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stats question...plz help

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hey..em sry not certain about ur ques :S :S..

do u know how to do oct/nov 2010 paper 62 quest 2 part i ??
 
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i meant that in some questions it is e.g
X>4
=Z>4.5-mean/s.d

why....will answer ur question 2morwow..Inshallah..
 
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at ks
when there is an equal sign with the >,< then u take the largest range eg
x<4
u take
z<4.5
but when there is no equal sign u take the least range eg
x>6
it will be
x>6.5
 
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ohhhhhhh
sorry if u wanted to know WHEN to use the normal approximation then it is wen
np>5
and
npq>5
 
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ahmed t said:
ohhhhhhh
sorry if u wanted to know WHEN to use the normal approximation then it is wen
np>5
and
npq>5

when they r BOTH >5...or 0.5...can u tell me plz..thx
 
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@ Samy:

to find Σ(x − 50):
We know Σ(x)=824. and the total number of students is 16. So Σ(x − 50)= Σ(x) - (16 x 50) = 824 - 800 = 24.

to find Σ(x − 50)^2:
we know the formula for variance, so substitute the values in it. So you get (Σ(x − 50)^2)/16 - (Σ(x − 50) / 16)^2 = (6.5)^2
using simple algebra, we get Σ(x − 50)^2 = (42.25 + (24/16)^2) x 16 = 712.

:)
 
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ks136 said:
ahmed t said:
ohhhhhhh
sorry if u wanted to know WHEN to use the normal approximation then it is wen
np>5
and
npq>5

when they r BOTH >5...or 0.5...can u tell me plz..thx
yes wen they are both >5
or if they ask u to use an appropriate approximation
 
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ks136 said:
http://www.xtremepapers.me/CIE/index.php?dir=International%20A%20And%20AS%20Level/9709%20-%20Mathematics/&file=9709_w09_qp_61.pdf

Q2...why had we taken boundaries???
are u sure ut talkin about q2 cause u dont take boundaries there
 
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well np is>5
and so is npq
so you would use the approximation method, unless clearly in the question it says that it is normally distributed then there is no need of using an approx
 
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no its because it clearly states that it is a normal distribution whereas q2 does not
 
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Amna said:
@ Samy:

to find Σ(x − 50):
We know Σ(x)=824. and the total number of students is 16. So Σ(x − 50)= Σ(x) - (16 x 50) = 824 - 800 = 24.

to find Σ(x − 50)^2:
we know the formula for variance, so substitute the values in it. So you get (Σ(x − 50)^2)/16 - (Σ(x − 50) / 16)^2 = (6.5)^2
using simple algebra, we get Σ(x − 50)^2 = (42.25 + (24/16)^2) x 16 = 712.

:)


hey thanku so much...but y did u have to multiply 16 with 50?? for the first one??
and can u tell the formulae of variance?? isnt it Σ(x − x(mean))^2)/n.??
 
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