VCE Methods Question Thread!

[Jenny_2108]

Consider the graph y=x³. If a question asks for the domain in which the graph is an increasing function, would x=0 be included?

included

I was under the impression that strictly increasing means including x=0 but just increasing does not include x=0.
Can anyone else confirm?

yep, you are right
I was obsessed by strictly increasing these days

[TrueTears]

Consider the graph y=x³. If a question asks for the domain in which the graph is an increasing function, would x=0 be included?

included

I was under the impression that strictly increasing means including x=0 but just increasing does not include x=0.
Can anyone else confirm?

yep, you are right
I was obsessed by strictly increasing these days

If unsure, consult wiki for definitions, then apply: http://en.wikipedia.org/wiki/Monotonic_function#Monotonicity_in_calculus_and_analysis

[Special At Specialist]

If unsure, consult wiki for definitions, then apply: http://en.wikipedia.org/wiki/Monotonic_function#Monotonicity_in_calculus_and_analysis

That's too complicated for me to understand.

I would've assumed that "increasing" and "strictly increasing" meant the same thing (that f'(x) > 0), except "strictly increasing" placed more emphasis on the fact that it MUST increase (hence why it is so "strict" about that) and simply staying the same (ie. f'(x) = 0) does not count.

[TrueTears]

all you really need to read is :

If the order ≤ in the definition of monotonicity is replaced by the strict order <, then one obtains a stronger requirement. A function with this property is called strictly increasing. Again, by inverting the order symbol, one finds a corresponding concept called strictly decreasing. Functions that are strictly increasing or decreasing are one-to-one (because for x not equal to y, either x < y or x > y and so, by monotonicity, either f(x) < f(y) or f(x) > f(y), thus f(x) is not equal to f(y)).

i'm sure thats accessible, so essentially you have the right idea

[BubbleWrapMan]

it's funny how that's different to what VCAA defines as strictly (in/de)creasing

[hongs-]

The continuous random variable X has a normal distribution with mean 14 and standard deviation 2
If the random variable Z has the standard normal distribution, then the probability that X is greater than 17 is

A Pr(Z>3)
B Pr(Z<2)
C Pr(Z<1.5)
D Pr(Z<-1.5)
E Pr(Z>2)

I dont understand why the answer is D, can someone help me please?

[BubbleWrapMan]

The Z score for that is 1.5, so it's Pr(Z > 1.5) which is the same as Pr(Z < -1.5) by symmetry

[polar]

[hongs-]

Oh okay, thankyou!!

[TrueTears]

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.

Nice 

Btw TT, I let and solve for p but didn't get

Where did I do wrong?
















But


Therefore


How can I prove (1)=(2)? Did I do somewhere wrong? 

And what does this notation mean?

sorry i didn't see this post, apologies for the late reply

nice that you went through all the algebra, you are correct in every step - good job

i think you are just missing something quite obvious,

hence the equality of (1) and (2)

[dfgjgddjidfg]

how do you do the last question in VCAA 2011 exam 1?

[polar]

10(a)


10(b)


10(c)


10(d)

[pi]

Personally, I felt 10(d) was a bit of an unfair q for methods-only students

*shrugs*

[Jenny_2108]

But in the other post, Hutcho said

is the same as , but instead of addition, it's multiplication.

Shouldn't it be

Edit: typo

[BubbleWrapMan]

It's n + n + n + ... + n, n times. So there's n ns.