A level Statistics doubt??Post your doubts here!

[panoramafolks]

DUde A star i can also access to the MS....but i need an explanation :/

can u post paper link and question number??

 

[Kangxidi]

can u post paper link and question number??

I gave a worked example for him one post back = o see if you can get the proper question so I can help him solve it

 

[panoramafolks]

I gave a worked example for him one post back = o see if you can get the proper question so I can help him solve it

ok

 

[abruzzi]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s12_qp_62.pdf
I need some help with question 6(v) of this paper. I understand how to use the combination method to find the probability.. could anyone please explain me the other method shown in the marking scheme?
Thanks

 

[Kangxidi]

It is very straightforward really, I in fact understand the probability one and can't understand the combination way.

He took 4 biscuits without replacement. Find that wrapped is EXACTLY 2.

Therefore, it can be in .. (W = wrap) (U = unwrap)

P(U) * P(W) * P(U) * P(W)

Or

P(W) * P(U) * P(W) * P(U)

Or

P(W) * P(W) * P(U) * P(U)

So on.

Counting the amount of ways this can be done, it is 4C2 ways.

Take any of the arrangement above ^, multiply it by 4C2 give you the answer.

 

[Kangxidi]

Things to note, remember that they are not independent and therefore the probability changes each time, be careful.

 

[sufyanalizaib]

3alaikumissalam.

Work out the P(X > mean + sd)

I can't remember which year that is, but I will just give you a quick example.

If the question is X~N(10,5^2) (10 mean and 5 sd) and you are looking for lengths MORE THAN 1 sd from the mean, that means you need to work out P(X > 10 +5)

That is P(X>15).

Whatever answer you get, multiply it by 1000.

Hope that helps.

thanx yar...but i need a little explanation here..question 6 part 2

 

[A star]

Things to note, remember that they are not independent and therefore the probability changes each time, be careful.

this makes binomial expansion invalid. yup took me 3 minutes to notice what i find odd in this question

It is very straightforward really, I in fact understand the probability one and can't understand the combination way.

He took 4 biscuits without replacement. Find that wrapped is EXACTLY 2.

Therefore, it can be in .. (W = wrap) (U = unwrap)

P(U) * P(W) * P(U) * P(W)

Or

P(W) * P(U) * P(W) * P(U)

Or

P(W) * P(W) * P(U) * P(U)

So on.

Counting the amount of ways this can be done, it is 4C2 ways.

Take any of the arrangement above ^, multiply it by 4C2 give you the answer.

naa i found latter one easy

 

[Bebeskii]

1st of all when converting binomial to normal :

1. np and npq should be greater than 5 to be suitable to normal dist.

2. X~N(np ,npq) will ne the distribution of X.

3. Use continuity correction as we have to be accurate :

if p(X>5) ... We will calculate p(X>5.5)..... if its p(x<5) we will calculate p(X<4.5)
if its p(X≥5)... it will be p(X≥4.5) .... if its p(X≤ 5 ).... it will be p(X≤5.5).....
4. then solve remember npq is variance so under root it to calculate s.d while substituting it in formula to find Z

thank you ♥

 

[A star]

thank you ♥

one thing wrongly said
1) np and nq should be greater than 5 to be suitable to normal dist.

 

[A star]

thanx yar...but i need a little explanation here..question 6 part 2

actually its a rule that within 1 sd of the average 2/3 of the values lie
within 3sd 99 percent of the values
within 2sd 95 % of the values
use this rule

 

[abruzzi]

It is very straightforward really, I in fact understand the probability one and can't understand the combination way.

He took 4 biscuits without replacement. Find that wrapped is EXACTLY 2.

Therefore, it can be in .. (W = wrap) (U = unwrap)

P(U) * P(W) * P(U) * P(W)

Or

P(W) * P(U) * P(W) * P(U)

Or

P(W) * P(W) * P(U) * P(U)

So on.

Counting the amount of ways this can be done, it is 4C2 ways.

Take any of the arrangement above ^, multiply it by 4C2 give you the answer.

Lol.. I understand the combination way better.

Thanks for you help. Just one more thing, is the no. of ways 4C2 because you are choosing 2 wrapped biscuits from 4?

 

[Kangxidi]

Lol.. I understand the combination way better.

Thanks for you help. Just one more thing, is the no. of ways 4C2 because you are choosing 2 wrapped biscuits from 4?

Yes Oh I'm surprised people find the (personally chinese) combination method easier wow glad to see the different approaches

 

[A star]

Yes Oh I'm surprised people find the (personally chinese) combination method easier wow glad to see the different approaches

lol

 

[Alice123]

find the no of ways 4boys and 5 girls can be arranged in a row if
a) a particular boy and a particular girl must be next to each other
b)the 1st n last persons are of the same sex
syed1995 littlecloud11 please help

 

[A star]

find the no of ways 4boys and 5 girls can be arranged in a row if
a) a particular boy and a particular girl must be next to each other
b)the 1st n last persons are of the same sex
syed1995 littlecloud11 please elp

a) 8! x 2!
b) 4 x5 x 7!

 

[Alice123]

a) 8! x 2!
b) 4 x5 x 7!

Explain

 

[Kangxidi]

*EDIT* star explained it first, we have different approaches don't look at mine haha I'm weak in permutations

a) Imagine you have 4 boys.

You will have B B B B

And you have 5 girls.

It says that you have to make them alternating (BGBGBGBG) and so on, because a boy must be next to a girl (who in turn has to be next to a boy and so on)

Let's "insert" the girls into the line of boys we have, we will notice that we can "insert" them in 5 different places.

| B | B | B | B |

| = place where we can "insert" the girls.

Since we have only 5 girls (which perfectly fits the slot) we can use 5P5 (or 5!) for the girls.

Since there are 4 different boys, we can use 4!

Therefore 5! * 4!

[There is a simpler explanation on grouping them and stuff, but I prefer explaining this way as it is much easier to be told over the net ]

b) We have two cases for this, the first case being that it's both a boy, and obviously second is that both a girl.

Let's look at first case:

B _ _ _ _ _ _ _ B and a remainder of 5 girls and 2 boys. Since it does not matter how they are arranged inbetween the boys, and that each person are different, we can use 7!.

Second case:

G _ _ _ _ _ _ _ G and a remainder of 3 girls and 4 boys. Same concept applies, 7!.

Therefore 7! + 7! or 7! * 2

Correct me if I'm wrong, again.

 

[A star]

Explain

so much anger
a) keep one boy and one girl as one unit you will have 8 blocks left which can change places hence 8!
and then boy and girl can also change place hence 2!
multiply 8! x 2!
b) take one side as noy and other as girl
only one boy can be at one side hence 4C1 and on side one girl 5C1
and then take factorial of children left makes
4 x5 x 7!

 

[felicia tan]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf
can anyone solve 5(i) ? I get 7.717 for the SD but marking scheme gave 6.38