about (iii) what did you get ??
hmm I dont quite remember, though i might.. in case you tell me your answer :P
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i got 9030 but i'm pretty sure it's wrong !! i think the right answer was around 4000 :-\
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2/3 is right.. i got that too :) and since i could not find the new s.d i just sed it was the same as the original.. lol what were ur values for the new s.d
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I think 7.94 was the correct answer !! do you remember the answer for question 7 (iii) ??
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could you remind me the question. i forgot what it was lol
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I think it was in how many ways can the man or the women be on the committee but not both
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As far as i remember, the info they gave was:
If there are 10 men and 9 women. What are the ways of choosing 6 people if there are atleast 4 men and 1 woman.
iii) If one particular man or one particular women are on the committee but not both.
Answer I got:
(9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230
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are you sure this is correct ???and By the way what do you expect the grade boundaries for an A to be for this variant ??
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i have no idea if this is correct.. :) i would like someone who is sure about their answer to comment on if i did something wrong :) and i think the threshold should vary between 36-39 out of 50
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I predict between 35-38 i think everyone found it harder than any other year .. but I hope we all get an A ;)
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Am sorry...................i've done so badly in the last question that I can't confirm any answers in it. :(
Yeah it was kinda hard for everyone. So i expect the threshold to decrease upto 30-35.
I really hope to get an A in that paper though it's not very likely due to the number of mistakes. :(
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As far as i remember, the info they gave was:
If there are 10 men and 9 women. What are the ways of choosing 6 people if there are atleast 4 men and 1 woman.
iii) If one particular man or one particular women are on the committee but not both.
Answer I got:
(9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230
I realy dont remeber wt I got, but i am confident of my working out, By the way, for the 1st part regarding how many combinations for the committe having at least 4m and at least 1 w??
it meant that only possible combinatins are 5m and 1 w OR 4m and 2 w
so my answer was 9828 ways is that correct??
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hey guyss how about Questions 5 and 6 any of you remember their answers???
which bitts you found tricky???
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hey guyss how about Questions 5 and 6 any of you remember their answers???
which bitts you found tricky???
That was the normal and binomial distribution, right?
Unfortunately I can't remember the values I got :(
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do u guys remember the actual question for the last question includin all its parts?
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do u guys remember the actual question for the last question includin all its parts?
Here you are :
7. A committee of 6 people, which must contain at least 4 men and at least 1 women, is to be chosen from 10 men and 9 women.
i) Find the number of possible committee that can be chosen. [3]
ii) Find the probability that one particular man, Albert and one particualr woman, Tracey are both on the committee.[2]
iii) Find the number of possible committees that include either Albert or Tracey but not both. [3]
iv) The committee that is chosen of 4 men and 2 women. They queue up randomly in a line for refreshments. Find the probability that the women are not next to each other in the queue. [3]
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These were my answers:
i) (10C4*9C1*14C1) + (10C4*9C2) + (10C5*9C1) = 36288
ii) (9C4*8C0) + (9C3*8C1) / 28728 = 1/36
iii) (9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230
iv) 6!-(5!*2!) / 6! = 2/3
Can anyone confirm my answers :)
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And i think i got 60/75 for P1 and 43/50 for S1 ... what do you think my grade will be, based on the threshold you guys consider to be used :)
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The permutation and combination for p63 was hard.. 10 marks for that, i only manage to do the first question, 8 marks gone.. other question was quite okay, except the first one, which distribution is used to model weight of students..? i answered stem and leaf, most of my friends answered normal distribution cos of the "distribution" on the question, 2 marks for that..
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These were my answers:
i) (10C4*9C1*14C1) + (10C4*9C2) + (10C5*9C1) = 28728
ii) (9C4*8C0) + (9C3*8C1) / 28728 = 1/36
iii) (9C3*8C1*13C1) + (9C3*8C2) + (9C4*8C1) + (8C1*9C4) + (8C0*9C5) = 13230
iv) 6!-(5!*2!) / 6! = 2/3
Can anyone confirm my answers :)
i dunno realy why you've done it like that;;;;; since there must be 6 ppl in total and they said a min. of 4 m and 1 w.. I have done it like that:
possible combinations are// 5 men and 1 women OR 4 men and 2 Women
therefore the answer would be :
10C5*9C1 + 10C4 * 9C2 =9828 ways...so i dunno know which one of us is correct.
