VCE Methods Question Thread!

[Hutchoo]

basically variance is the average squared deviation from the mean. it's just a random number that tells someone how much a particular data set varies. so by definition, Var(X) = E((X-u)^2), where u is the mean. we can use this to derive the other formula:

I know I'm being pretty pedantic here, for vce what you have said is perfectly fine, however I just wanted to clarify the use of the word "average" that you used, it is very important to distinguish between an average and an expected value, in general, they are not the same. In fact what you should say is the "variance is the expected value of squared deviations from the mean." In general "average" is not the same as "expected value". If you used the word "average" then your sentence should be "the sample variance is the average squared deviation from the sample mean". Although this doesn't mean much in VCE, but the implications are huge for beyond VCE purposes (which I think you have the mathematical maturity understand ). There is a huge difference between sample variance and actual variance (that is, the definition of variance from ). That is, whenever we talk about averages we are always talking about the sample variances, sample means, sample standard deviations, sample medians etc. You may ask, what is a "sample"? What is a "population"?

I will try show you the difference.

Let S denote the population set, that is S could include the income of every household on the planet earth, or the age of every dog in australia and so on.

A sample (or more formally, a random sample http://en.wikipedia.org/wiki/Random_sample) is where you take a subset of S, ie, the income of every household in Australia, or the age of every dog in victoria etc.

Now the variance, denoted by implies that we know the actual POPULATION variance, however in reality, we will NEVER know the population variance, it is impossible to survey every household in the world to ask for their income. Thus we often "estimate" the population variance with the sample variance (the correct term here to use is that the sample variance is an ESTIMATOR of the population variance). Thus the best approximation we can arrive at to approximate is from (where s^2 is the conventional notation for sample variance).

Now note the difference

BUT where is the SAMPLE mean (different from the POPULATION mean denoted by ) and n is the sample size

In fact can be shown to be an "biased" estimator of the population variance, thus to correct for this bias we often divide by n-1. So is the sample variance.

TT, simply a beast (I now fully understand by what you meant last time)

[Bhootnike]

hey guys
had some questions:

1. Given that x = 3 is an approximate solution to the equation , use
to find a better apprximation of the solution.

2.
I've forgotten the short cuts, so was wondering what was the quick way to differentiate

3. Find the values of x for which increases as x increases, and hence find the max value for

4.

a) find min, value of x.  -- all goo d here. x =0 ,y =0
b) hence show that for all real x.

thanks!

[brightsky]

1. f(3) = 27 - 27 + 3 - 2 = 1. take this as your new old 'y-value'. construct a tangent to the curve at x = 3. let h = some small negative number and approximate f(3 + h) using the tangent. this approximate solution should be more accurate than x = 3.
2. product and chain rule. 2*d/dt[(t+1)(t-5)^2] =1* (t-5)^2+(t+1)*2(t-5) = (t-5)^2 + 2(t+1)(t-5)
3. let f(x) = x^2 + 2x -5. the original function is increasing when f'(x) > 0. 2x + 2 > 0, x > -1. there is no max value because the original function just gets larger as x increases.
4. well min value occurs when x = 0, y = 0, meaning that the min vertical distance between the two functions is 0 (they touch at x = 0). hence the inequality holds.

[generalkorn12]

For absolute value functions, I'm still having trouble understanding when, we split them into its hybrid functions, how do we know when to edit the domains of each 'sub function' and when to not edit them?

[paulsterio]

In the exam, can I just use Var(X)= E(X^2) - [E(X)]^2 without expanding as above?

Yes, I have never had to use any other form of variance in my life

[TrueTears]

A multiple-choice test consists of eight questions and three answers to each question (of which only one is correct). If a student answers each question by rolling a balanced die and checking the first answer if he gets 1 or 2, the second answer if he gets 3 or 4, and the third answer if he gets 5 or 6, what is the probability that he will get exactly four correct answers?

Extension: He picks answer 1 if he rolls a 1 or 2. Picks answer 2 if he rolls 3 or 4 or 5. Picks answer 3 if he rolls a 6.

