Stat v. 62

[littlecloud11]

62 was very easy, so I'd say the threshold will be 40 or over.

 

[littlecloud11]

hmm then i will loose three marks or may b even four
if i used correct formulae but my IQ was incorrect i can get ecf right?
any one remember value of c cause if that is correct then i can get near 45

6.77

 

[A star]

6.77

you sure?

 

[A star]

62 was very easy, so I'd say the threshold will be 40 or over.

was easy but many people did mistakes in q4 and lost many marks :/

 

[Ryan123]

we can discuss all the answers? i remember many of them would like to know if they are correct

 

[A star]

we can discuss all the answers? i remember many of them would like to know if they are correct

what was ur value of c?

 

[Ryan123]

hm

what was ur value of c?

hmm around 6.78 or something like that it was c=1.881*r(and i cant remmeber the s.d value clearly but it was around 3.61 i think)

 

[littlecloud11]

you sure?

pretty much.
I did the back calculations.
the z value for c was 1.881, which gave a probability of .97
1.881- (1-1.881) = .97 - 1 + .97
total probability = .94 or 94%

 

[A star]

pretty much.
I did the back calculations.
the z value for c was 1.881, which gave a probability of .97
1.881- (1-1.881) = .97 - 1 + .97
total probability = .94 or 94%

Oops i took z value as -1.881 thats y my answer is wrong :/
well that reduces my total to 43

 

[Ryan123]

you guys the remember the value of n and k? is it 144 and 6? and abt the permutation 2nd and third is it 4!*6!*2!*3!(for no.ii) and for no.iii is it (12!-9!*4!)?

 

[littlecloud11]

you guys the remember the value of n and k? is it 144 and 6? and abt the permutation 2nd and third is it 4!*6!*2!*3!(for no.ii) and for no.iii is it (12!-9!*4!)?

yes, yes and Yes!

 

[A star]

you guys the remember the value of n and k? is it 144 and 6? and abt the permutation 2nd and third is it 4!*6!*2!*3!(for no.ii) and for no.iii is it (12!-9!*4!)?

yes yes and idr the third

 

[Ryan123]

yes, yes and Yes!

for the last question of X there was 3 possibilities 0 1 2 right?

 

[A star]

for the last question of X there was 3 possibilities 0 1 2 right?

yup

 

[littlecloud11]

for the last question of X there was 3 possibilities 0 1 2 right?

Yup.

 

[A star]

Yup.

can you write down the process for calculation of the value of c
still cannot figure out y i lost marks in that part

 

[Ryan123]

can you write down the process for calculation of the value of c
still cannot figure out y i lost marks in that part

P(445-c<X<445+c)=0.94(i think mean was 445 and s.d was 3.6 or 3.61 i think)
P(445-c-445/3.61<Z<445+c-445/r)=.94
P(-c/3.61<Z<c/3.61)=.94
2P(Z<C/3.61)-1=.94
P(Z<C/3.61)=(1+.94)/2
P(Z<C/3.61)=.97
c/3.61=1.881
so c=6.79

 

[A star]

if i only did till here
P(445-c<X<445+c)=0.94(i think mean was 445 and s.d was 3.6 or 3.61 i think)
P(445-c-445/3.61<Z<445+c-445/r)=.94
P(-c/3.61<Z<c/3.61)=.94
2P(Z<C/3.61)-1=.94
do you think i wil get half the mark

 

[littlecloud11]

can you write down the process for calculation of the value of c
still cannot figure out y i lost marks in that part

P <(445+c)
P >(445-c)
(445+c -445/3.6) - (445-c-445/3.6)
= c/3.6 - (-c/3.6)
Let the probability for c/3.6 be 'x', then the probability for -c/3.6 is '-x'
so x- (-x) = .94
x- 1 + x =.94
x = .97
z value for .97 is 1.881
So, c/3.6 =1.881
c= 6.77

 

[Ryan123]

if i only did till here
P(445-c<X<445+c)=0.94(i think mean was 445 and s.d was 3.6 or 3.61 i think)
P(445-c-445/3.61<Z<445+c-445/r)=.94
P(-c/3.61<Z<c/3.61)=.94
2P(Z<C/3.61)-1=.94
do you think i wil get half the mark

i think they will cut 2 marks if you did till that but not sure though. I am not very familiar with their marking.