brightsky's Maths Thread

[BlueSky_3]

From what I've been told, you just note what I said before, and its pretty straightforward since you will have a large number divided by that number squared and so as x approaches infinity that fraction approaches 0 leaving you with y=3....

[brightsky]

From what I've been told, you just note what I said before, and its pretty straightforward since you will have a large number divided by that number squared and so as x approaches infinity that fraction approaches 0 leaving you with y=3....

actually yeah stupid question, figured it out eventually. thanks muchly!

[kamil9876]

anyway, my main query is, would a function like y=x^2 be increasing in the interval [0,infty) or (0,infty)? vcaa (i think) seems to think the former, while kilbaha thinks the latter.

Both statements are true, in fact I think someone even mentioned on the forum sometime ago that VCAA does in fact use the definition the rest of the world does " A function is increasing if the following condition holds "whenever with then " ". In your case, this condition is satisfied for both of the sets.

It's true that differentiation can be used to determine many things like maxima,mimina, increasing,decreasing etc. but it's important to NOT make things like "decreasing" synonymous to "gradient<0"

As a general precaution: Whenever you want to use differentiation to determine some stuff, be careful of the endpoints (for example maxima/minima)

[brightsky]

k thanks kamil

also wondering, were students studying methods back in 2005 allowed calculators into exam 1?

[pi]

also wondering, were students studying methods back in 2005 allowed calculators into exam 1?

Yep, non-CAS though (and one scientific calc if they wanted, too)

[brightsky]

also wondering, were students studying methods back in 2005 allowed calculators into exam 1?

Yep, non-CAS though (and one scientific calc if they wanted, too)

cheers

[brightsky]

just to confirm: the phrase 'along the y-axis' in the context of transformations is synonymous with 'from the x-axis', right? so if one were to dilate y=2(2-x)^5 'along the y-axis', the equation of the image would be y=(2-x)^5 not y=2(2-2x)^5, as a putatively trustworthy source seems to think?

[b^3]

Yes, 'along the y-axis' is the same as saying 'from the x-axis', so you would be correct.

[brightsky]

Yes, 'along the y-axis' is the same as saying 'from the x-axis', so you would be correct.

k thanks

[brightsky]

how does the variance of a hypergeometric distribution compare with the variance of a correspnding binomial distribution? that is, which one is greater and which smaller?

[TrueTears]

that is, which one is greater and which smaller?

oh so that's what you meant, anyways as i explained in the pm, they are actually asymptotically the same for a large N (keeping n fixed), it doesn't really make sense to ask which one is "greater" and "smaller" how do you define "greater"? Plus there are too many parameters which are variable, you cannot really determine which is "larger" (in the normal sense of "large") by just looking at the expressions.

What matters most in most of these distributions are asymptotic properties as most practical processes have large sample sizes.

[brightsky]

thanks for that!

the original question was:

A box contains a number of Chocolates and Toffees and a random selection of the sweets is made. If A is the number of chocolates in the sample when a sample is taken without replacement, and B is the number of chocolates in the sample when a sample is selected with replacement:

Is Var(A) > Var(B) or is Var(A) < Var(B)?

i plugged random numbers into the calculator and the cases i tried seemed to suggest that it's the former, though the answers say it's the latter. no rigorous explanation was given as to why though.

could you help me with this? cheers.

[TrueTears]

Oh i see, the question actually isn't stated very well, but anyways:

if we assume population is large, i would argue the variances are equal.

if we assume small population size (which is what I think the question is alluding to) then:

in the pm i said: the variance of Y if Y is hypergeometric is given by: n(r/N)[(N-r)/N][(N-n)/(N-1)] = n(r/N)[1-r/N][(N-n)/(N-1)]

now let p = r/N (ie, the probability of success in the population, since r = success in population and N = population size, just use the classical definition of probability to obtain p) and we have np(1-p)*[(N-n)/(N-1)]

note that [(N-n)/(N-1)] -> 1 if N -> infinity, so the above quantity is always smaller than np(1-p) the variance of a binomial random variable for a small N

[brightsky]

Oh i see, the question actually isn't stated very well, but anyways:

if we assume population is large, i would argue the variances are equal.

if we assume small population size (which is what I think the question is alluding to) then:

in the pm i said: the variance of Y if Y is hypergeometric is given by: n(r/N)[(N-r)/N][(N-n)/(N-1)] = n(r/N)[1-r/N][(N-n)/(N-1)]

now let p = r/N (ie, the probability of success in the population, since r = success in population and N = population size, just use the classical definition of probability to obtain p) and we have np(1-p)*[(N-n)/(N-1)]

note that [(N-n)/(N-1)] -> 1 if N -> infinity, so the above quantity is always smaller than np(1-p) the variance of a binomial random variable for a small N

love ya tt! thanks very much!

[TrueTears]

love ya too babe