was last year's methods exam 1 considered easy?

[sailinginwater]

Was last year's methods exam 1 considered easy?
Since a lot of people said it was easy but there were a lot of questions that showed up that were completely new and didn't show up in any of the 2006-2016 methods exam 1's
Opinions on this?

[Sine]

Was last year's methods exam 1 considered easy?
Since a lot of people said it was easy but there were a lot of questions that showed up that were completely new and didn't show up in any of the 2006-2016 methods exam 1's
Opinions on this?

I thought it was fine but evidently from the grade distribution the state found it quite hard. An A+ was 34.5/40 which is lower than the usual A+ which is around 36-37/40.

However, 2016 the A+ was 34/40 (usually considered the hardest year) so last year was easier than the previous year.

[S_R_K]

Difficulty is subjective.

The grade distribution was similar to 2016's, which was found more difficult than previous years'. Going by the examiner's report, some questions were well answered, others were poorly answered – pretty typical.

There was really only one "dodgy" question on the 2017 exam, the last one, where you asked to find a tangent which turns out to be at the endpoint of a graph – normally methods students would regard this as undefined. Unsurprisingly, the average mark on the last part of this question was 5%. No doubt that contributed a lot to the relatively low A+ cutoff.

[sailinginwater]

I thought it was fine but evidently from the grade distribution the state found it quite hard. An A+ was 34.5/40 which is lower than the usual A+ which is around 36-37/40.
Did you reckon it was harder than the 2006-2016 methods exam 1's since there were a lot of new questions?

However, 2016 the A+ was 34/40 (usually considered the hardest year) so last year was easier than the previous year.

[S_R_K]

What are the "new" questions you have in mind? Obviously there was some new content on the study design from 2016 (ie. sampling proportions and statistical inference), and last year was the first time it appeared on Exam 1...

[sailinginwater]

What are the "new" questions you have in mind? Obviously there was some new content on the study design from 2016 (ie. sampling proportions and statistical inference), and last year was the first time it appeared on Exam 1...

Like you mentioned, the last question, and question 5c, 6ab, 7, 8c
These didn't show up in any of the exam 1's from 2006-2016

[Lear]

If they pulled out the same questions or same types of questions everyone would be smashing it.

[S_R_K]

Like you mentioned, the last question, and question 5c, 6ab, 7, 8c
These didn't show up in any of the exam 1's from 2006-2016

Well, I don't remember all of the exams off the top of my head, but I don't think any of those questions assessed anything that is beyond a well prepared student, or that is beyond the study design. I'd also be very surprised if answering any of those questions required a technique / concept that hadn't been assessed in previous exams.

There's probably little point in explaining each question. I think the general thing to say is that you shouldn't approach exams by thinking that the only thing VCAA will ask of you are slightly reworded versions of questions asked in previous years (perhaps with different numbers). VCAA will throw curveballs to check whether students have a good enough understanding to apply concepts / skills to unfamiliar questions.

If you found some of those questions very unfamiliar, then perhaps you should be talking to your teacher / fellow students about getting practice with answering unfamiliar questions and getting experience with thinking laterally to find the right technique.

[sailinginwater]

Thanks guys for the opinions

I did question 5c on the exam and did P(FSF) + P(FFS)

= (3/5)*(2/5)*(3/5)+(3/5)*(3/5)*(2/5)

= 36/125

But the answers did P(FS) + P(FFS) and they got 48/125

But how come they did P(FS) when the question clearly states that a maximum of three attempts can be made? Shouldn’t it be P(FSF)?
But how can we assume it when the question doesn't actually say that?

[S_R_K]

Thanks guys for the opinions

I did question 5c on the exam and did P(FSF) + P(FFS)

= (3/5)*(2/5)*(3/5)+(3/5)*(3/5)*(2/5)

= 36/125

But the answers did P(FS) + P(FFS) and they got 48/125

But how come they did P(FS) when the question clearly states that a maximum of three attempts can be made? Shouldn’t it be P(FSF)?

You have to think about the context here. Once he successfully logs on, he stops attempting to log on.

[sailinginwater]

But the question never says that?
So how can we assume

[S200]

But the question never says that?
So how can we assume

Well I wouldn't try to log on if I was already logged on. Would you?

Once an action succeeds, there isn't really any point going back to the start and trying again is there?

[sailinginwater]

Using the above logic I worked out part a as follows
a. (3/5)^1  = 3/5
Since if he doesn't log in correctly on the first attempt then he would not try to log in again?
part b
I did (2/5)^1 = 2/5
Since if he logs in correctly on the first attempt then he wouldnt attempt to log in again
But I got both part a and b wrong even though I followed the logic that was mentioned in part c
Can someone please explain how I got part a and b wrong even though I used the same logic as part c?
I hate probability
I'm feeling so overwjelmed, is there actually any way to still get 30-35 raw in methods when there's only one month before the exam and I'm struggling on basic exam 1 questions?

[S200]

Using the above logic I worked out part a as follows
a. (3/5)^1  = 3/5
Since if he doesn't log in correctly on the first attempt then he would not try to log in again?

Not quite. This question is asking for P(0). i.e: (FFF)

part b
I did (2/5)^1 = 2/5
Since if he logs in correctly on the first attempt then he wouldnt attempt to log in again

This question asks you to find the probability of any of the three being a success. This is most easily found by going 1-P(0)

I hate probability
I'm feeling so overwhelmed, is there actually any way to still get 30-35 raw in methods when there's only one month before the exam and I'm struggling on basic exam 1 questions?

And yeah. Probability is my Achilles as well. bloody sucks.

[sailinginwater]

Part a is just P(0),

Part b is 1-P(0) because if he logs in on any attempt, then yeah, he's in.
Can you please explain why part a is p(0) and not (3/5)^1 since if is not successful on the first attempt then he would not try again?
And why isn't part b just (2/5)^1=2/5?
These questions are really doing my head in

And yeah. Probability is my Achilles as well. bloody sucks.