Mathematics: Post your doubts here!

[mominzahid]

View attachment 10533
Can u explain this??

yes please someone explain this.. :/

 

[mukki]

I still dont get it.. :S
ok forget that question im too dumb to understand it lol..:/
can u please explain this how we find the standard deviation?

the area less then the upper quartile q3 is 75% =.75 Hence P(X<63)=.75
standardize and ul get P(z<63-51/s.d) =.75
dont have the normal table right now but this should do it find the value of fi inverse of .75 and equate it z=63-51/s.d

 

[Khan_971]

Can anyone help me with this question...It's about permutation= =2011 OCT statistic 9709/62-------Q2
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf

Tree diagram. You do know conditional probablility right. It will be P( It was in his pencil case {.7} / he finds it when he looks for it {.7 + .3*.2} )

 

[mominzahid]

the area less then the upper quartile q3 is 75% =.75 Hence P(X<63)=.75
standardize and ul get P(z<63-51/s.d) =.75
dont have the normal table right now but this should do it find the value of fi inverse of .75 and equate it z=63-51/s.d

Ahh... that i understoood...
thanks bro..
can u also please explain the question posted above by

View attachment 10533
Can u explain this??

???

 

[Khan_971]

yes please someone explain this.. :/

View attachment 10533
Can u explain this??

Part a i. Well duh there's no repeating. so just 4! = 24 arrangements.
Part a ii. Well U have to work on this. an odd number will have 1,3,5 in the end. 5 or 6 in the 1st digit and any of the 4 as 2nd digit. use this to write down possible combinations.

b. NOT next to each other has many possibilities, but to make it easy to can 1 - (P that they are together). Total no. of Permutations are: 6!
If u make a box diagram to solve it, meaning boxes containing this numbers. it will be easier. try making a box containing 4 and 5, another containing the rest (4 divisions in it though!!) so the 4,5 box can have 5 positions.
At any of the positions, the rest will have arrangements of 4! in which they are together.
AND it is also possible for it to be 4,5 or 5,4 (2 arrangements)
therefore arrangements: 5*4!*2 = 240
Therefore Prob. that they will be together: 6! /240
Subtract from 1 and see if it is correct.

 

[smartangel]

Q2. You should draw a tree diagram for this. Then you will understand. See, the questions if you analyze asks for everything except Males NOT watching the kids. Hence add up all probabilities except Males watching. it is 5/6. if you try to answer numerically, it usually doesnt come right.

Q3. A factory is making 3m to 5m ropes in the ratio 4:1.
part i) now you can draw a table in which 1st row is labelled 3m and 5m. 2nd row the probability of that 1 rope being 3 or 5m.

Lengths .........3m.....5m
Prob. ....... 4/5..... 1/5
And u know how to find E(x) and Var from this table.

Part ii) they can be 3,5 or 5,3 since 2 ropes of different lengths. Hence, (4/5 x 1/5) x 2 (since 2 arrangements)

part iii) the only possibility is a combination of 3+3+5. it can be 5,3,3 or 3,5,3. Therefore, (4/5 * 4/5 * 1/5) * 3!/2! (since 3 objects but 2 are the same)

can you plz check if my tree diagram is right..not getting the ans still..

 

[Khan_971]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
Q3..plz tell me how do we calculate other values of probabilitty of x?

Draw a probability distribution table. Since the rest except 0 have equal probability, let it be x
Let X be the random variable and P(X) being its probability

X....-2...-1...0..........1...2...3...4...5
p(x).x...x....1/10......x....x...x...x..x

Since total probability is always 1, 7x + 1/10= 1
7x = 9/10
Therefore x = 9/70

 

[sunnyexo]

Can U help me with this question...It's about permutation= =2011 OCT statistic 9709/62-------Q2
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf

 

[hassam]

help

 

[Khan_971]

can you plz check if my tree diagram is right..not getting the ans still..

Correct!
look the the question. they asked females or anyone who watches or both! (both already comes in when you add P(female)+ P(anyone who watches since it means both genders). U can add these up OR
since the only people left out are people who DONT watch, their probability = 5/30. Subtract from 1 to get the answer.

