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STEP III 2012 Discussion Thread

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Reply 180

msmith2512

Original post by mikelbird

my attempt at question 6

Could you check your arithmetic in the last part. I think (-2x) * x^2 = -2 x^3.

So y^2 = (1-x^3)/x

Reply 181

shamika

Original post by DFranklin

Is the variance at the end straightforward? I can see how you do the rest of it, but (without actual calculation) it feels the algebra for the variance is going to be horrific.

I think I agree with you - that wasn't not fun at all (I made a substitution to turn the integral of

E[|X|^2]

into a standard normal and then an integration of parts. I got...

Spoiler

(In case anyone is wondering why anyone would know this, this sort of problem comes up in real life when trying to work out how much reinsurance someone should buy. Isn't actuarial stuff fun? 13 (ii) has definitely come up in a past actuarial exam :smile:

)

Are Lognormal distributions covered in A-Levels? A nice question for someone to try:

If

f_X (x)

is the pdf of a LogNormal

(\mu,\sigma^2)

distribution, show that

\int_0^\infty \! x^k f_X(x) \, \mathrm{d} x = \mathrm{exp}(k\mu+\frac{1}{2}k^2\sigma^2)

(edited 11 years ago)

Reply 182

Lord of the Flies

Original post by msmith2512

Could you check your arithmetic in the last part. I think (-2x) * x^2 = -2 x^3.

So y^2 = (1-x^3)/x

You're forgetting a term:

(-2x)x^2-(-2x)y^2+(-2x)^2x+2=0

-2x^3+2xy^2+4x^3+2=0

2x^3+2xy^2+2=0

y^2=-\dfrac{1+x^3}{x}

Reply 183

MrDD

Original post by msmith2512

Could you check your arithmetic in the last part. I think (-2x) * x^2 = -2 x^3.

So y^2 = (1-x^3)/x

indeed (-2x) * x^2 = -2 x^3, but then it gets added to (-2x)^2*x to make 2x^3

i agree with mikelbird

i liked this question, but didn't feel confident about my solutions in the exam. i'm pleased to say i got the sketch right for the p=-2x case, but frustratingly missed the restriction on the values of x when y=0

Reply 184

Dirac Spinor

Original post by shamika

I think I agree with you - that wasn't not fun at all (I made a substitution to turn the integral of

E[|X|^2]

into a standard normal and then an integration of parts (although I still maintain its not difficult as such, just very long). I got...

Spoiler

(Are Lognormal distributions covered in A-Levels? The result for the Lognormal generalizes nicely for all central moments and actually is a much easier result to derive. A nice question for someone to try:

If

f_X (x)

is the pdf of a LogNormal

(\mu,\sigma^2)

distribution, then

\int_0^\infty \! x^k f_X(x) \, \mathrm{d} x = \mathrm{exp}(k\mu+\frac{1}{2}k^2\sigma^2)

I don't think they are on A levels anymore. I only know them from statistical mechanics.

Reply 185

B Jack

Original post by ben-smith

I don't think they are on A levels anymore. I only know them from statistical mechanics.

They were referred to in one of the past papers for the OCR MEI S4 module, as part of the queston under probability/moment generating functions; but it wasn't really covered in the module

Reply 186

TheMagicMan

Original post by MrDD

i agree there probably isn't a perfectly fair solution, but it is certainly interesting to speculate what the least unfair solution is.

one solution might be to take candidates 5 best solutions from questions 2-13 and multiply total by 1.2

anyone got any better ideas?

They will have to come up with something, but it will be unfair to someone. They'll probably end up putting more weight on step ii.

This was posted from The Student Room's iPhone/iPad App

Reply 187

shamika

Original post by B Jack

...

Original post by ben-smith

...

http://en.wikipedia.org/wiki/Log-normal_distribution (all you really need to answer the question is the very first property in the article)

Reply 188

desijut

I'm honestly surprised that Q1 was this poorly attempted... What about the phrasing threw people off? I still dont get it

Reply 189

Lord of the Flies

Original post by desijut

I'm honestly surprised that Q1 was this poorly attempted... What about the phrasing threw people off? I still dont get it

"Use the above result to show that [...]" when the initial expression was anything but a "result"? Also, without defining

z

in some way the question makes little sense...

Reply 190

Stray

I can't see a one-size-fits-all solution based on a numerical adjustment that is sufficiently fair.

Those who successfully guessed a suitable substitution for the missing part and completed the question did more work than was intended to be rewarded with the 20 marks. Should they get a bonus?

There will be people who spent 20 minutes on Q1, moved on unsatisfied and then were niggled with possible guesses for the missing part throughout the next couple of hours. Our brains don't let go of things just because we've 'decided' to switch to a different question - that trait isn't necessarily a bad one in a mathematician!

Perhaps an optional "have another go" paper should be offered for those who feel they were adversely affected, or maybe they'll separate the grading of papers from numerical scores for those who tackled Q1. Of course that is still unfair on those who considered Q1 but didn't put pencil to paper - only following Silkos's own advice that one should spend some time thinking about the question before diving in.

It'll be interesting to see what they come up with - it'll have to be transparent given how much rests on a single grade here. I can imagine the air in the admissions offices is a little blue...

Reply 191

fruktas

Original post by desijut

I'm honestly surprised that Q1 was this poorly attempted... What about the phrasing threw people off? I still dont get it

There was completely no mention of z whatsover, apart from stating dz/dx = ....

Reply 192

fruktas

I would imagine a lot of people are going to appeal if they do not get in due to their score in iii. I know I would!

Reply 193

MitchBriggles

I found this paper really hard, it didn't have any of the things I normally go for, was rooting for number theory, combinatorics or integration. I did 3 all the way, 1 apart from a sketch, and 2 scraps.

I wasted far too much time on question 1, normally being quite fond of differential equations, hopefully I'll get some credit for that. I think I'm probably talking 70s to 90s, depending on lots of things. So a comfortable 1, but almost definitely not an S. :frown:

Hopefully getting an S in STEP II will be enough...

Reply 194

desijut

Original post by Lord of the Flies

"Use the above result to show that [...]" when the initial expression was anything but a "result"? Also, without defining

z

in some way the question makes little sense...

Original post by fruktas

There was completely no mention of z whatsover, apart from stating dz/dx = ....

Fair enough, I honestly thought it made sense, maybe the phrasing was off but i thought it was obvious what you were meant to do with that.

If they give extra marks to people that struggled on that question, then it's unfair on those that got through the question

Reply 195

Tobedotty

Original post by desijut

What if you get extra marks too? :tongue:

Reply 196

desijut

Original post by Tobedotty

What if you get extra marks too? :tongue:

Then it's ok :biggrin:

Then i should definately get a 1.... Right now, im on the border of 1/2 with a strong 1 in II, so if i dont get in and it's by a few marks, then it could be (if people are given extra marks) those few extra marks raising the grade boundaries a bit.

Reply 197

msmith2512

Original post by desijut

It would seem to me that an option would be to reduce each grade boundary by, say, 10. This wouldn't affect people who managed to do Q1 - in fact their grade may go up. And this would compensate people who struggled with or were put off by Q1.

The only issue with this would be it would create more S's, 1's and 2's so more people will meet their offers so fewer pool places.