ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: costargh on August 06, 2008, 05:11:40 pm
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If you have the following data:
Mean 94.6
Std Dev. 45
Lower limit 0
Upper Limit 180
... can we find out what percentage of people would get over
a) 152
b) 154
c) 156
d) 158
e) 160
f) 162
g) 164
h) 166
i) 168
j) 170
i)172
j)174
????
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dude, you did this last year :P
=normcdf(X,\infty,\mu,\sigma))
calculator work
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I know I did this last year. I tried to work it out but I wasnt getting the right answers.. I even did it on my CAS calculator
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Can you do one for me so I get the gist how to do the rest?
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LOL LOL LOL COSTARGH!!! :D
This is for the accounting mid year isn't it?? ROFL
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... nooo LOL
I know it can be worked out in methods and I think I'm doing it the right way and I probz am getting the right answer... but for over 152 I got 7.2% ... is that right?
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TIstat.normcdf(152,
,94.6,45)=0.101056....
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... nooo LOL
I know it can be worked out in methods and I think I'm doing it the right way and I probz am getting the right answer... but for over 152 I got 7.2% ... is that right?
LOL very funny :P hahahaa
Anyways...hmm... if you are doing it correctly...then shouldn't over 152 be 10%...
like it says on the distribution table... O_o
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Yeh I know... thats why I thought I was getting it wrong... anywyas what i was doing wrong was puttin upper limit at 180... why is it infinity instead of 180? Is it cause some people dont sit the exam?
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ooooh can u work out the rest now and post them?
wait...can this be done on ti84 ? or only on CAS?
Omg, now i really want our SOM now...lol
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Well I think something is wrong.... cause apparently 2.9% of people got 180/180 which is total bullshit....
lol and apparently 2.6% of people got 91/90 LOL
I think it might have something to do with the unassessed people? or something else. Mao? lol
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Yeh I know... thats why I thought I was getting it wrong... anywyas what i was doing wrong was puttin upper limit at 180... why is it infinity instead of 180? Is it cause some people dont sit the exam?
its infinity because it's a normal distribution. but then, to make it realistic, it should be done with conditional probability (and not to infinity, my bad :P )
=\frac{normcdf(152,180,94.6,45)}{normcdf(0,180,94.6,45)}\approx 7.57%)
mmm, according to the stats given here, 95.3% of people get realistic scores? :P
[ps: yeah, sorry my bad. i didn't take into account there was a maximum and minimum. do it with conditional probability]
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Well I think something is wrong.... cause apparently 2.9% of people got 180/180 which is total bullshit....
lol and apparently 2.6% of people got 91/90 LOL
I think it might have something to do with the unassessed people? or something else. Mao? lol
LooooL wtf ??
hmm...awww so it doesn't work properly?
i got all excited for a sec :(
Oh wells....
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Well I dont get it. Why wouldnt 100% of people get a realistic score?
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Well I dont get it. Why wouldnt 100% of people get a realistic score?
imperfection in the normal distribution (they can only try to *fit* people to the curve), hence it'll be approximate, but not perfect.
that and, some people didn't sit the exam (not assessed)
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Hmmm ok... so we can't really find out what percent of people got 172/180 or higher?
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you can:
)
conditional probability! =D
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So whats the answer.... lol... Mao, you know I am mathematically retarded. :P
And just to ease my mind that its correct could you do it for like percentage of people that got 178/180?
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172+: 1.45%
178+: 0.32%
(another problem with discrete data... do we count a score such as 170 to be 168~170? or 170~172? either ways anomaly at ends...)
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THANKS MAOOO
:)
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Hah, you should also remember these:
1σ (s.d.) away from the mean on both sides is 68% of values
2σ is 95%
3σ is 99%
They can ask you this in a tech free exam.
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Hah, you should also remember these:
1σ (s.d.) away from the mean on both sides is 68% of values
2σ is 95%
3σ is 99%
They can ask you this in a tech free exam.
if you use this in the tech free exam you'd get it wrong :P
3σ is 99.7% in VCELand
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Damn study designs.
If you're gonna round to 2 s.f., why not KEEP it that way! :P
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This question is specifically for Mao considering he has been able to give me an answer for the other ones.
166/180. top what %?
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Pr(X > 166)
You do not need to include an upper X < 180 bound as you are saying 'EVERYONE above 166'
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You can argue that there should be noone above 180 though.
However, I did it this way because it seemed that VCAA was using the same method:
 \approx 0.10)
Putting an upper bound would have yielded approximately 7.8% instead.