 

[mominzahid]

Part a i. Well duh there's no repeating. so just 4! = 24 arrangements.
Part a ii. Well U have to work on this. an odd number will have 1,3,5 in the end. 5 or 6 in the 1st digit and any of the 4 as 2nd digit. use this to write down possible combinations.

b. NOT next to each other has many possibilities, but to make it easy to can 1 - (P that they are together). Total no. of Permutations are: 6!
If u make a box diagram to solve it, meaning boxes containing this numbers. it will be easier. try making a box containing 4 and 5, another containing the rest (4 divisions in it though!!) so the 4,5 box can have 5 positions.
At any of the positions, the rest will have arrangements of 4! in which they are together.
AND it is also possible for it to be 4,5 or 5,4 (2 arrangements)
therefore arrangements: 5*4!*2 = 240
Therefore Prob. that they will be together: 6! /240
Subtract from 1 and see if it is correct.

for aii the MS says

3 digit odd 500+ = 4 ways
3 digit odd 600+ = 3 × 2
= 6 ways
4 digit odd 1000+ = 4 ways
4 digit odd 3000+ = 4 ways
4 digit odd 5000+ = 4 ways
4 digit odd 6000+ = 6 ways
OR 4 digit odd, last digit in 3 ways,
2
nd
to last in 3 ways, 2
nd
in 2 ways
first in 1 way = 18
Total = 28 ways
can u explain this?

 

[Jublie]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w10_qp_63.pdf
Quesion no. 6 part 3 plz how did they do this narking scheme main tricky answer hai plz clarify me soon tommorrow is my papers.

 

[usmiunique]

When ever you get a number that isnt a whole number you round it off to the next number. Thats what i was told and that is what i follow and so far havent come across any problems.

For your first example of 10.5 you use the 11th vaule and 5.25 you use the 6th vaule and for the next one the 16th value

thnx.. but in the markscheme, the median is given as 19(11th value). the LQ is given as 10(mean of the 5th and 6th values) and the UQ is given as 24(the mean of the 16th and 17th value)...
WHY??

 

[Khan_971]

Can U help me with this question...It's about permutation= =2011 OCT statistic 9709/62-------Q2
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf

2 Possibilities of outcome, Head or tail.
since twelve coins, arrangements = 2*2*2*2*2*2*2*2*2*2 = 2^12

part 2 is binomial expansion. Let X be no. of heads.
and since it is unbiased Prob. of getting heads (p) or tails (q)

20C7 * (.5)^7 * (.5)^5

 

[Wanzi21]

Part a i. Well duh there's no repeating. so just 4! = 24 arrangements.
Part a ii. Well U have to work on this. an odd number will have 1,3,5 in the end. 5 or 6 in the 1st digit and any of the 4 as 2nd digit. use this to write down possible combinations.

b. NOT next to each other has many possibilities, but to make it easy to can 1 - (P that they are together). Total no. of Permutations are: 6!
If u make a box diagram to solve it, meaning boxes containing this numbers. it will be easier. try making a box containing 4 and 5, another containing the rest (4 divisions in it though!!) so the 4,5 box can have 5 positions.
At any of the positions, the rest will have arrangements of 4! in which they are together.
AND it is also possible for it to be 4,5 or 5,4 (2 arrangements)
therefore arrangements: 5*4!*2 = 240
Therefore Prob. that they will be together: 6! /240
Subtract from 1 and see if it is correct.

yes thank u actually how to do the 1st part.. just the next 2 parts.. thanks again

 

[smartangel]

Correct!
look the the question. they asked females or anyone who watches or both! (both already comes in when you add P(female)+ P(anyone who watches since it means both genders). U can add these up OR
since the only people left out are people who DONT watch, their probability = 5/30. Subtract from 1 to get the answer.

YAYYY I GOT IT thankss !

 

[smartangel]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s11_qp_61.pdf
Q5 b..no idea what to do in this one!!

 

[parthrocks]

hey can anyone plzz help me with this
:"

 

[Khan_971]

yes thank u actually how to do the 1st part.. just the next 2 parts.. thanks again

first part is direct 4!. 2nd part u gotta write down all possible combinations. I already showed the third part

 

[sunnyexo]

2 Possibilities of outcome, Head or tail.
since twelve coins, arrangements = 2*2*2*2*2*2*2*2*2*2 = 2^12

part 2 is binomial expansion. Let X be no. of heads.
and since it is unbiased Prob. of getting heads (p) or tails (q)

20C7 * (.5)^7 * (.5)^5

Why 20C7??? Still don't get it.....The marking scheme uses 12!/7!5!......