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These are questions from my review packet for IB math test in couple days... I'm worried because anything related to probability + statistics = I fail and anything very random = I fail. Plz help me do well! Thx
Question 1
Andrew shoots 20 arrows @ a target. He has probability of .3 of hitting the target. All shots = independent of next.
Z = # of arrows hitting target.
1. Find mean + S.D of X
2. Find P(x=5)
3. << = Equal to or less than >> = Equal to or greater than.
Find P(4<<X<<8)
Bill also shoots arrows @ a target w/ probability of .3 of hitting the target. (independent events)
4. Calculate probability that Bill hits the target for the first time on his 3rd shot.
5. Calculate the minimum # of shots required for the probability of @ least 1 shot hitting the target to exceed .99
Question 2
A = ( 2, -1, 0 )
B = (3, 0, 1)
C = 1, -1, 2)
D = 6, -7, 2)
Area Triangle ABC = (Sqrt 14 / 2)
Line L is perpendicular to Plane ABC and passes through point A.
Cartesian equation of Plane ABC = -2i + 3j - k
Point D lies on L.
Find the volume of pyramid ABCD.
V = 1/3 B * H = Pyramid.
Question 3.
Theta = x
Let z = cos x +isin x
1. Find z^3
2. Use de Moivre's theorem to show that:
Cos3x = 4(cosx)^3-3cosx and
sin3x = 4(sinx)^3-4sinx
Using Result, Prove (sin3x - sinx) / (cos3x + cosx) = tanx
  
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United States24700 Posts
How will us doing your math problems for you without you trying help? What exactly are you looking for?
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Hmm I got a 3 on my Math HL test, no college credit for me!
GL HF.
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These are the problems where I have no idea of how to approach.
I don't need answer.
telling me how to arrive to answer is all I really need.
Ancestral - how was Paper 1 for you? and what was questions mainly do you remember?
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United States24700 Posts
For question 3 I think they are talking about when you do the following:
e^ix = e^ix
e^(nix) = e^(nix)
e^(nix) = (e^(ix))^n
expand, set real and imaginary parts equal
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I got q1, I just used Binomial theorem. For Q2, I don't see
Cos3x = 4(cosx)^3-3cosx can be related to de Morives.
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I think 4 is (0.3) x (0.3) x (1 - 0.3) = (0.3) x (0.3) x (0.7) = ...
x3 I think...makes sense, right?...
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United States24700 Posts
On May 05 2008 15:41 kdog3683 wrote:
I got q1, I just used Binomial theorem. For Q2, I don't see
Cos3x = 4(cosx)^3-3cosx can be related to de Morives.
On May 05 2008 15:34 micronesia wrote:
For question 3 I think they are talking about when you do the following:
e^ix = e^ix
e^(nix) = e^(nix)
e^(nix) = (e^(ix))^n
expand, set real and imaginary parts equal
http://en.wikipedia.org/wiki/De_Moivre's_formula#Derivation
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Question 2
A = ( 2, -1, 0 )
B = (3, 0, 1)
C = (1, -1, 2)
D = (6, -7, 2)
Area Triangle ABC = (Sqrt 14 / 2)
Line L is perpendicular to Plane ABC and passes through point A.
Cartesian equation of Plane ABC = -2i + 3j - k
Point D lies on L.
Find the volume of pyramid ABCD.
V = 1/3 B * H = Pyramid.
...If Line L is perpendicular to ABC's plane, then the absolute value of the distance from D to A is the pyramid's height. So just find that distance, then find the triangle made by A, B, and C (that's the base), and then you get to plug those two values into the equation (V = 1/3 B * H). =3
( (6 - 2)^2 + ((-7) - (-1))^2 + (2 - 0)^2 )^(1/2) = distance from D to A = (56)^0.5
Then...Err...Oh, yeah. xO
The area of the triangle (ABC) = the base = (1/2) the magnitude of the vector (-2i + 3j - k)
base = (1/2) ((-2)^2 + (3)^2 + (-1)^2)^(0.5) = (1/2) (14)^0.5 = (1/2) x (square root of 14)
Pyramid's volume = (1/3) (0.5 x (14)^0.5) (56)^0.5 = ...14/3... I'm not sure if this is right... xD I almost forgot what to do after getting the pyramid's height...
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Question 1
Andrew shoots 20 arrows @ a target. He has probability of .3 of hitting the target. All shots = independent of next.
Z = # of arrows hitting target.
1. Find mean + S.D of X
2. Find P(x=5)
3. << = Equal to or less than >> = Equal to or greater than.
