A level Statistics doubt??Post your doubts here!

[sma786]

oh no ummm... ok c d table on d ms it says no. of matches(x) above n no. of teams (f) below so wat happens is dat only two teams make it to a fifth match so u have two below 5! uh tell me if i cudnt clear u still!

You go thanks!

 

[Anika Raisa]

You go thanks!

no problem at all thnx to u actuali .... i didnt do dis sum b4 bt i got 2 do it only 4 ya!
N it was seriously a challenge!

 

[Anika Raisa]

sma786 here wat u do is 1st take midvalues of the range as x! now square x! then multiply x^2 wid d its complementary frequency (aka frequency of d range!).... now add all the f*x^2 s n divide the result by total frequency then subtract square of the mean from dis result n then square root the final result! n there u go u got d sd!!!

 

[sma786]

sma786 here wat u do is 1st take midvalues of the range as x! now square x! then multiply x^2 wid d its complementary frequency (aka frequency of d range!).... now add all the f*x^2 s n divide the result by total frequency then subtract square of the mean from dis result n then square root the final result! n there u go u got d sd!!!

i got it so i deleted the post :$ thanks though

 

[Anika Raisa]

i got it so i deleted the post :$ thanks though

Ah dats xplains it dats y i cudnt quote it! Na its k!!!

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_6.pdf
Question 3, (i), why did they take 5C4 ? why not 1-5C5 ?

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf
Question 3, (i), why did they take 5C4 ? why not 1-5C5 ?

Here u r told to find the probability of exactly 4 dices landing on green face so u consider P(X=x) but if u were told at most four dice lands on green face then u wud have considered.... P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) which is same as 1-5C5!
Cud i be clear enough! :~

 

[sma786]

pe

Here u r told to find the probability of exactly 4 dices landing on green face so u consider P(X=x) but if u were told at most four dice lands on green face then u wud have considered.... P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4) which is same as 1-5C5!
Cud i be clear enough! :~

perfect, means they are saying of exactly 4, so we take distribution of EXACTLY 4, that is 5C4

 

[Anika Raisa]

pe
perfect, means they are saying of exactly 4, so we take distribution of EXACTLY 4, that is 5C4

yupp.

 

[sma786]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_6.pdf
Question 7, a (i), why not do this : 3C1*3C1*5C1/11C3 ?!

 

[Kumkum]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf
Question 7, a (i), why not do this : 3C1*3C1*5C1/11C3 ?!

ok i'll try to explain
first of all you cannot divide by 11C3 because ur not asked to find any probabilities only the number of ways.
ur combinations are ok except it should be 3C1*3C0*5C1
why is it 3C0?
u see in the question they say the captain must be included, so there is no need of choosing a person from the 3 mid-fielders, it means the captain is already in the team. so now u have to choose two ppl, one who plays in defence and the other who plays in forward (since u need to choose 3 ppl to collect a gold medal)
therefore number of ways = 3C1 * 3C0 * 5C1 = 15
u dnt need 3C0 since it is 1...
hope i've helped nd didn't confuse u more

 

[sma786]

ok i'll try to explain
first of all you cannot divide by 11C3 because ur not asked to find any probabilities only the number of ways.
ur combinations are ok except it should be 3C1*3C0*5C1
why is it 3C0?
u see in the question they say the captain must be included, so there is no need of choosing a person from the 3 mid-fielders, it means the captain is already in the team. so now u have to choose two ppl, one who plays in defence and the other who plays in forward (since u need to choose 3 ppl to collect a gold medal)
therefore number of ways = 3C1 * 3C0 * 5C1 = 15
u dnt need 3C0 since it is 1...
hope i've helped nd didn't confuse u more

Got it, thanks

 

[Kumkum]

Got it, thanks

no prob

 

[prekshya]

http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s05_qp_6.pdf 6 .ii.help please..

 

[A n g e l a ♥]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_61.pdf3iib and 6jj please

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf 6 .ii.help please..

i cant even get the right answer for the first part -__-
i am getting 0.0208
z1 = (1.82-1.9)/1.92
z2 = (1.92-1.9)/1.92
z2-z1

this is wat i did. is it not right??

 

[daredevil]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf 6 .ii.help please..

i get the idea for the second bar as take this :
(1.9-b) < X < (1.9+b)

find the expressions for z1 and z2

equate z2-z1 with 0.8

0.8 is ur probability as they have given 80% and that means 0.8 if we write it as deimals... i didnt get the corect answer for he first part so i didnt give this one a shot. try this if u can do it then tell me how u did all of it. thanks

i seem to undertand wat to do in the question but at the end of it i cant get the answer and how that come around i have no freaking idea!! the examiner is not going to give me marks for my thinking skills for crying out loud!! :O *frustrated*

 

[prekshya]

i cant even get the right answer for the first part -__-
i am getting 0.0208
z1 = (1.82-1.9)/1.92
z2 = (1.92-1.9)/1.92
z2-z1

this is wat i did. is it not right??

it is...just subtract the negative z with 1 and u get the answer and thnx for d help

 

[prekshya]

it is...just subtract the negative z with 1 and u get the answer and thnx for d help

ok 6.i .
1.82-1.9/0.15 < x < 1.92-1.9/0.15
-0.53333<x<0.1333
so.. 0.13333-(-0.533)
0.1333-[1-0.5333]..put the values and u get the ans as 0.256 but i didnt get y the ans is as (0.256)^4 ..it may b bcoz we have to find probability of all four tyres on a car....is that true??

 

[daredevil]

ok 6.i .
1.82-1.9/0.15 < x < 1.92-1.9/0.15
-0.53333<x<0.1333
so.. 0.13333-(-0.533)
0.1333-[1-0.5333]..put the values and u get the ans as 0.256 but i didnt get y the ans is as (0.256)^4 ..it may b bcoz we have to find probability of all four tyres on a car....is that true??

ooo i did the stupidest mistake evvaar!! i toook 1.92 as standard deviation i dont know y :O >__<