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March 29, 2024, 06:33:47 am

Author Topic: VCE Methods Question Thread!  (Read 4802688 times)  Share 

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matthewzz

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Re: VCE Methods Question Thread!
« Reply #18105 on: August 31, 2019, 08:29:34 pm »
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The normal distribution is a continuous probability distribution. Properly, it has a Probability Density Function of \({\displaystyle {\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}}\). This means that to calculate probabilities of a random variable that is normally distributed, you integrate \({\displaystyle {\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}}\) with appropriate bounds. Here’s where your CAS comes in: there is no (elementary) integral for the normal distribution, \({\displaystyle {\frac {1}{\sqrt {2\pi \sigma ^{2}}}}e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}}\), so your CAS has a function that numerically approximates this integral.

This function is called \(\mathtt{normCDF}\) - for the cumulative distribution function of the normal distribution. A cumulative distribution function gives you the probability that a random variable takes on values less than equal to a given number. Graphically, using \(\mathtt{normCDF}\) with a lower bound of, say \(a\), would correspond to the area of the normal distribution to the left of the point \(a\). Note that is not quite the 'area to the left of a probability’.

As an example, if \(X\sim\mathcal{N}(2,2^2)\) then \(\Pr(X<0.5)=\int_{-\infty}^{0.5}{\displaystyle {\frac {1}{\sqrt {2\pi \cdot 2^{2}}}}e^{-{\frac {(x-2)^{2}}{2\cdot2^{2}}}}} \, dx \approx 0.2266\). So you would specify a lower bound of negative infinity and an upper bound of 0.5 to the \(\mathtt{normCDF}\) function.

Great! Thanks so much!

milanander

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Re: VCE Methods Question Thread!
« Reply #18106 on: September 03, 2019, 09:21:30 am »
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Having a bit of a trouble with question 9 (c) of the 2017 VCAA exam (tech-free):

My understanding of the question is that we must find point B (which I found to be (1,0), I guess the diagram is a bit off?), then find the tangent to the graph at (1,0). Except if I did that then I would be calculating the tangent at its endpoint which isn't allowed. So B can't be at (1,0) because it can't be at its endpoint except there is no other coordinate B can be?

Plz help!
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colline

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Re: VCE Methods Question Thread!
« Reply #18107 on: September 03, 2019, 03:19:32 pm »
+1
Having a bit of a trouble with question 9 (c) of the 2017 VCAA exam (tech-free):

My understanding of the question is that we must find point B (which I found to be (1,0), I guess the diagram is a bit off?), then find the tangent to the graph at (1,0). Except if I did that then I would be calculating the tangent at its endpoint which isn't allowed. So B can't be at (1,0) because it can't be at its endpoint except there is no other coordinate B can be?

Plz help!

I am almost certain the writers of the exam made a mistake there. The question doesn’t seem right 🤔 Point B has to be the endpoint which doesn’t work.

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DrDusk

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Re: VCE Methods Question Thread!
« Reply #18108 on: September 03, 2019, 03:48:52 pm »
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Having a bit of a trouble with question 9 (c) of the 2017 VCAA exam (tech-free):

My understanding of the question is that we must find point B (which I found to be (1,0), I guess the diagram is a bit off?), then find the tangent to the graph at (1,0). Except if I did that then I would be calculating the tangent at its endpoint which isn't allowed. So B can't be at (1,0) because it can't be at its endpoint except there is no other coordinate B can be?

Plz help!

Oh wait I just took the time to look at it. Yeah that is a bit dodgy.
« Last Edit: September 03, 2019, 05:51:47 pm by DrDusk »

colline

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Re: VCE Methods Question Thread!
« Reply #18109 on: September 03, 2019, 07:42:04 pm »
+1
RE: 2017 Exam 1 Question 9 c

Have a look at the 2017 methods post-exam discussion thread on ATAR Notes. By the looks of it, that question threw most people off.

I had a look at the Examiners' Reports and it seems that VCAA just decided to work it out using a  ""hypothetical situation""  where you can have a tangent at an endpoint. Would've been nice for them to specify that in the question though. :-/
« Last Edit: September 03, 2019, 07:44:20 pm by colline »

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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18110 on: September 03, 2019, 09:07:58 pm »
+2
My understanding of the question is that we must find point B (which I found to be (1,0), I guess the diagram is a bit off?).

The diagram labels the angle as \(\theta\). It doesn't need to represent \(45^\circ\).

Except if I did that then I would be calculating the tangent at its endpoint which isn't allowed. So B can't be at (1,0) because it can't be at its endpoint except there is no other coordinate B can be?
I am almost certain the writers of the exam made a mistake there. The question doesn’t seem right 🤔 Point B has to be the endpoint which doesn’t work.

While the tangent (first-order approximation) to the graph of \(f\) at \(x=1\) is not totally defined, the question only ever refers to the line through \(BC\) being tangential to the graph, which, in the context of the question, clearly uses a well-defined left derivative.

