VCE Methods Question Thread!

[Evolio]

Hi.
From the NHT 2019 Exam 2, question 2 e, I was wondering what domain we use when we are solving the equations? Because I keep getting general solution.

[S_R_K]

Hi.
From the NHT 2019 Exam 2, question 2 e, I was wondering what domain we use when we are solving the equations? Because I keep getting general solution.

There might be a more elegant way to do it, but I'd observe that the period of v2 is 14, so I'd solve v1(60) = v2(60) with 0 ≤ k ≤ 14.

[wildsivanna]

helloo  , does anyone have detailed written solutions to the 2019 NHT exam 2? (other than VCAA's) if not, at the very least.... for the MC?

thanks heaps! 

[JR_StudyEd]

Can you confirm if my methods of working out and the answers are correct for the following question?

the question

Malaysia (MAL) is a highly tactical team and puts a lot of emphasis on how far their players can 'Header' a ball. The distance that the MAL players can Header the ball is normally distributed with a mean of 18 metres and standard deviation of 3 metres.

a. Find the probability, correct to 3 decimal places, that a MAL player can Header at least 20 metres (2 marks)

my working and my answer

I did X ~ N (18,3²)
Using the normCdf function, I got 0.252.

b. 8 of the MAL players are selected at random and asked to header the ball at training. Find the probability correct to 3 decimal places that:

i. Only the first two players Header the ball at least 20 metres. (2 marks)

my working and my answer

X ~ Bi(8, 0.252).
Using the binomCdf function, I got 0.638

ii. Exactly 2 of the players Header the ball at least 20 metres. (2 marks)

my working and my answer

X ~ Bi(8, 0.252)
Using the binomPdf function, I got 0.311

iii. At least 4 of the players Header the ball at least 20 metres, given the first player chosen Headers the ball at least 20 metres (2 marks)

my working and my answer

Using the conditional probability formula, I got 0.117.

The below section is the source of my uncertainty and confusion. (It doesn't help that it's badly worded)

c. Find the minimum number of MAL players required to ensure that the probability that at least one of them can Header the ball at least 20 metres is at least 0.98. (3 marks)

I started by doing Pr(X≥1) = 0.98 but then my calculator exposed my ignorance. Makes no sense, doesn't it? Now I don't know what to do for this particular question.

Hopefully I typed up enough information to enable you to help me out! Thanks!

[S_R_K]

I believe your answers to a and b.ii. are correct.

I don't think your answer to b. i. is correct. This is because the question asks for the probability that only the first two players head the ball at least 20 metres. This should be 0.252^2 * (1 – 0.252)^6 = 0.011. Using the binomial CDF function with n = 8, 2 ≤ x ≤ 8 gives the probability that at least 2 of the eight players heads the ball at least 20 metres, and this player(s) could be the first through to the eighth player.

For b.iii., we are given that the first player heads the ball 20 m. So that means at least four players in total head the ball 20 m if at least three of the next seven head the ball 20 m. So just find the probability that at least three out of seven players head the ball 20 m, using binomcdf(7,0.252,3,7). You can use the conditional probability formula for this, but you need to be careful that you set it up correctly - I was unable to get anything like your answer, so I'm not sure what you've done wrong here.

For part c. I would observe that the complement to at least one player heading the ball at least 20 metres is no player heading the ball at least 20 metres. So you should find the number of players required for the probability of no 20 m headers to be 0.02. This is just solve(0.02 = (1 – 0.252)^n, n), and then round up to the next integer (rounding down would give a probability of no headers that is greater than 0.02).

[Evolio]

There might be a more elegant way to do it, but I'd observe that the period of v2 is 14, so I'd solve v1(60) = v2(60) with 0 ≤ k ≤ 14.

Cool, thank you!

[sarah15]

Hi! I am having trouble with questions in probability where you have to find the median of a hybrid function. Specifically, when there is a second equation other than 0. How do you know which equation to integrate?
Thanks!

