A level Statistics doubt??Post your doubts here!

[Anika Raisa]

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

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

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_62.pdf
Question 2, (ii), we found p as 0.854 in part i, so in part (ii), why did they take q as 0.854?

Usually q is considered as the probability of failure. In 1 we found out the we found the probability of pencils with length greater than 10.9cm.But in two we are dealing with choosing pencils which are less than 10.9cm.So our p here should be the probability of pencil with length less than 10.9 which would be 1-our answer in i . n our answer in 1 is our q!
Was that helpful enough?

 

[sma786]

Usually q is considered as the probability of failure. In 1 we found out the we found the probability of pencils with length greater than 10.9cm.But in two we are dealing with choosing pencils which are less than 10.9cm.So our p here should be the probability of pencil with length less than 10.9 which would be 1-our answer in i . n our answer in 1 is our q!
Was that helpful enough?

great got it, how do guys have such good brains?

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_62.pdf
Question 2, (ii), we found p as 0.854 in part i, so in part (ii), why did they take q as 0.854?

Usually q is the probability of failure. n p is the probability of success!
Here in i we found out the prob of pencil with length greater thn 10.9cm. However we r to deal with pencils less than 10.9 cm in ii. So our p wud be 1-ans in i n our q wud be the ans in i . So if we considered the value of p to be same in both i n ii then we wud end up getting wrong ans 4 part ii as we would be finding probability of choosing 2 pencils greater than 10.9cm from 6 rather than less!
Hope u got that now!

 

[Anika Raisa]

great got it, how do guys have such good brains?

haha nt really! Just pray dat the real xam goes well! Its pleasure to hv helped u!

 

[salvatore]

Hello

I need help with question no. 5(iii) of the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

Why is p=0.03 and not 0.97? syed1995

 

[iFuz]

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

P62 or P61?

 

[syed1995]

Hello

I need help with question no. 5(iii) of the following paper:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

Why is p=0.03 and not 0.97? syed1995

SD=6.38
Mean=20..

P(X>32)
P(z>1.8809)
P(z<-1.8809) = 1-Phi(1.8809)
P(z<-1.8809) =1- 0.97
P(z<-1.8809) = 0.03

hence p=0.03

 

[Champ101]

It is known that, on average, 2 people in 5 in a certain country are overweight. A random sample of 400 people is chosen. Using a suitable approximation, find the probability that fewer than 165 people in the sample are overweight.

help me out guys. Its a binomial distribution which is then approximated by using normal distribution. Right? But how do we start.

 

[Sarah_3420]

Are there any notes or something i can get for statistics :S i am really lost and scared i have no hope anymore :/

 

[felicia tan]

can anyone help me to solve 3b?http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

 

[Anika Raisa]

can anyone help me to solve 3b?http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s07_qp_6.pdf

Let the random variable X be normally distributed where with u(meu) n s(sigma) as mean n standard deviation respectively....
Nw we need to find
P((u+s)<X<(u-s))

Now standarising
we have
p{((u-s-u)/s)<Z<((u-s-u)/s)}
so, P(-1<Z<1)
so phi(1)-phi(-1)
=phi 1 -[1-phi(-1)]
={2phi(1)}-1
=(2*0.8413)-1
= .6826

therefore for 1 observation the p is .6862
so, for 800 it is .6826*800
=546 (Ans)

Hope i cud help!!!

 

[Anika Raisa]

Are there any notes or something i can get for statistics :S i am really lost and scared i have no hope anymore :/

http://people.richland.edu/james/lecture/m170/

 

[Anika Raisa]

P62 or P61?

Both aka all variants !

 

[Anika Raisa]

This is the question syed1995 Anika Raisa

C over here u got p(A)=1/3 in i
n P(B)=5/9 in ii

now A intersection B means that the score differ by 3 or more and also there product is greter than 8

so P(A intersection B)=6/36

not 0 so A n B are not mutually exclusive!!
Btwn dis is frm on06 or mj06...

 

[daredevil]

C over here u got p(A)=1/3 in i
n P(B)=5/9 in ii

now A intersection B means that the score differ by 3 or more and also there product is greter than 8

so P(A intersection B)=6/36

not 0 so A n B are not mutually exclusive!!
Btwn dis is frm on06 or mj06...

thankks alot.... that P(A intersection B) was wat i was missing on.. like i knew it had to be there .... but cudnt figure where it was -_-

 

[felicia tan]

t

Let the random variable X be normally distributed where with u(meu) n s(sigma) as mean n standard deviation respectively....
Nw we need to find
P((u+s)<X<(u-s))

Now standarising
we have
p{((u-s-u)/s)<Z<((u-s-u)/s)}
so, P(-1<Z<1)
so phi(1)-phi(-1)
=phi 1 -[1-phi(-1)]
={2phi(1)}-1
=(2*0.8413)-1
= .6826

therefore for 1 observation the p is .6862
so, for 800 it is .6826*800
=546 (Ans)

Hope i cud help!!!

thank you very much! I can do it now

 

[Anika Raisa]

It is known that, on average, 2 people in 5 in a certain country are overweight. A random sample of 400 people is chosen. Using a suitable approximation, find the probability that fewer than 165 people in the sample are overweight.

help me out guys. Its a binomial distribution which is then approximated by using normal distribution. Right? But how do we start.

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

 

[Anika Raisa]

np!

t
thank you very much! I can do it now

 

[Anika Raisa]

thankks alot.... that P(A intersection B) was wat i was missing on.. like i knew it had to be there .... but cudnt figure where it was -_-

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