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    :woo:

    And S3 wouldn't be S3 without it clashing with another module, and this year it clashes with C2. :woo: :party:

    Here are some resources, inspired by resources from the thread of the previous year.

    Madasmaths - awesome website for loads of things for all of your exams, with a few topics that come up in S3.

    PhysicsAndMathsTutor - link to loads of useful things.

    FMSP revision page - some videos to watch about the specification and a specific set of exam questions.

    If there is demand I can start an S4 thread as well..!

    And if you have any questions, fire away!


    2015 S3 PAPER FROM EDEXCEL'S WEBSITE:

    QP


    MS:



    Model Solutions courtesy of Zacken
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    Thanks for making the thread, please could you explain why on question 2a of this paper https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf the bias can be plus or minus a/2 as I thought the bias would only be positive? Thanks
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    (Original post by economicss)
    Thanks for making the thread, please could you explain why on question 2a of this paper https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf the bias can be plus or minus a/2 as I thought the bias would only be positive? Thanks
    I think it might just be the mark scheme being lenient.
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    (Original post by aymanzayedmannan)
    I think it might just be the mark scheme being lenient.
    I see, thank you! I don't suppose you could explain how to do question 2c please https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf I really can't get my head around it! Thanks
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    In. :-)
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    (Original post by economicss)
    Thanks for making the thread, please could you explain why on question 2a of this paper https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf the bias can be plus or minus a/2 as I thought the bias would only be positive? Thanks
    It's very simple! You know that X is an unbiased estimator of the parameter \alpha and the data given to you has been obtained from {X}, so the data is also biased. You must now find the mean, \bar{X}, from the data given by using the formula \bar{X}= \frac{1}{n}\sum x and since you want to find the unbiased estimator value of \alpha, you use the function of Y. Can you take it from here?
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    (Original post by aymanzayedmannan)
    It's very simple! You know that X is an unbiased estimator of the parameter \alpha and the data given to you has been obtained from {X}, so the data is also biased. You must now find the mean, \bar{X}, from the data given by using the formula \bar{X}= \frac{1}{n}\sum x and since you want to find the unbiased estimator value of \alpha, you use the function of Y. Can you take it from here?
    Many thanks! I don't suppose you could explain question 2c on that paper aswell please? https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf Thank you
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    Could anyone explain please how you would know what significance level to use in question 6c, is it concluded as not significant because it's below the critical value of 11.345 obtained in part a? https://drive.google.com/folderview?...&usp=drive_web Thanks
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    (Original post by economicss)
    Many thanks! I don't suppose you could explain question 2c on that paper aswell please? https://869d950bf149eefd5cd2652b4001...%20Edexcel.pdf Thank you
    I'm not sure if you have solved this yet, but think about the links between the parts.

    (Original post by economicss)
    Could anyone explain please how you would know what significance level to use in question 6c, is it concluded as not significant because it's below the critical value of 11.345 obtained in part a? https://drive.google.com/folderview?...&usp=drive_web Thanks
    What an awful/ambiguous question

    I would go with the 1% given in part a).
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    (Original post by HT12345)
    Can anyone help me with this question (part b) on goodness of fit?

    A pesticide was tested by applying it in the form of a spray to 50 samples of 5 flies. The numbers of dead flies after 1 hour were then counted with the following results:

    Number of dead flies: 0, 1, 2, 3, 4, 5
    Frequency 1, 1, 5, 11, 24, 8

    a) Calculate the probability that a fly dies when sprayed. (Answer: 0.72)
    b) Using a significance level of 5%, test to see if these data could be modelled by a binomial distribution.
    What have you tried?
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    Ooh Im taking this. S3 aint too bad. S4 is a boring piece of ****.


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    What's S2 like?
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    (Original post by Zacken)
    In. :-)
    Have a look at this question what is ur answer to 7d?
    Name:  ImageUploadedByStudent Room1458990637.213495.jpg
Views: 691
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    7d i disagree with the book answer.
    Lets say -1/2X is outside the sum(they shidve used brackets tbh).



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    If Z=X1+X2+X3 where X1,2,3 have the same distribution
    Which method is correct
    Var(z)=Var(3Xi)=9Var(Xi)
    Var(z)=Var(X1)+Var(X2) etc


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    (Original post by physicsmaths)
    Have a look at this question what is ur answer to 7d?
    Name:  ImageUploadedByStudent Room1458990637.213495.jpg
Views: 691
Size:  152.8 KB
    7d i disagree with the book answer.
    Lets say -1/2X is outside the sum(they shidve used brackets tbh).



