Statistics P6 : post your doubts here!

[yashi30]

NOV 07 P6 NUMBER 1 AND JUNE 2012 P62 NUMBER 3 ??? PLEASE HELP ME WITH ALL THE CALCULATIONS.

November 2007 number 1
(i)n=24, Σ(x-a)=-73.2 , Σ(x-a)^2=2115

Mean = Σ(x-a)/n +a
8.95 = -73.2/24 + a
a= 8.95 + 73.2/24 = 12

(ii) SD = square root of ( Σ(x-a)^2/n -( Σ(x-a)/n)^2)
SD= square root of ( 2115/24-(-73.2/24)^2) = 8.88(3sf)

 

[bamteck]

June 2004 P6 nO 5
(i) For a three-course meal,
nO of choices=( 3C1 X 5C1*2 X 3C1 ) =90
From the starter course, you have to choose 1 from 3 so 3C1
From main course, you have to choose 1 from 5 so 5C1, but it's mentioned that main courses are either served with new potatoes of french fries, so from your choice of main course, you either have new potatoes or french fries, total = 5C1*2
from dessert course you have 1 choice from 3 so 3C1
add all and you get the answer.

(ii) If a customer chooses the
(a)Starter and Main course
number of choices= 3C1*5C1*2=30
(b) Starter and Dessert Course
Number of choices= 3C1 X 3C1= 9
(c) Main course and dessert
Number of choices= 5C1*2 X 3C1 =30

total number of choices= 30+30+9= 69

Thank you

 

[yashi30]

Thank you

you're welcome.

 

[momo_khan]

can any one help me with question 6 part 2 from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_61.pdf

question 2 from -----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_62.pdf

question 2 and question 6 part two from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_63.pdf

does anyone have some good notes for statistics ?

 

[yashi30]

can any one help me with question 6 part 2 from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_61.pdf

question 2 from -----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_62.pdf

question 2 and question 6 part two from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_63.pdf

does anyone have some good notes for statistics ?

for the 1st one, more than 1SD from mean, either X=μ+SD or X=μ-SD
so you have to find P(μ-SD< X < μ+SD)
if you standardise(by subtracting μ and dividing by SD), you'll get P(-1< Z < 1)= A . to obtain the answer you multiply A by 1000

 

[yashi30]

can any one help me with question 6 part 2 from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_61.pdf

question 2 from -----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_62.pdf

question 2 and question 6 part two from ----> http://olevel.sourceforge.net/papers/9709/9709_s12_qp_63.pdf

does anyone have some good notes for statistics ?

for June 2012 P62 number 2
Y can take values 0,2,4
P(Y=0)= P(2,2) or P(4,4) or P(6,6)
= (0.5*0.5)+(0.4*0.4)+(0.1*0.1)=0.42
P(Y=2)= P(2,4) or P(4,2) or P(4,6) or P(6,4)
=(0.5*0.4)x2 +(0.4*0.1)x2=0.48
P(Y=4)= P(2,6) or P(6,2)
= (0.5*0.1)x2=0.1
you just need to draw your probability density function.

 

[momo_khan]

jazakAllah !
omo you sound like a genius to me !
thank you very much
can i ask some more questions?

 

[momo_khan]

question 5 and 7 from -------> http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf

question 6 from -----> http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf

question 2part 1 , 6 from -----> http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_63.pdf

question 3 part 3 from -----> http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_61.pdf

question 5 part 2 and 3 from ----> http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_62.pdf

thanks in advance.
can anyone lend me their statistics (p6) notes ??

whitecorp yashi30
@everyone

 

[yashi30]

question 5 and 7 from -------> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

question 6 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf

question 2part 1 , 6 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_63.pdf

question 3 part 3 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf

question 5 part 2 and 3 from ----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_62.pdf

thanks in advance.
can anyone lend me their statistics (p6) notes ??

whitecorp yashi30
@everyone

November 2011P61 number 5
(i) Let X be the random variable representing weight of letters
X~N(20, SD^2)
P(μ-12< X < μ+12)= 0.94
standardising, P(-12/SD < Z < 12/SD)=0.94
let a= 12/SD
P(-a< Z < a)=0.94
2*phi(a) -1=0.94
phi(a)=o.97
reading normal table in reverse order, you obtain a=1.881
therefore, SD=12/a= 12/1.881=6.3795=6.38(3sf)

(ii) Calculate P(X>13) by standardising( u already have mean and SD from previous part)

(iii) You can model it by a binomial with n=7, p=0.94 and q=0.06
find P(atleast two)= 1-[P(X=0)+P(X=1)]

 

[bamteck]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w05_qp_6.pdf

HELP NEEDED for no. 6
yashi30

 

[momo_khan]

November 2011P61 number 5
(i) Let X be the random variable representing weight of letters
X~N(20, SD^2)
P(μ-12< X < μ+12)= 0.94
standardising, P(-12/SD < Z < 12/SD)=0.94
let a= 12/SD
P(-a< Z < a)=0.94
2*phi(a) -1=0.94
phi(a)=o.97
reading normal table in reverse order, you obtain a=1.881
therefore, SD=12/a= 12/1.881=6.3795=6.38(3sf)

(ii) Calculate P(X>13) by standardising( u already have mean and SD from previous part)

(iii) You can model it by a binomial with n=7, p=0.94 and q=0.06
find P(atleast two)= 1-[P(X=0)+P(X=1)]

jazakAllah khair

 

[yashi30]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_6.pdf

HELP NEEDED for no. 6
yashi30

(i)Let X be the profit made.
profit= amount of money gained - amount of money paid= $3-$1= $2

(ii)For profit = $0
it means that he got $1 after he hit the target on the fourth or fifth throw, right?
so P(X=0)= P(WWWR) or P(WWWWR) where W= he didn't hit the target and R= he hit the target
therefore P(X=0)= (4/5*4/5*4/5*1/5)+(4/5*4/5*4/5*4/5*1/5)=0.18432=0.184

(iii) do the same for the other profit
define X as the profit gained
then X takes values of -1,0,2,4 and calculate the probabilities as I did in the previous part
is it okay?