The question is perhaps a little unfair though. I believe they should've just extended the domain of \(f\) just a bit further to the right to avoid this confusion though.
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milanander

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Re: VCE Methods Question Thread!
« Reply #18111 on: September 04, 2019, 06:19:12 pm »
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The diagram labels the angle as \(\theta\). It doesn't need to represent \(45^\circ\).

While the tangent (first-order approximation) to the graph of \(f\) at \(x=1\) is not totally defined, the question only ever refers to the line through \(BC\) being tangential to the graph, which, in the context of the question, clearly uses a well-defined left derivative.

The question is perhaps a little unfair though. I believe they should've just extended the domain of \(f\) just a bit further to the right to avoid this confusion though.

Sorry but I'm still a bit confused. Does that mean that with the current domain, we have to assume that it is possible to have a tangent at point B because its coordinates weren't explicitly defined in the question? (hope the question makes sense, thanks!)
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DrDusk

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Re: VCE Methods Question Thread!
« Reply #18112 on: September 04, 2019, 06:56:26 pm »
+1
Oh what, I was confused about something completely different. I found it weird how the line is actually tangent at B but the image shows it as tangent to another point lol.

JR_StudyEd

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Re: VCE Methods Question Thread!
« Reply #18113 on: September 08, 2019, 05:15:36 pm »
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What's the difference between Binomial pdf and Binomial cdf, and in what situations would I use these functions?
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RuiAce

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Re: VCE Methods Question Thread!
« Reply #18114 on: September 08, 2019, 05:24:19 pm »
+1
What's the difference between Binomial pdf and Binomial cdf, and in what situations would I use these functions?
The cdf gives you \( P(X\leq x)\). Notice how there's a less than or equal to.

Whereas the pdf gives you \( P(X=x)\). Here we just have 'equal to'.
« Last Edit: September 08, 2019, 05:25:57 pm by RuiAce »

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #18115 on: September 08, 2019, 08:00:15 pm »
+3
What's the difference between Binomial pdf and Binomial cdf, and in what situations would I use these functions?

Let \(X\sim \text{Bi}(n,p)\).

If you want to calculate \(\Pr(X=a)\), you could do so using \(\texttt{binomPdf}\) with the following inputs: \begin{align*}\texttt{Num Trials, n:}&\ \ n\\ \texttt{Prob Success, p:}&\ \ p\\ \texttt{X Value:}&\ \ a\end{align*}
If you want to calculate \(\Pr(a\leq X\leq b)\), you could do so using \(\texttt{binomCdf}\) with the following inputs: \begin{align*}\texttt{Num Trials, n:}&\ \ n\\ \texttt{Prob Success, p:}&\ \ p\\ \texttt{Lower Bound:}&\ \ a\\ \texttt{Upper Bound:}&\ \ b\end{align*}
« Last Edit: September 09, 2019, 08:54:48 am by AlphaZero »
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ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #18116 on: September 12, 2019, 04:35:54 pm »
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Hey! I just had a question related to probability.

So lets say that E(X) (the mean) is 2 for a function, and The variance, is 0.84.

Then, to find E(2/3X), I simply times 2 by 2/3, which will give 4/3.

So, in order to find the variance that corresponds with E(2/3X), do I just simply multiply 0.84 by 2/3?

Sorry if this question doesnt make sense!

Would really appreciate any help though :)

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Re: VCE Methods Question Thread!
« Reply #18117 on: September 12, 2019, 04:50:31 pm »
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Hey! I just had a question related to probability.

So lets say that E(X) (the mean) is 2 for a function, and The variance, is 0.84.

Then, to find E(2/3X), I simply times 2 by 2/3, which will give 4/3.

So, in order to find the variance that corresponds with E(2/3X), do I just simply multiply 0.84 by 2/3?

Sorry if this question doesnt make sense!

Would really appreciate any help though :)
The formulas are different. Your first answer is correct, because \(\operatorname{E}(aX)=a\operatorname{E}(X)\).

But for the variance, we instead have \( \boxed{\operatorname{Var}(aX) = a^2\operatorname{Var}(X)} \). You need to multiply by the square of that extra constant factor. (In your case, it would be multiplication by 4/9 instead of 2/3.)

ArtyDreams

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Re: VCE Methods Question Thread!
« Reply #18118 on: September 12, 2019, 04:56:29 pm »
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The formulas are different. Your first answer is correct, because \(\operatorname{E}(aX)=a\operatorname{E}(X)\).

But for the variance, we instead have \( \boxed{\operatorname{Var}(aX) = a^2\operatorname{Var}(X)} \). You need to multiply by the square of that extra constant factor. (In your case, it would be multiplication by 4/9 instead of 2/3.)

Right I understand now. Thank you so much!!!

persistent_insomniac

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Re: VCE Methods Question Thread!
« Reply #18119 on: September 12, 2019, 07:30:39 pm »
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When we are approximating the distribution of the sample proportion to find probabilities, do we just assume that p hat = x/n is the same as the probability of success of a particular event?
e.g.) "Find the approx probability that in the next 50 tosses of a fair coin, the proportion of heads observed will be less than or equal to 0.46).
So to find the mean and sd of p hat by using the formula, do we assume pr(sucess) = 1/2 is the same as x/n??