A level Statistics doubt??Post your doubts here!

[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??

yeah thats it... the probabiility of tyre having this pressure is 0.256 so of all 4 it will be 0.256*0.256*0.256*0.256
[its an AND case so all prob will mulitply]
that is how the answer is (0.256)^4

 

[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??

yaar i got the answer up until getting
(b/0.15)=+/- 1.281 but after that i'm not getting the limits... i mean we have to cross multiply here ryt??
did u get this far?

 

[prekshya]

frm where d

yaar i got the answer up until getting
(b/0.15)=+/- 1.281 but after that i'm not getting the limits... i mean we have to cross multiply here ryt??
did u get this far?[/quote
i didnt get...i mean frm where that 1.281 came frm?

 

[daredevil]

frm where d

look wen u have the probability u look at the distribution table and look for this number.
look the prob is given 0.8 ryt?
and wen u solve z2-z1 for this case u come up with this: -

[(fie)(b/0.15)] - [1 - (fie)(b/0.15)] = 0.8
solving this above eq u get: -
2[(fie)(b/0.15)] = 1.8

{the 1 on LHS goes to the RHS and is added to 0.8
then divide both sides by 2 and u get: -
[(fie)(b/0.15)] = 0.9

take fie inverse and u have:
(b/0.15) = (fie inverse) 0.9

u get fie inverse by looking it up in that distribution table we have for the Z probabilities.
in this case the value close to 0.9000 is 1.281 so u take that and bcz the +/- will both be the same answer u take +/-1.281

got it?

 

[syed1995]

anyone got questions.. ?

 

[Alice123]

anyone got questions.. ?

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_6.pdf
no 4iii with explanation please

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_6.pdf
no 4iii with explanation please

Umm I don't understand this one very well... if you get the answer tag me as well..!

 

[littlecloud11]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_6.pdf
no 4iii with explanation please

3 of the tenors refuse to stand next to and of the basses right? So one of the tenors must always stand between them. Therefore you can only move the other 3 tenors around. So for them it's 3!. All the bases can be moved around as you please, so it's 4! for them. Now imagine the 4 tenors and the 4 basses in a line. The first 4 can be tenors and the next 4 can be basses, or, the first 4 can be basses and the next tenors. So that's two more ways.
Total number of ways = 3!* 4! *2 = 288

Hope this helps!
syed1995

 

[syed1995]

3 of the tenors refuse to stand next to and of the basses right? So one of the tenors must always stand between them. Therefore you can only move the other 3 tenors around. So for them it's 3!. All the bases can be moved around as you please, so it's 4! for them. Now imagine the 4 tenors and the 4 basses in a line. The first 4 can be tenors and the next 4 can be basses, or, the first 4 can be basses and the next tenors. So that's two more ways.
Total number of ways = 3!* 4! *2 = 288

Hope this helps!
syed1995

Thanks a lot littlecloud11 .. Honestly leaving statistics for 6-7 days for P1 and other AS papers has made me forget some of statistics or maybe it's that I am sleepy right now

Best of Luck for the papers littlecloud11 .. May Allah grant you the grades you desire and make you successful in life

 

[littlecloud11]

Thanks a lot littlecloud11 .. Honestly leaving statistics for 6-7 days for P1 and other AS papers has made me forget some of statistics or maybe it's that I am sleepy right now

Best of Luck for the papers littlecloud11 .. May Allah grant you the grades you desire and make you successful in life

Same here actually. I did stat just a few days back but now that I'm doing it again I seem to have forgotten so much.

All the very best for your papers too. May Allah grant you sucess. =)

 

[syed1995]

Same here actually. I did stat just a few days back but now that I'm doing it again I seem to have forgotten so much.

All the very best for your papers too. May Allah grant you sucess. =)

Thanks!

 

[shafayat]

hello peeps! I don't at all understand y they do +/- in the markscheme and I don't even get 7.2 as my SD .. Please if someone can help

 

[littlecloud11]

hello peeps! I don't at all understand y they do +/- in the markscheme and I don't even get 7.2 as my SD .. Please if someone can help

3i) z value for .9 = 1.282
as .9 is for probability of >5.2
z= -1.282
so, 5.2 - 3σ/σ = -1.282
σ = 7.24

ii) For this part the range is 1 standard deviation from the mean:
so the range of z value should be -1<z<1
ϕ1 - (1-ϕ1)
= .8413 - 1 +.8413
= .683

Number of observations = .683* 800 = 546.4

 

[Alice123]

3 of the tenors refuse to stand next to and of the basses right? So one of the tenors must always stand between them. Therefore you can only move the other 3 tenors around. So for them it's 3!. All the bases can be moved around as you please, so it's 4! for them. Now imagine the 4 tenors and the 4 basses in a line. The first 4 can be tenors and the next 4 can be basses, or, the first 4 can be basses and the next tenors. So that's two more ways.
Total number of ways = 3!* 4! *2 = 288

Hope this helps!
syed1995

thanks

 

[fishfish]

hhi guys, i cant get this simple question on normal distribution.
X~N(0,1)
P(Z > u ) = 0.8496
i dont get how its done, i mean if we minus 0.8496 by 1, then it wont even show on the table :/
helelplp

 

[syed1995]

hhi guys, i cant get this simple question on normal distribution.
X~N(0,1)
P(Z > u ) = 0.8496
i dont get how its done, i mean if we minus 0.8496 by 1, then it wont even show on the table :/
helelplp

Can you give me a link to the question?


I think it will be ...
P(Z>u) = 0.8496
P(Z<u)= 0.1504
u=inversePhi(0.1504)
u=-invPhi(1-0.1504)
u=-invPhi(0.8496)
u=-1.034

 

[fishfish]

Can you give me a link to the question?


I think it will be ...
P(Z>u) = 0.8496
P(Z<u)= 0.1504
u=inversePhi(0.1504)
u=-invPhi(1-0.1504)
u=-invPhi(0.8496)

its in my S1 maths book, no link. but you did it correctly.
but like why did you reverse the sign in the second step? you first subtracted 1, then what did you do?

 

[syed1995]

its in my S1 maths book, no link. but you did it correctly.
but like why did you reverse the sign in the second step? you first subtracted 1, then what did you do?

Well .. It's a rule like

P(Z>u) = P(Z<-u)

And P(Z<-u) = 1-Phi(u)

1-Phi(u) = 0.8496
Phi(u)=0.1504
u= invPhi(0.1504)

Now there's another rule that .. If the inverse phi is less than 0.5 .. we take it negative and minus it from 1.. like phi(0.250) = -phi(1-0.250) so that's what I used in this question...

so u= -invPhi(1-0.1504)
u=-invPhi(0.8496)

 

[fishfish]

Well .. It's a rule like

P(Z>u) = P(Z<-u)

And P(Z<-u) = 1-Phi(u)

1-Phi(u) = 0.8496
Phi(u)=0.1504
u= invPhi(0.1504)

Now there's another rule that .. If the inverse phi is less than 0.5 .. we take it negative and minus it from 1.. like phi(0.250) = -phi(1-0.250) so that's what I used in this question...

so u= -invPhi(1-0.1504)
u=-invPhi(0.8496)

OHH, okayyy! thank you so much, i get it now

 

[syed1995]

OHH, okayyy! thank you so much, i get it now

No problem mate! helping you .. cleared one of my concepts as well.. so thank you