A level Statistics doubt??Post your doubts here!

[Champ101]

Hey solved it but i m not sure if the ans is ryt!! Hope i cud help!!

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I was confused but you helped me out. Thankyousomuch. Appreciated.

 

[Anika Raisa]

I was confused but you helped me out. Thankyousomuch. Appreciated.

NP! Feels good that i cud help! BOL 4 xam!N pray i cn do well in xam!

 

[Anika Raisa]

Some really gud note available on Statistics here:
http://people.richland.edu/james/lecture/m170/

 

[daredevil]

happens! BOL 4 xam! N pray 4 me 2!!!

yeahhh it does happen ryt?!! :| and that sucks wen it happens..... i hope i dont have that moment in the exam room X(
yeah BOL! i will IA ... u pray for me too

 

[Anika Raisa]

yeahhh it does happen ryt?!! :| and that sucks wen it happens..... i hope i dont have that moment in the exam room X(
yeah BOL! i will IA ... u pray for me too

Sure i wil pray to! N everything wil be fine Xam hall in shaa Allah!

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf
ques num 1 plz

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prekshya nxt tym post ur stat probs here please!

Hope i was helpful!!!

 

[sma786]

Among so many types of maths n Statistics xam later thn d other math xams it seems the statistics problems stay completely ignored in the thread:Mathematics: Post your doubts here!
So this is for the candidates taking both S1-Paper 6 n also for candidates taking S2-Paper 7

yeah so true, stats is completely ignored

 

[A star]

thats cause many people dont have stats. yeah good job for the thread

 

[prekshya]

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prekshya nxt tym post ur stat probs here please!

Hope i was helpful!!![/quote]
ok and thank u so much!!

 

[Champ101]

hey guys the only thing in stats i'm stuck at is probability+permutation combination. How do i cope up with them?

The answers i get is itself a probability don't know but i need more confidence. Please someone help me out!!!!

 

[prekshya]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_6.pdf ques num 3 :'( i even dont know this....help please

 

[Champ101]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf ques num 3 :'( i even dont know this....help please

hi prekshya,
3(a) I'm sure its a binomial distribution question,
where X is the random variable
n =5 ,
probability of getting a green face=p= 1/4 (which is the probability of success)

The question asked for P(X=4), so just use the formula, which gives
5C4 *(1/4)^4*(3/4)

i hope this helps!

 

[sma786]

I love this thread lol
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 4, (iii), wtf?

 

[Champ101]

continued->
3(b) well this question is quite tedious. In this case also, n=5, p=1/4 hence q=1-p=3/4
We are given 5 chances out of which we have to see how many of them lands on the green face. It could be all of them or none of them or 1 of them.
When x=0, which means none of the dice lands on the green face, then P(X=0) = 5C0*(1/4)^0*(3/4)^5
When x=1, which means only one dice lands on the green face, then P(X=1) =5C1*(1/4)^1*(3/4)^4
And likewise we calculate upto x=5, and draw up a table.

i hope you get it

 

[Champ101]

I love this thread lol
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 4, (iii), wtf?

wow it feels great after helping someone, wait i'm on it

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 6, (iii), how did they get a 5th match in the table? shouldn't 4th match be the last?

 

[Champ101]

I love this thread lol
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 4, (iii), wtf?

uff, finally got it
Its 3c2*(5c1+2c1)

In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples AND 1 other person, who may OR may not be married?

first condition : out of 3 married couples, we need two so selection = 3c2
second condition: the other 1 person may OR may not be married (look at that keyword) so you need to add. out of 5 unmarried you can select any one, so selection : 5c1, now consider the remaining unmarried couple of which only 1 couple left, so we can select any one of the two married person, therefore selection: 2C1

ahh im tireedd of stats :\

 

[Champ101]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s06_qp_6.pdf
Question 6, (iii), how did they get a 5th match in the table? shouldn't 4th match be the last?

32->16->8->4->2 (so 5 matches)

Thus 16 teams play in the second round, 8 teams play in the third round, AND SO ON, until 2 teams play in the final round.

i hope you get it DD

 

[sma786]

uff, finally got it
Its 3c2*(5c1+2c1)

In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples AND 1 other person, who may OR may not be married?

first condition : out of 3 married couples, we need two so selection = 3c2
second condition: the other 1 person may OR may not be married (look at that keyword) so you need to add. out of 5 unmarried you can select any one, so selection : 5c1, now consider the remaining unmarried couple of which only 1 couple left, so we can select any one of the two married person, therefore selection: 2C1

ahh im tireedd of stats :\

great, genius you are thankyou boy !

 

[sma786]

32->16->8->4->2 (so 5 matches)

Thus 16 teams play in the second round, 8 teams play in the third round, AND SO ON, until 2 teams play in the final round.

i hope you get it DD

na na, in the markscheme its different, i did the same, but see what they did here : http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s06_ms_6.pdf