VCE Methods Question Thread!

[barydos]

How do you find the the answer for question b?

Im getting 0.0961 which is not the right answer?

Part b)
Pr(x>2) + Pr(x=2)*Pr(x>0)
so binomialCDf(3,10,10,0.05) + binomialPDf(2,10,0.05)*binomialCDf(1,10,10,0.05)
should give you 0.04145

... if you were referring to this question of course haha

[barydos]

Sorry to barge, but I've got a probability related question (binomial)
(Image removed from quote.)

It's part c) which has got me stumped! D:

So, we know that;
p = 0.05
n = 10

So, presuming that you are getting this A and B ok, I'll move to C.

 So, the probability that two batches are tested is if the first one fails. This occurs when Pr(X=2).

Pr(X=2) = nCr(10,2)(0.05)^2(0.95)^8 = 0.07463
Now, we know that if there are any faulty articles in the 2nd batch, it fails. So there will never be a 3rd batch. There is also always 1 batch tested.

So, we know that the Pr(one batch) + Pr(2 batches) = 1, so Pr(one batch) = 0.92537. Drawing up a table, and working out the mean of a discrete distribution, you get;

Now, E(X) = (0.92537)(1) + (0.07463)(2) = 1.07463 batches.

SHIT, articles, not batches. Damn it. Well, we know that one batch is equal to 10 articles, so

E(X)articles = 10.7463

May be wrong, but you get the idea. You can make a discrete table and work out the mean from there. A similar question is on the Q2 of 2010 VCAA Methods.

hold on now, I was trying out similar methods, but I cannot get the actual answer which is 10.702, do you know what would cause this slight difference?

[Hancock]

Well, everything hinges on the fact that Pr(X=2) = 0.07436

Could you check that on your calculator? Otherwise, dropping of significant figures may have caused the issue.

[barydos]

Well, everything hinges on the fact that Pr(X=2) = 0.07436

Could you check that on your calculator? Otherwise, dropping of significant figures may have caused the issue.

so Pr(x=2) = 0.0746347...... => 'z' value
from there i did

z*20 + (1-z)*10 = 10.7463...
so using exact values gives 10.7463, im pretty sure this textbook usually has its answers calculated via exact value methods like what i did :S

[b^3]

Well, everything hinges on the fact that Pr(X=2) = 0.07436

Could you check that on your calculator? Otherwise, dropping of significant figures may have caused the issue.

so Pr(x=2) = 0.0746347...... => 'z' value
from there i did

z*20 + (1-z)*10 = 10.7463...
so using exact values gives 10.7463, im pretty sure this textbook usually has its answers calculated via exact value methods like what i did :S

Theres nothing wrong with what Hancock has done as far as I can see, and I just dug out the extended respone solutions for the methods textbook. It does it the way Hancock did it (which makes sense), the answer given there is 10.7463, so the textbook answer at the back of the book is wrong.

[barydos]

Well, everything hinges on the fact that Pr(X=2) = 0.07436

Could you check that on your calculator? Otherwise, dropping of significant figures may have caused the issue.

so Pr(x=2) = 0.0746347...... => 'z' value
from there i did

z*20 + (1-z)*10 = 10.7463...
so using exact values gives 10.7463, im pretty sure this textbook usually has its answers calculated via exact value methods like what i did :S

Theres nothing wrong with what Hancook has done as far as I can see, and I just dug out the extended respone solutions for the methods textbook. It does it the way Hancook did it (which makes sense), the answer given there is 10.7463, so the textbook answer at the back of the book is wrong.

omg that clears everything up, wasted so much time on this haha
thanks b^3 , and thanks again Hancook

[Hancock]

Awesome. Felt like I screwed up something there. Cheers guys, glad to be able to help.

[dinosaur93]

1.Let A and B be events that Pr (A) = 0.6, Pr (A U B) = 0.8 and Pr (A|B) = 0.6
Find Pr (B).


2. Having a mean of 20 and standard deviation of 4 where X is the normal random variable and Z is the random variable with a standard normal distribution Find b such that Pr 9X .240 = Pr (Z < b)


3. On a multiple choice exam, each question will have 5 alternatives where only i given of which is correct. For each question, candidates gain 4 mark for a correct answer but Raoul is able to elimiate 2 of the incorrect answers from the alternatives given before guessing from those remaining. Given that the exam contains 24 questions, calaculate his explected total mark?


