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January 03, 2026, 12:33:30 am

Author Topic: VCE Methods Question Thread!  (Read 5863651 times)  Share 

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Special At Specialist

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Re: VCE Methods Question Thread!
« Reply #1155 on: October 31, 2012, 02:40:22 pm »
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Ans is E
x -> x+2 means translation 2 units in neg x direction -> (-2)
y-> 5y+2 means dilation by 5 parallel y axis and translation 2 units in positive y direction -> x5 then +2

I still don't understand Ennjy or ClimbTooHigh. We start off with y = f(x)
Now we have y' = 5 + 2f(x + 2)
So we have:
y' = 5 + 2y
x' = x + 2
The transformed function has been translated +2 units (in regards to x). So why not option E, but with a +2 instead of a -2 on the top of the 2x1 matrix?
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d3stiny

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Re: VCE Methods Question Thread!
« Reply #1156 on: October 31, 2012, 03:12:04 pm »
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You got your ' mixed up. It should be  y' = 2 + 5f(x' + 2) from y = f(x)
So y' = 2 + 5y and x = x' + 2
Then rearrange for the matrix
ClimbTooHigh's method is a bit more logical IMO, not saying yours is wrong/bad
Also the (x+2) should alert you to a translation in the negative dir anyway.
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BubbleWrapMan

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Re: VCE Methods Question Thread!
« Reply #1157 on: October 31, 2012, 03:41:01 pm »
+1
It's hard to mix up x and x' if you write the whole transformation out. Start with the fact that



Pick the transformation you think is the answer, then apply it



Solve for the original variables




Substitute these into the original equation



And then rearrange. You should get the transformed function specified in the question, if you applied the correct transformation.

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Jenny_2108

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Re: VCE Methods Question Thread!
« Reply #1158 on: October 31, 2012, 08:12:18 pm »
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why not option E, but with a +2 instead of a -2 on the top of the 2x1 matrix?

I thought in the earlier post, ClimbTooHigh already explained everything in details?  :-\

We have and

The transformed expression can be rearranged to give

In this form we can see that and , i.e. and

so you can see x-2=x', thus in transition matrix, in x direction it will be -2

Anyway, I told you it was the shortcut I did in my head for MCQ so that I don't have to spend time re-arranging and writing everything down. If you do in short ans, you have to re-arrange like ClimbTooHigh did
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Re: VCE Methods Question Thread!
« Reply #1159 on: October 31, 2012, 08:18:22 pm »
+1
The proportion of faulty items in a large population is "p". A sample of 8 items is randomly chosen from a large population. Find the value of "p" for which the probability that a sample of 8 containing exactly 3 faulty items is maximal.
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Re: VCE Methods Question Thread!
« Reply #1160 on: October 31, 2012, 08:42:39 pm »
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Solve for p
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Re: VCE Methods Question Thread!
« Reply #1161 on: October 31, 2012, 08:56:26 pm »
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The proportion of faulty items in a large population is "p". A sample of 8 items is randomly chosen from a large population. Find the value of "p" for which the probability that a sample of 8 containing exactly 3 faulty items is maximal.
Solve for p

Or, a much simpler method is to use E(X) = np:

3 = 8p
p = 3/8
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Re: VCE Methods Question Thread!
« Reply #1162 on: October 31, 2012, 08:58:56 pm »
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owned haha
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Re: VCE Methods Question Thread!
« Reply #1163 on: October 31, 2012, 09:35:16 pm »
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The proportion of faulty items in a large population is "p". A sample of 8 items is randomly chosen from a large population. Find the value of "p" for which the probability that a sample of 8 containing exactly 3 faulty items is maximal.
Solve for p

Or, a much simpler method is to use E(X) = np:

3 = 8p
p = 3/8
um not sure how you got E(X)=3?
or did i miss something?
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availn

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Re: VCE Methods Question Thread!
« Reply #1164 on: October 31, 2012, 10:31:30 pm »
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um not sure how you got E(X)=3?
or did i miss something?

It's all in the question. They want to know what p is, when "the probability that a sample of 8 containing exactly 3 faulty items is maximal". Maximal probability is just their way of saying mean or expected value, which is 3.
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Re: VCE Methods Question Thread!
« Reply #1165 on: October 31, 2012, 10:46:44 pm »
+2
Solve for p
ClimbTooHigh's method is very nice and I'll elaborate further on the interesting aspects of it.

Basically what he did was find the "maximum likelihood" parameter for the binomial distribution.

What that means is that he has selected the parameter p which maximises the probability of the event occuring given the data set. This method of likelihood approach is very flexible and can be applied to a range of distributions. Consider the normal distribution, one can maximise jointly the probability of a certain event by maximising the probability with respect to mu and sigma, although the difference here is that you will be maximising with respect to 2 parameters, thus you will need to partial differentiate etc

One can actually show that the maximum likelihood estimator of the parameter p in a binomial distribution is always given by , this is derived as follows:

Assume we have a sample of X_i's which are identically and independently distributed binomial random variables.

The pmf is thus given by , we now construct what is known as the likelihood function which is a function of the parameter p in this case:

Now if we take logs on each other we get:



Now if we take the first order derivative with respect to p:



Set this to 0 and solve for p yields:

as required.

Furthermore, it is interesting to note that if we define X to be a bernoulli random variable, then the maximum likelihood estimator for p is just simply the sum of the bernoulli random variables, divided by the sample space, which makes intuitive sense too. Furthermore, it is easily shown that this estimator is unbiased, consistent, efficient, all in all, it is the "best" estimator for p.

Note that not every distribution has a maximum likelihood estimator for some parameters in it, and this method is quite powerful to yield unbiased and consistent estimators of parameters.

You can read more about it here: http://en.wikipedia.org/wiki/Maximum_likelihood
« Last Edit: October 31, 2012, 10:49:37 pm by TrueTears »
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TrueTears

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Re: VCE Methods Question Thread!
« Reply #1166 on: October 31, 2012, 10:52:49 pm »
+1
um not sure how you got E(X)=3?
or did i miss something?

It's all in the question. They want to know what p is, when "the probability that a sample of 8 containing exactly 3 faulty items is maximal". Maximal probability is just their way of saying mean or expected value, which is 3.
Incorrect, what you said here is by coincidence the same matter, but in general maximal does not imply expected value at all.

It is easily shown why it works here because:

As shown above, is an unbiased estimator of the mean, now

What you really did was actually used the unbiased estimator for p to arrive at the maximal probability.
« Last Edit: October 31, 2012, 10:55:19 pm by TrueTears »
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Re: VCE Methods Question Thread!
« Reply #1167 on: October 31, 2012, 11:52:02 pm »
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I really need to study probability more, shit's interesting
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Re: VCE Methods Question Thread!
« Reply #1168 on: November 01, 2012, 01:20:13 am »
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sure is, probability theory is my fave branch of maths lol
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Re: VCE Methods Question Thread!
« Reply #1169 on: November 01, 2012, 05:07:08 pm »
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TWO bowls each contain 10 pieces of fruit. In a pink bowl there are 5 oranges and 5 bananas, in the the purple bowl there is one orange and 9 bananas,


for each bowl, find the prob that two pieces of fruit chosen at random will both be bananas, if the first piece is not replaced before the second piece of fruit is chosen???


AND one bowl is chosen at random and from it one piece of fruit is chosen at a random, Find the prob. that the piece of fruit chosen is a banana?