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

1. (Original post by Damien_Dalgaard)
I was getting you confused with ten of them, you names are similar.

I think I did some of your papers before, great website sir!
It will be far more useful if you do a degree with a high mathematical content.

Sorry many people confuse our names (I was here first)
2. (Original post by TeeEm)
It will be far more useful if you do a degree with a high mathematical content.

Sorry many people confuse our names (I was here first)
Ah thanks, hmmn newb mistake on my part.

If anyone has access to the edexcel s3 pearson book, can you explain to me please on page 4 a) at the end of the question it goes from P(-1.44<Z<1.44) to 1.44 - - 1.44

Why does it do this and how is the final answer obtained.

Thanks.
3. (Original post by Damien_Dalgaard)
Ah thanks, hmmn newb mistake on my part.

If anyone has access to the edexcel s3 pearson book, can you explain to me please on page 4 a) at the end of the question it goes from P(-1.44<Z<1.44) to 1.44 - - 1.44

Why does it do this and how is the final answer obtained.

Thanks.
I do not I am afraid
4. (Original post by TeeEm)
I do not I am afraid
Apologies I should have quoted this separtely.
5. If anyone has access to the edexcel s3 pearson book, can you explain to me please on page 4 a) at the end of the question it goes from P(-1.44<Z<1.44) to 1.44 - - 1.44

Why does it do this and how is the final answer obtained.

Thanks.
6. (Original post by Damien_Dalgaard)
Apologies I should have quoted this separtely.
no worries
7. (Original post by Damien_Dalgaard)
If anyone has access to the edexcel s3 pearson book, can you explain to me please on page 4 a) at the end of the question it goes from P(-1.44<Z<1.44) to 1.44 - - 1.44

Why does it do this and how is the final answer obtained.

Thanks.
Its a typo. They meant phi(1.44).
So,
P(-1.44<Z<1.44)=P(z<1.44)-(1-P(z<1.44))
= phi(1.44)-1+phi(1.44)
= 2phi(1.44) - 1
8. (Original post by Xin Xang)
Its a typo. They meant phi(1.44).
So,
P(-1.44<Z<1.44)=P(z<1.44)-(1-P(z<1.44))
= phi(1.44)-1+phi(1.44)
= 2phi(1.44) - 1
Thank you very much for the clear explanation.

I get P(z<1.44) - (1- P(z<1.44)
0.9251 - (1-0,9251)
=0.9251-0.0749
=0.8502.

Can I ask do we leave answers for z values to 3 or 4 d.p.
9. (Original post by Damien_Dalgaard)
Thank you very much for the clear explanation.

I get P(z<1.44) - (1- P(z<1.44)
0.9251 - (1-0,9251)
=0.9251-0.0749
=0.8502.

Can I ask do we leave answers for z values to 3 or 4 d.p.
The safe answer would be at least 3d.p if it isn't specified.
10. has anybody attached the paper?
11. (Original post by Xin Xang)
The safe answer would be at least 3d.p if it isn't specified.
Last last question, do we have to use phi notation or we can just use the P( x < z) format and then just sub in the phi values as we go.
(Original post by TeeEm)
has anybody attached the paper?
which paper lol?
12. (Original post by Damien_Dalgaard)
which paper lol?
Crap

I thought I was in the FP1 thread

13. (Original post by TeeEm)
Crap

I thought I was in the FP1 thread

lolek, that was a nice paper I think.

Will have to wait to results day srs
14. (Original post by Damien_Dalgaard)
Last last question, do we have to use phi notation or we can just use the P( x < z) format and then just sub in the phi values as we go.

which paper lol?
I don't think they even expect us to use the phi notation. Use whatever notation suits you. It won't cost you marks.
15. (Original post by Xin Xang)
I don't think they even expect us to use the phi notation. Use whatever notation suits you. It won't cost you marks.
Thank you.
16. (Original post by Damien_Dalgaard)
Thank you.
Its kl.
17. Could someone have a look at Q3 on June 2010 and explain how you do part a? It's only two marks and seems incredibly simple but I can't remember what to do in that situation and the mark scheme isn't helpful in explaining why they do that
18. (Original post by Iridann)
Could someone have a look at Q3 on June 2010 and explain how you do part a? It's only two marks and seems incredibly simple but I can't remember what to do in that situation and the mark scheme isn't helpful in explaining why they do that
We have X - N(w, 0.5^2)

We are interested in P(W - 0.6 < X < W + 0.6)

But, and if you don't know why this is drawing a diagram of the areas under the normal represented by this probability would help,

This is the same as P(X < W + 0.6) - P(X < W - 0.6)

Now converting into z, for w + 0.6 we have (w + 0.6 - w)/0.5 and (w - 0.6 -w)/0.5

Giving 1.2 and - 1.2 respectively

So we have P(Z < 1.2) - P(Z < -1.2)

But, and again a diagram will illustrate this, P(Z < -1.2) = 1 - P(Z < 1.2)

So we require, rearranging, 2P(Z < 1.2) - 1

And the table does the rest
19. (Original post by 13 1 20 8 42)
We have X - N(w, 0.5^2)

We are interested in P(W - 0.6 < X < W + 0.6)

But, and if you don't know why this is drawing a diagram of the areas under the normal represented by this probability would help,

This is the same as P(X < W + 0.6) - P(X < W - 0.6)

Now converting into z, for w + 0.6 we have (w + 0.6 - w)/0.5 and (w - 0.6 -w)/0.5

Giving 1.2 and - 1.2 respectively

So we have P(Z < 1.2) - P(Z < -1.2)

But, and again a diagram will illustrate this, P(Z < -1.2) = 1 - P(Z < 1.2)

So we require, rearranging, 2P(Z < 1.2) - 1

And the table does the rest
Ah thanks very much, something wasn't clicking there haha.
20. Does anyone know how strict the mark scheme is on working out?
Is it acceptable to state the result of this formula without doing individual calculations for each row (I do it on a calculator)?.

When combining variables, I noticed that the textbook includes steps which I tend to skip over (implicit imo). For example, 'Let W= X1 + X2 +Y1 ....', and drawn out mean and variance calculations. Is it acceptable to simply state the new distribution correctly (which implies that you have a full understanding of the concept)?

I just find with S3 that I answer 6 mark questions with 2-3 lines of working, which I'm hoping is adequate. I've had a look at some mark schemes but the notes were brief regarding how much 'working' is expected.

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