VCE Methods Question Thread!

[knightrider]

Say for continuous distributions.

say we know 

how come we can then say

why does this work ?

[keltingmeith]

Say for continuous distributions.

say we know 

how come we can then say

why does this work ?

... I'm honestly confused by the question. What's Z?

[knightrider]

... I'm honestly confused by the question. What's Z?

Z represents the standardised  score

like howcome we can say

[odeaa]

Z represents the standardised  score

like howcome we can say

Z is the standard score, so a score of 25 must correspond to .37 standard deviations from the mean of x. I find drawing it out to be helpful

Edit:I'm on my phone so can't figure out exactly what you are asking without latex, but hopefully that answers the question

[Adiamond]

Hello everyone!
I'm having lots of trouble with a question from the chapter and i'm wondering if any of you would be able to help me.

The question is attached and I don't understand how the book got their answer for C) i.
(Their answer is 0.78) I got the answer as being 0.731707

[Adiamond]

Hello everyone!
I'm having lots of trouble with a question from the chapter and i'm wondering if any of you would be able to help me.

The question is attached and I don't understand how the book got their answer for C) i.
(Their answer is 0.78) I got the answer as being 0.731707

Wow i'm stupid, i was working it out as if they had to win two sets in a row, but they only have to win two sets overall
My bad

[cosine]

Are we required to know the mutually exclusive, independent and such probability formulas for the exam?

[Adiamond]

Are we required to know the mutually exclusive, independent and such probability formulas for the exam?

I would suggest learning them anyway just in case, they pop up here and there.

[keltingmeith]

Are we required to know the mutually exclusive, independent and such probability formulas for the exam?

Independence is on the formula sheet, so yes. Otherwise, for mutually exclusive, you should be able to use/derive it. Essentially, if A and B are mutually exclusive, then P(A n B)=0, so P(A u B)=P(A) + P(B) - P(A n B)=P(A) + P(B) - 0. So, using logic to get back to it shouldn't be too hard of an endeavour.

[cosine]

Independence is on the formula sheet, so yes. Otherwise, for mutually exclusive, you should be able to use/derive it. Essentially, if A and B are mutually exclusive, then P(A n B)=0, so P(A u B)=P(A) + P(B) - P(A n B)=P(A) + P(B) - 0. So, using logic to get back to it shouldn't be too hard of an endeavour.

Alright cheers brah!

What does mutually exclusive mean? Does it just mean that the events will never occur at the same time? For example, they will never have elements within each other, they are exclusive? idk haha..

[Adiamond]

Alright cheers brah!

What does mutually exclusive mean? Does it just mean that the events will never occur at the same time? For example, they will never have elements within each other, they are exclusive? idk haha..

You pretty much hit the nail on the head, It means that the is no intersection between even A and event B.

[cosine]

Oh, cheers. xD

Where is the independent formula on the summary sheet?

[odeaa]

Oh, cheers. xD

Where is the independent formula on the summary sheet?

Kinda parallel to the bolded word "probability", under the quotient rule and approximation formula (Its Pr(A u B)=Pr(A) + Pr(B) - Pr(A n B) if thats what you meant)

Also, I think you need to know that Pr(A n B)=Pr(A)*Pr(B) for independant events. Not a hard formula to remember, and pretty intuitive if you forget it

[StupidProdigy]

It's not explicitly on it. However it can be derived from the Pr(a|b) formula. Pr(a|b)=(Pr(a and b))/Pr(b). Multiply through by Pr(b)===>Pr(a and b)=Pr(a|b)*Pr(b). Note that Pr(a|b) just equals Pr(a) since the probabilities are independent (not conditional). So we have Pr(a and b)=Pr(a)*Pr(b). All sweet?

[cosine]

All good, cheers. xD