[Sine]

Hi! I am having trouble with questions in probability where you have to find the median of a hybrid function. Specifically, when there is a second equation other than 0. How do you know which equation to integrate?
Thanks!

Basically you want to choose the one you believe will result in the least work. So if you think more than 50% of the area is under one question integrate that. But if you end up with something like 0.44 you would just try the other one to get 0.50 or just find the value that produces 0.06 so the total area to the left/right is 0.50.

[Evolio]

Hi! I am having trouble with questions in probability where you have to find the median of a hybrid function. Specifically, when there is a second equation other than 0. How do you know which equation to integrate?
Thanks!

So, usually you would work from the lower domain to the upper domain which is using the first branch first.
So, first you find the probability of the first branch by integrating that equation, using the numbers 4 and 0. If it is more than 0.5, you know that the median lies in that branch as that is the only branch there. So, using the CAS (I'm assuming this is a CAS active question), you would press solve type the integral, with m at the top of the integral and 0 at the bottom making it equal to 0.5 and , m. This will give you the median value.
However, if the median is less than 0.5 when calculating the probability, then you know that the median is in the second branch as 0.5 is higher than whatever probability you get. This is how you would solve it. You would type the integral using the first function and its domain+ the second function k(12-x) and putting 4 and m on the side of the integral, making this equal to 0.5 and solving for m.

I hope this makes sense and it answered your question!

EDIT: Sine beat me to it.

[sarah15]

Basically you want to choose the one you believe will result in the least work. So if you think more than 50% of the area is under one question integrate that. But if you end up with something like 0.44 you would just try the other one to get 0.50 or just find the value that produces 0.06 so the total area to the left/right is 0.50.

So, usually you would work from the lower domain to the upper domain which is using the first branch first.
So, first you find the probability of the first branch by integrating that equation, using the numbers 4 and 0. If it is more than 0.5, you know that the median lies in that branch as that is the only branch there. So, using the CAS (I'm assuming this is a CAS active question), you would press solve type the integral, with m at the top of the integral and 0 at the bottom making it equal to 0.5 and , m. This will give you the median value.
However, if the median is less than 0.5 when calculating the probability, then you know that the median is in the second branch as 0.5 is higher than whatever probability you get. This is how you would solve it. You would type the integral using the first function and its domain+ the second function k(12-x) and putting 4 and m on the side of the integral, making this equal to 0.5 and solving for m.

I hope this makes sense and it answered your question!

EDIT: Sine beat me to it.

Thank you! I think I get it now. So would 8 be the correct answer for that question?

[Evolio]

I got m=5.072.
Did you add the first part of the function as well when making it equal to 0.5?

[sarah15]

I got m=5.072.
Did you add the first part of the function as well when making it equal to 0.5?

Oops, my mistake! I got it now. Thanks again

[forsande]

Help plzz!

Solve for x: log2(x) - log2([rootx]-1)=2
I only got up to x=4([rootx]-1), dunno where to go from there

thxx

[DrDusk]

Help plzz!

Solve for x: log2(x) - log2([rootx]-1)=2
I only got up to x=4([rootx]-1), dunno where to go from there

thxx

I'm assuming what you'e done so far is correct by the way. There is a little trick you can do.
So we have



Standard solving a quadratic from there =)

Alternatively you can actually factorize it straight away from where you left off, but I understand that it may sometimes look a bit weird so it's much clearer to make a substitution like this.

Personally I would factorize it straight away with the square root.

[forsande]

I'm assuming what you'e done so far is correct by the way. There is a little trick you can do.
So we have



Standard solving a quadratic from there =)

Alternatively you can actually factorize it straight away from where you left off, but I understand that it may sometimes look a bit weird so it's much clearer to make a substitution like this.

Personally I would factorize it straight away with the square root.

Thanks, but where does y^2 come from (also, what's the rationale behind doing this method?)