core help...please :) URGENT

[spinaway]

i know this may seem simple...but how do you find the mean of a boxplot?

MC question 4 (CORE) in 2008 VCAA messed me around  ...and the rest of the exam was easy

[Wezanator123]

How'd you do question 3 in MC Core 2008 VCAA exam 1 ?

[chelseaFC]

You can determine whether the mean will be smaller than or greater than the median by lookin at the distribution and whether outliers exist. Eg) If values are bunched on the left side of the box plot the mean is likely to be less than the median if I'm not mistaken

[Deank]

Q3 is asking about the upper 25% of the boxplot...?

[spinaway]

90 seconds = Q3 which means 25% of people spent more than 90 seconds moving down the aisle.
 there are 79 customers so (79/100)*25=19.75..and 20 is the closest.

any help with 4? haha

[Deank]

You can determine whether the mean will be smaller than or greater than the median by lookin at the distribution and whether outliers exist. Eg) If values are bunched on the left side of the box plot the mean is likely to be less than the median if I'm not mistaken

Do you have anything to back that up? because going off the answer and itute.com's answer, it states that the median is less than the mean (Mean > Median), because it's positively skewed, you're saying that if the data is positively skewed, that the mean is going to be less than the median. (Mean < Median)

[Dominatorrr]

i know this may seem simple...but how do you find the mean of a boxplot?

MC question 4 (CORE) in 2008 VCAA messed me around  ...and the rest of the exam was easy

In an approximately symmetric distribution without outliers, the mean and median will be much the same. In this case the distribution is positively skewed with outliers meaning that these outliers at the extreme end of the data will increase the mean massively, so the median will be less than the mean.

[Deank]

Good job Dom.

[harlequinphoenix]

Well, means are affected by outliers much more than the median. And there are lots of outliers occurring among higher values, so the answer is A because the median is likely to be less than the mean, just by looking at it you can tell.

[Wezanator123]

Q4:

If the boxplot has a:

Symmetric distribution: Median = Mean value

Positively skewed distribution: Median < Mean value

Negatively skewed distribution: Median > Mean value

So... in Q4. Since its a positively skewed distribution the median will be less than the mean

so... the ans is A) and if you check the assessors report its right

[chelseaFC]

Q4:

If the boxplot has a:

Symmetric distribution: Median = Mean value

Positively skewed distribution: Median < Mean value

Negatively skewed distribution: Median > Mean value

So... in Q4. Since its a positively skewed distribution the median will be less than the mean

so... the ans is A) and if you check the assessors report its right

Wouldn't a positive skew leave the mean less than the median as the values are bunched at the Lower end. In this scenario it's the outliers that make the most effect rather than the skew? I'm confused

[Deank]

Q4:

If the boxplot has a:

Symmetric distribution: Median = Mean value

Positively skewed distribution: Median < Mean value

Negatively skewed distribution: Median > Mean value

So... in Q4. Since its a positively skewed distribution the median will be less than the mean

so... the ans is A) and if you check the assessors report its right

i know this may seem simple...but how do you find the mean of a boxplot?

MC question 4 (CORE) in 2008 VCAA messed me around  ...and the rest of the exam was easy

In an approximately symmetric distribution without outliers, the mean and median will be much the same. In this case the distribution is positively skewed with outliers meaning that these outliers at the extreme end of the data will increase the mean massively, so the median will be less than the mean.

Question is, is this only true for distributions with outliers, or is it for all distributions? thinking logically, it would have to be for both wouldn't it?, regardless both mean and median are affected by outliers, but the only way it would be for distributions with outliers only is if we assume the data doesn't change much in terms of median as it does in terms of mean?

[Dominatorrr]

How'd you do question 3 in MC Core 2008 VCAA exam 1 ?

There are 79 customers, if you look at the boxplot you can see that customers that spent 90 seconds or above is Q3 (Quartile 3) or above.

Now, between your minimum value and Q1 lies 25% of your data, between Q1 and Median lies 25%, between Median and Q3 lies 25% of your data and between Q3 and your max value lies the remaining 25% of your data.

SO, 25% of 79 = 19.75 ~ 20 = Answer B

[Deank]

Q4:

If the boxplot has a:

Symmetric distribution: Median = Mean value

Positively skewed distribution: Median < Mean value

Negatively skewed distribution: Median > Mean value

So... in Q4. Since its a positively skewed distribution the median will be less than the mean

so... the ans is A) and if you check the assessors report its right

Wouldn't a positive skew leave the mean less than the median as the values are bunched at the Lower end. In this scenario it's the outliers that make the most effect rather than the skew? I'm confused

If we have all the data grouped up at the left end, and one major outlier to the right, the mean will be affected greater than the median will for example, if we have 1,2,3,4,5,6 and 50, the median is still very close to the left end, and this outlier doesn't greatly affect the position of the median, whereas this 50 changes the mean from 21/6 (3.5) to 28.14.

[Dominatorrr]

Q4:

If the boxplot has a:

Symmetric distribution: Median = Mean value

Positively skewed distribution: Median < Mean value

Negatively skewed distribution: Median > Mean value

So... in Q4. Since its a positively skewed distribution the median will be less than the mean

so... the ans is A) and if you check the assessors report its right

i know this may seem simple...but how do you find the mean of a boxplot?

MC question 4 (CORE) in 2008 VCAA messed me around  ...and the rest of the exam was easy

In an approximately symmetric distribution without outliers, the mean and median will be much the same. In this case the distribution is positively skewed with outliers meaning that these outliers at the extreme end of the data will increase the mean massively, so the median will be less than the mean.

Question is, is this only true for distributions with outliers, or is it for all distributions? thinking logically, it would have to be for both wouldn't it?, regardless both mean and median are affected by outliers, but the only way it would be for distributions with outliers only is if we assume the data doesn't change much in terms of median as it does in terms of mean?

All distributions, regardless of outliers or not. Just remember, if it's not approx. symmetric then the mean won't equal the median. Median is only slightly affected by presence of outliers; mean is affected massively by an extreme outlier.