    Posted from TSR Mobile
    This is how I done it

    \displaystyle S = \sum_{i=0}^{3}Y_{i}-\frac{1}{2}X = Y_{1}+Y_{2}+Y_{3}-\frac{1}{2}X

    \displaystyle \mathrm{Var}\left ( S \right )=\mathrm{Var}\left ( Y_{1}+Y_{2}+Y_{3}-\frac{1}{2}X \right ) =\mathrm{Var}\left ( Y_{1} \right )+\mathrm{Var}\left ( Y_{2} \right )+\mathrm{Var}\left ( Y_{3} \right ) +\mathrm{Var}\left ( \frac{1}{2}X \right ) = 3\cdot \mathrm{Var}\left ( Y \right ) +\frac{1}{4}\mathrm{Var}\left ( X \right )=3\left ( 3^{2} \right )+\frac{2^{2}}{4}=27+1=28

    Is it alright?
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    (Original post by aymanzayedmannan)
    This is how I done it

    \displaystyle S = \sum_{i=0}^{3}Y_{i}-\frac{1}{2}X = Y_{1}+Y_{2}+Y_{3}-\frac{1}{2}X

    \displaystyle \mathrm{Var}\left ( S \right )=\mathrm{Var}\left ( Y_{1}+Y_{2}+Y_{3}-\frac{1}{2}X \right ) =\mathrm{Var}\left ( Y_{1} \right )+\mathrm{Var}\left ( Y_{2} \right )+\mathrm{Var}\left ( Y_{3} \right ) +\mathrm{Var}\left ( \frac{1}{2}X \right ) = 3\cdot \mathrm{Var}\left ( Y \right ) +\frac{1}{4}\mathrm{Var}\left ( X \right )=3\left ( 3^{2} \right )+\frac{2^{2}}{4}=27+1=28

    Is it alright?
    Yeh it is correct
    But why can't
    y1+y2+y3=3y1 or y2,3that's where my mistake was



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    (Original post by physicsmaths)
    If Z=X1+X2+X3 where X1,2,3 have the same distribution
    Which method is correct
    Var(z)=Var(3Xi)=9Var(Xi)
    Var(z)=Var(X1)+Var(X2) etc


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    If Z = X_1 + X_2 + X_3 \neq 3X_i then \text{Var}(Z) = \sum_{i} \text{Var}(X_i) \neq \text{Var}(3X_i)

    There's a difference between X_1 + X_2 + X_3 (3 samples being taken) rather than 3X (scaling the distribution by 3).
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    (Original post by Zacken)
    If Z = X_1 + X_2 + X_3 \neq 3X_i then \text{Var}(Z) = \sum_{i} \text{Var}(X_i) \neq \text{Var}(3X_i)

    There's a difference between X_1 + X_2 + X_3 (3 samples being taken) rather than 3X (scaling the distribution by 3).
    That last line was what I was looking for, cheers mate.



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    (Original post by physicsmaths)
    That last line was what I was looking for, cheers mate.



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    Once you know that, life becomes a hell of a lot easier, that's how you can derive distributions of the mean for n samples.

    You take n samples \sum_{i}^n X_i then scale it by \frac{1}{n} so we have:

    \displaystyle 

\begin{equation*}E(\bar{X}) = E\left(\frac{X_1 + \cdots X_n}{n}\right) = \frac{1}{n}(\mu + \cdots + \mu) = \frac{1}{n}(n\mu) = \mu \end{equation*}

    So the expected value of the mean is the same as the expect value of the distribution.

    You can do something similar with variance:

    \displaystyle 

\begin{equation*}\text{Var}{\bar  {X}} = \text{Var}\left(\frac{X_1 + \cdots X_n}{n}\right) = \frac{1}{n^2}(\sigma^2 + \cdots + \sigma^2) = \frac{n\sigma^2}{n^2} = \frac{\sigma^2}{n}\end{equation*  }

    This shows that the more samples you take, the smaller your variance becomes - which is what we expect.
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    (Original post by physicsmaths)
    Yeh it is correct
    But why can't
    y1+y2+y3=3y1 or y2,3that's where my mistake was



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    There's a little note on page 3 which says it can't be equal to that but I don't really know of a proof.. There's a proof here though, but I don't get it :laugh:

    Generally \displaystyle \mathrm{Var}\left ( \sum_{i=1}^{n}X_{i} \right ) =\sum_{i=1}^{n}\mathrm\left \mathrm{Var} ( X_{i} \right )

    Zacken do you know why they do it like this?

    EDIT: sorry, I'll read above

    EDIT II: thank you, that gives me a more intuitive understanding of it.
 
 
 
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