WitCH 33: Below Average

We’re not actively looking for WitCHes right now, since we have a huge backlog to update. This one, however, came up in another context and, after chatting about it with commenter Red Five, there seemed no choice. The following 1-mark multiple choice question appeared in 2019 Exam 2 (CAS) of VCE’s Mathematical Methods. The problem was to determine Pr(X > 0), the possible answers being

A. 2/3      B. 3/4      C. 4/5      D. 7/9      E. 5/6

Have fun.

Update (04/07/20)

Who writes this crap? Who writes such a problem, who proofreads such a problem, and then says “Yep, that’ll work”? Because it didn’t work, and it was never going to. The examination report indicates that 27% of students gave the correct answer, a tick or two above random guessing.
We’ll outline a solution below, but first to the crap. The main awfulness is the double-function nonsense, defining the probability distribution \boldsymbol{f} in terms of pretty make the same function \boldsymbol{p}. What’s the point of that? Well, of course \boldsymbol{f} is defined on all of \boldsymbol{R} and \boldsymbol{p} is only defined on \boldsymbol{[-a,b]}. And, what’s the point of defining \boldsymbol{f} on all of \boldsymbol{R}? There’s absolutely none. It’s completely gratuitous and, here, completely ridiculous. It is all the worse, and all the more ridiculous, since the function \boldsymbol{p} isn’t properly defined or labelled piecewise linear, or anything; it’s just Magritte crap. 
To add to the Magritte crap, commenter Oliver Oliver has pointed out the hilarious Dali crap, that the Magritte graph is impossible even on its own terms. Beginning in the first quadrant, the point \boldsymbol{(b,b)} is not quite symmetrically placed to make a 45^{\circ} angle. And, yeah, the axes can be scaled differently, but why would one do it here? But now for the Dali: consider the second quadrant and ask yourself, how are the axes scaled there? Taking a hit of acid may assist in answering that one.
Now, finally to the problem. As we indicated, the problem itself is fine, its just weird and tricky and hellishly long. And worth 1 mark. 
As commenters have pointed out, the problem doesn’t have a whole lot to do with probability. That’s just a scenario to give rise to the two equations, 
1) \boldsymbol{a^2 \ +\ \frac{b}{2}\left(2a+b\right) = 1} \qquad      \mbox{(triangle + trapezium = 1).}
2) \boldsymbol{a + b = \frac43} \qquad           \mbox{( average = 3/4).}
The problem is then to evaluate
*) \boldsymbol{\frac{b}2(2a + b)} \qquad \mbox{(trapezium).}
or, equivalently, 
**) \boldsymbol{1 - a^2 \qquad} \mbox{(1 - triangle).}
The problem is tricky, not least because it feels as if there may be an easy way to avoid the full-blown simultaneous equations. This does not appear to be the case, however. Of course, the VCAA just expects the lobotomised students to push the damn buttons which, one must admit, saves the students from being tricked.
Anyway, for the non-lobotomised among us, the simplest approach seems to be that indicated below, by commenter amca01. First multiply equation (1) by 2 and rearrange, to give
3) \boldsymbol{a^2 + (a + b)^2 = 2}.
Then, plugging in (2), we have 
4) \boldsymbol{a^2 = \frac29}.
That then plugs into **), giving the answer 7/9. 
Very nice. And a whole 90 seconds to complete, not counting the time lost making sense of all the crap. 

24 Replies to “WitCH 33: Below Average”

  1. Defining f(x) to be equal to p(x) between -a and b and 0 elsewhere seems redundant. Why not simply say that p is the probability density function?

    Further, p(x) is never defined other than via the graph, so we don’t technically know that it’s pieced together from two linear functions. The test should state/mathematically-imply in the definition of p that p is pieced together from two linear functions.

    1. Thanks, edder. Yes, whatever the intrinsic (de)merits of the question, the framing is appalling in the ways you indicate.

  2. It’s a curious question, because the only probabilistic part is noting that the area under the given graph must be 1, hence the average value given shows that a+b = 4/3. Then it’s just geometry and algebra. The area, as a triangle on the left and a trapezoid on the right, is a^2 + (b/2)(2a+b) = a^2 + ab + b^2/2. Since this is equal to one, we multiply by 2 to obtain 2a^2 + 2ab + b^2 = 2, and note that the right hand side can be rewritten to obtain a^2 + (a + b)^2 = 2. Since a+b = 4/3, the answer is 1 – a^2 = 7/9.

