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Edexcel S3 - Wednesday 25th May AM 2016

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

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

(edited 7 years ago)

Scroll to see replies

Reply 1

economicss

Thanks for making the thread, please could you explain why on question 2a of this paper https://869d950bf149eefd5cd2652b400160ceb38ca5c7.googledrive.com/host/0B1ZiqBksUHNYVGVDMmVJYzVnRnc/June%202014%20(IAL)%20MS%20-%20S3%20Edexcel.pdf the bias can be plus or minus a/2 as I thought the bias would only be positive? Thanks :smile:

Reply 2

Ayman!

Original post by economicss

I think it might just be the mark scheme being lenient.

Reply 3

economicss

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://869d950bf149eefd5cd2652b400160ceb38ca5c7.googledrive.com/host/0B1ZiqBksUHNYVGVDMmVJYzVnRnc/June%202014%20(IAL)%20QP%20-%20S3%20Edexcel.pdf I really can't get my head around it! Thanks :smile:

In. :-)

Original post by economicss

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?

(edited 8 years ago)

Reply 6

economicss

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://869d950bf149eefd5cd2652b400160ceb38ca5c7.googledrive.com/host/0B1ZiqBksUHNYVGVDMmVJYzVnRnc/June%202014%20(IAL)%20QP%20-%20S3%20Edexcel.pdf Thank you :smile:

Reply 7

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?id=0B3tgJoiYtxdOfkp0Q1EwT05XczdZcm0tMTZWZ25mN2NVcHNFZXdfSlVCeTk2NXNpeVBpaTA&usp=drive_web Thanks :smile:

Reply 8

Kevin De Bruyne

Original post by economicss

I'm not sure if you have solved this yet, but think about the links between the parts.

Original post by economicss

What an awful/ambiguous question :s-smilie:

I would go with the 1% given in part a).

Reply 9

Kevin De Bruyne

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? :h:

Reply 10

username1162972

Ooh Im taking this. S3 aint too bad. S4 is a boring piece of ****.

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What's S2 like?

Original post by Zacken

In. :-)

Have a look at this question what is ur answer to 7d?
ImageUploadedByStudent Room1458990637.213495.jpg

7d i disagree with the book answer.
Lets say -1/2X is outside the sum(they shidve used brackets tbh).

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Reply 13

username1162972

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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Reply 14

Ayman!

Original post by physicsmaths

Have a look at this question what is ur answer to 7d?
ImageUploadedByStudent Room1458990637.213495.jpg

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?

Reply 15

username1162972

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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Reply 16

Zacken

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

Posted from TSR Mobile

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).

Reply 17

username1162972

Original post by Zacken

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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Reply 18

Zacken

Original post by physicsmaths

That last line was what I was looking for, cheers mate.

Posted from TSR Mobile

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:

Unparseable latex formula:

\displaystyle [br]\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:

Unparseable latex formula:

\displaystyle [br]\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.

(edited 8 years ago)

Reply 19

Ayman!

Original post by physicsmaths

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

Posted from TSR Mobile

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: