VCE Methods Question Thread!

[SocialRhubarb]

Hey, I'll take what I can get.

[lzxnl]

Did that in a spesh SAC; I rounded something incorrectly because my mind wasn't working properly and my answer was correct, but I still dropped a mark because of that. That was really stupid. Pathetically stupid.

[Damoz.G]

I've been told the same thing as nliu1995 - that even if your answer is correct, you don't get marks if you show incorrect working. Showing no working is fine in a 1 mark question, but you lose marks for incorrect working regardless of the validity of your answer, even in 1 mark questions.

The same applies for questions worth multiple marks. If you show incorrect working and manage, fortuitously, to obtain the correct answer, you still don't get any answer marks if it appears you obtained it with an incorrect method.

Not sure about the consequential point. I think you can get method marks working from an incorrect assumption but not answer marks.

]I haven't been told that.

As far as I've been told, as long as the answer is there, the examiners don't care about what else is there in the answer.

If that's the case, then rub out all your working out. LOL

[Stevensmay]

This is from a spec paper but I believe will still apply.

There were several cases where incorrect working fortuitously led to a correct answer. Students are reminded that in
such cases, the mark for the final answer will not be awarded if the answer is not supported by relevant and correct
working.

From a methods paper.

As stated in the instructions, students must show appropriate working for questions worth more than one mark

In this case I imagine appropriate = correct and relevant.

[zhe0001]

An oil slick is found to radiate outwards at a rate modelled by,
t≥1
where t is the time measured in hours and r is the radius of the slick in metres. If the slick is 16 metres wide after one hour, how long, to the nearest hour, will it take to be 100 metres wide?

I keep getting t=88.36 h, but answer in book is 27.

Please help

[Phy124]

:

[clıppy]

Can I get some help on this one?

The heights of boys in year 12 are normally distributed with a mean of 175cm and a standard deviation of 10cm. it is known that Pr(Z>0.5) = 0.3
Using the information above, find the probability that a boy selected at random is greater than 170cm

[b^3]

HINT: , transform your distribution to a standard normal distribution (i.e. the one), work out the equivalent probability using the first equation. Then if needed draw out a small sketch of the normal distribution curve, and look at how you can use symmetry to find corresponding areas.

symmetry

i.e. for the symmetry part, if you know that , then , so

[clıppy]

HINT: , transform your distribution to a standard normal distribution (i.e. the one), work out the equivalent probability using the first equation. Then if needed draw out a small sketch of the normal distribution curve, and look at how you can use symmetry to find corresponding areas.

symmetry

i.e. for the symmetry part, if you know that , then , so

Ahhh thanks b.
I just couldn't make that connection with the symmetry of the distribution

[itsdanny]

Hello,

When having to evaluate a definite integral, are you allowed in VCE to have infinity at either of the terminals and evaluate it from there? Or must you define a limit before being being able to evaluate it? 

For example, when having to find an unknown variable for a probability density function, by utilising the property that the area under the density function must be 1, if the domain of the PDF is from a 5 to infinity, can you have infinity as one of the terminals? Or logically must you define a limit before doing so?

[Hancock]

In reality, you should put a limit out of the front of the integral, and change the terminal to the variable of the limit, but in VCE, I'm fairly sure you can keep the infinity in the terminals.

[Damoz.G]

Hello,

When having to evaluate a definite integral, are you allowed in VCE to have infinity at either of the terminals and evaluate it from there? Or must you define a limit before being being able to evaluate it? 

For example, when having to find an unknown variable for a probability density function, by utilising the property that the area under the density function must be 1, if the domain of the PDF is from a 5 to infinity, can you have infinity as one of the terminals? Or logically must you define a limit before doing so?

No, you would need to apply a limit at the start of an integral instead of doing that.

[BubbleWrapMan]

No, you would need to apply a limit at the start of an integral instead of doing that.

You wouldn't have to be as rigorous as that in methods. Not that using limit notation is that rigorous, but that's sort of the point I guess - simply slapping the limit sign in front of it doesn't really change what you're doing. Whether you use limit notation or abbreviate it with terminals, either can be done without really knowing what a limit is. So it would be pretty stupid of VCAA to prefer one over the other.

I don't think VCAA has actually ever required limit notation in any of their answers. Take for example question 2b from 2011 exam 2 (in particular the solutions in the assessment report). This is the only instance where it might have been appropriate (that I can remember), but they use the abbreviated from.

[clıppy]

Another probability question for those that can help me out. I usually get confused on finding compliment intersections when independent events are involved.

Consider the independent events A and B where Pr(A|B) = 3/4 and Pr(B) = Pr(A' n B'). Find Pr(B)

[lzxnl]

Another probability question for those that can help me out. I usually get confused on finding compliment intersections when independent events are involved.

Consider the independent events A and B where Pr(A|B) = 3/4 and Pr(B) = Pr(A' n B'). Find Pr(B)

Pr(A' n B')
Means what is both not in A and not in B. Everything not in this set is either in A or B. The complement of this, the union of A and B, is thus given by 1-Pr(B)
Independent events => Pr(A|B)=Pr(A). Pr(A U B) = 1-Pr(B) from above
Which is equal to Pr(A)+Pr(B)-Pr(A intersect B)=1-Pr(B)
Pr(A intersect B)=Pr(A)Pr(B) due to independence.
Pr(A)+Pr(B)-Pr(A)*Pr(B) = 1- Pr(B)
2Pr(B)-Pr(A)*Pr(B)=1-Pr(A)
Pr(B)(2-Pr(A))=1-Pr(A)
Now using Pr(A)=3/4
Pr(B)(2-3/4)=1/4
Pr(B)=1/5