VCE General & Further Maths Question Thread!

[Cosec]

Can anyone explain how you get the answer in Question 5 of vcaa Exam 1-2005? Thankyou!

I cheated and looked at the answers but the way i got the correct answer is by putting the x variables, the number of mobile phones in list 1 on my cas, the y variables in list 2 whcih is the frequency, set it up and did a one variable analysis. Giving us a mean for the x variable of 1.1298701, which is (a) 1.13. As to why this is the case i have no clue, why you cant just find the mean of the x variables. If euler or alike can explain that it would be much appreciated!

[AngelWings]

Can anyone explain how you get the answer in Question 5 of vcaa Exam 1-2005? Thankyou!

Hey guys.
Reality is: you just have to take in the actual table properly.
One variable you need to omit. Try and have a guess as to which on it is!

Spoiler

Text messaging use is kind of useless in this one.

Then figure out the direction the no. of mobile phones are going:

Spoiler

Since they're the columns, check out the totals at the bottom of the table.

Then multiply the x value (no. of phones in a household) with the y value (the amount of households with that no. of phones.)
e.g. if there were 10 households with 1 phone, go 10 x 1 and so forth

Spoiler

so:
0 x 34
1 x 78
2 x 30
3 x 12

Then add them together:

Spoiler

0 + 78 + 60 + 36 = 174 phones

Note that this is the amount of phones all together...
so we have one last step: find the mean.
How do we usually do that?

Spoiler

mean = (sum of values)/ (no. of values there are)

So therefore we take our total and divide by the amount of households there are.

Spoiler

174 phones / 154 households = approx. 1.12987...

... and that's exactly how the answer arises. Thus the answer is A.

I hope that all made sense. Please tell me if that's not.

[Cosec]

Hey guys.
Reality is: you just have to take in the actual table properly.
One variable you need to omit. Try and have a guess as to which on it is!

Spoiler

Text messaging use is kind of useless in this one.

Then figure out the direction the no. of mobile phones are going:

Spoiler

Since they're the columns, check out the totals at the bottom of the table.

Then multiply the x value (no. of phones in a household) with the y value (the amount of households with that no. of phones.)
e.g. if there were 10 households with 1 phone, go 10 x 1 and so forth

Spoiler

so:
0 x 34
1 x 78
2 x 30
3 x 12

Then add them together:

Spoiler

0 + 78 + 60 + 36 = 174 phones

Note that this is the amount of phones all together...
so we have one last step: find the mean.
How do we usually do that?

Spoiler

mean = (sum of values)/ (no. of values there are)

So therefore we take our total and divide by the amount of households there are.

Spoiler

174 phones / 154 households = approx. 1.12987...

... and that's exactly how the answer arises. Thus the answer is A.

I hope that all made sense. Please tell me if that's not.

Makes sense from a methods kids point of view, but never got taught it nor did i ever see it in the textbook (essentials)?

[AngelWings]

Makes sense from a methods kids point of view, but never got taught it nor did i ever see it in the textbook (essentials)?

Actually, it's a Core thing. You should have learnt this if you have done the Core Module. If you haven't, I'd be glad to teach you this properly when I'm rested up.

[chansena]

Hi,

Could someone please help me with this trig question. I keep getting the wrong answer to the text book.



Thanks !! 

[keltingmeith]

Hi,

Could someone please help me with this trig question. I keep getting the wrong answer to the text book.

Thanks !! 

Try drawing a diagram - the answer should be 8sin(angle). Sounds like a pretty silly flying fox, though...

[scarletmoon]

How do you find IQR from a dot plot? In VCAA 2013 Exam 2 question 2b, I not sure how they got their IQR and Q1.

Why can't you solve this question by typing the values into the CAS (e.g "70" in column 1 and then "1" in column 2), finding the mean and standard deviation that way and going from there?

Also how do you find median in a dot plot? I'm stuck on VCAA 2012 exam 2 question 1a (LOL)
And I'm stuck on 1b as well

[MightyBeh]

How do you find IQR from a dot plot? In VCAA 2013 Exam 2 question 2b, I not sure how they got their IQR and Q1.

Why can't you solve this question by typing the values into the CAS (e.g "70" in column 1 and then "1" in column 2), finding the mean and standard deviation that way and going from there?

Also how do you find median in a dot plot? I'm stuck on VCAA 2012 exam 2 question 1a (LOL)
And I'm stuck on 1b as well

Questions 1 and 2, in order:

Q1,a:
The question is asking for how many times the variable 'very high' appears in the set. Since it's in a pretty (and cheap to print) black and white bar chart, you can just eyeball it from the top of the bar to the frequency (in count, because it doesn't have a % sign) in the vertical axis.

Just in case what I'm trying to say isn't clear, I made a picture. <3

Spoiler

Here my lovely dotted line shows us that the 'number of countries that have a very high development index' is 31.

Q1,b: Here, they're asking for the percentage of the sum of 'low' and 'medium' development index values.