    I have several points to make:

    1, This seems like quite a lot of work for 1 point.

    2, I have no idea of the pedagogy behind this. What understanding are the examiners intending to test, and how does this question actually test this understanding? (Mind you, this query could be made of many VCE maths exam questions.)

    3, My solution above has b being independent of a, whereas in the picture it seems that b < 2a. If you start by assuming this, and obtain the same answer, are you wrong? But then, as a multiple choice question there is no way of checking working.

    4, Multiple choice questions, unless very carefully designed, have no place in mathematics assessment. They reinforce the idea that mathematics is about “answers” and “getting it right”, instead of demonstrating reasoning and argument.

    5, It’s a dog’s breakfast of a question: a mish-mash of ideas, and really really poorly examined. A perfect example of a question which should not be multiple choice.

    You could probably go deeper into the setup, as edderiofer has done above, and pull apart some of the sloppy definitions. But as we all know, the only way to make sense of VCAA mathematics is to take a general level of sloppiness on board and so give everybody a large measure of doubt.

    1. Thanks, amca01, and I agree entirely. I would just shorten your point 5: A perfect example of a question which should not be.

  3. @edderiofer @amca01 – Agreed; Further,wrt sloppiness, the diagram is a sketch not a graph, in the conventional sense that a graph is a representation of the function in Cartesian coordinates with respect to linearly-scaled orthogonal axes. If correctly drawn, the lengths of the abscissa and ordinate of (b,b) would be in the same ratio as the length of the abscissa of (-a,0) and 0.5*length of ordinate of (0,2a). ie Point (b,b) should lie on a line through the origin with half the slope of the left-hand side arm of the graph.

  4. Alternatively, an astute student could realize that as the answer is independant of b, they can choose any value they like, say b=1. Then a=1/3 from which the answer is immediate. And all my points still stand, with the extra possibility of confirming the perception that mathematics is really all about learning a mass of nifty tricks.

    1. I think I see what you’re getting at here… and will now go and try a few values to see. I suspect that if it were not for the left hand part of the graph, this might work since b acts as a scale factor.

      But it does raise a valid point about areas under a graph more generally, since Methods examiners have been known (at those “meet the assessors” sessions) to suggest that “guessing a function” is sometimes valid on MCQs.

  5. By the way, talking about VCAA Exam questions ….. At the risk of being off-topic, I’m sure everyone has heard the wonderful news:

    VCAA have at long last made the marking scheme for its exams public. Finally. Unfortunately, it’s not freely available. You have to pay the MAV a sum of money to get access. Apparently the VCAA Assessors are no longer bound by confidentiality and can disclose the marking scheme through the MAV VCAA Exam Solutions. A cosy arrangement where the Assessors get to make extra money, MAV make money and teachers get the VCAA marking scheme. Everyone’s happy.

    1. Late note: It turns out the VCAA exam solutions did NOT provide the marking scheme after all. What a let-down. The solutions simply give screenshots of each exam question and what it was worth – far short of a marking scheme by any reasonable definition.

      The MAV advertisement was marketing hyperbole (not to be confused with hyperbolae – which are no longer on the course) and was subsequently amended.

  6. Hi, JF, and everyone. I’ll post on the MAV-VCAA solutions thing shortly. Please leave thoughts and comments for that post.

  7. Marty et al,
    Curious to hear your thoughts on the 2011 (non-)Mathematical Methods Exam 1, Question 7, regarding the flipping of a biased coin.
    I was just browsing some exams to see how they “examine” the Binomial Theorem and came across this doosey (especially part b. and the allowance of p=0.
    Shitty question and shitty scenario if you ask me.
    If it deserves it, a new post on this question would be awesome – titled “Flipping Crap” or something more cognisant of the scenario?
    Keen to hear everyone’s thoughts.

    1. Hi, Steve. I agree that p = 0 being a permitted solution is pretty screwy, and the question should obviously have been arranged to avoid this issue. I read the report as confessing, in the predictable mealy-mouthed manner, that they screwed up. I read it that they allowed students to “cancel out” p (i.e. give just the one solution for p), as long as this cancelling was “supported”.

      It’s also worth pointing out that the solution in the report is much muddier than need be, because, like everybody else, the examiners have forgotten the value of defining and working with q = 1 – p.

      Is there anything else that bugs you about the question?

