VCE General & Further Maths Question Thread!

[clarke54321]

The answer is wrong.

As you suggested, the answer should be an increase of 150%

Thanks Peanut Butter 

[knightrider]

For this question attached.

how do you find q1 and q3.

Because the median is the average between the dots i have circled in red ?

so which dot do i use as the median then?( in order to allow me to find q1 and q3) ?

[Peanut Butter]

For this question attached.

how do you find q1 and q3.

Because the median is the average between the dots i have circled in red ?

so which dot do i use as the median then?( in order to allow me to find q1 and q3) ?

When two values are the median, you take the value that is halfway between both of them. Therefore, your median would be 2.

So instead of circling both the medians, put a line between them. Then, just count the dots and find your Q1 and Q3 as usual

[knightrider]

When two values are the median, you take the value that is halfway between both of them. Therefore, your median would be 2.

So instead of circling both the medians, put a line between them. Then, just count the dots and find your Q1 and Q3 as usual

Thanks Peanut Butter 

[knightrider]

For the distribution attached .

the answer says The outliers are at 5, 8, 36 and 40.

However when you check for the lower fence you get 6.

therefore 8 shouldn't be a outlier right?

so the outliers should be 5,36 and 40  right?

[Peanut Butter]

For the distribution attached .

the answer says The outliers are at 5, 8, 36 and 40.

However when you check for the lower fence you get 6.

therefore 8 shouldn't be a outlier right?

so the outliers should be 5,36 and 40  right?

Yes, the outliers should be at 5, 36 and 40.

The box plot is incorrect. You might also notice that the upper fence should be at 30, however the whisker seems to extend further than this. This question is completely wrong

[knightrider]

Yes, the outliers should be at 5, 36 and 40.

The box plot is incorrect. You might also notice that the upper fence should be at 30, however the whisker seems to extend further than this. This question is completely wrong

Thanks for clarifying  Peanut Butter 

the further book literally has so many errors !!

also when we mention these outliers can we say their values directly?

or do we have to say that they are possible outliers and then state their values ?

[Peanut Butter]

Thanks for clarifying  Peanut Butter 

the further book literally has so many errors !!

also when we mention these outliers can we say their values directly?

or do we have to say that they are possible outliers and then state their values ?

You have used the formula to prove that the values are outliers.

You can just say something like: the data contains outliers at 5, 36 and 40

Does that answer your question?

[knightrider]

You have used the formula to prove that the values are outliers.

You can just say something like: the data contains outliers at 5, 36 and 40

Does that answer your question?

yep it does thanks 

[knightrider]

How would you go about answering these types of questions as attached?

the answer says b) On average, brand A batteries have longer and less variable lifetimes.

Like how do they determine the above answer ?

[Darth_Pepe]

How would you go about answering these types of questions as attached?

the answer says b) On average, brand A batteries have longer and less variable lifetimes.

Like how do they determine the above answer ?

Brand A has a higher median than Brand B. Which means on average they have longer lifetimes. ( I should be quoting medians here..this is why i didnt get a 50 in further lol)
Brand B has more variability than A. Roughly Brand B has variations from 18-52 hours compared to A's 22-50 hours. Thus on average Brand A batteries have longer and less variable lifetimes.

Always express the answer with numerical proof.. vcaa made a point about that sometime ago.

[knightrider]

Brand A has a higher median than Brand B. Which means on average they have longer lifetimes. ( I should be quoting medians here..this is why i didnt get a 50 in further lol)
Brand B has more variability than A. Roughly Brand B has variations from 18-52 hours compared to A's 22-50 hours. Thus on average Brand A batteries have longer and less variable lifetimes.

Always express the answer with numerical proof.. vcaa made a point about that sometime ago.

Thanks Darth_Pepe  nice explanation !!

[snow257]

What is the question asking you to do if it says 'compare'? E.g. Compare the mean and standard deviation to the median and IQR for the sample. Btw, I haven't done further maths 1/2.

[MightyBeh]

What is the question asking you to do if it says 'compare'? E.g. Compare the mean and standard deviation to the median and IQR for the sample. Btw, I haven't done further maths 1/2.

Probably just a short comment on the differences - the mean/median are measures of centre and in a symmetrical distribution they should be about the same. A mean significantly different from the median usually implies a skew and/or outliers. Standard deviation with the IQR seems a bit odd though, since the numbers don't really line up (1sx is 34%, IQR is 50%). A comment on whether the mean or median is more appropriate for the data is mostly optional but a good idea most of the time.

[Darth_Pepe]

What is the question asking you to do if it says 'compare'? E.g. Compare the mean and standard deviation to the median and IQR for the sample. Btw, I haven't done further maths 1/2.

Comparing mean and median allows you to know if they are negatively skewed or positively skewed. I cant remenber right now but i think these are how you know:
Mean = median: Data is symmetrical
Mean > median: Positive skewed
Mean < median: Negative skewed.

Comparing standard deviation and IQR allows you to know which has more deviation. This type of question is not common, but the mean and median is. Be sure to use numerical proof Such as The mean (21.3) is larger than the median (20.7) which means that the data is positive skewed.