Thursday, November 10, 2016

Predictions are not necessarily bad if they don't turn out to be true


This was in The Guardian:

 Oh dear!  This again.  I have no drum to bang for pollsters and they may well have messed things up completely, but this paragraph is fairly hopeless. 

If we have a prediction, with a probability attached, it is natural to look at the biggest probability and to make use of this.  If there is a “15% chance that Donald Trump will win” then this is deemed to be unlikely.  But unlikely isn’t impossible.  By saying that the statement was ‘wrong’ we would also be saying that if a total of 6 comes up when you roll two dice then you must have cheated!

This crops up over and over.  An example is the (in-)famous prediction by the Met Office in 2009 of a ‘barbecue summer’.  Here is how the Daily Mail website refers to it:


The April statement says sun is “odds on” and that it is “likely” to be warmer.  That it didn’t happen doesn’t make the prediction necessarily “wrong”, just that a less likely scenario prevailed this time around.  (I am not sure it makes sense to make such a high profile prediction where you think it is only 2/3 likely to happen, though.)  

To be clear, I am not saying that either of these predictions are definitely “right”, but just that a scenario that is deemed to be unlikely can occur without the probabilities being wrong.

FFT
As teachers we also see this with FFT grades.  These grades used to be presented as percentages.  They would show that, for a pupil with a similar background (eg KS2 results, etc) a certain proportion of the pupils went on to get an A*, A, B, C, D, etc grade in each GCSE subject.  

Nowadays only the median version is given, and this is shown as _the_ FFT grade for the pupil.  In the new system these are given to the nearest third of a grade. 

If a  pupil has B+ as their FFT grade what this really means is that about half of children with a similar background to that pupil will get a high grade B or above, and half will get a high grade B or below.  It might well be the case that a small fraction actually get a high grade B.  If that pupil doesn’t get a high grade B that doesn’t mean that over- or under-achieved, or that their FFT grade was ‘wrong’, just that it didn’t happen. 

[The initial article also forms part of a Quibans.  Find it here.]

Sources:

Friday, November 04, 2016

Aliens vs Barca

I usually enjoy the weekly puzzle on the Weekly Newsletter from Chris Smith (find him on Twitter @aap03102 – he will happily add you to his list of subscribers).

The puzzle in the most recent edition was a corker:


I loved this because there are so many ways of doing it.  I won’t quibble over whether two ways really are ‘different’ or not and will list as many as I and various students have come up with.  (Yr 11 thoroughly enjoyed it too!)

In all diagrams CN is labelled as a.  All explanations are sketched rather than presented as full proofs.

1) Similar Triangles

Triangles SNM and CNS are similar (look at the angles).  This means that a/180 = 180/135
Rearranging gives a = 240 and the diameter is 375.

2) Pythagoras 1



In triangle SNM the hypotenuse is 225. 
Angle MSC is a right angle (angle in a semi-circle), so we have two further right-angled triangles: CNS and MSC.  We can use Pythagoras in both of them:
(a+135)2 = b2 + 2252
b2 = a2 + 1802
Eliminate b2 and solve to get a=240.

A brief excursus – 3,4,5 triangles.
The numbers that Chris chose had some meaning (this puzzle appeared in newsletter number 375 and the answer is … 375).  It is also nice that they are numbers I recognise (usually in the form of angles).  135 = 3x45 and 180 = 4x45, which means we have a 3,4,5 triangle, so the hypotenuse is 5x45 = 225.  In my handwritten notes for these solutions I made the arithmetic easier by using e = 45 and calling the sides 3e, 4e and 5e.

3) Pythag 2
Draw in a radius (labelled r):


Do Pythagoras on triangle ONS and work out r.  Then double it.

4) Pythag 3
This is really the same as version 3 but with different labels.


Use Pythagoras on the triangle ONS

5) Intersecting Chord Theorem
This is related to version 1.



The intersecting chord theorem states that CN x NM = SN x NU
This gives 135a = 180 x 180

6) Area of a triangle


Looking at the big triangle, the area is ½ (a+135) x 180
If we treat CS as the base then it is also ½ b x 225.
Equate these and then use Pythagoras on CNS (a^2 + 180^2 = b^2) as a second equation and solve for a.


7) More similar triangles – using cos.


OT is the perpendicular bisector of SM.
We know SM = 225, so TM is half of that, 112.5.  Cos(M) = 135/225 (from triangle SNM).  But in triangle OTM we have cos(M) = 112.5/r

8) Trig 1



Work out the size of angle α (inverse tan of 180/135).
Angle OSM = α (isosceles triangle)
Angle NSM = 90 – α
Angle OSN = α – (90 – α)
So angle OSN = 2α – 90.
Use this angle, trigonometry and the side SN to work out either ON or OS.

9) Trig 2


We know what angle α is (using trig in triangle SNM) and we know that SM is 225 (Pythag).  We also know that angle CSM is a right angle (angle in a semi-circle).  Use trig in that triangle to get the diameter.

10) Trig 3
We could also find the length of CS using trig (as in version 9) and can then use Pythagoras to find CM.

11) Similar Triangles again



Triangle SNM is similar to triangle CSM. 
225/135 = diameter / 225.

How many different bits of maths did we use?
Many of these methods overlap, but it is rather nice that there are so many different aspects of maths that are involved.  It was also good (with Yr 11) to be able to focus on methods and explanation, rather than on finding the answer.

I think this can involve:
  • Circle theorems and other angle rules
  • Pythagoras
  • Trigonometry
  • Area of a triangle
  • Similar triangles



Many thanks to Chris for a great problem!