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same method as before, 2.88%
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172+: 1.45%
178+: 0.32%
(another problem with discrete data... do we count a score such as 170 to be 168~170? or 170~172? either ways anomaly at ends...)
You count 170 as 169-171 if your discrete data is only every 2nd number which I think you're suggesting there.
But in reality people can get 171/180 so you would count 170 as ~ 169.5-170.5
i lernt dis in mth1030 so it iz 1052.7% more accurate den vce infoz lol.
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Well, I don't know which figure is correct.
So far I've been told it is
around 5%
around 7.8&
around 2.88%
:S
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Well, I don't know which figure is correct.
So far I've been told it is
around 5%
around 7.8&
around 2.88%
:S
7.8% is what you get if you used Mao's method to evaluate scores above 152. However, VCAA quoted that scores above 152 corresponded to the top 10% (which would be using the simpler method).
You get 5.5% using VCAA's method for scores above 166. You get 2.8% using Mao's method.
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Ok, so we can presume that VCAA's method is the one most applicable here yeh?
So, 5.5% corresponds to a SS of around 42 you would think?
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Why are you using an exam mark to figure that out? Just use the scaling reports. They tell you the mean and standard deviation (usually around 30 and 7 respectively), which makes it fairly easy to calculate.
A study score of 42 should be approximately the top 4.1% of students in the subject.
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i think he's trying to predict his SS from his exam score
coblin: are you sure that's VCAA's method? by that, about 2.98% of people achieve full score [or more], and 2.60% of people get 182? =\
The normal distribution is definitely truncated: http://en.wikipedia.org/wiki/Truncated_normal_distribution
the probability density function of a truncated distribution [from wikipedia]:
=\frac{\frac{1}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)}{\Phi\left(\frac{a-\mu}{\sigma}\right)-\Phi\left(\frac{b-\mu}{\sigma}\right)})
}{\sigma\cdot normcdf(a,b,\mu,\sigma)})
hence, the cumulative function would be:
=\int_{\alpha}^{\beta}f(x;\mu,\sigma,a,b,)\; dx = \frac{1}{normcdf(a,b,\mu,\sigma)}\int_{\alpha}^{\beta} \frac{\phi\left(\frac{x-\mu}{\sigma}\right)}{\sigma}\; dx = \frac{normcdf(\alpha,\beta,\mu,\sigma)}{normcdf(a,b,\mu,\sigma)})
tada~ conditional probability~
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For the sake of clarity, my method (conditional probability) [with suggestion from caramel to implement each discrete data range to be from -0.5 to +0.5]:
for discrete set of data 
that gives:
0 - 100%
25 - 95.6%
50 - 85.2%
75 - 67.5%
100 - 44.9%
110 - 35.8%
120 - 27.4%
130 - 20.0%
140 - 13.7%
150 - 8.70%
160 - 4.87%
162 - 4.23%
164 - 3.64%
166 - 3.08%
168 - 2.56%
170 - 2.08%
172 - 1.63%
174 - 1.22%
176 - 0.83%
178 - 0.48%
and finally for amusement:
180 - 0.15%
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Thanks Mao, but then why would VCAA do it such that the top 10 % is 152 or above. Why would they use that way of calculating the distribution if according to their data, people get above the total marks allocated for the exam. lol
This is really annoying lol. Thanks for the effort guys
Finally, with a score of 83/90 (ie. half way between cut off A+ and maximum mark allocation) (where cut off A+ is 76/90 and max score is 90/90), would anyone like to predict mathematically what study score I would be looking at if I did proportionately well on exam 2 (ie. disregard the score on the exam and just take into account the percentile rank)
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Why would they use that way of calculating the distribution if according to their data, people get above the total marks allocated for the exam. lol
I do not believe that's what happens. I think the bell curve is truncated at 0,180 (I defined it to be -0.5 to 180.5 since 0 and 180 are valid scores)
as for what VCAA actually do in detail, I am unsure. But I am do think that their model does not include people with unrealistic scores (well, I hope at least)
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Hmm VCAA is confusing
Finally, with a score of 83/90 (ie. half way between cut off A+ and maximum mark allocation) (where cut off A+ is 76/90 and max score is 90/90), would anyone like to predict mathematically what study score I would be looking at if I did proportionately well on exam 2 (ie. disregard the score on the exam and just take into account the percentile rank)
Just incase I edited while you were replying and missed this. Lol don't feel obliged, just pointing it out
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Mao, if you calculate it, you will get 10% for an upper bound of infinite, and 7.8% for an upper bound of 180 (lower bound being 152).