Statistics P6 Help needed

[fatima 007]

anyone giving s1 in the morning session?

 

[Jaf]

According to my teacher, the mark scheme for the question I asked is wrong. Splendid!

 

[fatima 007]

mark schemes are never wrong! when teachers cant find a solution to the question and they cant thinking of a way to get the right answer, they usually say the following ^

 

[Knight]

(b)(ii).........plz
thanx.......

 

[Nouman Shafique]

(b)(ii).........plz
thanx....... View attachment 10633

5! * 4! = 2880

 

[Jaf]

mark schemes are never wrong!

Did you make them? Have you checked ALL of them? Are you an examiner?
If the answer to anyone of the question is no, then that is a pretty bold and perhaps asinine remark to make.

Over the years there have been numerous mistakes in the mark schemes. Some were misprints, but many were (and are) actual mistakes. There have been mistakes in actual questions! There have been huge blunders in the marking (ever heard of successful rechecks?). We need to stop thinking the people on the CIE end are an infallible team of experts. They make mistakes too.

Another one that I'm pretty sure of in Statistics is in the last question in this paper; the last part. Even though the answer remains the same (with the correct answer) but the working in the mark scheme is missing something.
The examples are countless. I've seen at least 2 deleted questions because they were wrong and at least one MCQ for which they have credited 2 answers in A levels alone.

So yes, I could be wrong about this question and so can the teachers I had this checked from. But then, so can the examiners.

Good luck for your exam tomorrow.

 

[aloha]

how do we know whether 2 events are mutually exclusive or independent???????????????

 

[Knight]

5! * 4! = 2880

the answer is right......but can you plz explain it.....
thanx

 

[Nouman Shafique]

how do we know whether 2 events are mutually exclusive or independent???????????????

for indepented...P(A intersection B)=P(A) * P(B)
For Mutually Exclusive evetns P(A U B)= P(A) + P(B)

 

[Nouman Shafique]

the answer is right......but can you plz explain it.....
thanx

quite hard to xplain here.. well since 5 boys are acting like a single pack and now we got arrangements of this pack and the 3 girls, the pack can come anywhere so 4!(3 girls + Pack), now in the pack there can be different arrangements too, since 5 boys in the pack so 5!, so overall 5! *4!

 

[Knight]

how do we know whether 2 events are mutually exclusive or independent???????????????

If A and B are two events
Then they are:
Independent if P(A and B)=P(A) * P(B)
Mutually Exclusive if P(A or B)=P(A) + P(B) and P(A and B)=0

Hope you like it...........

 

[Knight]

Part three(iii)...........plz

 

[hamza_sheikh182]

hey can anyone help me with this doubt

question 3 part 3..... how do we solve this?​

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hey can anyone help me with this doubt

question 3 part 3..... how do we solve this?​

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  9709_s11_qp_61.pdf
  
  

  File size:​

  94.3 KB​

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[Jaf]

Part three(iii)...........plzView attachment 10647

A number is even when its last digit is even.
So we fix the even numbers at the end of the numbers, one by one.
_ _ _ _ _ _ 2 - here the number of arrangements is 6!

_ _ _ _ _ _ 6 - here the number of arrangements is 6!/2!

_ _ _ _ _ _ 8 - here the number of arrangements is 6!/2!.

So total, 6! + (6!/2!) x 2 = 1440 numbers

 

[Knight]

hey can anyone help me with this doubt ​

​

question 3 part 3..... how do we solve this?​

Attached Files:

• ​
  9709_s11_qp_61.pdf
  ​
  File size:​
  94.3 KB​
  
  
  Views:​
  8​

hey can anyone help me with this doubt ​

​

question 3 part 3..... how do we solve this?​

Attached Files:

​

• ​
  9709_s11_qp_61.pdf
  ​
  File size:​
  94.3 KB​
  
  
  Views:​
  8​

a=1 because when a=1 P(-a<_X<_2a)=P(-1<_X<_2)
P(-1<_X<_2)=(9/70) + (1/10) + (9/70) + (9/70)
=17/35

 

[Knight]

help plz in third part.........hurry........

 

[applepie1996]

help plz in third part.........hurry........ View attachment 10671

(i) it will be 8C2 as there askin us to punch 2 holes in the 8 holes of each column
(ii)max 4 holes can be punched so 8C1+8C2+8C3+8C4=162
(iii)there are 4 columns and each column as worked out in part (ii) have 162 ways of getting the holes punched so it will be 162x162x162x162=162^4

 

[Knight]

(i) it will be 8C2 as there askin us to punch 2 holes in the 8 holes of each column
(ii)max 4 holes can be punched so 8C1+8C2+8C3+8C4=162
(iii)there are 4 columns and each column as worked out in part (ii) have 162 ways of getting the holes punched so it will be 162x162x162x162=162^4

thanxxxx.......

 

[applepie1996]

thanxxxx.......

anytym

 

[sonu1996]

help needed urgently. 9709_w11_qp_62 Question 7 (ii), (iii) & (iv) Kindly help.