4. For a stnadard normal variable Z, Pr (0 Z c) = 0.3 and Pr (Z d) = 0.3 The value of Pr (Z d| c) is ___?


5. The random variable X us normally distributed with a  mean of 10 and a variance of 4. If Pr (8 Z 12) = 1 - 2 Pr (Z c) where Zi is a stnadrad normal distribution then c is ___?

[barydos]

1.Let A and B be events that Pr (A) = 0.6, Pr (A U B) = 0.8 and Pr (A|B) = 0.6
Find Pr (B).


2. Having a mean of 20 and standard deviation of 4 where X is the normal random variable and Z is the random variable with a standard normal distribution Find b such that Pr 9X .240 = Pr (Z < b)


3. On a multiple choice exam, each question will have 5 alternatives where only i given of which is correct. For each question, candidates gain 4 mark for a correct answer but Raoul is able to elimiate 2 of the incorrect answers from the alternatives given before guessing from those remaining. Given that the exam contains 24 questions, calaculate his explected total mark?


4. For a stnadard normal variable Z, Pr (0 Z c) = 0.3 and Pr (Z d) = 0.3 The value of Pr (Z d| c) is ___?


5. The random variable X us normally distributed with a  mean of 10 and a variance of 4. If Pr (8 Z 12) = 1 - 2 Pr (Z c) where Zi is a stnadrad normal distribution then c is ___?

1. As you can see, Pr(A) = Pr (A|B) = 0.6 so they are independent.
Hence Pr(A) = Pr(A n B)/Pr(B)
or Pr(A)Pr(B) = Pr (A n B)
Pr(A) + Pr(B) - Pr(A n B) = Pr (A U B) , let Pr(B) = x
0.6 + x - 0.6x = 0.8
0.4x = 0.2
x = 0.5
so Pr(B) = 0.5

2. I'm assuming you're asking for Pr(X >24) = Pr (Z< b) find b.
Pr(Z<b) = Pr(Z >-b) by symmetry, so Pr(X>24) = Pr( >-b)
so  finding the z-score...
(24-20)/4 = 1
so -b = 1
or b = -1 !!!

3. so really there are 3 options, so pr(correct) = 1/3, pr (incorrect) =2/3
im assuming, correct = +4 marks, incorrect = 0 marks
so expected, 24*(4*1/3) = 32 marks ... yeh im guessing this one? correct me if i'm wrong someone

4. so we're going to assume c>d from the info given
so Pr (Z>=d|=<c) = Pr(d=< Z =< c)/Pr(Z=< C)
drawing out the standard norm distr, you can see that
Pr(Z =< c) = 0.8
and Pr(Z=<d) = 0.7
so answer is 0.1/0.8  or 1/8

5. sd = 2
c = 12 from inspection? idk

[#1procrastinator]

Two questions related to the binomial distribution

The probability that a full forward in Aussie Rules football will kick a goal from outside the 50-metre line is 0.15. If the full forward has 10 kicks at goal from outside the 50-metre line, find the probability that he will
a) kick a goal every time
b) kick at least one goal
c) kick more than one goal, given that he kicked at least one goal

I've gotten a) and b) but don't know to do c)


Also, this one:

An examination consist of 25 multiple choice questions. Each question has four possible answers. At least 13 correct answers are required to pass the exam. Suppose the student guesses the answer to each question
a) What is the probability that the student guesses exactly 13 questions correctly?
b) What is the probability the student will pass the exam?

I've got a) but don't know how to do b). I think I have to find the probability of the student getting anything over 13 but don't know how to compute without having to calculate individually the probability for each score and summing it

[barydos]

Two questions related to the binomial distribution

The probability that a full forward in Aussie Rules football will kick a goal from outside the 50-metre line is 0.15. If the full forward has 10 kicks at goal from outside the 50-metre line, find the probability that he will
a) kick a goal every time
b) kick at least one goal
c) kick more than one goal, given that he kicked at least one goal

I've gotten a) and b) but don't know to do c)


Also, this one:

An examination consist of 25 multiple choice questions. Each question has four possible answers. At least 13 correct answers are required to pass the exam. Suppose the student guesses the answer to each question
a) What is the probability that the student guesses exactly 13 questions correctly?
b) What is the probability the student will pass the exam?

