2019 NHT Exam 1 Solutions (draft)

[S_R_K]

Here's my attempt at solutions to 2019 NHT Exam 1. Please post corrections / suggestions for improvements. Hopefully Exam 2 is not too far away...

https://www.dropbox.com/s/k0j2zvoc1yfed9e/2019SM1-nht-w_SOLUTIONS.pdf?dl=0

[BKmaps]

I think Question 2b should be 0.03 because Pr(Z<-2) = 0.025 which rounds up to 0.03.

Also units are missing for 5b), 6) and 9b). I believe that when VCAA makes an exam relatively easy they start withdrawing marks like that.

That is also an interesting way to solve question 9a) btw, very cool! I'd think the only way to solve that was using the tan(x+y) formula, but you've proved me wrong lol.

[Tau]

I think Question 2b should be 0.03 because Pr(Z<-2) = 0.025 which rounds up to 0.03.

Also units are missing for 5b), 6) and 9b). I believe that when VCAA makes an exam relatively easy they start withdrawing marks like that.

That is also an interesting way to solve question 9a) btw, very cool! I'd think the only way to solve that was using the tan(x+y) formula, but you've proved me wrong lol.

One can only speculate, but that’d seem excessively harsh to take off marks for units when none are mentioned in the question and they weren’t specifically asked for.

Regarding 9a, it actually kinda is the tan(x+y) approach! Using the double angle formula for tan i.e where x=y so tan(x+y)=tan(2x)

[S_R_K]

I think Question 2b should be 0.03 because Pr(Z<-2) = 0.025 which rounds up to 0.03.

Also units are missing for 5b), 6) and 9b). I believe that when VCAA makes an exam relatively easy they start withdrawing marks like that.

That is also an interesting way to solve question 9a) btw, very cool! I'd think the only way to solve that was using the tan(x+y) formula, but you've proved me wrong lol.

Thanks for your comments.

3b: To three decimal places, Pr(Z < –2) = 0.0227, hence to two decimal places it is 0.02. This is the most accurate answer. However, the question asks for an approximate probability, so in this case using Pr(–2 < Z < 2) = 0.95 is accepted, which gives 0.03 as the approximate answer to the question.

With respect to 5b, 6, 9b (and your comment also applies to 8, which asks for an arc length), there is no requirement to write "units" or "units^2" or whatever.

For 9a, I think the compound angle approach is more straightforward, because it avoids any need to mention why one of the roots to the quadratic should be rejected.