VCE Methods Question Thread!

[Floatzel98]

In the context of normal distributions I'm a bit confused by these 2 transformations:

Spoiler

and

The first one takes a standard normal distribution to a general normal distribution and the second one takes a general normal distribution back to the standard normal distribution? Is that correct?

The standard normal distribution would just have the density function:



While the general form would have the density function:

I'm just having a bit of trouble interpreting my book and its giving me trouble on a few questions. Also is it worth trying to understand a lot of the derivations of mean, variance, steady state probabilities etc for probability? I would usually try to understand things better but I'm struggling a lot in probability and skipping a lot of that stuff.

[Floatzel98]

If i have a z score of 1.5, does that mean I lie 1.5 standard deviations away from the mean?

[Orb]

If i have a z score of 1.5, does that mean I lie 1.5 standard deviations away from the mean?

Yes

Remember that the normal distribution curve is based on mean = 0 and sd = 1

[qwerty101]

if you look at the lines for the two highlighted bits i dont get how it happens.

It seems as if they have gone on their calc and used prob= 0.01  mean = 0 and Sd = 1, to get the -2.3264, but why?, i understand the equalling part, but only if i understand why they did the above things.

From the question it self we have a mean = x   and sd = 4 so why the 0 and 1 rather than x and 4?

Thanks!!

[Zealous]

if you look at the lines for the two highlighted bits i dont get how it happens.

It seems as if they have gone on their calc and used prob= 0.01  mean = 0 and Sd = 1, to get the -2.3264, but why?, i understand the equalling part, but only if i understand why they did the above things.

From the question it self we have a mean = x   and sd = 4 so why the 0 and 1 rather than x and 4?

Thanks!!

I'll explain their steps in a different way:

, then use an inverse normal to find the z-score on a standard normal distribution (mean=0, dev=1) such that 99% of the data lies before it.

You'll get a z-score of 2.32635. What this z-score is saying is that on a standard normal distribution, you'd need to be 2.32635 standard deviations away from the mean in order to have 99% of the data lie before you.

Therefore, we can use the z-score formula to find the mean given a value of 17 has a z-score of 2.32635.



So we can't start straight away and use the normal distribution with mean=x and sd=4 because we don't know the mean yet. Instead, we have to use the standard normal distribution then apply the result to find the mean.

[qwerty101]

Thanks,

so why do we need to use the standard normal distribution (mean=0, dev=1) when we have different mean (or same if mean ends up as 0) and different standard deviation?

Or is it sort of like a method of comparison, comparing to the standard normal distribution '17' gets z-score of 2.32635 - sorry if i repeated you.

Thanks

[shivaji]

Heffernan Exam 2, Question 2)g)iii)

When using the transition matrix, why is that when the question asks for TENTH year, we have to raise the transition matrix to the power 9, and not 10?

Thanks

[keltingmeith]

Heffernan Exam 2, Question 2)g)iii)

When using the transition matrix, why is that when the question asks for TENTH year, we have to raise the transition matrix to the power 9, and not 10?

Thanks

You have not listed the year this paper was from, and not all of us have a collection of Heffernan papers on us - could you please attach the whole question for us?

[_fruitcake_]

Heffernan Exam 2, Question 2)g)iii)

When using the transition matrix, why is that when the question asks for TENTH year, we have to raise the transition matrix to the power 9, and not 10?

Thanks

Maybe it is because the inital state (S) is after one year. Thus it is S^1 , so T^9 x S^1 would give you T^10.

[shivaji]

Maybe it is because the inital state (S) is after one year. Thus it is S^1 , so T^9 x S^1 would give you T^10.

ohh is that why, cheers.

You have not listed the year this paper was from, and not all of us have a collection of Heffernan papers on us - could you please attach the whole question for us?

soz forgot

[Orb]

ohh is that why, cheers.

soz forgot

Because after multiplying by the transition matrix, you get the result for the 2nd year. Multiplying it 9 times get you the result for the 10th year

[AndyCau]

hey, got a question from a maths methods exam paper
QUESTION 27) a) If Pr(A) = 0.7 and Pr(B) =0.9, what is the greatest value PR(A|B_ can have? (2 marks)


b) If A and B are mutually exclusive events, what can be said about Pr(B|A)? (1 mark)

thanks in advance 

[The Mathemagician]

A hint for part a - think about the largest value of that is possible (think about why this value should be maxmised as well). You could use a Karnaugh Map for this.
A hint for part b - if A and B are mutually exclusive, what can you say about ? The definition of is on the formula sheet.

If you're still stuck I've provided more explanation below.

Answers

A) As the highest value of is 0.7 because you cannot have negative probabilities. In other words, for this sample space if B occurs A also occurs. Then you use the formula . Maximising will maximise
B) Mutually exclusive means that hence . In other words, mutually exclusive events cannot occur at the same time, so if A has occured B will not occur.

[knightrider]

How would you solve this equation for x?

[knightrider]

How would you differentiate ?