I've got a) but don't know how to do b). I think I have to find the probability of the student getting anything over 13 but don't know how to compute without having to calculate individually the probability for each score and summing it

1. c)
Pr(X>1|X>=1) = Pr(X>1 n X>=1)/Pr(X>=1) = Pr(X>1)/Pr(X>=1)
=0.4557/0.8031
=0.5674... hope that's right! haha

2.b) so Pr of answer right = 0.25
X~Bi(25,0.25) i think that's the notation? for binomial where n = 25 and p = 0.25
so we're looking for Pr(X>=13) which is binomialCDf(13,25,25,0.25) = 0.00337 I think, is that right?

-----------------------------------------------------
okay my question!
suppose we have cos(2pi*t)and sin(4pi*t/3)
how do you find a general solution for when these two expressions are equated (aka cos(2pi*t)=sin(4pi*t/3)
an alternative form i've found is cos(2pi*t)=cos(4pi*t/3 - pi/2)

but the general solution i can't find, THANKS in advance!

[barydos]

I wanted to ask a question about the solutions for the 2006 VCAA exam 2.
There's a question that asks to find the y-value of the graph y=3-e^x-e^-x when x=-0.5
The question asks to round the value to two decimal places. The answer I got was 0.744748 and when rounded to two decimal places it should equal 0.75 right?
The solutions say that the right answer is 0.74. Furthermore, no marks were given if the answer given was 0.75.

I think what you're doing is rounding the 3rd decimal up to make it 0.745
which in turn makes it 0.75 to 2 decimal places.

but when doing to 2 decimal places, you should just look at the 3 decimal digits: 0.744
completely ignore the rest so you wont make the mistake. here you can see it rounds down to 0.74

[bonappler]

Yeah it was a pretty easy exam until I came to that question and lost around 3 marks. I guess I have to go back to the basics.

[#1procrastinator]

1. c)
Pr(X>1|X>=1) = Pr(X>1 n X>=1)/Pr(X>=1) = Pr(X>1)/Pr(X>=1)
=0.4557/0.8031
=0.5674... hope that's right! haha

2.b) so Pr of answer right = 0.25
X~Bi(25,0.25) i think that's the notation? for binomial where n = 25 and p = 0.25
so we're looking for Pr(X>=13) which is binomialCDf(13,25,25,0.25) = 0.00337 I think, is that right?

Yeah that's correct, thanks a lot....I'm cramming everything on probability and statistics for my SAC this week, lol. Two chapters left

[brightsky]

Two questions related to the binomial distribution

The probability that a full forward in Aussie Rules football will kick a goal from outside the 50-metre line is 0.15. If the full forward has 10 kicks at goal from outside the 50-metre line, find the probability that he will
a) kick a goal every time
b) kick at least one goal
c) kick more than one goal, given that he kicked at least one goal

I've gotten a) and b) but don't know to do c)


Also, this one:

An examination consist of 25 multiple choice questions. Each question has four possible answers. At least 13 correct answers are required to pass the exam. Suppose the student guesses the answer to each question
a) What is the probability that the student guesses exactly 13 questions correctly?
b) What is the probability the student will pass the exam?

I've got a) but don't know how to do b). I think I have to find the probability of the student getting anything over 13 but don't know how to compute without having to calculate individually the probability for each score and summing it

1. c)
Pr(X>1|X>=1) = Pr(X>1 n X>=1)/Pr(X>=1) = Pr(X>1)/Pr(X>=1)
=0.4557/0.8031
=0.5674... hope that's right! haha

2.b) so Pr of answer right = 0.25
X~Bi(25,0.25) i think that's the notation? for binomial where n = 25 and p = 0.25
so we're looking for Pr(X>=13) which is binomialCDf(13,25,25,0.25) = 0.00337 I think, is that right?

-----------------------------------------------------
okay my question!
suppose we have cos(2pi*t)and sin(4pi*t/3)
how do you find a general solution for when these two expressions are equated (aka cos(2pi*t)=sin(4pi*t/3)
an alternative form i've found is cos(2pi*t)=cos(4pi*t/3 - pi/2)

but the general solution i can't find, THANKS in advance!

make use of the formula cosx - cosy = 2sin((x+y)/2)sin((x-y)/2)