WitCH 25: U Turn

Our second (and last for now) NHT WitCH is due to the ever-vigilant John the Merciless (who shall, to begin, hold his fire …). It comes from the 2019 Exam 1 of Specialist Mathematics (calculator-free):

The examination “report” gives the answers as: (a) (51,65); (b) 0.02, 0.03 accepted.

Have fun.

8 Replies to “WitCH 25: U Turn”

  1. \1. The number of cars making a U-turn per day would likely better fit a Poisson distribution than a normal distribution (which would allow for a negative number of cars making a U-turn!).

    \2. Under what conditions would one know the population standard deviation but not the population mean?

    \3. The correct number of standard deviations (for a normal distribution) is 1.96, which is not an integer.

    \4. In part (b), the question switches from the number of cars making U-turns to the number of U-turns made; a car can make multiple U-turns and we’re not given the standard deviation of the number of U-turns made per day.

    \5. Given the large numbers, it’s unlikely that the students are using a statistical table, but a calculator instead. In that case, since the probability for part (b) is ~0.02275, I don’t see why an answer of 0.03 should be accepted (unless there’s some rounding errors going on somewhere, which I highly doubt).

    (Forgive the backslashes before the list numbers here; apparently this site’s commenting system tries to strip them out?)

    1. Hi edderiofer. Great observations.

      Expanding on comment 1: An additional reason why Poisson would likely be a better fit is that the number of cars is obviously a discrete random variable (why would whoever wrote this question say that a discrete random variable follows a continuous distribution??) But the Poisson distribution is not on either the Methods or Specialist courses so cannot be used here. In fact, discrete random variables are not on the Septic Maths course, so why would the writer even go near them (particularly when there are so many continuous random variables to build a context around)??

      Expanding on comment 3: There is precedent on past exams (although not in the Study Design) that 2 rather than 1.96 is the desired approximation in Exam 1. In which case, the answer of 0.025 would be expected. Except that only two decimal place accuracy is asked for …. Why not just ask for three dp rather than forcing students to wonder whether 0.03 or 0.02 is the intended answer?? But no matter according to the Examiners non-Report, let’s just accept multiple answers because the writer screwed up. Again. (The answer of 0.03 is obviously wrong, but who cares!?).

      Expanding on comment 4: Yes, an unintended subtle change of random variable, but no matter …. the intent is to say that now we do know the mean of the normal distribution that the discrete random variables (take your pick here, because apparently they’re interchangeable) follow.

      Conclusion: More evidence (as if more was needed) that the exam setting panel and the exam vetters are incompetent when it comes to statistics. And in this case does not even understand the Study Design. If this question appeared on any commercial exam I was vetting, I would be strongly recommending deletion and a new question written.

      1. In fact, it is an absolute disgrace that 0.03 is accepted for part (b). And it’s an example of not knowing whether the critical value of 2 or 1.96 or both needs to be memorised.

        \displaystyle \Pr(Z < -2) is required. So in the first instance we get an approximate answer of 0.025. But VCAA wants an answer \displaystyle correct (my emphasis) to two decimal places. So what to do? Round up or round down? As a student sitting the exam, I don't know that VCAA will stealthily (*) decide to accept either answer. I have to make a decision (unknown to me, I don’t! (**)). What do I do? I'll lose 1 mark – 2.5% – for making the wrong decision after having demonstrated a clear understanding of the required calculation (**).

        So I need to know the more accurate critical value of 1.96. Because -2 is less than -1.96 and so \displaystyle \Pr(Z < -2) will be a little bit LESS than \displaystyle \Pr(Z < -1.96) and so I must make the decision to round DOWN. The correct answer to this question is 0.02. The 'answer' 0.03 is blatantly wrong (***).

        Are we required to teach students this sort of decision making process? Who knows? The Report for this exam suggests we don't. But do we take the risk that this sets a precedent for all exams from 2020 onwards? Including 2022 Exam 1 Q3 part (a) ….

        Has this 2019 NHT Exam Report set a precedent that answers to 2022 Exam 1 Q3 part (a) rounded up or down will be accepted? Marty, get ready to add this to the horror *ahem* I mean error list with the additional observation that VCAA never learns.

        * Because VCAA can't admit its mistake of not asking for an answer correct to 3dp. (But see ***)

        ** Little do I know that I won't lose any marks because VCAA will accept either decision. My genuine and reasonable anxiety is all for nothing!

        *** Furthermore, we get the more accurate answer of 0.0227501 from a CAS and this clearly rounds down to 0.02, correct to 2 dp.
        But we note that this answer does not round to 0.025 if VCAA \displaystyle had asked for an answer correct to 3dp. So it’s impossible in Exam 1 to give an answer \displaystyle correct to 3dp … So what’s the solution? Maybe we yell at VCAA and say:
        “Don’t ask this sort of question unless you give data that can be used to answer it correctly (like you did in the 2021 Exam 1 Question 3 – why can’t you do this \displaystyle every year!)”

        Or maybe VCAA should simply have said
        “Give your answer to three decimal places”
        with none of this “correct to …” crap.

    2. Hi again, edderiofer. Expanding further on some of your comments:

      Comment 2: That the variance is known is usually an artificial convenience for the purposes of the question (particularly given that the t-distribution is not part of the course). However, this is not always the case. For example, suppose “…. that we have a machine that is designed to produce parts of a specified length. Because of imperfections in the process however, the true length of a part will be a random variable. The variance of the distribution may be due to inherent factors in the machine, which may remain fairly stable over time. In this case, the variance may [be] known from historical data to a high degree of accuracy. The mean, on the other hand, may be set by adjusting the machine and hence may change to an unknown value fairly frequently because of vibrations or other factors.”
      (Source: https://store.fmi.uni-sofia.bg/fmi/statist/education/Virtual_Labs/interval/interval2.html).

      But I doubt these considerations apply in the given question.

      Comment 1: Using a normal distribution to model random variables that do not take on ‘unreasonable’ values (eg. negative values) is not a problem. For example, the distribution of heights can be modelled using a normal distribution even though it allows for negative heights. In these cases the probability of the rv having ‘unreasonable’ values is so small that it can be ignored and the model is OK. The real trouble with using a normal distribution in this question is that you have a discrete random variable and from the get-go you’re told that it follows a continuous distribution ….

  2. Hi,

    For those interested in the use of the Poisson Distribution .

    This link provides a reasonable summary


    I recall counting number of cars passing a lamppost in 30 sec intervals on a country road in the first year statistics a while back as we didn’t have access to current military accidents in Prussia

    The question assumes a normal distribution and I would only gives a mark for 0.025 or 0.02 to 2dp to last part

    Steve R

    1. Indeed. I suspect a few of those Prussian soldiers that were kicked in the head might be writing VCAA exams.

      Yes, those are the two main things that caught my eye in this question:
      1. Inappropriately treating a discrete random variable as continuous (I mean seriously, of all the continuous random variables to choose from, the Prussians have to pick a discrete random variable …)
      2. The required two decimal accuracy (leading to the dodgy acceptance of 0.03 as an answer).

  3. Thank you all for your comments. (And I’m sorry for the issues with comment editing. One day I’ll try to figure out which plug-in makes symbols and latex work nicely.) I’ll update this WitCH, and all the WitCHes, as soon as I catch up posting on current crap.

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