      1. Hi marty, I admit that I did not read the report prior to my submission of the comment, however after reading it, I don’t see evidence of them of them confessing – far from it. Do you have a different version of the report compared to the one I’m reading?

        All I can see is “The cancelling out of p was rarely supported” (evidence that they were looking to “trip” students up with algebraic tricks) and “Many students incorrectly assumed p could not be zero” (which doesn’t bode well with the fact that the word “heads” specifically appeared, by my count, five times throughout the course of the question, without reference to any vagiue possibility of there existing a double-tailed coin.

        I wonder what the “Meet the Assessors” event was like in 2012 (for the 2011 exams) – I can only imagine there would’ve been an outcry – which of course would have by and large fallen on deaf ears. To quote JF, they’re just “too DuLL” to realise what the gvdl they’re doing, and it’s students and (a fair portion of) teachers that suffer.

        1. Hi, Steve. I’m reading the same report as you. I could be wrong and, in any case, it would have been more accurate for me to have written “grudgingly conceding” rather than “confessing”. There is no doubt that the examiners intended to require p = 0 as a solution. There is also no doubt that the question is screwed for having forced students to contemplate this implausible solution and, um, flip a coin to include it or not.

          But here’s my thinking. The examiners wrote “The cancelling out of p was rarely supported”. So, what if the cancelling was “supported”? What does that mean, and what then?

          The point is, if one is permitting p = 0 as a solution then one cannot cancel p, end of story; no “support” is possible. But the “support” is only worth commenting upon, as was done, if it is relevant to the grading. For me, this suggests that the examiners permitted a student to cancel p and to leave out p = 0, as long as the student also wrote something like “p cannot be 0”.

          Again, I could be wrong. And the more important point is that an exam shouldn’t include such a screwed question and, when an exam includes such a screwed question, the examiners should be straight-forward and honest about having written a screwed question, and the examiners should be straight-forwarded and honest about how they graded their screwed question. None of which should be that hard, but all of which appears to be well nigh impossible for VCAA clowns.

          1. 100% agreed, marty. It’s a shit question, it’s a screwed question, no amount of “comments” provided on their Assessor “Reports” can hide that fact.
            I really do feel for the students that have to face this crap year after year – I was in their shoes not that long ago, and now I’m on the ‘other side’ of the teacher’s desk, so it’s hard to fathom that my students, every single year, will just have to keep throwing shit at the wall and hope that something sticks.

            Anyway, sorry for derailing your original post, it was the most recent Methods – related one so it was the best I could come up with.

            Thanks for your thoughts & insights.

            1. No worries, Steve. I didn’t think the question was worth a separate post, but it was a close call. And, adding it to this post was as good a place as any. You can always email me if you want to suggest I post on something, or if you want comment but aren’t sure where is best.

    2. I’m more interested in the current number of amendments to the reports. So many amendments to so many reports! One can only speculate on how many more amendments there would be if the reports came out any sooner than the 6 or so months we typically have to wait.

      And we only see the date of the latest current amendment – there may have been numerous previous amendments.

      Yeah, allowing p = 0 in this context is a cheap trick. If you want to test whether students do things like x^2 – x = 0 therefore x – 1 = 0 therefore x = 1 there are much better ways of doing this.

      1. JF, I don’t think permitting p = 0 was intended to be a cheap trick. It smells to me like the examiners thought up a cute algebraic question for a biased coin and, unfortunately, the most natural variations will have p as a factor and thus p = 0 as a solution. They were then either too obtuse to realise that p = 0 is really stupid for a coin, or they were too stubborn or too lazy to modify the question.

        1. Yeah. Even a cheap trick requires intelligence.

          And as Steve said, all those heads in the earlier parts of the question occurring with a non-zero probability … It really sets the students up for failure. Besides, who ever heard of a two-tailed coin. (A two-headed coin I could almost believe). I can almost hear the VCAA clowns now:

          “Hey DuLL, let’s toss a coin to decide whether or not we fix the question. Heads we fix, tails we don’t.”
          “Great idea, AllMac. We can use my brand new two-tailed coin.”
          (“Hmmm … I think I’ll use it to make all my decisions on whether or not to give a shit about what the teachers think”)

          1. Spot-on, J(n)F. Bunch of clowns running the show.
            And as the saying goes, hire clowns, get dodgy outcomes… Or something along those lines.

            Almost forgot about this question until I went trawling through past exams looking for Binomial Probability exam question examples.

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