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March 19, 2024, 02:06:16 pm

Author Topic: VCE Methods Question Thread!  (Read 4795772 times)  Share 

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Sine

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Re: VCE Methods Question Thread!
« Reply #15165 on: August 18, 2017, 08:45:00 pm »
+1
Hi

For part c, calculating the variance, here http://imgur.com/a/Rx2WE

Why is my method wrong? The correct answer is 400 but the answer made it in terms of the random variable X before evaluating
You are fine most likely just the answer is wrong in the textbook.

VanillaRice

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Re: VCE Methods Question Thread!
« Reply #15166 on: August 18, 2017, 09:05:44 pm »
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Hi

For part c, calculating the variance, here http://imgur.com/a/Rx2WE

Why is my method wrong? The correct answer is 400 but the answer made it in terms of the random variable X before evaluating
The statement Var(X + Y) = Var(X) + Var(Y) is only true if X and Y are independent. In this question, Y = 2X + 1 and U = 10 - 3X. Since both Y and U are dependent of the same variable (X), they are not independent.

That is why you get a different answer when you substitute Y and U to get an expression in terms of X.

Hope this helps :)
« Last Edit: August 18, 2017, 09:10:23 pm by VanillaRice »
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Sine

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Re: VCE Methods Question Thread!
« Reply #15167 on: August 18, 2017, 09:39:56 pm »
+1
The statement Var(X + Y) = Var(X) + Var(Y) is only true if X and Y are independent. In this question, Y = 2X + 1 and U = 10 - 3X. Since both Y and U are dependent of the same variable (X), they are not independent.

That is why you get a different answer when you substitute Y and U to get an expression in terms of X.

Hope this helps :)
my bad this is why ^

Var(V) = Var (Y + 2U) = Var(2X+1 + 2(10-3X)) = Var(2X+1+20-6X) = Var(21-4X) = 16Var(X) = 16 * 25 = 400

Basically this is why you don't go Var(2X + 2X ) = 4Var(X) + 4Var(X) = 8Var(X)
but you go Var(4X) = 16Var(X)

clarke54321

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Re: VCE Methods Question Thread!
« Reply #15168 on: August 19, 2017, 01:50:33 pm »
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Hey!

Is this type of question still applicable to the new study design? In the examiner's report, transition matrices are used. Is there another method I can use?

Thanks  :)
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Sine

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Re: VCE Methods Question Thread!
« Reply #15169 on: August 19, 2017, 01:58:38 pm »
+4
Hey!

Is this type of question still applicable to the new study design? In the examiner's report, transition matrices are used. Is there another method I can use?

Thanks  :)
Yeah I think last year I just skipped that question LOL. Well you could probably do it without transition matrices - maybe something like conditional probability because that is basically what a transition matrix includes. Would probably take too long so I don't think it is necessary to do.

alternatively you can learn transition matrices for FUN  ;)
« Last Edit: August 19, 2017, 02:00:20 pm by Sine »

Syndicate

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Re: VCE Methods Question Thread!
« Reply #15170 on: August 19, 2017, 02:01:52 pm »
0
Hey!

Is this type of question still applicable to the new study design? In the examiner's report, transition matrices are used. Is there another method I can use?

Thanks  :)

which year is this from?
2017: Chemistry | Physics | English | Specialist Mathematics | Mathematics Methods
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Sine

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Re: VCE Methods Question Thread!
« Reply #15171 on: August 19, 2017, 02:03:23 pm »
+3
which year is this from?

It is 2009 exam 2 (I haven't re-re-read the question yet LOL)

clarke54321

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Re: VCE Methods Question Thread!
« Reply #15172 on: August 19, 2017, 02:28:39 pm »
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Yeah I think last year I just skipped that question LOL. Well you could probably do it without transition matrices - maybe something like conditional probability because that is basically what a transition matrix includes. Would probably take too long so I don't think it is necessary to do.

alternatively you can learn transition matrices for FUN  ;)

No worries, thanks for the advice! Haha I did transitional matrices for Further last year, so could probably work it out :)
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Lavar Big BBB Balls

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Re: VCE Methods Question Thread!
« Reply #15173 on: August 19, 2017, 06:43:33 pm »
0
http://imgur.com/a/TFxmR

For part c, just to confirm that the answer is saying that the claim is questionable because the percentage of a battery lasting over 100 hours is over 90%, not exactly 90%?


Sine

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Re: VCE Methods Question Thread!
« Reply #15174 on: August 19, 2017, 07:01:00 pm »
+5
http://imgur.com/a/TFxmR
For part c, just to confirm that the answer is saying that the claim is questionable because the percentage of a battery lasting over 100 hours is over 90%, not exactly 90%?
yes it causes doubt and it is quite questionable because the probability of obtaining a sample where p^ = 0.848 or less is extremely small. Question b) really just shows how unlikely you are to obtain a p^ of 0.848 or "worse" given the manufacturers claims.

Of course they could be saying the truth but got extremely unlucky.

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Re: VCE Methods Question Thread!
« Reply #15175 on: August 19, 2017, 11:06:57 pm »
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Thanks brethren!

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Re: VCE Methods Question Thread!
« Reply #15176 on: August 20, 2017, 11:38:40 am »
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Hey guys, is there any way to derive something in the form of 1/(a-bx^2) by hand? I've tried putting it into my CAS and i get an inverse tan with negative square roots so this seems unlikely.

Thanks in advance!!
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Sine

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Re: VCE Methods Question Thread!
« Reply #15177 on: August 20, 2017, 12:26:12 pm »
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Hey guys, is there any way to derive something in the form of 1/(a-bx^2) by hand? I've tried putting it into my CAS and i get an inverse tan with negative square roots so this seems unlikely.

Thanks in advance!!






This is what the derivitive looks like. When integrating you will get something with inverse tans and square roots.

TheCommando

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Re: VCE Methods Question Thread!
« Reply #15178 on: August 20, 2017, 01:34:07 pm »
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This is what the derivitive looks like. When integrating you will get something with inverse tans and square roots.
For the third step, why does tthat power of negative one go to the front and the power of negative 2 appears

Sine

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Re: VCE Methods Question Thread!
« Reply #15179 on: August 20, 2017, 01:38:46 pm »
+5
For the third step, why does tthat power of negative one go to the front and the power of negative 2 appears
It is using the rule

**Bring the power to the front and reduce the power by 1**
However the inside of the function is not x so I must also take that into account by using the chain rule hence why I multiply by the derivitive of the inside.
Remember the chain rule is basically derivitive of outside multiplied by the derivitive of the inside.