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Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691]

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Done almost every past paper and am pretty confident with everything except adding variances (even though its so basic!)
In Q4 of the June 2014 paper you had to work out A=B+4C-3D. I did this correctly by doing:
1^2Var(B)+4^2Var(C)+3^2Varc(D) So that was all fine.

However in question 7b of the June 2012 paper you have 6 Male employees and 4 Female employees. So i thought the Variance would be 6^2Var(M)+4^2Var(F) but its 6Var(M)+4Var(F). How comes the 6 and 4 aren't squared in this question?
Original post by kidder007
Done almost every past paper and am pretty confident with everything except adding variances (even though its so basic!)
In Q4 of the June 2014 paper you had to work out A=B+4C-3D. I did this correctly by doing:
1^2Var(B)+4^2Var(C)+3^2Varc(D) So that was all fine.

However in question 7b of the June 2012 paper you have 6 Male employees and 4 Female employees. So i thought the Variance would be 6^2Var(M)+4^2Var(F) but its 6Var(M)+4Var(F). How comes the 6 and 4 aren't squared in this question?


Because the 6 male employees are independent to each other when you calculate the variance of independent items like(X1+X2+......X6) u need to just mutiply the number of items with variance to give out the new one
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Original post by kidder007
Done almost every past paper and am pretty confident with everything except adding variances (even though its so basic!)
In Q4 of the June 2014 paper you had to work out A=B+4C-3D. I did this correctly by doing:
1^2Var(B)+4^2Var(C)+3^2Varc(D) So that was all fine.

However in question 7b of the June 2012 paper you have 6 Male employees and 4 Female employees. So i thought the Variance would be 6^2Var(M)+4^2Var(F) but its 6Var(M)+4Var(F). How comes the 6 and 4 aren't squared in this question?


If A= C1 + C2 + C3 + C4
(A sample of 4), then Var(A) = 4*Var(C)

If A= 4C
(4 * one sample), then Var(A) = 4^2*Var(C)
Original post by shloke123
These high grade boundaries coupled with either one dodgy question or a couple of silly errors (like arithmetic errors) is making that A* very annoying :angry:


True, it is a bit like c1 - not very hard but a task of not dropping stupid marks! 69/75 was an A* last year 64 /75 was A so very little for maneovre
Original post by bobo19966
Because the 6 male employees are independent to each other when you calculate the variance of independent items like(X1+X2+......X6) u need to just mutiply the number of items with variance to give out the new one


Thank you to both you and Starter, i knew it shouldn't be to difficult to get my head around.
For SRCC when do I know whether it's a two tailed test or not?
Original post by nayilgervinho
For SRCC when do I know whether it's a two tailed test or not?


Are you showing the existence of some correlation? Then its two tail
Are you showing a positive or a negative correlation? Then its one tail
Original post by Navo D.
Are you showing the existence of some correlation? Then its two tail
Are you showing a positive or a negative correlation? Then its one tail


:smile:
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For Q 6b 2009, why does the stock comment about the "central limit theorem enabling the assumption..." not apply?
5c June 2008 - how do we use the random number tables? Literally no idea
Original post by tazza ma razza
True, it is a bit like c1 - not very hard but a task of not dropping stupid marks! 69/75 was an A* last year 64 /75 was A so very little for maneovre


I know man, praying for a decent paper tomorrow where i dont cock up easy marks haha
Original post by STATER
For Q 6b 2009, why does the stock comment about the "central limit theorem enabling the assumption..." not apply?


CLT doesn't count as an assumption. As you said it is a theorem that enables an assumption but in those type of questions they aren't looking for CLT. Most of the time anyway. I think there was one question in the old syllabus that had a marking point for that. If I were you I would just go ham and write every assumption I can think of.
Original post by STATER
For Q 6b 2009, why does the stock comment about the "central limit theorem enabling the assumption..." not apply?


because the central limit theorem is not an assumption, its a theorem, but there are assumptions that have to be there for it to be applied, like that the two samples are independent, and the variance of population = variance of sample for large n
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Original post by mmms95
because the central limit theorem is not an assumption, its a theorem, but there are assumptions that have to be there for it to be applied, like that the two samples are independent, and the variance of population = variance of sample for large n


I did not phrase it too well. In 95% of these questions, you'd say you assume that the sample means are normally distributed given that the sample is large enough (by the central limit theorem). So why is that not acceptable in this case?
Original post by STATER
I did not phrase it too well. In 95% of these questions, you'd say you assume that the sample means are normally distributed given that the sample is large enough (by the central limit theorem). So why is that not acceptable in this case?


I think the assumptions is to make you to apply the central limit theorem to calculate the test statistic out . You are using the central limit theorem to calculate the test statistic so that the question is asking what assumption to make you use CLT
Original post by STATER
I did not phrase it too well. In 95% of these questions, you'd say you assume that the sample means are normally distributed given that the sample is large enough (by the central limit theorem). So why is that not acceptable in this case?




EDIT: actually, i know why, you say the mean's are normally distributed when they ask for the importance/significance of CLT
(edited 8 years ago)
When do we have to change intervals to 0.5 below and 0.5 above? Continuous? Also only if the distribution does not cover all possible values?

Eg

Continuous: 0-3, 3-4.... Would not need to change,

Continuous: 0-3, 4-6... Would need to change,

Discrete: never change


Is this correct?


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Original post by mmms95
EDIT: actually, i know why, you say the mean's are normally distributed when they ask for the importance/significance of CLT


But it is still an assumption? To be able to carry out the hypothesis test you must assume a normal distribution.
(edited 8 years ago)
Original post by adorablegirl1202
5c June 2008 - how do we use the random number tables? Literally no idea


Number the samples of range for example (1-100) then use random number table start at a position (for example start at 8613) then the question is usually ask you look across or look down (in this example look down). Using the first 3 digits as the range is 1-100. Then ignore the number is greater than 100 when looking down then select the allocate numbers
June 10 Paper

Q3a) A woodwork teacher measures the width, w mm, of a board. The measured width, X mm,is normally distributed with mean w mm and standard deviation 0.5 mm. Find the probability that X is within 0.6 mm of w.

I don't understand why in the mark scheme they do: 2 x 0.8849 - 1 = 0.7698. Can someone explain please?

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