Mathematics: Post your doubts here!

[hassam]

its w11 qp 62

 

[Diamona151294]

Question 7 (iii)

70 days at random, how many would have temperature more than 3 times the mean.

So first we calculate the probability, that a single day has, that its minimum temperature is more than 3 times the mean.


So, p(X > 3u )

p(Z > (3u - u)/2u )

p(Z > 2u/2u)

p( Z > 1 )
1 - P(Z<1)

after looking from the table:
1 - 0.8413
0.1587

So the probability of a single day having a minimum temperature more than 3 times the mean:
p = 0.1587

n, number of days = 70.

Expected value, = np, = 70(0.1587) = 11.109,
Therefore 11 days.

thanku!! got it.... ugh i just dont know how im suposed to conc on every single word in the paper.....:/

 

[smzimran]

its w11 qp 62

Im not getting this method but i managed to do it with the other method mentioned in the mark scheme!
That 6! * (7! / 3!)

 

[alphabravocharlie]

Im not getting this method but i managed to do it with the other method mentioned in the mark scheme!
That 6! * (7! / 3!)

Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!

 

[Diamona151294]

um in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?

 

[smzimran]

Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!

I dont think so!

 

[alphabravocharlie]

I dont think so!

Why?

 

[smzimran]

its w11 qp 62

Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!

I got it hassam:
They are first taking 8! / 3! to find arrangements in which two Gs are grouped together and the third is placed randomly
Then they are subtracting 2 occasions when this third G may come besides the group of Gs
First possibility :
G (GG) ............
(GG) G ............
That is why the 2 is present,
So it is (8! / 3!) - 2 * (7! / 3!) [grouping the third G with the rest two in subtraction]

 

[smzimran]

Why?

The answer is coming negative!
* If i did the calculation right!

 

[alphabravocharlie]

The answer is coming negative!
* If i did the calculation right!

i get 5040

 

[hassam]

well this is so damn cnfusing

 

[smzimran]

i get 5040

Well then my calculation must have been wrong but i cant seem to get the calculations right!

 

[tasnim007]

its just you

LOL

good luck for exam

 

[hassam]

"Then they are subtracting 2 occasions when this third G may come besides the group of Gs
First possibility :
G (GG) ............
(GG) G ............"
now this bit i cant understand...@smzimran wat i was thinking that.....i got 840 arrangements in which the 3Gs were 2gether and 6720 arrangements in which 2Gs and 3Gs were 2 gether......so just thought of subtracting them

 

[mukki]

well this is so damn cnfusing

its very easy bro just think of it like this
32 teams play the first match but 16 get knocked out so hence the teams who have played only A SINGLE MATCH IS 16. then these 16 teams who survived play a second match of which 8 got knocked out hence 8 teams have played exactly two match. the remaining 8 teams play a third match in which 4 of them get knocked out hence 4 teams have played exactly three matches. Then out of the remaining 4 teams 2 get knocked out in the 4th match hence 2 teams have played exactly 4 matches. For the 5th match there are only two teams remaining hence only two teams have played exactly 5 matches

 

[hassam]

oohhh got it......

 

[helloboy]

um in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?View attachment 10654View attachment 10655

i had the same question!! this one's bugging me since yesterday

 

[mukki]

i had the same question!! this one's bugging me since yesterday

here check this out

 

[zain123]

m.s allows 0.94 too I think

 

[usmiunique]

um in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?View attachment 10654View attachment 10655

The probability of a letter being >12g above the mean is 0.94+(0.06/2)=0.94+0.03=0.97
What u are taking is 0.94 which is the probablity of the letters being within 12g of the mean but what u have to take is the probabality of the letters being MORE than 12g above the mean!