Mathematics: Post your doubts here!

[sea_princess]

can somone please explain question 5 :a (ii)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w09_qp_61.pdf

 

[omar hazem]

please how I know I have to use the continuity correction in the normal distribution problems??

 

[hm12]

please how I know I have to use the continuity correction in the normal distribution problems??

whenever they use the word USING SUITABLE APPROXIMATION in the question

 

[reina81]

in MJ10 p6 variant 2, Q 7 the last part (v)---- don't get it.

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_62.pdf
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_ms_62.pdf

 

[omar hazem]

whenever they use the word USING SUITABLE APPROXIMATION in the question

ok but there is a problem in nov 11 variant 2 or 3 i think they say random sample and in the mark scheme they used the continuity correction !!

 

[hm12]

ok but there is a problem in nov 11 variant 2 or 3 i think they say random sample and in the mark scheme they used the continuity correction !!

i cant check it right now busy with my chem exam tomorrow when i am free i will try and figure it out and explain

 

[DragonCub]

can somone please explain question 5 :a (ii)
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_61.pdf

Between 5000 and 6000, so the first digit must be 5.
Then, allowed repetition of number, the following three digits each can have any one number from 1 to 6. That is, they each have 6 options. The total number is then
6 × 6 × 6 = 216

 

[DragonCub]

in MJ10 p6 variant 2, Q 7 the last part (v)---- don't get it.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_62.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_ms_62.pdf

[number without pink next to green] = [total number] - [number with pink next to green]
total number, as calculated in (iii), is 9P3 = 504
number with pink next to green can be done this way:
- find the number of choices that has both the pink and green cards, which is just, with the pink and green removed, choosing 1 extra from 7.
1 × 1 × 7C1 = 7
- find the number of arrangements of the three chosen cards that have the pink next to green.
2! × 2! = 4
So overall with pink next to green there are 7 × 4 = 28 arrangements
504 - 28 = 476 arrangements without pink next to green.

 

[reina81]

[number without pink next to green] = [total number] - [number with pink next to green]
total number, as calculated in (iii), is 9P3 = 504
number with pink next to green can be done this way:
- find the number of choices that has both the pink and green cards, which is just, with the pink and green removed, choosing 1 extra from 7.
1 × 1 × 7C1 = 7
- find the number of arrangements of the three chosen cards that have the pink next to green.
2! × 2! = 4
So overall with pink next to green there are 7 × 4 = 28 arrangements
504 - 28 = 476 arrangements without pink next to green.

thanks

 

[omar hazem]

i cant check it right now busy with my chem exam tomorrow when i am free i will try and figure it out and explain

ok thats fine, and good luck for tomorrow exam

 

[GoodRobot]

please tell me some rules regarding reading reverse table in normal distribution.i get so confused..

 

[omar hazem]

please any help when we use the continuity correction in solving the normal distribution problems, for example in june 11 paper 62 there is a question using normal approximation and in the mark scheme they used the continuity correction !!!!

 

[DragonCub]

please any help when we use the continuity correction in solving the normal distribution problems, for example in june 11 paper 62 there is a question using normal approximation and in the mark scheme they used the continuity correction !!!!

Continuity correction is used when we approximate a discrete variable using a continuous variable. Like using X~N to approximate X~B, or X~N approximate X~Po

 

[omar hazem]

i really don't understand it well, i mean the question terms in the past papers
when i know from these terms that i have to solve using continuity correction??

 

[sea_princess]

can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)

- Each of the digits 1, 1,, 2,3, 3, 4, 6,is written on a separarte card. The seven cards are then laid out in a row to form a 7-digit number. How many of these 7-digit numbers is divisible by 4?( ans:300)

-The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
a) how many possible selections are there of four letters?( ans :12)
b)how many arrangements are there of four letters? ( ans:115)

thanx in advance

 

[DragonCub]

can someone please explain how to solve these questions:
-eight cards are selected with replacement from a standard pack of 52 playing cards with 12 picture carsd, 20 odd cards and 20 even cards.
a)how many different sequences of eight cards are possible?( ans:5.346 X 10 ^13
b)how many of the sequences in part (a) will contain three picture cards, three odd- numbered cards and two even numbered cards?(ans: 3.097 X10 ^12)

Are you sure that for (a) the answer is 5.346E^13 ?
Choosing 8 from 52, sequence included, so that is permutation, 52P8, and I got 3.034E^13.

For (b) I also got a different result. The procedure is like choosing 3 from 12, 3 from 20 and 2 from 20, then put them in sequence. So it's first combination, 12C3 × 20C3 × 20C2, then permutation, 8P8 which is 8!.
I got 12C3 × 20C3 × 20C2 × 8! = 1.92E^12

 

[sea_princess]

Are you sure that for (a) the answer is 5.346E^13 ?
Choosing 8 from 52, sequence included, so that is permutation, 52P8, and I got 3.034E^13.

For (b) I also got a different result. The procedure is like choosing 3 from 12, 3 from 20 and 2 from 20, then put them in sequence. So it's first combination, 12C3 × 20C3 × 20C2, then permutation, 8P8 which is 8!.
I got 12C3 × 20C3 × 20C2 × 8! = 1.92E^12

yeah the answers are like that , it's the cambridge statistics 1 book by Steve Dobbs adn JAne Miller
in (a) the steps were written as ( 52 C 1)^8 , but when I used a similar method for b , it didn't work

 

[ppaayas]

Hello!
Can someone please explain how the following question is solved. And care to explain any properties involved. Thank you.
1 The random variable X is normally distributed and is such that the mean µ is three times the standard
deviation σ. It is given that P(X < 25) = 0.648.
(i) Find the values of µ and σ.
(ii) Find the probability that, from 6 random values of X, exactly 4 are greater than 25.

 

[ppaayas]

Help!

 

[mukki]

-The letters of the word POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
a) how many possible selections are there of four letters?( ans :12)
b)how many arrangements are there of four letters? ( ans:115)

First write down the number of times an alphabet is being repeated
P=1 O=1 E=2 S=5 (these are 4 options and not 9)
a) select - all different alphabets first i.e chose 1 from each 4 options above(p,o,e,s) =4c4
- all same alphabets i.e (ssss)= 1c1 (as it is S which is 4 times u have to select it only)
- 2 same alphabets , 2 same alphabets (ss,ee in this case both E and S occur two time so select from E and S) =2c2
- 2 same alphabets , 2 different alphabets (ss/ee and 2 different , in this case u either select E or S and then 3 alphabets remain from which u have to select)= 2c1 *3c2
- 3 same alphabets , 1 different (sss and 1 from remaining 3 , since S occur three times u have to select it and 1 from remaining 3) = 1c1 * 3c1
sum them all up u get 12
b) its very simple once uve done the combinations just apply permutation to each part above
- all different= 4c4 *4!
- all same = 1c1 * 4!/4!
-2same,2same =2c2*4!/(2!*2*)
-2 same 2 different =2c1*3c2* 4!/2!
-3 same 1 different =1c1 *3c1* 4!/3!
add em ul get 115...............
hope ive helped