 

[yashi30]

question 5 and 7 from -------> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf

question 6 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf

question 2part 1 , 6 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_63.pdf

question 3 part 3 from -----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf

question 5 part 2 and 3 from ----> http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_62.pdf

thanks in advance.
can anyone lend me their statistics (p6) notes ??

whitecorp yashi30
@everyone

November 2011 P62 number 6
(i) Let X be the random variable 'number of boys who choose music'
model it as a binomial ie X~Bin(n,p) where n=6 and p=0.35 q=0.85(0.75+0.1)

P(X<3)= P(X=0)+ P(X=1) + P(X=2)= 0.85^6+ 6C1*0.15*0.85^5+6C2*0.15^2*0.85^4=0.95266=0.953(3sf)

(ii) consider the conditional probability where
A(required): a boy
B(given): drama student

P(A/B) = P(AnB)/ P(B)
P(B)= P(drama student)= P(girls doing drama) + P(boys doing drama)= (0.4*0.55)+(0.6*0.1)=0.28
So, P(A/B)=P(boy doing drama)/P(drama student)= (0.6*0.1)/0.28=0.21428=0.214

Let X be the random variable 'a boy doing drama'
X~Bin(5,0.214)
P(atleast 1 boy)= 1-P(no boy)=1-[(1-0.214)^5]=0.70055=0.701(3sf)

 

[momo_khan]

could you explain me when to use suitable approximation . recent years do not mention when to use it ....can you guide me when to for example in paper nov 2010 question 6 mark scheme used 35.5 and 26.5 as limits. i simply used 35 and 27 and i got the wrong answer .

yashi30

XPFMember

 

[yashi30]

could you explain me when to use suitable approximation . recent years do not mention when to use it ....can you guide me when to for example in paper nov 2010 question 6 mark scheme used 35.5 and 26.5 as limits. i simply used 35 and 27 and i got the wrong answer .

yashi30

XPFMember

Consider the random variable X with X~Bin(100,0.2)
if you had to calculate P(X<50) which is equal to P(X=0,1,2.......,49) right? so this is a very long and tedious calculation.
In this case if product np and nq are both greater than 5 then X~N(np,npq)
then you proceed by making your continuity correction
(i) greater than and/or less or equal to: you add 0.5
(ii) less than and/or greater or equal to: you deduct 0.5

in the example i gave you
np=100*0.2=20 nq=100(1-0.2)=80 Var(x)=npq= 100*0.2*0.8=16
since both np and nq>5, X~N(20,16)
P(X>50)=P(X>50.5) ------> when you apply continuity correction, you add 0.5 in this case)
standardising, P(Z> 50.5-20/square root of 16)= P(Z>7.625)

this is only an example, you won't be able to solve it because the normal table does not go till that value obtained, but the principle is the same. you get it?

 

[momo_khan]

Consider the random variable X with X~Bin(100,0.2)
if you had to calculate P(X<50) which is equal to P(X=0,1,2.......,49) right? so this is a very long and tedious calculation.
In this case if product np and nq are both greater than 5 then X~N(np,npq)
then you proceed by making your continuity correction
(i) greater than and/or less or equal to: you add 0.5
(ii) less than and/or greater or equal to: you deduct 0.5

in the example i gave you
np=100*0.2=20 nq=100(1-0.2)=80 Var(x)=npq= 100*0.2*0.8=16
since both np and nq>5, X~N(20,16)
P(X>50)=P(X>50.5) ------> when you apply continuity correction, you add 0.5 in this case)
standardising, P(Z> 50.5-20/square root of 16)= P(Z>7.625)

this is only an example, you won't be able to solve it because the normal table does not go till that value obtained, but the principle is the same. you get it?

oohkay ..

jazakAllah khair

you explain really well mashaAllah are you a maths teacher by any chance ? :O

 

[yashi30]

oohkay ..

jazakAllah khair

you explain really well mashaAllah are you a maths teacher by any chance ? :O

Haha,no i'm a student taking my A level for the first time.

 

[momo_khan]

Haha,no i'm a student taking my A level for the first time.

what subjects ? are you done with AS?

 

[bamteck]

(i)Let X be the profit made.
profit= amount of money gained - amount of money paid= $3-$1= $2

(ii)For profit = $0
it means that he got $1 after he hit the target on the fourth or fifth throw, right?
so P(X=0)= P(WWWR) or P(WWWWR) where W= he didn't hit the target and R= he hit the target
therefore P(X=0)= (4/5*4/5*4/5*1/5)+(4/5*4/5*4/5*4/5*1/5)=0.18432=0.184

(iii) do the same for the other profit
define X as the profit gained
then X takes values of -1,0,2,4 and calculate the probabilities as I did in the previous part
is it okay?

Yeah thank you loads

 

[yashi30]

what subjects ? are you done with AS?

A level: Maths, Chemistry, Physics
As level: Biology and General Paper(over)