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Statistics P6 : post your doubts here!

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A level: Maths, Chemistry, Physics
As level: Biology and General Paper(over)
i got a B in biology AS :( actually i was the only one from my whole school who got a B according to my teacher . best of luck with your other subjects :) thank you for your help :) May Allah reward you for this .
 
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i got a B in biology AS :( actually i was the only one from my whole school who got a B according to my teacher . best of luck with your other subjects :) thank you for your help :) May Allah reward you for this .

You're welcome... Good Luck too..:)
 
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hey guys. please can someone help me with coding data.. i have a problem with that..
9709_w10_qp_62 , question 2
 
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hey guys. please can someone help me with coding data.. i have a problem with that..
9709_w10_qp_62 , question 2
to find Σ(x-50) means that from all the numbers/variables you have collected,subtract 50 from each one.
Lets say you have the number 51,52,53 , Σx=51+52+53 and Σ(x-50)=1+2+3
so to find Σ(x-50) just find the sum of number and subtract (50*16) because you are subtracting 50 16 times as there are 16 numbers
can someone help me with oct/nov 9709_w10_qp_63, question 4

I dont understand what is Σ(x − 60) = 245 ???? can someone explain to me?
 
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Q1 i)
If the mean is 8.95,then Σ(x-a)/n + a = mean
So 8.95= -73.2/24 + a,then just solve for a :)
ii)
standard deviation = sqr root ( Σ(x-a)^2/n -(Σ(x-a)/n)^2)
THis can be found in the formula booklet

q7 ii) P of taking a red paper clip from Box B is either to put a white clip from A into B and then take a red one from B or to put a red clip from A and put it into B.
You have already found out the probability of the first option in part i.
So,P of selecting red from A and transferring it to B and selcting a red from B is (5/6)*(8/10)
Find this probability and since it there are two options add this to you answer from part i
iii)
This conditional proabability states p(clip a is red and clip B is red)/P(clip B is red)
P(clip a is red and b is red)-This was found in part ii-(5/6)*(8/10)
P(clip b is red) -this is the answer of part ii

iv)
When they mean number of times red is taken it means to say how many times you pick a red.Ie transferring a red to box b,and chosing a white clip from box B,means one time
So the random variable x can be 0-(WW) 1-(WR or RW) 2-(RR)
okay,for the first option WW-P of chosing white from A is (1/6) and P of chosing white from B will be(3/10)
for the second option (WR or RW) -do the same thing as above and add the two probabilities
same goes for the third
Hope it helped :)
 
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Since the survey will definately include 1 house of style C and 1 from B,there are two houses remaining to chose from
Eliminating Style C and B houses you have 7 different kinds of houses to chose from.
So 1 of style B gives 2C1
1 of style C gives 3C1
and for the remaninig your'e left with 7C2
Putting these together you need 1 of B and 1 of C and two of something else other than B or C
so ----> 3C1 * 2C1 * 7C2
Sorry im quie confused here, arent all the houses in style A the same???

Also, what are the Conditions for a situation to be modelled be a normal distribution?

Thanks
 
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Q1 i)
If the mean is 8.95,then Σ(x-a)/n + a = mean
So 8.95= -73.2/24 + a,then just solve for a :)
ii)
standard deviation = sqr root ( Σ(x-a)^2/n -(Σ(x-a)/n)^2)
THis can be found in the formula booklet

q7 ii) P of taking a red paper clip from Box B is either to put a white clip from A into B and then take a red one from B or to put a red clip from A and put it into B.
You have already found out the probability of the first option in part i.
So,P of selecting red from A and transferring it to B and selcting a red from B is (5/6)*(8/10)
Find this probability and since it there are two options add this to you answer from part i
iii)
This conditional proabability states p(clip a is red and clip B is red)/P(clip B is red)
P(clip a is red and b is red)-This was found in part ii-(5/6)*(8/10)
P(clip b is red) -this is the answer of part ii

iv)
When they mean number of times red is taken it means to say how many times you pick a red.Ie transferring a red to box b,and chosing a white clip from box B,means one time
So the random variable x can be 0-(WW) 1-(WR or RW) 2-(RR)
okay,for the first option WW-P of chosing white from A is (1/6) and P of chosing white from B will be(3/10)
for the second option (WR or RW) -do the same thing as above and add the two probabilities
same goes for the third
Hope it helped :)

Thank you loads :)
 
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Q6 part one plz!
 

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It'd be a great help if anyone can help me on explaining the mark scheme in ON2010 P63 question no. 6(iii). -- http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_63.pdf

I know someone has posted the 'explanation', but I still couldn't understand why the students only have 5 ways when arranged? Someone can explain with a further explanation--because I still don't get it :(

I know what you mean, the question is worded quite awkwardly so don't blame yourself for not getting it.
Its basically asking you to take the total number of way for the passengers to be seated, and since it uses the word and you then multiply those ways and divide them by the result in part (i).
Hence, there being 3 front row seats for Mrs. Brown --> 3P1.... For Mrs. Lin, she must sit behind a student, and for her to do so she cannot be in the front and since there are 10 seats available for this to be true --> 10P1.... now for the STUDENTS, Mrs. Lin must be behind 1 student, which means this student must be infront of her, which leads to only 1 possibility for that student to be seated. BUT we do not know which student this is, since there are 5 students and we need to choose 1, we do ------> 5P1 =5.

Hope this helps
 
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Oh my, now I understand where the 5 comes from!
Thanks a lot for your explanation! :D
 
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