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Edexcel S3 - Wednesday 25th May AM 2016

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Reply 60
Original post by ServantOfMorgoth
Are there like ebooks around? I don't have money to buy a textbook for a few weeks

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There are but I can't endorse them, being a mod and all. :tongue:
Original post by SeanFM
I don't understand why you're doing that :s-smilie: the sum that you're asked to find is given and you just have to compute it.





I'm not sure :redface: for Edexcel I found that the book was all you really need.


Do you know if there's an ebook text for it?

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Original post by Zacken
There are but I can't endorse them, being a mod and all. :tongue:


Can you endsore them in PM/WhatsApp, Kik? Lol

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Original post by ServantOfMorgoth
Do you know if there's an ebook text for it?

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I've never looked :tongue: sounds dodgy, but good luck :tongue:
Original post by SeanFM
I've never looked :tongue: sounds dodgy, but good luck :tongue:


:sad:
Original post by SeanFM
I've never looked :tongue: sounds dodgy, but good luck :tongue:


What's the name of the book you use?

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Original post by ServantOfMorgoth
What's the name of the book you use?

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The textbook :tongue: I have it from A-level but I don't bring it to uni. I find that googling topics (eg F distribution test) gives these uni notes from America which are usually fine.
Y'all on this site are so difficult :rolleyes:
Original post by SeanFM
I don't understand why you're doing that :s-smilie: the sum that you're asked to find is given and you just have to compute it.

Are you trying to use

Sorry, I really can't figure it out I was using the sum of the x values minus the mean all squared divided by 6, what should I be using? Thanks :-)


I'm not sure :redface: for Edexcel I found that the book was all you really need. I guess it's worth googling individual topics if you don't have the textbook.


Sorry, I really can't figure it out I was using the sum of the x values minus the mean all squared divided by 6, what should I be using? Thanks :-)
Original post by economicss
Sorry, I really can't figure it out I was using the sum of the x values minus the mean all squared divided by 6, what should I be using? Thanks :-)


Maybe it's the sum notation that's the problem.

i=16XiXˉ6\sum_{i=1}^6 \frac{X_i - \bar{X}}{6} is the sum (since each term in the sum is divided by 6, you can just take that outside of the sum to give 16i=16XiXˉ\frac{1}{6}\sum_{i=1}^6 X_i - \bar{X}.

So what does i=1 mean that you do, and how do you find the sum of that?

It's missing the squares, but you get the idea.
(edited 7 years ago)
Original post by SeanFM
Maybe it's the sum notation that's the problem.

i=16XiXˉ6\sum_{i=1}^6 \frac{X_i - \bar{X}}{6} is the sum (since each term in the sum is divided by 6, you can just take that outside of the sum to give 16i=16XiXˉ\frac{1}{6}\sum_{i=1}^6 X_i - \bar{X}.

So what does i=1 mean that you do, and how do you find the sum of that?

It's missing the squares, but you get the idea.


It's finally clicked, thanks so much!
Could something please explain question 3 a and 3 b in the S3 June 2010 paper:

3. 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.
(a) Find the probability that X is within 0.6 mm of w.
The same board is measured 16 times and the results are recorded.
(b) Find the probability that the mean of these results is within 0.3mm of w.

The answers are: for a - 0.7698, for b - 0.9836.
Reply 72
Original post by astarsian
Could something please explain question 3 a and 3 b in the S3 June 2010 paper:

3. 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.
(a) Find the probability that X is within 0.6 mm of w.
The same board is measured 16 times and the results are recorded.
(b) Find the probability that the mean of these results is within 0.3mm of w.

The answers are: for a - 0.7698, for b - 0.9836.


What about it confuses you? Do you have any thoughts of your own?
Original post by Zacken
What about it confuses you? Do you have any thoughts of your own?


I managed to get the right answer, using confidence intervals but the mark scheme uses a normal variable with mean 0, which I don't get.
Reply 74
Original post by astarsian
I managed to get the right answer, using confidence intervals but the mark scheme uses a normal variable with mean 0, which I don't get.


Do you agree that XN(w,0.52)X \sim N(w, 0.5^2)? You want the probability that Xw<0.6|X - w| < 0.6 i.e: the difference between X and w is 0.6 mm.

But Y=XwN(ww,0.52)YN(0,0.52)Y = X - w \sim N(w-w, 0.5^2) \Rightarrow Y \sim N(0, 0.5^2) then you want P(Y<0.6)P(|Y| < 0.6).
Original post by Zacken
Do you agree that XN(w,0.52)X \sim N(w, 0.5^2)? You want the probability that Xw<0.6|X - w| < 0.6 i.e: the difference between X and w is 0.6 mm.

But Y=XwN(ww,0.52)YN(0,0.52)Y = X - w \sim N(w-w, 0.5^2) \Rightarrow Y \sim N(0, 0.5^2) then you want P(Y<0.6)P(|Y| < 0.6).


Oh I understand it now, as w doesn't have a normal distribution it's a constant.

Thank you :smile:)
Reply 76
Original post by astarsian
Oh I understand it now, as w doesn't have a normal distribution it's a constant.

Thank you :smile:)


Yep, pretty much. That's why it's only the mean affected and not the variance, since a constant has no variance.

You're welcome.
Please could someone explain the working on exercise 3e question 13b https://doc-10-0g-docs.googleusercontent.com/docs/securesc/tv01ac8rh9u7v3n9tei94valah3mgfnn/37oq44i1eqkg6mghbup4014cni4q0mav/1459692000000/01042298237316186224/02734711105213058970/0B1ZiqBksUHNYbDRYNUQ0ZktzNGs I understand how to find the value of z, but from there I don't understand what is shown? Thanks :smile:
Reply 78
Original post by economicss
Please could someone explain the working on exercise 3e question 13b https://doc-10-0g-docs.googleusercontent.com/docs/securesc/tv01ac8rh9u7v3n9tei94valah3mgfnn/37oq44i1eqkg6mghbup4014cni4q0mav/1459692000000/01042298237316186224/02734711105213058970/0B1ZiqBksUHNYbDRYNUQ0ZktzNGs I understand how to find the value of z, but from there I don't understand what is shown? Thanks :smile:


Again, can't seem to open the link. Is it on a private site or privately shared?
Original post by Zacken
Again, can't seem to open the link. Is it on a private site or privately shared?


Sorry, it shouldn't be, is from solutions bank It's exercise 3E question 13b on page 44 of the textbook :smile: Thanks

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