VCE General & Further Maths Question Thread!

[Peanut Butter]

Well, according to maths quest, what i said above is correct because an outlier lies outside the interval of the attached picture.

What book are you using?

I use Essentials, and my book just says less than/greater than but not equal to? Ahhh!!

I have attached a photo.

[BakedDwarf]

I use Essentials, and my book just says less than/greater than but not equal to? Ahhh!!

I have attached a photo.

Yes, that's correct. What maths quest and essentials are saying is the same thing.

An outlier is a data point that is outside this interval
Q1 - 1.5 x IQR <= x <= Q3 + 1.5 x IQR

which is the same thing as an outlier that is within these intervals:
x < Q1 - 1.5 x IQR  and  x > Q3 + 1.5 x IQR

[Peanut Butter]

Yes, that's correct. What maths quest and essentials are saying is the same thing.

An outlier is a data point that is outside this interval
Q1 - 1.5 x IQR <= x <= Q3 + 1.5 x IQR

which is the same thing as an outlier that is within these intervals:
x < Q1 - 1.5 x IQR  and  x > Q3 + 1.5 x IQR

Isn't Maths Quest saying that an outlier can be, for instance, the value of Q1 - 1.5IQR, whereas Essentials is saying that would not be an outlier as it has to be less than (and not equal) to that value?

Sorry I'm really confused now :/

[BakedDwarf]

Isn't Maths Quest saying that an outlier can be, for instance, the value of Q1 - 1.5IQR, whereas Essentials is saying that would not be an outlier as it has to be less than (and not equal) to that value?

Sorry I'm really confused now :/

No, maths quest is not saying that. It is saying that an outlier must be less than (and not equal to) Q1 - 1.5IQR, which is what essentials is saying.

Basically, just understand this:
An outlier must be less than (but not equal to) Q1 - 1.5IQR or greater than (but not equal to) Q3 + 1.5IQR

[Peanut Butter]

No, maths quest is not saying that. It is saying that an outlier must be less than (and not equal to) Q1 - 1.5IQR, which is what essentials is saying.

Basically, just understand this:
An outlier must be less than (but not equal to) Q1 - 1.5IQR or greater than (but not equal to) Q3 + 1.5IQR

Oh right, sorry!

I totally misread the photo you attached and I assumed that x was the outlier (whereas it's the complete opposite haha).

Thanks for your help!! And sorry for all the confusion!

[Escobar]

the answer is B
does that mean when considering average rate of change, we just look at the number (as if it had | | around it) and ignore the sign (positive/negative)
in this case it asked for maximum rate & the answer was when it was decreasing fastest

[Peanut Butter]

(Image removed from quote.)
the answer is B
does that mean when considering average rate of change, we just look at the number (as if it had | | around it) and ignore the sign (positive/negative)
in this case it asked for maximum rate & the answer was when it was decreasing fastest

Yes, you just ignore the sign

The rate of change is basically just asking where the distance between one point and another is the greatest - this distance can be increasing or decreasing.

Hope this makes sense!!

[I am a unicorn]

Could someone please explain how to do this question?
(Graph is attached)

Using the 68–95–99.7% rule, the standard deviation for temperature is closest to
A. 1 °C
B. 2 °C
C. 3 °C
D. 4 °C
E. 6 °C

Spoiler

Answer is B

I thought we were meant to use the formula 'standard deviation=range/4' but it turns out that was wrong...

Thank you!

[MightyBeh]

I thought we were meant to use the formula 'standard deviation=range/4' but it turns out that was wrong...

I thought the same when I did that question! My teacher told me to just do it graphically and mark the graph with the standard deviations given to you. Here:

Spoiler

I've only marked out a standard deviation of 2°C to save some time, but here you can see it fits the normal distribution's bell curve almost perfectly, whereas the others would be significantly over or under fitting (from memory).

[AngelWings]

I thought we were meant to use the formula 'standard deviation=range/4' but it turns out that was wrong...
Thank you!

I thought the same when I did that question! My teacher told me to just do it graphically and mark the graph with the standard deviations given to you.

This was VCAA 2010 Exam 1 Core Q3 wasn't it?
Well, I'm just going to put this out there (and please don't hate me for saying this, I do this with good intention), but anything with standard deviation isn't always going to work out nicely - there's often exceptions. The curve you have here isn't exactly symmetrical, so you can't use that formula. (That formula only works in some cases. Be careful.)
 
It's possible to do this mathematically by inputting the data here into your CAS calculator's spreadsheet pages and getting the statistics (menu + 4 + 1 +1 on the colour CAS), but, it will take longer than the graphical version that MightyBeh has used. There are a number of other ways in which you could do this via calculator, some of which are not discussed in Further. With a sensible guess, you can eliminate many of the options, such as E (too big).

Changing the subject now, you guys should also note the following:

• Exam covers have been out for a while. See: Exam 1 and Exam 2 for Further covers.
• VCAA 2011 Exam 1 Core Q12 reappeared last year in another form. Only 16% improvement across these years (10% in 2011 --> 26% in 2014). If you're wondering how to do it, please inbox me and I'll teach you how to tackle the 2014 one.

[I am a unicorn]

I thought the same when I did that question! My teacher told me to just do it graphically and mark the graph with the standard deviations given to you. Here:

Spoiler

I've only marked out a standard deviation of 2°C to save some time, but here you can see it fits the normal distribution's bell curve almost perfectly, whereas the others would be significantly over or under fitting (from memory).

It's possible to do this mathematically by inputting the data here into your CAS calculator's spreadsheet pages and getting the statistics (menu + 4 + 1 +1 on the colour CAS), but, it will take longer than the graphical version that MightyBeh has used. There are a number of other ways in which you could do this via calculator, some of which are not discussed in Further. With a sensible guess, you can eliminate many of the options, such as E (too big).

Thanks MightyBeh and AngelWings!

[Berimbolo King]

Could someone please explain how to do this question?
(Graph is attached)

Using the 68–95–99.7% rule, the standard deviation for temperature is closest to
A. 1 °C
B. 2 °C
C. 3 °C
D. 4 °C
E. 6 °C

Spoiler

Answer is B

I thought we were meant to use the formula 'standard deviation=range/4' but it turns out that was wrong...

Thank you!

My teacher told me when dealing with large data sets standard deviation= range/6.
Its not in the textbook and there was a similar question on my sac.

[belle123]

NUMBER PATTERNS

the sum of the first four terms of a geometric sequence has the value -30
the sum of the first and third terms is 30

the value of the common ratio for this sequence is:

a) -3
b) -2
c) -1
d) 0
e) 1


thanks in advance

[Berimbolo King]

NUMBER PATTERNS

the sum of the first four terms of a geometric sequence has the value -30
the sum of the first and third terms is 30

the value of the common ratio for this sequence is:

a) -3
b) -2
c) -1
d) 0
e) 1


thanks in advance

I got b) -2
Equation 1 a(r^(4)-1)/(r-1)=-30
Equation 2 a+a(r)^(3-1)=30
Solve for a and r
a=6 and r=-2

Sorry i hope thats clear dunno how to use latex yet :/

[belle123]

I got b) -2
Equation 1 a(r^(4)-1)/(r-1)=-30
Equation 2 a+a(r)^(3-1)=30
Solve for a and r
a=6 and r=-2

Sorry i hope thats clear dunno how to use latex yet :/

yeah that does make sense!
thank you so much