VCE Methods Question Thread!

[I\'m Not A Robot]

you could also think about it as needing 3/7 if that's easier

Ohhhh, so pretty much its stating that since 1 has already done it (header over 20 metres) we only need 3/7 more people. If so, how would we show that in terms of working out? i still dont understand what the intersection of x>4 and x=1 would be?
Would it be easier to create a tree diagram and find all the scenarios that way??

[MB_]

Hey, could someone please take a look at this question, it really confused me.

BACKGROUND:
The questions below came from a question that involved normal distributions first. The question asked us to find the probability of a player doing a "header" and obtaining over/atleast 20 meters, where the mean is 18 and Sd is 3. Working this out, i got the answer of 0.252
The question then transitioned to binomial and the questions were like this...

X~Bi(8,0.252) i.e. the number of trials is 8 and POS is 0.252.
8 players are each called upon randomly to header the ball...
1. What is the probability that at least 4 of the players header the ball at least 20 meters given that the first player chosen headers the ball at least 20 meters.
2. What is the probability of only the first 2 players headering the ball at least 20 meters.
3. What is the probability of exactly 2 players headering the ball at least 20 meters.

These questions all seem kind of confusing, i don't know where to start with the first question...
and aren't the 2nd 2 pretty much asking the same thing...
when i asked my teacher, he just said to 'think about only the first 2 players' (in relation to Q2).

Could someone please explain to me how this works.

There is a subtle difference between questions 2 and 3. Question 2 asks the probability of only the first two players heading the ball 20 meters whereas Question 3 asks the probability of exactly two players heading the ball 20 meters but this does not have to be the first two players, it could be the third and seventh or the second and sixth.

So to work out question 2, its the probability of the first two heading it at least 20 meters and the next 6 not heading it that far which is (0.252)^2*(1-0.252)^6

Question 3 uses the binomPDF function as there are 26 ways in which exactly 2 players header the ball at least 20 meters

Hopefully this helps!

[MB_]

Ohhhh, so pretty much its stating that since 1 has already done it (header over 20 metres) we only need 3/7 more people. If so, how would we show that in terms of working out? i still dont understand what the intersection of x>4 and x=1 would be?
Would it be easier to create a tree diagram and find all the scenarios that way??

If you know that the first person headers it at least 20 meters then all you need to find out is the probability that at least 3 of the remaining 7 header it over 20 meters. So its just P(X>3) over 7 trials and you can work this out using the binomCDF function with bounds 3 and 7.

[trendsetter123ftw]

I make sure I understand the concepts well
I ask my teacher for help whenever I need to
I do practice sacs and chapter review questions before each sac
And yet I'm only getting 75% in my sacs
Where the hell am I going wrong?
I want to get 90% in exam 1 and 2, but I just don't know what the hell is wrong with my revision. Any help would be appreciated

[I\'m Not A Robot]

If you know that the first person headers it at least 20 meters then all you need to find out is the probability that at least 3 of the remaining 7 header it over 20 meters. So its just P(X>3) over 7 trials and you can work this out using the binomCDF function with bounds 3 and 7.

Ohhh okay, THANK YOUUU!! So I'm not really meant to use the whole intersection between (X>4 and X=1) / (X=1) formula thing. What you said makes heaps more sense, cus i was confused about the whole formula thingy. And...

As for the question about the only the first 2 headering over 20 meters... I'm meant to kind of work that out on paper and not through BinomPDF right??

Thanks heaps again though.

[I\'m Not A Robot]

I make sure I understand the concepts well
I ask my teacher for help whenever I need to
I do practice sacs and chapter review questions before each sac
And yet I'm only getting 75% in my sacs
Where the hell am I going wrong?
I want to get 90% in exam 1 and 2, but I just don't know what the hell is wrong with my revision. Any help would be appreciated

Based on what you have said, the only conclusion i can come to is that you are making 'silly' mistakes. My best advice would be to just try and iron out any mistakes that people may sometimes make negligently. Such mistakes could include things like notation, or even things like using 'x' in your working out instead of another variable that was stated to be used. .e.g. The question is y=4t^2... and in your working out and answer, you use the variable 'x' instead of the required t.

I know that i lost a mark or two for both of my previous SACs for making mistakes like that. Or even not presenting your answer in the way the question wants you too.

[trendsetter123ftw]

Based on what you have said, the only conclusion i can come to is that you are making 'silly' mistakes. My best advice would be to just try and iron out any mistakes that people may sometimes make negligently. Such mistakes could include things like notation, or even things like using 'x' in your working out instead of another variable that was stated to be used. .e.g. The question is y=4t^2... and in your working out and answer, you use the variable 'x' instead of the required t.

I know that i lost a mark or two for both of my previous SACs for making mistakes like that. Or even not presenting your answer in the way the question wants you too.

I don't make silly mistakes, I just always get stumped on some questions in the sac, since the way they're worded?

[MB_]

Ohhh okay, THANK YOUUU!! So I'm not really meant to use the whole intersection between (X>4 and X=1) / (X=1) formula thing. What you said makes heaps more sense, cus i was confused about the whole formula thingy. And...

As for the question about the only the first 2 headering over 20 meters... I'm meant to kind of work that out on paper and not through BinomPDF right??

Thanks heaps again though.

Technically it's still using the intersection but you have to think about what the intersection means in the context of the question.

The second question is a Bernoulli trial where you have either a success with probability \( p \) or a fail with probability \( 1-p \) over only 1 trial and this can be worked out by hand.

