EDEXCEL GCE S1 June 2014 (R) Paper Q5c- HELP! Watch

emx_eco
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There is a table for the question and I cant copy and paste its format on here so here is the link- it's Q5c that I am stuck on:
http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
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Dragolien
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The upper class boundary is the midpoint of the upper limit of the class in question and the lower limit of the next class. In this instance, the upper limit is 4 minutes, and the lower limit of the next is 5 minutes. The midpoint is 4.5 minutes, and that is the boundary.

The tallest bar of the histogram is the one that represents the group with the largest frequency density. Let's work out all of them to make sure.

2-4 :: 15/3 = 5
5-6 :: 9/2 = 4.5
7 :: 6/1 = 6
8 :: 24/1 = 24
9-10 :: 14/2 = 7
11-15 ::12/5 = 2.4

8 has the largest freq. density, so its height is 6cm. The next largest is 9-10, with a density of 7. If a density of 24 is represented by 6cm, then each cm is worth a density of 4. So a density of 7 is represented by a bar of 1.75cm height.
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Dragolien
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To estimate the number of people between 3.5 and 7, we add up the bars in that range.

The bar 2-4 actually represents the group that waited for 1.5-4.5 minutes, since it is rounded to the nearest minute. A third of the bar is between 3.5 and 7, so we take 15/3 = 5.
Then we add the entire 5-6 bar, which is 9.
And finally since the 7 bar is actually 6.5-7.5, and half of that is inside 3.5-7, we take 6/2 = 3.
Add them up to get 17, which is the answer for c).
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Dragolien
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To find the median of the group, we need to find the exact middle of the data. To do so, we need to add the frequencies until we get to 40.

If we take the frequencies of 2-7, we end up with a total of 30. The next group has 24 people, taking us to 54. So we know that the median is in the 8 group.

We only need 10 people out of that 24. So we take 10/24 = 5/12 or 0.41666...
This is multiplied by one because the 8 group only represents a range of 1 minute.

Add that to 7.5 (because 7.49999... rounds to 7) to get 7.91666....
This is the median.

The same logic is used to calculate Q1 and Q3, but you need to find the 20th and the 60th instead.
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Dragolien
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If Q3 - Q2 < Q2 - Q1, then it is negatively skewed. Otherwise if Q3 - Q2 > Q2 - Q1, then it is positively skewed. If they are equal, there is no skew.

Hope that I helped.
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emx_eco
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(Original post by Dragolien)
To estimate the number of people between 3.5 and 7, we add up the bars in that range.

The bar 2-4 actually represents the group that waited for 1.5-4.5 minutes, since it is rounded to the nearest minute. A third of the bar is between 3.5 and 7, so we take 15/3 = 5.
Then we add the entire 5-6 bar, which is 9.
And finally since the 7 bar is actually 6.5-7.5, and half of that is inside 3.5-7, we take 6/2 = 3.
Add them up to get 17, which is the answer for c).
how did you work out that from 3.5 minutes it was a third of the 1.5-4.5 minute group?
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Dragolien
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(Original post by emx_eco)
how did you work out that from 3.5 minutes it was a third of the 1.5-4.5 minute group?
4.5 - 1.5 = 3. This is the range of the group.
4.5 - 3.5 = 1. This is the part of the group that is between 3.5 and 7.

1/3 is a third.
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emx_eco
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(Original post by Dragolien)
4.5 - 1.5 = 3. This is the range of the group.
4.5 - 3.5 = 1. This is the part of the group that is between 3.5 and 7.

1/3 is a third.
Ohhh I get it now Really appreciate all your help
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Dragolien
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(Original post by emx_eco)
Ohhh I get it now Really appreciate all your help
No problem, I'm happy to help.
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