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i dunno realy why you've done it like that;;;;; since there must be 6 ppl in total and they said a min. of 4 m and 1 w.. I have done it like that:
possible combinations are// 5 men and 1 women OR 4 men and 2 Women
therefore the answer would be :
10C5*9C1 + 10C4 * 9C2 =9828 ways...so i dunno know which one of us is correct.
ur talking about part i right? if u are, i agree what u did but if u see i also added + (10C4*9C1*14C1). This was because u can ALSO have 4 men, 1 woman and any 1 other, either man or woman.
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Hmm............am not sure myself.
Just wait for some time. I'll ask Astar to clear it up for us. ;)
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ok thankx.. do let us know :) and By the way DK, as i asked b4, what do u think my grade wil be based on the marks i gave you.
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ok thankx.. do let us know :) and By the way DK, as i asked b4, what do u think my grade wil be based on the marks i gave you.
Ooh..yeah. Sorry i forgot about it. :-[
I managed better in p1 but worst in S1. So from my point of view, the threshold is not going to decrease much in p1 but a bit more in S1.
Though am not sure koz it will depend on the general trend and how students all across the world have performed.
But you'll sure be getting a B or an A.
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that jus made my day so much worse.. "But you'll sure be getting a B or an A." .. a B.. noooooooooo :( i NEED an A in order to get into uni! :(
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that jus made my day so much worse.. "But you'll sure be getting a B or an A." .. a B.. noooooooooo :( i NEED an A in order to get into uni! :(
Hey am not sure about it.
I've calculated your percentage from the points you think you'll be getting.
If it is the case then 82.4% will get you and A. ;)
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oh ok... what is the lowest overall percentage to get an A generally.. 80?
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oh ok... what is the lowest overall percentage to get an A generally.. 80?
Usually it is 80. But it can decrease upon difficulty of paper. ;)
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can you guys remember the answer for question five the one about the normal distribution ??
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can you guys remember the answer for question five the one about the normal distribution ??
Hmm.............nope. Sorry can't even remember the question. :-[
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well i remember there were two cars a zodiac with mean 320 and s.d 21.6 and the other car has mean 350 and I think the s.d was the same as the first one !! any memories ?? :P
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well i remember there were two cars a zodiac with mean 320 and s.d 21.6 and the other car has mean 350 and I think the s.d was the same as the first one !! any memories ?? :P
Ooh...........yeah.
The other car was Ganmor I think. And for the second part we needed to find d, right?
d was a small value, can't remember exactly though. :-[
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i got 4.96 or sumthin like that for d
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i got d= 4.96 but in (i) did you subtract the answer from 1 or not ??
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i got d= 4.96 but in (i) did you subtract the answer from 1 or not ??
Yeah.......i think I got that too. :)
Hmm............don't know :-[
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hehehe okii thanks anyway and don't worry about the paper i'm sure you'll do great just focus on the coming exams for now ;)
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hehehe okii thanks anyway and don't worry about the paper i'm sure you'll do great just focus on the coming exams for now ;)
Anytime :)
I hope so :)
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ur talking about part i right? if u are, i agree what u did but if u see i also added + (10C4*9C1*14C1). This was because u can ALSO have 4 men, 1 woman and any 1 other, either man or woman.
I'm not sure of this of course, but I don't think your answer is right. I got 9828 ways. 4 men, 1 woman and 1 man/woman- this doesn't make sense. It has to have a minimum of 4 men, right? This means that the only possible combinations are 4 men-2 women and 5 men-1 woman. So the answer should be (10C4*9C2)+(10C5*9C1)= 9828
I also got 2/3 for the last answer and I don't remember my other answers to that question, but I can probably work them out later and check them with you guys.
I got 8/11 for that 6 mark question, I'm SO HAPPY that other people got that answer too! That question was extremely confusing, you had to think about it for quite a while before attempting it. This was a tough S1 paper by comparison, so threshold should be atleast 35.
My new mean and standard deviation were 52.something and 7.something, if I remember right.
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Ahh ! I didn't know about this thread ! :(
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Click here (https://studentforums.biz/math-146/s1-doubts-here!!!!/) to see Astar's answers concerning the last number of p62. ;)
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can i get ans of cie maths 62 for oct/nov 2010
i need answer of all questions
1.
2.
3.
4.
5.
6.
7.
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can i get ans of cie maths 62 for oct/nov 2010
i need answer of all questions
1.
2.
3.
4.
5.
6.
7.
Am sorry but we are not the marking scheme. We are not sure of our own answers.
This paper has been thoroughly discussed in this thread. I would advice you to take the time and read the posts here. Some of our answers might coincide with yours. ;)