[jacquieg]

Hi all,
I've attached a screenshot of a question I got a bit confused with, I thought it would be easy but I'm not sure of what i'm doing wrong.
I chose to do the integral of -3x + 1/3x^3 between -3 and 0 (I got the intersection points alright) then I subtracted the integral of -1/3x^3 + 3x between 3 and 0. I thought this is correct because i'm minusing the area under the X axis which would be negative. I know i could do 2 times either of those areas but I want to do it the long way so I know for next time when I get a graph that isn't symmetrical.
Should i just use modulus signs or what?
What am i doing wrong?

thanks everyone

[I\'m Not A Robot]

Hi all,
I've attached a screenshot of a question I got a bit confused with, I thought it would be easy but I'm not sure of what i'm doing wrong.
I chose to do the integral of -3x + 1/3x^3 between -3 and 0 (I got the intersection points alright) then I subtracted the integral of -1/3x^3 + 3x between 3 and 0. I thought this is correct because i'm minusing the area under the X axis which would be negative. I know i could do 2 times either of those areas but I want to do it the long way so I know for next time when I get a graph that isn't symmetrical.
Should i just use modulus signs or what?
What am i doing wrong?

thanks everyone

Hey, ive attached the required integrals for the area.

[Dais_Deorum]

Hi guys, heres just a question thats a bit ambiguous...

The time taken to eat at a particle restaurant, T, is normally distributed with a mean of 2.25 hours and a variance of 45 minutes.
a) What time range will 95% of customers dine for? Give answers expressed in nearest whole number of minutes.

So when you convert hours to minutes and use a mean of 135 minutes & variance 45 minutes, the range comes out at about 122 - 148.
But when you convert the variance to 0.75 hours, and do the working out, it comes to 31 - 239 minutes. The answer is the latter option, but it says that students would be notified in reading time "Variance equals 3/4 hours". I'm just wondering why this would be the case, and further, why don't answers come out to be the same? I worked out that if you use var(T)=3/4 hours, the sd comes out bigger than the variance, and the opposite when you use 45 minutes as the variance. I figure this has something to do with number base systems, but idk. I asked the teacher, but he said to always use the units that are stated in the mean.

Any explanation would be appreciated!

Cheers

[jacquieg]

SOMEONE EXPLAIN CONFIDENCE INTERVALS TO ME PLEASE

[Bell9565]

SOMEONE EXPLAIN CONFIDENCE INTERVALS TO ME PLEASE

Okay so in non-mathematical terms, you have done your sample and you've found for instance, 4 out of the 20 year 12 students in Victoria (so 0.2 as a proportion which would be your p-hat) in the sample do methods. Now, as this is a sample, it's not fully representative of the population therefore there is a chance that in reality, more than or less that 20% of year 12 students do methods. So a confidence interval is a range based on the collected data which you're 95% sure the real p value lies in.
As you probably know about the 68/95/99.7 rule I'm not gonna go into it but pretty much a 95% confidence interval is ±1.96 standard deviations away from the mean.

Mathematically, this is done by as follows
(p̂ - 1.96√(p̂q̂/n), p̂ + 1.96√(p̂q̂/n))
Where p̂ - the sample proportion
            q̂ - (1 - p̂)
            n - sample size
           √(p̂q̂/n) - the standard deviation

In the situation I had above,
as p̂ = 0.2, the sd =  √(0.2 x 0.8 / 20) = 0.0894427191
Therefore,
(0.2 - 1.96√(0.2 x 0.8/20), 0.2 + 1.96√(0.2 x 0.8/20))
(0.1105572809, 0.2894427191) is your confidence 95% interval, meaning you are 95% sure that the population parameter is between these two values.
If you want it in term of x, just multiply everything by 20 once finished.

I hope that helps 

[yourdavid]

can someone help me with question 2. theres no worked examples in my book for it.
this is methods units 3 & 4

2. Sammy and Guido play racquetball each week. In any one game, Sammy has the probability of
winning of 0.7. The outcome of any one game between the two is independent of the outcome of the
previous games played.
a. If the play 8 games, what is the probability, to four decimal places, that Sammy wins 5 out of 8
games?
b. What is the probability, to four decimal places, that Guido wins at most 6 of the 8 games?
c. What is the probability, to four decimal places, that Sammy only wins the second, fourth, fifth,
seventh and eighth games?
d. Compare the value obtained in part (a) with that of part (c). Explain the result.
e. How many games do they need to play so that Guido has at east 80% chance of winning at least
one game?

[Sine]

can someone help me with question 2. theres no worked examples in my book for it.
this is methods units 3 & 4

2. Sammy and Guido play racquetball each week. In any one game, Sammy has the probability of
winning of 0.7. The outcome of any one game between the two is independent of the outcome of the
previous games played.
a. If the play 8 games, what is the probability, to four decimal places, that Sammy wins 5 out of 8
games?
b. What is the probability, to four decimal places, that Guido wins at most 6 of the 8 games?
c. What is the probability, to four decimal places, that Sammy only wins the second, fourth, fifth,
seventh and eighth games?
d. Compare the value obtained in part (a) with that of part (c). Explain the result.
e. How many games do they need to play so that Guido has at east 80% chance of winning at least
one game?

what is your working out so far or thoughts about the question?

Basically any probability question is possible to do even if you haven't learn the content and formulas as long as you think about the question logically and how it would work it real life.