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Statistics S1 post your doubts here

Haya Ahmed

Haya Ahmed

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_Ahmad said:
Sorry for my bad handwriting.
For the second part of this question you can also use normal distribution to find probability as you have got the (sd in the first part)
Click to expand...
Can I know for the second part why is the probability of success 0.15 not 0.85 ?
 
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_Ahmad

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Haya Ahmed said:
Can I know for the second part why is the probability of success 0.15 not 0.85 ?
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since the probability we want is of P(X>77) ie 0.15
because from the normal distribution curve you can see that the area is same for both (X greater than 77) and (X less than 75) and they are at same distance from mean so the probability is same.
Or as i told, since you have got your sd you can find the probability of success ie X greater than 77 using normal distribution.
 
Haya Ahmed

Haya Ahmed

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_Ahmad said:
since the probability we want is of P(X>77) ie 0.15
because from the normal distribution curve you can see that the area is same for both (X greater than 77) and (X less than 75) and they are at same distance from mean so the probability is same.
Or as i told, since you have got your sd you can find the probability of success ie X greater than 77 using normal distribution.
Click to expand...
there is no other way to do this excluding the graph thingy method ? .. btw I got it but Doing the method where we have to get the values from the probability distribution table is that fine ?!
 
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Haya Ahmed said:
there is no other way to do this excluding the graph thingy method ? .. btw I got it but Doing the method where we have to get the values from the probability distribution table is that fine ?!
Click to expand...

yes, use the same method as in part i the difference only is finding the P(x greater than 77) by using the sd you found in part i and you will end up with the same thing finding values from table
 
axetreme.O

axetreme.O

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Talha Irfan said:
The precise way of justifying a normal approximation is to establish the binomial in the question and show that np>5 and nq>5 where n is no. of the random variable (e.g. days) p is probability of success and q is probability of failure
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(y)
Thanks
 
Talha Irfan

Talha Irfan

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B

Browny

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IGCSE13 said:
You are not required to find the mean and standard deviation in this question but if you are asking in general then the mean can be found by creating a CF table from the graph since you have the upper bounds on the x axis and the respective CF on the y axis then you can work out the frequencies ,class boundaries and midpoints
I have no idea how to calculate the sd sorry
Click to expand...

That's ok.
Thanks for your help.:)
 
Haya Ahmed

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My doubt is just a confusion in the permutation combination how do we calculate for items not together .. what is the difference between using the way that
we make them together and subtract them from the arrangement of all (method 1).. or
multiplying by for eg we have 9 places and 4 items should not be together so 9P4 ?! (method 2)

.. why sometimes I get this wrong !!

for example here we use (method 2) in Q6 (iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_62.pdf


and here we use (method 1) in Q7 (iii)
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s13_qp_63.pdf


So can someone "Please" Explain what is the difference between this 2 methods and when to use each method .. thanks in Advance !! :)
 
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_Ahmad

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Haya Ahmed said:
My doubt is just a confusion in the permutation combination how do we calculate for items not together .. what is the difference between using the way that
we make them together and subtract them from the arrangement of all (method 1).. or
multiplying by for eg we have 9 places and 4 items should not be together so 9P4 ?! (method 2)

.. why sometimes I get this wrong !!

for example here we use (method 2) in Q6 (iii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_62.pdf


and here we use (method 1) in Q7 (iii)
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s13_qp_63.pdf


So can someone "Please" Explain what is the difference between this 2 methods and when to use each method .. thanks in Advance !! :)
Click to expand...

In the both questions they use the same method (method 2).

ok here is an example

Lets say there are 6 different consonants and 4 different vowels in a letter

when they ask all 4 vowels should not be next to each other then use method 1
because three vowels can be next to each other

But when they ask no two vowels next to each other than use method 2

Hope you got it.
 
Talha Irfan

Talha Irfan

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_Ahmad said:
Can anyone please explain Q6 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w10_qp_61.pdf
and
Q5 from http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w05_qp_6.pdf
Click to expand...
First Query :

The scene is going on such that there are 4 holes shown and they're telling you different combination of identical and non-identical pegs to calculate number of ways they can be arranged.

i) 6 non identical pegs : four of them to be arranged within those holes = 6C4 (choose 4) * 4! (arrange four of them) = 360

ii) 2 Blue 1 Orange 1 Yellow so no choice has to be made as there are four pegs and four holes so BBYO ; arrange distinctly = 4!/2! = 12

iii) Now a pair of pegs are given of each colour, this part requires all distinct. So total 6 different coloured pegs that can be chosen = 6C4 * 4! = 360

iv) Three different means, for e.g. BBGO, OOGR ; which implies that there must be a pair of a single colour present , Let me call for pairs so from 6 pairs I need one
6C1 * (5C2 (5 different colours remaining and they both needs to be singly selected in order to achieve the required given condition and two are left as other two has already been selected of the same colour)) * 4!/2! (arrangement) = 720

v) Let us recount the possibilites

All different : DONE = 360
Three different : DONE = 720
Two different : Possible? Yes. BBYY so again from 6 pairs two pairs selection = 6C2 * 4!/(2!*2!) = 90
All same : Possible? eg YYYY : NO

Add all of them up : 1,170 ways
 
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_Ahmad

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Haya Ahmed said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_62.pdf

Q7 (iii)
Please someone explain the question and answer :(
Click to expand...

Find P(X less than 34.6 ) for both symphonies separately
For Beethoven’s Sixth Symphony you can use normal distribution to find the probability
For Beethoven’s Fifth Symphony they have given P(X greater than 34.6) =0.05 so you can find P(X less than 34.6 ) just by 1-the P(X greater than 34.6)
=1-0.05=0.95
at last multiply both the probabilities (since we are playing both the symphonies and we want both of the time less than 34.6)
 
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