[brightsky]

1) i don't think the die actually affects the probability at all, since you still have 1/3 chance of choosing each answer. thus answer is ncr(8,4)*(1/3)^(4)*(2/3)^4 = 1120/6561 = 0.1707

2) now you have 1/3 chance of choosing first answer, 1/2 chance of choosing second, and 1/6 chance of choosing third. each answer has 1/3 chance of being right, and 2/3 chance of being wrong. so the probability that you will get a right answer is 1/3*1/3 + 1/2*1/3 + 1/6*1/3 = 1/3. so answer still appears to be 0.1707..?!

[TrueTears]

1) i don't think the die actually affects the probability at all, since you still have 1/3 chance of choosing each answer. thus answer is ncr(8,4)*(1/3)^(4)*(2/3)^4 = 1120/6561 = 0.1707

2) now you have 1/3 chance of choosing first answer, 1/2 chance of choosing second, and 1/6 chance of choosing third. each answer has 1/3 chance of being right, and 2/3 chance of being wrong. so the probability that you will get a right answer is 1/3*1/3 + 1/2*1/3 + 1/6*1/3 = 1/3. so answer still appears to be 0.1707..?!

correct!

Now can you prove in general that rolling the die has completely no influence on the probability of picking the correct answer?

[poorengish]

I need some help
Question 22 MC from a kilbaha practice exam 2007
It says an Ipod contains 250 songs. There are 100 by artist A, 90 by artist B and 60 by artist C. Three different songs are played in random order. The probability that the three songs are different is closest to...
A) 0.1037
B)0.2099
C)0.2602
D)0.03456
E)0.03498
I get 6*100/250*90/250*60/250 as there are 6 combinations when you can get three different artist, and the probability (I'm assuming) is with replacement.
My answer is 0.20736 which is close to B, but I still don't think that it is correct.
Can anybody work this problem out?

[TrueTears]

I need some help
Question 22 MC from a kilbaha practice exam 2007
It says an Ipod contains 250 songs. There are 100 by artist A, 90 by artist B and 60 by artist C. Three different songs are played in random order. The probability that the three songs are different is closest to...
A) 0.1037
B)0.2099
C)0.2602
D)0.03456
E)0.03498
I get 6*100/250*90/250*60/250 as there are 6 combinations when you can get three different artist, and the probability (I'm assuming) is with replacement.
My answer is 0.20736 which is close to B, but I still don't think that it is correct.
Can anybody work this problem out?

Hence B

[brightsky]

1) i don't think the die actually affects the probability at all, since you still have 1/3 chance of choosing each answer. thus answer is ncr(8,4)*(1/3)^(4)*(2/3)^4 = 1120/6561 = 0.1707

2) now you have 1/3 chance of choosing first answer, 1/2 chance of choosing second, and 1/6 chance of choosing third. each answer has 1/3 chance of being right, and 2/3 chance of being wrong. so the probability that you will get a right answer is 1/3*1/3 + 1/2*1/3 + 1/6*1/3 = 1/3. so answer still appears to be 0.1707..?!

correct!

Now can you prove in general that rolling the die has completely no influence on the probability of picking the correct answer?

Quite crude but anyhows...

Let a be the probability of choosing first answer, b be the probability of choosing second answer. So probability of choosing third is 1 - a - b. each answer has 1/3 chance of being correct. so probability of getting an answer correct is 1/3(a + b + 1-a-b) = 1/3, which is the same as if you randomly chose an answer. Thus no influence.

[HERculina]

Hey guys, need help with this question:
Find the values of q, where q is a real number for which the eqtn qx^4-5000qx^2 + 90 = 0 has more than one solution for x.
It's part bii (of q.3) of an extended response question from kilbaha exam 2 2007 but not sure if the previous details had any relevance to this part.

[Jenny_2108]

Hey guys, need help with this question:
Find the values of q, where q is a real number for which the eqtn qx^4-5000qx^2 + 90 = 0 has more than one solution for x.
It's part bii (of q.3) of an extended response question from kilbaha exam 2 2007 but not sure if the previous details had any relevance to this part.

let x^2=u then use the quadratic formula

You already did trial exam I havent done any 

[HERculina]

^ ohhhh why didn't I think of that! Thanks Jenny
nah this was homework so I was forced to do it but, it seems that I do need all the practise I can get

[generalkorn12]

This is just something that's been confusing me.

1. When it comes to the constant 'c' in Integrated functions, if we had to divide both sides by two or something like that, would the 'c' remain as 'c' (which my teacher says is the answer) or is it, 2c?

2. √(x^2 ) does that equal, x or |x|?