Spoiler

substituting the values in gives us:

Simplifying one step at a time gives us:


(rounded to four decimal places)

However, it is also stated in the question: 'Write your answer, correct to the nearest percentage' (which is poorly phrased imo because I'm pretty sure 61.44% is the nearest percentage since it's already... you know, a percentage. What they mean is the nearest integer (whole number) percentage, but I digress.)
So your answer is 61%.

Q2,a: Interpretting the dot plot, fun stuff.
Mode = 78 (It's the tallest tower, therefore the most occurring)
Range = 9 (max - min, 79-70)

Q2.b:

Spoiler

the question is asking you to prove that the number 70 is an outlier.
So we need to find the quartiles and the IQR.

It is shown on the plot that n = 28, so:

Either on your calculator or in your head:

All you need to do to find the median is to count 14.5 places from the left, because the points in a dot plot are already in order.

another picture in case that didn't make sense

Because you need the 14.5th position, you add the 14th and the 15th positions and divide them by two:



The median split the data into two equal halves. That means that there are 14 values on either side.

So now we need to find Q1 and Q3 to calculate the IQR and lower fence.



Therefore, Q1 = 75




therefore, Q3 = 78

So the IQR = 78 - 75
IQR = 3

Now we can work out the lower fence using the following:

Just sub in our values and you get:



Now we need to explain why the country with a development index of 70 is an outlier for this group of countries.
The appropriate calculation is the lower fence calculation above (more specifically the calculation is the once where you sub in the values), and the explanation can simply be:
"70 is an outlier because it is less than 70.5".

Hope this helped!
(Also I'm sorry if the format is iffy, I don't usually mess with latex)

Edit: in answer to your question

Why can't you solve this question by typing the values into the CAS (e.g "70" in column 1 and then "1" in column 2), finding the mean and standard deviation that way and going from there?

This is univariate data (the only variable is development index), so you should only ever need to put data into one column. Also, the mean and standard deviation must take very value in the column(s) into account to be calculated, so just entering 70 and the count (1) wouldn't give you the correct values for them. I think. I have a very flimsy knowledge of cool CAS tricks so there still might be a way to do it like that.

Edit 2: Also if you're not comfortable doing a question like 2b by hand, with a question like that the whole set is displayed so you could pretty easily put it into your CAS and do a one variable stat analysis.

[keltingmeith]

Holy crap, go MightyBeh! This place really needs a lot more love, you rock.

Also, the mean and standard deviation must take very value in the column(s) into account to be calculated, so just entering 70 and the count (1) wouldn't give you the correct values for them. I think. I have a very flimsy knowledge of cool CAS tricks so there still might be a way to do it like that.

In this case, if you just entered 70 as your only bit of data, then the CAS should say you have a mean of 70 and standard deviation of 0 (which is pretty bogus. ). One data point is never enough to find a mean or sd.

[MightyBeh]

In this case, if you just entered 70 as your only bit of data, then the CAS should say you have a mean of 70 and standard deviation of 0 (which is pretty bogus. ). One data point is never enough to find a mean or sd.

Thought so, time to edit my post with that exact information so I look smart.

Seriously though, I love the maths subs. It kinda sucks that everywhere but Methods and Spesh seem to be ghost towns.

[RazzMeTazz]

Can you use the terms 'discrete' and 'continuous' to describe categorical data, or are these terms confined for use with 'numerical' data only?

Also what exactly does 'centre' and 'spread' of data mean?

[MightyBeh]

I don't think so. Even if it were logical to describe categorical data numerically, using terms like discrete and continuous would probably be confusing. Generally with categorical data, you describe it in terms of the dominant category; conventionally with percentages.

Centre is what divides your data into two halves. We use the mean and median. Not sure about the other textbooks, but I thought Essentials had a really neat explanation for the difference between the two:

Whereas the median lies at the midpoint of a distribution, the mean is the balance point of the distribution. For approximately symmetric distributions, both the median and mean will be approximately equal in value.

Spread refers to... how spread out your data is? The spread puts a number on how tightly (or loosely) your data is grouped together. We use measurements like the range, standard deviation and IQR to calculate it.

If that makes sense.

[Garden]

Hi all,

Regarding the following image, I'm inclined to classify the following box plot as slightly positively skewed on a SAC. Would it more closely resemble an approximately symmetric distribution? Is it appropriate to classify a box plot as slightly positively skewed?



Similarly so with this one, although I'm more confident that this is positively skewed.



Thanks!

[MagicGecko]

Hi all,

Regarding the following image, I'm inclined to classify the following box plot as slightly positively skewed on a SAC. Would it more closely resemble an approximately symmetric distribution? Is it appropriate to classify a box plot as slightly positively skewed?

(Image removed from quote.)

Similarly so with this one, although I'm more confident that this is positively skewed.

(Image removed from quote.)

Thanks!

I've never heard nor used the term 'slightly' positively skewed. From my knowledge, both box plots are positively skewed, mainly because the median line is more inclined to the left side. I get why you would think that the first one is approximately symmetric but the median line though.

[Garden]

Thanks for the clarification MagicGecko!