Find P(4<<X<<8)
Bill also shoots arrows @ a target w/ probability of .3 of hitting the target. (independent events)
4. Calculate probability that Bill hits the target for the first time on his 3rd shot.
5. Calculate the minimum # of shots required for the probability of @ least 1 shot hitting the target to exceed .99
I have no idea what level or sort of mathematics you'd be expected to use here, but this is how I'd solve it - mind you there could well be a faster way than this - I haven't done combinatorics for about 12 years! 0_o
Anyway :
If X is to be the number of hits out of 20, then the mean of X will be 6, since on average he hits 30% of the time - out of any given set of 20 shots, you expect that Andrew will average about six hits.
To get the standard deviation, you consider that there are only 20 possible outcomes of each trial - Andrew can hit the target only between 0 and 20 times out of 20. You need to calculate all of these to get a set of 20 probabilities to assign to each outcome. Since the probability for each hit is 0.3 (or each miss is 0.7), you can do this very quickly
p(0) = 0.7^20
p(1) = 0.3 + 0.7^19
p(2) = 0.3^2 + 0.7^18
etc.
This only gives the probability of one particular configuration of each, so you multiply by the total number of combinations for each event :
C_nk=n!/k!(n-k)!
with n=20 for k=0--20
(only one way to miss all 20 shots, but 20 ways to get one hit, 190 ways to get two hits, etc)
This should give you a list like this :
hits/20 -- prob. of one permutation - number of combinations - weighted probability
0 _ 0.7 _ 1 _0.000797923
1 _ 13.6 _ 20 _ 0.006839337
2 _ 13.2 _ 190 _ 0.027845873
3 _ 12.8 _ 1140 _ 0.071603672
4 _ 12.4_ 4845 _ 0.130420974
5_ 12 _ 15504 _ 0.178863051
6 _ 11.6 _ 38760 _ 0.191638983
7 _ 11.2 _ 77520 _ 0.164261985
8 _ 10.8 _ 125970 _ 0.11439674
9 _ 10.4_ 167960 _ 0.065369566
10_ 10 _ 184756 _ 0.030817081
11 _ 9.6 _ 167960 _ 0.012006655
12 _ 9.2 _ 125970 _ 0.003859282
13 _ 8.8 _ 77520 _ 0.001017833
14 _ 8.4_ 38760 _ 0.000218107
15_ 8 _ 15504 _ 3.73898E-05
16 _ 7.6 _ 4845 _ 5.00756E-06
17 _ 7.2 _ 1140 _ 5.04964E-07
18 _ 6.8 _ 190 _ 3.60688E-08
19 _ 6.4_ 20 _ 1.62717E-09
20_ 6 _ 1 _ 3.48678E-11
(last column plotted)
Here you see that six hits out of 20 gives the mean, as expected, at almost 20% probability of occurence.
To get the standard deviation now, just do it in the standard way for the set of data in the last column :
s=sqrt(1/N*SUM(x-mu)^2)
The 4th column data is already normalised, so N=1. mu is 6 from above, and x becomes the probability data in the 4th column - multiply by (column1 - 6)^2 and add them all up to get a standard deviation of 2.05
For questions 3 & 4, you can just read off the data :
p(x=5) = 17.9%
p(4<<x<<8) = 77.9%
Interesting to note that the last answer is a bit higher than that for a normal distribution (a gaussian) - decent speculation about why this is the case may be the sort of stuff that wins you extra points with your markers.
See if this gets things moving for you and I'll come back for the Bill questions if it seems worthwhile.
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this is overkill. Just use that the variance of a binomial distribution = npq (since a binomial distribution is the sum of k Bernoulli trials for some positive integer k.)
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Bahrain4949 Posts
Paper 2 is a real killer, can't wait for the exams next year ...
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On May 05 2008 22:28 sigma_x wrote:
this is overkill. Just use that the variance of a binomial distribution = npq (since a binomial distribution is the sum of k Bernoulli trials for some positive integer k.)
I did use the binomial distribution, I just didn't explicitly call it that - it produced the data in the plot. I guess you could just say sigma=sqrt(Npq), though, lol
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I don't have time to write out full answers, but here's an outline.
Question 1, Parts 1-3 use properties of a binomial distribution. Parts 4 and 5, use geometric distribution. Answer is 13 for part 5; solve for k, p+qp+q^2p+...+q^(k-1)p>0.99
Question 2, Only difficulty is to work out height, but given that D is perpendicular to the plane, this is fairly straight-forward. (Hint: Find AD)
Question 3, simple algebraic manipulation. write cis(3x) in two different ways. Note that your triple angle results are not written correctly. I assume this is just a typo. For the identity, start on the LHS and divide top and bottom by cos^3. Use sec^2=1+tan^2, and the answer will follow easily.
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EPT![[image loading]](http://www.geocities.com/the_soofman16/WTF.jpg)
![[image loading]](http://xs227.xs.to/xs227/08191/plot300.png)
