Hey!
Can you double check the methods exam 1 question 2 b solution.
I am getting pi/2.
I think solution missed out, the -1/sqrt(2) part of answer. Just double check pls.
cos(x+pi/4) = +- 1/sqrt(2)
x+pi/4 =pi/4
x=0
x+pi/4 =3pi/4
x=pi/2 --- first positive value of x
x+pi/4 =5pi/4
x=pi ------ 2nd solution
x=3pi/2 -- 3rd solution not first !! :)
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double check question 4 solutions as well pls.
"In a large forest, one quarter of all trees are infected with cinnamon fungus."
p=1/4, not 1/3 :)
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q7 solution needs to revised. ;) :) as well
Thanks so much - have passed this on for review. :)There were a couple of errors in exam 2 solutions as well.
Question 2
The function \(f\) cannot possibly have the domain \(\mathbb{R}\). For example, \(f(\pi/4)\) is undefined. In Q2b, the solution neglects the possibility \[\cos\left(x+\frac{\pi}{4}\right)=\frac{-1}{\sqrt{2}}.\] The correct answer is \[x_\text{min}=\frac{\pi}{2}.\]
Question 3
In the solution to Q3a, the notes section writes that the endpoint is \((-4,\ 5/4)\), which it isn't. The correct endpoint is \((-4,\ 9/8)\).
Question 4
The question gives \(p=1/4\) yet the solutions take \(p=1/3\) throughout the entire question. So, all the solutions are wrong for this question. Putting this to the side, the notation for the normal distribution is misused. For a normally distributed random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), the correct notation is \[X\sim\mathcal{N}(\mu,\ \sigma^2).\] Notice that the second parameter gives the variance, not the standard deviation. For Q4d, the correct solution yields \(\Pr(\hat{P}\geq 1/3)=\Pr(Z\geq 1)\), which is one of the only ways to give an exact answer without specifying an integral.
Question 5
Infinity is not a 'value' and I think it's a bad idea to treat it as such. In Q5b.i, just ask for the largest subset of \(\mathbb{R}_{\geq 0}\) for which \(h\) is defined. The wording in Q5b.ii can then be fixed by asking for the implied range of \(h\).
Question 6
Nothing wrong with the question other than the solution to Q6a is a bit overkill. The events \(B\) and \(A'\) are also independent, so \[\Pr(B\mid A')=\Pr(B)=p.\]
Question 7
Solution of Q7a has an arithmetic error. Although the method is correct, it's a bit overkill. Where \(R\) denotes 'rejected', we have \[\Pr(R)=1-\Pr(R')=1-\frac12\times \frac35\times \frac56=\frac34.\] If we denote \(R_i\) to mean 'rejected by test \(i\)', then the question asks for \(\Pr(R'\mid R_1')\), not \(\Pr(R_1'\mid R')\). Even if the question did ask for the latter, the solution is still incorrect since a toaster necessarily needs to pass the first test in order to not be rejected. \[\Pr(R'\mid R_1')=\frac{\Pr(R'\cap R_1')}{\Pr(R_1')}=\frac{\Pr(R')}{\Pr(R_1')}=\frac{1/4}{1/2}=\frac12.\]
Question 8
Solution for Q8b makes an algebraic slip despite giving the correct answer. It should have \(-54\), not \(+54\).
Question 9
The wording of Q9b is problematic. The mean value of a function is not the same as the mean of a random variable. The former means average value: \[\overline{f}=\frac{1}{1-0}\int_0^1f(x)\,\text{d}x,\] while the latter means \[\mu=\int_0^1x\,f(x)\,\text{d}x.\] So, one needs to define a random variable (\(X\), say) somewhere in the question. The solution for Q9b is also problematic since the second last line lacks the correct bracketing. The \(x\) needs to multiply the whole function, not just the \(\sin()\) term. That is, \[\mu=\int_0^1 x\big(\!\sin(2\pi x+k)+1\big)\,\text{d}x=\int_0^1 x\sin(2\pi x+k)\,\text{d}x+\int_0^1x\,\text{d}x.\] Solution to Q9c is also really overkill. Maximum \(\mu\) occurs when \(\cos(k)=-1\). Ie. \[\mu_\text{max}=\frac{\pi+1}{2\pi},\quad k=\pi.\]No calculus is required, so the marking scheme needs to be revised.
Are these exams adapted to the changed VCE curriculum?