2011 Further Maths Questions

[Streaker]

Haha, thanks onur369!
Could you or anyone else tell me how to do it manually? D:

Also, does anyone know what it means; standard deviation?

Standard deviation is basically the spread of the data around the mean. This relates to the 68-95-99.7% rule where, in a normal distribution (bell-shaped curve), 68% of data values can be found within one standard deviation of the mean, 95% of data values can be found within 2 standard deviations of the mean and 99.7% of data values can be found within 3 standard deviations of the mean.

If you're using the Essential textbook then there's a small section where it tells you how to calculate it manually but this process is very time-consuming and just plain silly in my opinion. You will never be expected to calculate the standard deviation manually in an exam.

Range/4. I think. Wait for confirmation.

Never use this formula. It's merely an estimation and even if a question asks you to estimate the standard deviation, don't use it. One of the exam questions last year asked students to find the standard deviation of a set of values by looking at a boxplot. Many students used the R/4 rule and got it wrong, there was another way to work it out.

If you get a simple question which asks you to find the standard deviation, then just use your calculator to find it. For the CAS calculator, enter your data values on a spreadsheet, highlight them all then press menu, 4,1,1. This will give you a wide variety of univariate statistics you can use, including standard deviation.

[ech_93]

thankyou streaker!

[BubbleWrapMan]

five finger summary

I lol'd.

[pHysiX]

To add to streaker's answer:

for univariate data:

enter your values into a list. eg: Type: {1,5,7,9}->L1.
That tells your calc to store the values 1, 5, 7, 9 into a list called L1.

If you have the TI-89:
Then press 2nd, Math, List or Stats (I forget), Standard Dev (something like that)

Reason for this post being that on the TI-89, it is much faster to use this technique compared to a spreadsheet; you just need to know your way around.
I did an entire uni statistics exam using this method for mean, median, mode, range, min, max, st dev etc etc. Very useful.

If i get a chance to muck around on a Classpad, I will post as well. I'd imagine it will be very similar.

[ech_93]

I was attempting the Core from VCAA's exam 1 and i got stuck on a couple of questions...
From the 2008 one, how do you do question 3? From a box plot how do you work out the number of people above the upper quartile (i assume thats what i am meant to be working out) ?

And from the 2010 exam, I'm not sure how to find the answer to question 7!? How do work out r from just looking at the graph?

The links to the exams are:
http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/further/pastexams/2008/2008furmath1-w.pdf

and

http://www.vcaa.vic.edu.au/vcaa/vce/studies/mathematics/further/pastexams/2010/2010furmath1-w.pdf

Thanks to anyone that can help!

[jasoN-]

for the 2008 one:
you need to know that IQR = 50% of values, <Q1 = 25%, >Q3 = 25%
so more than 90 seconds spent in the aisle of the supermarket is >Q3, so 25% of 79 = 19.75 ~= 20, B

2010 exam:
looking at the graph (and the dots specifically), you can tell that the correlation is fairly strong, but it's also not perfectly aligned on the line (hence r=1, or 0.98 approx in this case)
clearly the dots aren't spread out so far as to get a weak correlation (so 0.23 is out of the question)
the two other answers (negative ones) are clearly not right too, cos well, you know.
so answer is B!

[ech_93]

Thanks!
how do you do question 3 from the 2010 exam as well?

[jasoN-]

not really sure but the way i would do it is:
by looking at both the bar graph and box plot, the graph is approx. symmetrical
so just count the bars on either side of the mean (theres 7 on either side).
the last bar is pretty negligible due to its height, 3 standard deviations away from the mean is 99.7% of values.
Looking at the bar graph again, the two bars on the sides APPROX add up to 0.3% (well actually looks more like 1% lol but oh well)
we also know standard deviations are equally spaced, so 6 bars / 3 standard deviations = 2 (in this case 2 degree s.d)

also you could look at it this way. its not gonna be 1 degree, cos 3 s.d away from the mean is only 183 degrees, clearly wrong
2 degrees s.d makes sense, it covers most of the values, so a viable answer
3 degrees s.d goes way over the values

hence answer is once again, B

(lol this method is completely weird, not sure how they'd expect you to do this properly, but hope u understand)

[ech_93]

Thanks again, I understand what you mean but i was just hoping there was a method or something rather than just estimating.

[nightblaze]

i have a question; when your describing residual plots that has no clear pattern can you say that they are randomly scattered?
i think i remember someone saying that you can't say that they are scattered. i don't know if this is true?
how would you describe a residual plot if it had no pattern?

[mmonn1906]

15 + 1.5 x 5 = 22.5
BODMAS <- LOL

Rofl.

[WhoTookMyUsername]

15 + 1.5 x 5 = 22.5
BODMAS <- LOL

Rofl.

y is that funny? its all you have to learn for year 12 spesh?

[ech_93]

I have a question from the graphs and relations module...
F is farenheit and C is celsius and

Question. Calculate the temperature at which the Farenheit and Celsius temperatures are the same.
Answer. -40

[_avO]

Let F=C at temp T

T = 9/5T + 32

5/5T = 9/5T + 32
-4/5T = 32
T = -40

at Temp -40 is where F=C

[Hodgeyhodgey]

When using the TVM solver to work out something like 'the amount owing on a loan after x years', is it expected of us to write how we worked out the answer or do we just write the answer straight from the calculator?