How was your statistics paper 1.. (9709/62)

[Farhanchowd25]

Hey guys how was your xam..?? Lets nt discuss in details bt atleast match our answers.. whats say..!!?

 

[Yusuf Iqbal]

I find question 1 and 2 tough....the rest were kinda ok for me

 

[Natitah18]

I sat for paper 63. It was easier than June 2014 p63

 

[Natitah18]

Really? The questions for P63 were like the usual ones. The only difficult part is the Permutation question which carried 6 marks. Quite big for one question

 

[Kaiser Mahmud]

It was awful..

 

[Hafiz Ahmed pak]

Let's discuss about the paper here!

http://xat.com/DiscussionStatistics62

 

[Farhanchowd25]

http://xat.com/970962PaperDiscussion
Let's discuss about the paper here!

 

[Rehman Mallik]

I might get a D on this paper ..i left out number 6 completely..

 

[B!(+)(+)dy $k(_)!!]

1st - 48C43
2nd - 6! x 5!
- 6! x 7P4
3rd - 1/324
- 0.0212
4th - Tree diagram
- E(X) = 2 attempts
5th - P(X<8) was easy to find
- q was 7.74
- P(X<4 mean) = 1 - P(X<2)
6th - cumulative frequency graph
- h = 15.6 cm or 21.5 cm (not sure)
- SD = 6.01
7th - binomial distribution (4<=X<=6)
- Normal distribution P(X<114.5) dont remember
- np > 5 and nq >5

 

[Rehman Mallik]

for number 3... how did you get 1/324?... :/.... are you sure about answer... cause my answer is totally different.

 

[Dawarkazi]

for number 3... how did you get 1/324?... :/.... are you sure about answer... cause my answer is totally different.

He is right! It was 1/324

 

[Rehman Mallik]

He is right! It was 1/324

how did you get that?

 

[geni0usheeni]

for num 3 isnt (1,1,2,2) an option also?

 

[Feroz Tahir]

Just in case, we made correct tree diagram with little error in it and then draw Prob.Distr with Probabilities 1/3, 1/6, 1/2, and then calculated E(X) by using the values which gives 11/7 how many marks can I get? Out of 4 in treediagram I know I'll score 2-3. But what about in next part?

 

[geni0usheeni]

No 1+1+2+2 is 6 not 5

oh yhhh... nw i realise ths... Damn!

 

[Rehman Mallik]

Sum= 5 so either (1,1,1,2) or all other 4 ways and total was 6^4 so 4/1296 = 1/324

oh part 1 of question 3 was actually easy i see.. but i made a silly mistake :/ .. but how could you get 0.0212 in the second part?? my answer was 0.0121 :/ ??

 

[Rehman Mallik]

We use the anawer to be p and n=7, the want exactly 1 or 2 so 7c2(p)^2(q)^5 + 7c3(p)^3(q)^6

oooh damn. -.- i messed up with the calculation. nd Could you do question 4? I couldnt do it at all.. what did we have to do there? :/

 

[Rehman Mallik]

I solved it for you but how do i insert a pic on here? Im on my mobile tho

thanks a lot for that! i wish i could see the workout!..i am so confused about this math :/

 

[B!(+)(+)dy $k(_)!!]

for number 3... how did you get 1/324?... :/.... are you sure about answer... cause my answer is totally different.

Yuppp... I'm sure...!!

 

[B!(+)(+)dy $k(_)!!]

oooh damn. -.- i messed up with the calculation. nd Could you do question 4? I couldnt do it at all.. what did we have to do there? :/

You mean the tree diagram right?
---------------------------------------------------
x 1 2 3
P(X=x) 1/3 1/3 1/3
---------------------------------------------------
Therefore, E(X) = 1/3 + 2/3 + 3/3 = 2 attempts