A level Statistics doubt??Post your doubts here!

[Mecd66]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf
Question 7)ii)
if somebody can please explain this to me please.

 

[Anika Raisa]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w12_qp_61.pdf
( Q7 - c part ) I don't get it at all . brain not working ! please help me
Jiyad Ahsan

9 diff fruit pies
3 ppl ---- all gets odd no.

List of odd no. 1 3 5 7 _(9 not possible otherwise all will not get pies!)

Now , lets list possible ways the no. can be divided

1st per 2nd per 3rd per

1 3 5

1 5 3

1 7 1

7 1 1

1 1 7

3 1 5

3 3 3


(See that all combinations add to 9)

Now u c ms says:


1 1 7 = 9C1× 8C1× 7C7 (oe)× 3C1 =216 In above table:here c only 1 out of the three cn get 7 pies so 3C1.
1 3 5 = 9C1× 8C3× 5C5(oe)× 3! = 3024 In above table:here c either can hv 5 or 3 or 1 pie in any order so 3!
3 3 3 = 9C3× 6C3× 3C3 (oe) = 1680
In above table:here c its all getting 3 pie n order doesnt matter!
so you add all n get d final ans!
Hope i cud xplain! Though i myself is not much clear abt y we use 3C1 instd of 3! in the first part ...Bt dat was d explanation we were told @class . A little help there with this Alice123 or syed1995 ASTAR or any1!

 

[A star]

9 diff fruit pies
3 ppl ---- all gets odd no.

List of odd no. 1 3 5 7 _(9 not possible otherwise all will not get pies!)

Now , lets list possible ways the no. can be divided

1st per 2nd per 3rd per

1 3 5

1 5 3

1 7 1

7 1 1

1 1 7

3 1 5

3 3 3


(See that all combinations add to 9)

Now u c ms says:


1 1 7 = 9C1× 8C1× 7C7 (oe)× 3C1 =216 In above table:here c only 1 out of the three cn get 7 pies so 3C1.
1 3 5 = 9C1× 8C3× 5C5(oe)× 3! = 3024 In above table:here c either can hv 5 or 3 or 1 pie in any order so 3!
3 3 3 = 9C3× 6C3× 3C3 (oe) = 1680
In above table:here c its all getting 3 pie n order doesnt matter!
so you add all n get d final ans!
Hope i cud xplain! Though i myself is not much clear abt y we use 3C1 instd of 3! in the first part ...Bt dat was d explanation we were told @class . A little help there with this Alice123 or syed1995 ASTAR or any1!

it because two variables are the same 1 and 1 so only change that can make the difference is 7 so 3C1 shows
this 1 1 7 or
171 or
711

 

[Anika Raisa]

it because two variables are the same 1 and 1 so only change that can make the difference is 7 so 3C1 shows
this 1 1 7 or
171 or
711

Oh so when possibility is 1 5 3 we are doing 3! as d 3 digits are diff ryt!
Thank u!

 

[A star]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf Q5 part ii

 

[syed1995]

9 diff fruit pies
3 ppl ---- all gets odd no.

List of odd no. 1 3 5 7 _(9 not possible otherwise all will not get pies!)

Now , lets list possible ways the no. can be divided

1st per 2nd per 3rd per

1 3 5

1 5 3

1 7 1

7 1 1

1 1 7

3 1 5

3 3 3


(See that all combinations add to 9)

Now u c ms says:


1 1 7 = 9C1× 8C1× 7C7 (oe)× 3C1 =216 In above table:here c only 1 out of the three cn get 7 pies so 3C1.
1 3 5 = 9C1× 8C3× 5C5(oe)× 3! = 3024 In above table:here c either can hv 5 or 3 or 1 pie in any order so 3!
3 3 3 = 9C3× 6C3× 3C3 (oe) = 1680
In above table:here c its all getting 3 pie n order doesnt matter!
so you add all n get d final ans!
Hope i cud xplain! Though i myself is not much clear abt y we use 3C1 instd of 3! in the first part ...Bt dat was d explanation we were told @class . A little help there with this Alice123 or syed1995 ASTAR or any1!

Repetition is there in part i.

It becomes 3!/2! .. which is 3.

 

[A star]

guys and gals lets use this thread to discuss answers as well 24 hours after the paper

 

[A star]

guys and gals lets use this thread to discuss answers as well 24 hours after the paper

 

[suryashekhar]

The weights of letters posted by a certain business are normally distributed with mean 20 g. It is
found that the weights of 94% of the letters are within 12 g of the mean.
(i) Find the standard deviation of the weights of the letters.
w11 61 q5(i)

 

[A star]

The weights of letters posted by a certain business are normally distributed with mean 20 g. It is
found that the weights of 94% of the letters are within 12 g of the mean.
(i) Find the standard deviation of the weights of the letters.
w11 61 q5(i)

please see the answers posted before

 

[suryashekhar]

please see the answers posted before

hey man why do they take 0.97 instead of 0.94?

 

[vinit]

hey guys how do u do OCT/NOV 2011 variant 61 question 5?? my answers are completely different from the mark scheme!

 

[aniketjain]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf
1st question how do u calculate s.d (standard deviation)

 

[kiara15]

No problem at all! BOL 4 2mrw! Hope everything goes well!! #fingerscrossed
May Allah help us all while we take our exams Ameen!

ameen.. yeah fingers crossed

 

[x-gamer-x]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf
1st question how do u calculate s.d (standard deviation)

add square of all i.e. (48.3^2 + 55.2^2 + . . . 64.1^2) then divide this by 11 then subtract mean^2 and then take square root

 

[aniketjain]

add square of all i.e. (48.3^2 + 55.2^2 + . . . 64.1^2) then divide this by 11 then subtract mean^2 and then take square root

its not coming lyk tht ..... the value is coming in negative ... and square root of negative is not possible ....

 

[A star]

add square of all i.e. (48.3^2 + 55.2^2 + . . . 64.1^2) then divide this by 11 then subtract mean^2 and then take square root

u could have stated the sd formulae -_-

 

[A star]

its not coming lyk tht ..... the value is coming in negative ... and square root of negative is not possible ....

whats ur mean

 

[aniketjain]

add square of all i.e. (48.3^2 + 55.2^2 + . . . 64.1^2) then divide this by 11 then subtract mean^2 and then take square root

hey sorry i got it .....

 

[aniketjain]

thnk u